old.math.slu.czold.math.slu.cz/studmat/MatAnalIII-IV/MAIII-IV.pdfPrˇedmluva Tento text vznikl na za´kladeˇ pozna´mek ke kurzu pˇredna´sˇek z matematicke´ analy´zy 3 a 4, ktere´

Predmluva

Tento text vznikl na zaklade poznamek ke kurzu prednasek z matematicke analyzy 3 a 4,ktere vedl na Slezske univerzite prvnı autor, a poznamek k cvicenı z tehoz kurzu, kterevedl druhy autor.

Teoreticka cast je psana anglicky (prednasky probıhaly v tomto jazyce) a cvicenıv cestine.

Autori by chteli podekovat Michalovi Miklasovi za skvelou praci pri pocıtacovemzpracovanı obrazku.

Jako kazda publikace i tento text obsahuje tiskove a jine chyby, ackoliv bylo venovanovelke usilı se jich zbavit. Autori budou velice vdecni za libovolne korektury, ktere jemozno jim zaslat na emailove adresy:

[email protected] a [email protected]

Opava, zarı 2003 V. Averbuch a M. Malek

1

2

Contents

Introduction 7

1 Normed spaces 91.1 Norms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.2 Balls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.3 Norm topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.4 Equivalent norms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.5 Bounded sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.6 Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.7 Natural topology in Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.8 Bounded sets and compact sets in Rn . . . . . . . . . . . . . . . . . . . . 151.9 Uniqueness of the norm topology in Rn . . . . . . . . . . . . . . . . . . 161.10 Linear mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2 First derivative 212.1 Frechet and Gateaux derivatives . . . . . . . . . . . . . . . . . . . . . . 212.2 Computation Rule and directional differentiability. . . . . . . . . . . . . 232.3 Rules of differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.4 Partial derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.5 Finite-dimensional case . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.6 Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.7 Continuous differentiability . . . . . . . . . . . . . . . . . . . . . . . . . 332.8 Continuous partial derivatives . . . . . . . . . . . . . . . . . . . . . . . 34

3 Inverse Function Theorem 373.1 Lipschitz functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373.2 Banach spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.3 Contraction Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.4 Isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393.5 Inverse Function Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 403.6 Implicite Function Theorem . . . . . . . . . . . . . . . . . . . . . . . . 42

4 Higher derivatives 474.1 Multilinear mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474.2 Higher derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524.3 Rules of differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . 534.4 Higher partial derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . 554.5 Class C p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

3

4 CONTENTS

4.6 Symmetry of higher derivatives . . . . . . . . . . . . . . . . . . . . . . . 594.7 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 624.8 Taylor Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

5 Extreme Problems 695.1 Generalized theorem of Fermat . . . . . . . . . . . . . . . . . . . . . . . 695.2 Necessary and sufficient conditions of locmin . . . . . . . . . . . . . . . 695.3 Setting of the problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 725.4 General (non-smooth) problems: necessary condition of locmin . . . . . . 725.5 Smooth problems: sufficient conditions . . . . . . . . . . . . . . . . . . 755.6 Smooth problems: tangent cone to A = g−1(0) . . . . . . . . . . . . . . 765.7 Smooth problems: necessary condition . . . . . . . . . . . . . . . . . . . 795.8 Problems with equations and inequalities . . . . . . . . . . . . . . . . . . 80

6 Riemann integral in Rn 836.1 Partitions and cubes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 836.2 Riemann integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 846.3 Criterion of existence of Riemann integral . . . . . . . . . . . . . . . . . 856.4 Null sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 876.5 Oscillation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 896.6 Lebesgue theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 916.7 Jordan measurable sets . . . . . . . . . . . . . . . . . . . . . . . . . . . 946.8 Fubini Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

7 Partition of unity. Change of variables 997.1 Smooth indicators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 997.2 Partition of unity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1007.3 Partition of integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1027.4 Change of variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

8 Differential forms 1098.1 Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1098.2 Asymmetric tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1128.3 Differential forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1178.4 Exterion differentiation (operator d) . . . . . . . . . . . . . . . . . . . . 120

9 Stokes Formula for Chains 1239.1 Chains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1239.2 Integral over a chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1249.3 Stokes formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

10 Stokes Theorem for Manifolds 12910.1 Manifolds in Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12910.2 Tangent space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13110.3 Mapping between manifolds. Vector fields and forms . . . . . . . . . . . 13210.4 Manifolds with a boundary . . . . . . . . . . . . . . . . . . . . . . . . . 13510.5 Orientation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13510.6 Integral of a form on an oriented manifold . . . . . . . . . . . . . . . . . 13810.7 Stokes Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14010.8 Classical special cases . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

CONTENTS 5

11 Functions of complex variable 14511.1 Analytic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14511.2 Complex forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14711.3 Integrals of complex 1-forms . . . . . . . . . . . . . . . . . . . . . . . . 14911.4 Cauchy formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15111.5 Representation by series . . . . . . . . . . . . . . . . . . . . . . . . . . 15211.6 Elementary functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15511.7 Residues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

12 Ordinary differential equations 15912.1 Analytic setting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15912.2 Geometric setting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16112.3 Basic Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16312.4 Methods of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16712.5 Linear equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16912.6 Linear equations with constant coefficients . . . . . . . . . . . . . . . . . 173

Prıklady a cvicenı 177

References 224

Notations 227

IntroductionAnalysis 3-4 is a generalization of usual, one-dimensional analysis to the case of

arbirary finite dimensions.More precisely, instead of functions R→ R we consider functions Rn → Rm .In fact elements of finite-dimensional analysis were developed already in 18th century,

but an appropriate frame for this analysis are normed spaces. These spaces were introduced(independently and almost simultaneously (1916–1922))by A. Bennett, F. Riesz, H. Hahn,S. Banach, N. Wiener. The differentiation operator for mappings between normed spaceswas defined in 1925 by M. Frechet. (More “weak” notion of differentiability was definedearly (1913) by R. Gateaux, but for this kind of differentiability the chain rule is not valid.)This date can be consider as the birthdate of modern analysis.

* * *

Our course contains 12 Chapters. In Chapter 1 we study normed spaces (up to Chapter 5these spaces are allowed to be infinite-dimensional). In Chapter 2 we consider Frechet(and Gateaux) derivative. Chapter 3 is devoted to the most important theorem of analysis,Inverse Function Theorem (which can be equivalently reformulated as Implicit FunctionTheorem). As a tool for proving this theorem we prove at first so called ContractionLemma (this is the main tool also in the final Chapter 12). In Chapter 4 we study higherderivatives, up to Taylor formula. In Chapter 5 we give some applications of the theory tooptimization problems.

Starting from Chapter 6 we restrict ourselfs just by FINITE-dimensional case.In Chapter 6 we construct Riemann integral in Rn . Chapter 7 is devoted to two im-

portant technical results. In Chapter 8 we consider differential forms, which are in factgeneralizations of “length element”, “area element” and “volume element” of classical“old” analysis. Chapters 9 and 10 are devoted to the crown theorem of the theory, StokesTheorem, which is a generalization of different known results of “old” analysis (Eu-ler (1771), Green (1828), Ostrogradskij (1834), Stokes (1854)). For this end we definemanifolds in Rn .

The last two chapters are facultative and written in more compressed style. In Chap-ter 11 we apply Stokes Theorem for study of analytic complex function, and in Chapter 12we apply Contraction Lemma for proving of Existence and Uniqueness Theorem forordinary differential equations.

* * *

Some remarks on notations.

If we write, e.g., aA< b, this means “using A we conclude that a < b”.

Symbols ⊳ and ⊲ denote, resp., the beginning and the end of the proof. If we provesome “small” assertion inside the proof of a “great” one, we use symbols ⊳⊳ and ⊲⊲ forthis “small” proof, et-cetera.

“Exerc.” over e.g. an equation mark means that to prove this equation is an exercizefor the reader.

The reader has to remember that misprints are POSSIBLE and to use ever his commonsense.

7

8

Chapter 1

Normed spaces

1.1 Norms

Let X be a vector space over R. By a norm on (or in) X we mean a function ‖·‖ : X → Rwith the following properties:

(i) ∀x ∈ X... ‖x‖ ≥ 0 (positivity); ‖x‖ = 0⇔ x = 0 (non-degeneracy);

(ii) ∀x ∈ X ∀t ∈ R... ‖tx‖ = |t| ‖x‖ (positive homogenity); in particular ‖−x‖ = ‖x‖

(symmetry);

(iii) ∀x, y ∈ X... ‖x + y‖ ≤ ‖x‖ + ‖y‖ (subadditivity).

||x||

||y||

||x||

||y||

||x+y||x x+y

0 y

If we interpret ‖x‖ as the LENGTH of the vector x then theproperty (iii) expresses the triangle inequality (△-in.).

A normed space X is a vector space equipped with a norm.(X ∈ NS)

Examples.

1. (R, | · |);2. (Rn, ‖·‖p), where ‖·‖p is defined for 1 ≤ p < ∞ by theformula

‖x‖p := (|x1|p + · · · + |xn|p)1/p

(x = (x1, . . . , xn)).

For p = 2 we obtain the usual Euclidean length.

3. (Rn, ‖·‖∞), where‖x‖∞ := max {|x1|, · · · , |xn |} .

NB ‖x‖∞ = limp→∞ ‖x‖p .

4. ℓ2. This is the set of all sequences x = (x1, x2, . . .) of real numbers such that∑∞

i=1 x2i <

∞, with the norm defined so:

‖x‖2 :=∞∑

i=1

x2i .

9

10 CHAPTER 1. NORMED SPACES

5. C([0, 1]). This is the set of all continuous real-valued functions on [0, 1] equipped withthe norm

‖x‖ := maxt∈[0,1]

|x(t)| .

||x−y||

x

0 ||y|| y

||x||

Second Triangle Inequality. Let ‖·‖ be a norm in X . Then

∀x, y ∈ X... |‖x‖ − ‖y‖| ≤ ‖x − y‖ .

⊳ Without loss of generality we can assume that ‖x‖ > ‖y‖(since ‖x − y‖ = ‖y − x‖). We need to verify that ‖x‖−‖y‖ ≤‖x − y‖. But indeed

‖x‖ trick= ‖x − y + y‖ △-in.≤ ‖x − y‖ + ‖y‖ . ⊲

1.2 Balls

Let X be a normed space. Put for x ∈ X , r > 0

Br (x) := {y ∈ X | ‖y − x‖ ≤ r} (the closed ball with the center x and radius r );◦Br (x) := {y ∈ X | ‖y − x‖ < r} (the open ball with the center x and radius r );

For balls with center at 0 we write for short

Br := Br (0),◦Br := ◦Br (0).

Properties of balls. It is easy to verify (please!) that1

1) Br (x) = x + Br ;◦Br (x) = x + ◦Br ;

2) Br = r B1;◦Br = r

◦B1;

3)◦Br $ Br ;

4) if r1 < r2, then Br1 $◦Br2 ;

5)◦Br =

⋃(0<)α<r

◦Bα =

⋃α<r

Bα ;

6) Br =⋂α>r

Bα =⋂α>r

◦Bα;

NB Here and below we use the following notations:

A + B := {a + b | a ∈ A, b ∈ B} (A, B ⊂ X),

T A := {ta | t ∈ T, a ∈ A} (T ⊂ R, A ⊂ X).

In particular

x + A := {x} + A = {x + a | a ∈ A} (x ∈ X),

t A := {t}A = {ta | a ∈ A} (t ∈ R).

Notation. For balls in R we write I (“interval”) instead of B. For example

I1 = [−1, 1].

1In 3) and 4) we suppose that our normed space is non-trivial: X 6= {0}.

1.3. NORM TOPOLOGY 11

1.3 Norm topology

xG B (x)ε

Let X be a normed space. We define the topology τ generated by the norm so: a set G isopen if for each point x ∈ G there exists a ball Bε(x) that iscontained in G:

G ∈ τ :⇔ ∀x ∈ G ∃ε > 0 : Bε(x) ⊂ G.

The first assertion of the following theorem means that thisdefinition is correct.

Theorem 1.3.1.a) So defined τ is a topology.b) Open balls in X are open sets in this topology, and closed balls are closed sets.c) Both the open balls and the closed balls with the center at x are bases of neighbour-

hoods of x in this topology.

⊳ a) Gα ∈ τ ⇒⋃α Gα ∈ τ . Indeed if x ∈ ⋃Gα then x ∈ Gα0 for some α0, hence

Bε(x) ⊂ Gα0 for some ε > 0; a fortiori Bε(x) ⊂⋃

Gα.Further, G1,G2 ∈ τ ⇒ G1 ∩ G2 ∈ τ . Indeed, if x ∈ G1 ∩ G2, then x ∈ G1

and hence Bε1 ⊂ G1 for some ε1 > 0. Analogously Bε2 ⊂ G2 for some ε2 > 0. Putε := min(ε1, ε2). Then Bε(x) ⊂ G1 ∩ G2.

Thus τ is a topology.

y} }x

s

ε−s

B (x)ε

B (y)δ

b) Let us prove that◦Bε(x) ∈ τ . Let y ∈ ◦Bε(x). Then

s := ‖y − x‖ < ε.Take any δ > 0 such that

(1)δ < ε − s.

Thenz ∈ Bδ(y)⇒ ‖z − y‖ ≤ δ ⇒ ‖z − x‖

△-in.≤ ‖z − y‖︸︷︷︸

≤δ

+‖y − x‖︸︷︷︸=s

≤ δ + s(1)< ε ⇒ z ∈ ◦Bε(x),

which means that Bδ(y) ⊂◦Bε(x). Thus,

◦Bε(x) ∈ τ .

That (Bε(x))c ∈ τ can be proved analogously.c) If U is an open neighbourhood of x in τ , then (by our definition of τ ), Bε(x) ⊂ U

for some ε > 0; a fortiori◦Bε(x) ⊂ U . But

◦Bε(x) is an open set (by b)) and contains x

(obviously), so◦Bε(x) is an open neighbourhood of x , therefore Bε(x) is a (closed by b))

neighbourhood of x . All is proved. ⊲

By a norm topology we mean the topology generated by a norm.NB Any norm topology is Hausdorff. (Prove!)

Convergence and continuityConvergence in a normed space X means convergence in the topology generated by

the norm. It follows from the definitions that

xn Kn→∞ x ∈ X ⇔ ‖xn − x‖ K

n→∞ 0.


Continuity of a mapping f : X → Y (where X,Y are normed spaces) means continuityin topologies generated by the norms in X and Y . It follows from the definitions that f iscontinuous at a point x ∈ X if and only if (iff)

∀ε > 0 ∃δ > 0 :∥∥x − x

∥∥ ≤ δ ⇒∥∥ f (x)− f (x)

∥∥ ≤ ε (2)

(just as in usual analysis, only with ‖·‖ instead of | · |).For short we write (2) in the form

∥∥x − x∥∥→ 0⇒

∥∥ f (x)− f (x)∥∥→ 0,

or ∥∥ f (x)− f (x)∥∥ K‖x−x‖→0

0.

Theorem 1.3.2. Let X be a normed space. Then the norm ‖·‖ : X → R is a continuousfunction⊳ Continuity of ‖·‖ at a point x means that

∣∣‖x‖ −∥∥x∥∥∣∣ K‖x‖−‖x‖→0

0.

But the latter relation is true, since, by the Second Triangle Inequality,∣∣‖x‖ −

∥∥x∥∥∣∣ ≤ ‖x‖ −

∥∥x∥∥ . ⊲

1.4 Equivalent norms

Let ‖·‖1 and ‖·‖2 be two norms on a vector space X . We say that the norm ‖·‖1 is strongerthan the norm ‖·‖2, and write

‖·‖1 ≻ ‖·‖2 ,if the topology τ1 generated by ‖·‖1 is FINER than the topology τ2, generated by ‖·‖2:

‖·‖1 ≻ ‖·‖2 :⇔ τ1 ⊃ τ2.

Theorem 1.4.1. The following conditions are equivalent (TFAE):a) ‖·‖1 ≻ ‖·‖2 ;b) ∃r1, r2 > 0 :

◦B‖·‖1r1 ⊂

◦B‖·‖2r2 ;

c) ∃r1, r2 > 0 : B‖·‖1r1 ⊂ B‖·‖2r2 ;d) ∃r1, r2 > 0 : r1 ‖·‖1 ≥ r2 ‖·‖2. (Which means that ∀x ∈ X

... r1 ‖x‖1 ≥ r2 ‖x‖2.).

⊳ 1◦ (a)⇒ (b):

◦B‖·‖2r1

1.3∈ τ2τ1⊃τ2⇒ ◦

B‖·‖21 ∈ τ1def. of τ1⇒ ∃ε > 0 :

◦B‖·‖21 ⊃ B‖·‖1ε ⇒ ◦

B‖·‖21 ⊃ ◦B‖·‖1ε ;

thus, we can put r1 = ε, r2 = 1.2◦ (b)⇒ (c):

B‖·‖111.2=⋂

r>r1

◦B‖·‖1r

1.2=⋂

α>1

α◦B‖·‖1r1

(b)⊂⋂

α>1

α◦B‖·‖2r2

1.2=⋂

r>r2

◦B‖·‖2r

1.2= B‖·‖2r2 .

1.5. BOUNDED SETS 13

3◦ (c)⇒ (d): Let (c) is true. We need to verify that ∀x ∈ X... r2 ‖x‖1 ≥ r1 ‖x‖2. Without

loss of generality x 6= 0. We have ‖r1x/‖x‖1‖1 = r1 ⇛ r1x/‖x‖1 ∈ B‖·‖1r1

(c)⇛ r1x/‖x‖1 ∈

B‖·‖2r2 ⇛ ‖r1x/‖x‖1‖2 ≤ r2 ⇛ r1 ‖x‖2 ≥ r2 ‖x‖1.4◦ (d)⇒ (a): Let (d) is true. We need to verify that τ1 ⊃ τ2. Let U ∈ τ2 and let x be anarbitrary point in U . By the definition of τ2, for some ε > 0

B‖·‖2ε (x) ⊂ U. (1)

Now, ‖x‖1 < r1(d)⇒ ‖x‖2 < r2, which means that

◦B‖·‖1r1 ⊂

◦B‖·‖2r2 . Multiplying by ε/r2 we

obtain◦B‖·‖1εr1/r2

⊂ ◦B‖·‖2ε . And the translation by x yields (by the property 2 of balls, see 1.2)◦B‖·‖1εr1/r2

(x) ⊂ ◦B‖·‖2ε (x)(1)⊂ U . Thus U ∈ τ1. ⊲

Example. In ℓ2

‖·‖2 ≻ ‖·‖∞ , ‖·‖∞ 6≻ ‖·‖2 ,where

‖x‖∞ := supi∈{1,2,...}

|xi | (x = (x1, x2, . . .)).

(Prove!)

Equivalent normsWe say that two norms ‖·‖1 and ‖·‖2 on a vector space X are equivalent and write

‖·‖1 ∼ ‖·‖2 ,

if each norm is stronger than the other, that is, if they generate one and the same topology:

‖·‖1 ∼ ‖·‖2 :⇔ (‖·‖1 ≻ ‖·‖2 , ‖·‖2 ≻ ‖·‖1)⇔ τ1 = τ2.

B1

|| ||. 8

B1

|| ||.2

B1

|| ||.1

Theorem 1.4.2. In Rn

‖·‖1 ∼ ‖·‖2 ∼ ‖·‖∞⊳ 1◦ ‖x‖1 ≥ ‖x‖2 ≥ ‖x‖∞ (prove!), hence (byTheorem 1.4.1.) ‖·‖1 ≻ ‖·‖2 ≻ ‖·‖∞.2◦ n ‖x‖∞ ≥

√n ‖x‖2 ≥ ‖x‖1 (prove!), hence

‖·‖∞ ≻ ‖·‖2 ≻ ‖·‖1. ⊲

NB In fact ALL norms in Rn are equivalent (see1.9).

1.5 Bounded sets

A set A in a normed space X is called bounded if A is contained in the ball Br for somer > 0.NB A set A is bounded iff the number set {‖x‖ | x ∈ A} ⊂ R is bounded.

Example. Each ball (closed or open), with any center, is bounded. (Prove!)


Theorem 1.5.1. For equivalent norms the set of all bounded sets is one and the same, thatis, if ‖·‖1 ∼ ‖·‖2, then

A is bounded in ‖·‖1 ⇔ A is bounded in ‖·‖2 .

⊳ This follows from equivalence (a)⇔ (d) in Theorem 1.4.1. ⊲

Remark. The theorem suggests that boundedness can be expressed in terms of the TOPO-LOGY τ generated by the norm. And indeed A is bounded iff for each neighbourhood Uof zero in τ there exists δ > 0 such that δA ⊂ U .

1.6 Product

Let X1, . . . , Xn be normed spaces. The vector space X1× · · ·× Xn can be equipped withthe norm

‖(x1, . . . , xn)‖ := ‖ (‖x1‖X1, . . . , ‖xn‖Xn )︸︷︷︸∈Rn

‖p, (1)

where 1 ≤ p ≤ ∞, and ‖·‖p is the norm in Rn defined in 1.1. Just as in Theorem 1.4.2.,it can be verified that for p = 1, 2 and∞ we obtain equivalent norms. (In fact, the norms(1) are equivalent for ALL p ∈ [1,∞], since all norms IN Rn are equivalent, see 1.9.)Remark. The norm ‖·‖p in Rn is a special case of this construction (Rn = R× · · · ×R).NB The topology generated by the norms (1) coincides with the product topology inX1 × · · · × Xn , each X i being supplied with the topology generated by the norm.Criterion. Let X1, . . . , Xn be normed spaces. Then

(x1, . . . , xn)→ (x1, . . . , xn) ∈ X1 × · · · × Xn ⇔ ∀i...∥∥xi − xi

∥∥→ 0.

⊳ This follows at once from the definitions. ⊲

Theorem 1.6.1. For each normed space the algebraical operations

· : X × R→ X, (x, t) 7→ tx (multiplication)

and+ : X × X → X, (x, y) 7→ x + y (addition)

are continuous.⊳ 1◦

0 ≤∥∥tx − t x

∥∥ △-in≤∥∥tx − t x

∥∥+∥∥t x − t x

∥∥ = |t|︸︷︷︸K

t→t|t |

∥∥x − x∥∥+ |t − t|

∥∥x∥∥ K‖x−x‖→0|t−t |→0

0,

(t,x)

t,x)> >(

hence∥∥tx − t x

∥∥ → 0 as (x, t) 7→ (x, t). Thus, the multiplication iscontinuous.2◦

0 ≤∥∥(x + y)− (x + y)

∥∥ △-in.≤∥∥x − x

∥∥+∥∥y − y

∥∥ K‖x−x‖→0‖y−y‖→0

0,

hence∥∥(x + y)− (x + y)

∥∥→ 0 as (x, y)→ (x, y). Thus, the addition is continuous. ⊲

1.7. NATURAL TOPOLOGY IN RN 15

1.7 Natural topology in Rn

The topology generated by the equivalent norms ‖·‖1 , ‖·‖2 , ‖·‖∞ in Rn (see 1.4) iscalled the natural topology.NB We ever consider Rn with the natural topology.

Theorem 1.7.1. The natural topology τnat in Rn coincides with the product topology τprod(that is, the topology of the product R× · · · × R (n times), where R is equipped with itsstandard topology).

In particular the natural topology in R is its standard topology.⊳ For short consider the case n = 2.0◦ Since the neighbourhoods of x are translations by x of the neighbourhoods of 0 (byProperty 1 of balls, see 1.2), it is sufficient to verify that each neighbourhood of 0 in τnatcontains a neighbourhood of 0 in τprod and vice versa. Below the notation U ∈ Nbx meansthat U is a neighbourhood of x .1◦ Let U ∈ Nb0(τprod). Then by the definition of the product topology there exist ε1 >

0, ε2 > 0 such that

U ⊃ Iε1 × Iε2 ⊃ε:=min(ε1,ε2)

Iε × Iε︸︷︷︸={(x,y)||x |≤ε, |y|≤ε}

= B‖·‖∞ε ∈ Nb(τnat). O.K.

2◦ Let U ∈ Nb0(τnat). Then (since τnat is generated by ‖·‖∞) there exists ε > 0 such thatU ⊃ B‖·‖∞ε = Iε × Iε ∈ Nb0(τprod). O.K. ⊲

1.8 Bounded sets and compact sets in Rn

A set A ⊂ Rn is called bounded if it is bounded in one of the norms ‖·‖1 , ‖·‖2 , ‖·‖∞(then, by Theorem 1.5.1., it it bounded also in the two others; in essence it is boundednesswith respect to the natural topology, see Remark in 1.5).

Theorem 1.8.1. A set K ⊂ Rn is compact (in the natural topology) iff it is bounded andclosed.⊳ For simplicity of notations consider the case n = 2.0◦ We need the following important theorem of general topology:

Tichonov Theorem. The product∏

i X i of (arbitrary many) topologicalspa-ces (equipped with the product topology) is a compact space iff each X i is acompact space.

K

K1

K2

x1

x2

1◦ Let K be compact (in R2). Then K is closed (as a compact set in a Hausdorff topologicalspace). Now the projections K1 and K2 of K onto the axes x1and x2 are compact (in R) as the images of a compact set bycontinuous mappings. Hence, as is known from one-dimensionalanalysis, K1 and K2 are bounded. So there exists a > 0 such thatKi ⊂ Ia , i = 1, 2, . . . whence it follows that

K1 × K2 ⊂ Ia × Ia = B‖·‖∞a .

Thus K1×K2 is bounded. A fortiori K ⊂ K1×K2 is bounded.2◦ Vice versa, let K be a closed bounded set in R2. Then K ⊂ B‖·‖∞a for some a > 0.But B‖·‖∞a = Ia × Ia is compact by Tichonov Theorem, hence K is compact as a closedsubset of a compact set. ⊲


1.9 Uniqueness of the norm topology in Rn

Up to equivalence there exists just one norm in Rn — all norms in Rngenerate one andthe same topology:

Theorem 1.9.1. All the norms in Rn are equivalent.⊳ Let ‖·‖ be a norm in Rn . Show that ‖·‖ ∼ ‖·‖1.1◦ ‖·‖1 ≻ ‖·‖: Consider the canonical basis {e1, . . . , en} of Rn , (ei = (0, . . . , 0, 1,

i0, . . . , 0)). For any point x = (x1, . . . , xn) ∈ Rn it holds

‖x‖ = ‖x1e1 + · · · + xnen‖△-in.≤ ‖x1e1‖ + · · · + ‖xnen‖

= |x1| ‖e1‖ + · · · + |xn| ‖en‖≤

M :=max{‖e1‖,···,‖en‖}M(|x1| + · · · + |xn|) = M ‖x‖1 .

Hence, by Theorem 1.4.1., ‖·‖1 ≻ ‖·‖.2◦ ‖·‖ ≻ ‖·‖1: Consider the unit sphere S in the norm ‖·‖1:

S := {x ∈ Rn | ‖x‖1 = 1}.

This set is compact (in the natural topology). Indeed, S is obviously bounded, and S isclosed as the pre-image of the closed set {1} ⊂ R by the continuous mapping ‖·‖1 (seeTheorem 1.3.2.).

Now we claim that ‖·‖ is a continuous function on Rn (with the natural topology).Indeed, ‖·‖ is continuous with respect to the topology τ generated by ‖·‖ (once again byTheorem 1.3.2.) and τnat is FINER than τ by 1◦.

We conclude that ‖·‖ attains its MINIMAL value m on S, that is,

‖x‖1 = 1⇒ ‖x‖ ≥ m,‖x0‖ = m for some x0 with ‖x0‖1 = 1.

(1)

This value m must be greater than 0, since otherwise x0 = 0 and ‖x0‖1 = 0. It followsfrom (1) that

x0

B1

|| ||. 1

S

Bm

|| ||. ‖x‖1 > 1⇒ ‖x‖ = ‖x‖1︸︷︷︸>1

∥∥∥∥x

‖x‖1

∥∥∥∥︸︷︷︸≥m

> m.

Hence

‖x‖ ≤ m ⇒ ‖x‖1 ≤ 1,

that is,

B‖·‖m ⊂ B‖·‖11 .

Thus, by the same Theorem 1.4.1., ‖·‖ ≻ ‖·‖1. ⊲

1.10 Linear mappings

For a linear mapping l we usually write lx or l · x instead of l(x):

lx ≡ l · x ≡ l(x).

1.10. LINEAR MAPPINGS 17

The set of all linear mappings from a vector space X into a vector space Y is a vectorspace with respect to operations

(l + m)x := lx + mx, (tl)x := t (lx) (t ∈ R),

and we denote this vector space byL(X,Y ).

The vector subspace of all CONTINUOUS linear mappings, in the case where X and Y arenormed spaces, we denote by

L (X,Y ).

Theorem 1.10.1. Let X and Y be normed spaces and let l ∈ L(X,Y ). Then l is continuousiff l is continuous at 0.⊳ “Only if”: obvious.

“If”: Let l be continuous at 0, that is, ‖h‖ → 0⇒ ‖lh‖ → 0. Then for an arbitraryx ∈ X it holds

‖ l(x + h)︸︷︷︸=lx+lh

−lx‖ = ‖lh‖ K‖h‖→0

0,

which means that l is continuous at x . ⊲

Operator normLet X,Y be normed spaces. We define the norm of a mapping l ∈ L(X,Y ) as

‖l‖ := sup‖x‖≤1

‖lx‖ .

Very often one says “operators” for linear mappings, that is why this norm is usuallynamed operator norm. (Below we will see that this is really a norm.)

Example. For any k ∈ R the linear mapping R→ R, x 7→ kx has the norm |k|.Basic Inequality (BI). ∀l ∈ L(X,Y ) ∀x ∈ X

... ‖lx‖ ≤ ‖l‖ ‖x‖ .⊳ If x = 0 then our inequality is trivially true. If x 6= 0 then

‖lx‖ =∥∥∥∥l

(‖x‖ x

‖x‖

)∥∥∥∥ =∥∥∥∥‖x‖ l

x

‖x‖

∥∥∥∥ = ‖x‖∥∥∥∥l

x

‖x‖

∥∥∥∥︸︷︷︸≤

‖x/‖x‖‖=1‖l‖

≤ ‖l‖ ‖x‖ . ⊲

Criteria of continuity. Let l ∈ L(X,Y ). The following conditions are equivalent:

a) l is continuous;b) the image l B1 of the unit ball in X is bounded in Y ;

c) ∃k > 0 ∀x ∈ X... ‖lx‖ ≤ k ‖x‖ (the norm of lx admits an estimation linear in

‖x‖);d) ‖l‖ <∞ (the operator norm is finite).

⊳ (a)⇒(b): Since l is continuous at 0, there exists δ > 0 : l BXδ ⊂ BY

1 . Multiplying byδ−1 we obtain (by linearity of l) l BX

1 ⊂ BYδ−1 , which just means that the image l BX

1 isbounded.


(b)⇒(c): If l BX1 ⊂ BY

k , then (without loss of generality x 6= 0)

‖lx‖wlogx 6=0= ||

∈BYk︷︸︸︷

lx

‖x‖︸︷︷︸∈BX

1

|| ‖x‖ ≤ k ‖x‖ .

(c)⇒(d): If ‖lx‖ ≤ k ‖x‖ for all x , then sup‖x‖≤1 ‖lx‖︸︷︷︸≤k ‖x‖︸︷︷︸

≤1

≤ k, that is, ‖l‖ ≤ k.

(d)⇒(a): If ‖l‖ <∞ then 0 ≤ ‖lx‖ BI≤ ‖l‖ ‖x‖ K‖x‖→0

0, hence ‖lx‖ K‖x‖→0

0, that

is, l is continuous at 0. But then, by Theorem 1.10.1., l is continuous everywhere. ⊲

Remark. ‖l‖ = inf{k > 0 | ∀x ∈ X... ‖lx‖ ≤ k ‖x‖}.

Theorem 1.10.2. The mapping L (X,Y )→ R, l 7→ ‖l‖ is a norm.⊳ That ‖l‖ ≥ 0 is obvious. If ‖l‖ = 0 then ‖lx‖ = 0 for all x with ‖x‖ ≤ 1 and hence,by linearity of l, for all x , which means that l = 0. Further

‖tl‖ = sup‖x‖≤1

‖(tl)x‖ = sup‖x‖≤1

‖t (lx)‖ = |t| sup‖x‖≤1

‖lx‖ = |t| ‖l‖ .

At last

‖l1 + l2‖ = sup‖x‖≤1

‖(l1 + l2)x‖ = sup‖x‖≤1

‖l1x + l2x‖ ≤ sup‖x‖≤1

‖l1x‖︸︷︷︸≤‖l1‖ ‖x‖︸︷︷︸

≤1

+ ‖l2x‖︸︷︷︸≤‖l2‖ ‖x‖︸︷︷︸

≤1

≤ ‖l1‖ + ‖l2‖ . ⊲NB We EVER consider L (X,Y ) as a normed space with this norm.

The case X = Rn

Theorem 1.10.3. Any linear mapping from Rn (with the natural topology) into a normedspace Y is continuous:

L(Rn,Y ) =L (Rn,Y ).

⊳ Let l ∈ L(Rn,Y ). Each element x = (x1, . . . , xn) of Rn can be written as x1e1 + · · · +xnen , where {e1, . . . , en} is the canonical basis, so if we put lei =: ai , then, by linearityof l,

lx = x1a1 + · · · + xnan. (1)

In view of Theorem 1.10.1. it is sufficient to verify that l is continuous at 0, that is, thatx → 0⇒ lx → 0. But x → 0 means that all xi → 0 (since the natural topology coincideswith the PRODUCT topology), whence it follows (by continuity of algebraical operationsin a normed space, see Theorem 1.6.1.) that

x1a1 + · · · + xnan → 0.

Thus, by (1), lx → 0.

1.10. LINEAR MAPPINGS 19

(ANOTHER PROOF:‖lx‖ = ‖l(x1e1 + · · ·)‖ ≤ |x1|‖ le1︸︷︷︸

ai

‖ + · · · ≤ max ‖ai‖︸︷︷︸=:k

(|x1| + · · ·)︸︷︷︸=‖x‖1

≤ k ‖x‖1,

so l is continuous by Criterium (c) of continuity.) ⊲

Evaluation at a pointLet X,Y be vector spaces, and let h be a FIXED element of X . The evaluation at h (ordelta-function at h) is the mapping

evh ≡ δh : L(X,Y )→ Y, l 7→ lh.

This mapping is (obviously) LINEAR.

Theorem 1.10.4. Let X,Y be normed spaces. Then for each h ∈ X the evaluation at h isCONTINUOUS:

evh ∈L (L (X,Y ),Y ).

⊳ ‖evh‖ = sup‖l‖≤1

‖lh︷︸︸︷

evh l ‖︸︷︷︸BI≤ ‖l‖︸︷︷︸≤1

‖h‖

≤ ‖h‖, hence by Criteria (d) of continuity, evh is continuous. ⊲

“Lemma” from Functional AnalysisMappings into R are usually called functionals in the case where the “first” space isinfinite-dimensional. (The name “Functional Analysis” originates from this word.) Thevector space of all linear (resp., continuous linear) functionals on a given vector space(resp., normed space) X we shall denote by X ′ (resp., X∗):

X ′ := L(X,R), X∗ :=L (X,R).

Later we at least two times shall use the following:

Theorem 1.10.5. (“Lemma” from Functional Analysis) Let X be a normed space. Thenfor each vector x ∈ X there exists a functional l ∈ X∗ of the unit norm such that its valueat x is just the norm of x:

‖l‖ = 1, lx = ‖x‖ .⊳ We give the proof for X = Rn only. If x = 0 then we can take as l ANY functional ofthe norm 1. Let x 6= 0. Put (below ‖·‖ denotes ‖·‖2)

e := x

‖x‖ and ly := e · y (y ∈ Rn)

(where the latter point means scalar product). It is clear that ‖e‖ = 1, so

‖l‖ = sup‖y‖≤1

|e · y|︸︷︷︸≤

prop.of scal.prod.

‖e‖︸︷︷︸=1

‖y‖︸︷︷︸≤1

≤ 1, |le| = | e · e︸︷︷︸=1

| = 1.

Since e belongs to the unit ball, over which we take the supremum, we conclude that‖l‖ = 1.

At last

lx = e · x = x

‖x‖ · x =x · x‖x‖ =

‖x‖2‖x‖ = ‖x‖ . ⊲

Chapter 2

First derivative

2.1 Frechet and Gateaux derivatives

The classic definition of the derivative

f ′(x) := limh→0(h 6=0)

f (x + h)− f (x)

h

can be written in the form (below we drop for short “h 6= 0”)

r(h)

hK

h→00,

wherer(h) := f (x + h)− f (x)− f ′(x)h.

So we can reformulate the definition as follows: a function f : R→ R is differentiable ata point x if there exists a number l (= f ′(x)) such that f admits the representation

∀h ∈ X... f (x + h) = f (x)+ lh + r(h),

where r is a mapping R→ R, that satisfies the conditions r(0) = 0 and

r(h)

hK

h→00. (1)

Such a mapping we call small.A key point to generalize this definition is the idea that l can be considered as a

(continuous) LINEAR MAPPING R→ R:

l : R→ R, h 7→ lh

(we identify a number l with the linear function with the (slope) coefficient l). This leadsto the following definition:

A mapping f : X → Y between normed spaces X and Y is differentiable (in a givensense) at a point x ∈ X (notation: f ∈ Dif(x)) if there exists a continuous linear mappingl : X → Y such that f admits the representation

∀h ∈ X... f (x + h) = f (x)+ lh + r(h), (2)

21

22 CHAPTER 2. FIRST DERIVATIVE

where r is a mapping X → Y , that is SMALL (in this sense). There are two basic kinds ofsmallness for mappings between normed spaces:

A mapping r : X → Y is Frechet-small (F-small) if r(0) = 0 and

‖r(h)‖‖h‖ K

‖h‖→00; (3)

r is Gatteaux-small (G-small) if r(0) = 0 and

∀h ∈ X...

r(th)

tK

t→00 (t ∈ R). (4)

NB For X = Y = R both (3) and (4) are equivalent to (1) (verify!).Accordingly we speak about F-differentiability and G-differentiability. Very often we

drop the symbol “F”, so “differentiability” means ever “Frechet differentiability” and“ f ∈ Dif(x)” means “ f ∈ F- Dif(x)”.Remark. Both our differentiabilities do not depend on the choice of EQUIVALENT normsin X and Y . (Verify!)

The mapping l in the representation (2) is called the derivative of f at x and it isdenoted by f ′(x).

l 1.l

1 R

f(t) f (t)

Y

YExamples.

1. If X = R (“time”) then we can identify a linear mappingl : R → Y with the element l · 1 of Y and it is easy to see thatF-differentiability is equivalent to G-differentiability, and

f ′(t) · 1 = lim1t→0

f (t +1t)− f (t)

1t(the limit in Y ).

Below we shall denote the last limit by f (t) and call f (t)(which is a vector in Y ) the usual derivative of f at t (it is thevelocity of a point that moves in Y by the “law” f ).

2. Any CONSTANT function is differentiable everywhere, with zero derivative.⊳ If f ≡ c, then f (x + h) = f (x)+ 0.h+ 0, and 0 is small (in any reasonable sense!) ⊲

3. Any continuous linear mapping l is differentiable, and its derivative at each point isequal to this mapping itself:

l ′ ≡ l (that is, ∀x ∈ X... l ′(x) = l).

⊳ l(x + h) = lx + lh + 0. ⊲

4. The function f : Rn → R, f (x) = ‖x‖22 = x21 + · · · + x2

n (x = (x1, . . . , xn)) isdifferentiable everywhere, and

f ′(x) · h = 2x · h,

where the point to the right means scalar product. (Prove!)

1

0

5. Let the function f : R2 → R is equal to 1 on the right branch of theparabola {(t, t2)} WITHOUT the origin (t > 0) and is equal to 0 at all restpoints of the plane. Then f is G-differentiable at 0(= (0, 0)), with f ′(0) = 0(∈L (R2,R)), but is NOT F-differentiable. (Verify!)

2.2. COMPUTATION RULE AND DIRECTIONAL DIFFERENTIABILITY. 23

Theorem 2.1.1. F-differentiability implies G-differentiability, with the same derivative.⊳ It is sufficient to show that if r : X → Y is F-small, then r is G-small. Let r(0) = 0and ‖r(h)‖

‖h‖ K‖h‖→0

0. (5)

Then for each fixed h ∈ X \ 0 (for short we write X \ 0 instead of X \ {0})∥∥∥∥∥

r(t h)

t

∥∥∥∥∥ =

∥∥∥r(t h)∥∥∥

∥∥∥t h∥∥∥‖h‖ K

t→00,

since ‖t h‖ = |t|∥∥∥h∥∥∥ K

t→00 and hence, by (5),

∥∥∥r(t h)∥∥∥/∥∥∥t h

∥∥∥ Kt→0

0. But this just means

thatr(t h)

tK

t→00.

Thus, r is G-small. ⊲

Theorem 2.1.2. G-derivative is defined uniquely.⊳ We need to verify that if for a given x ∈ X

∀h ∈ X... f (x)+ l1h + r1(h) = f (x)+ l2h + r2(h), (6)

where l1, l2 ∈ L (X,Y ) and r1, r2 are G-small, then l1 = l2, that is, for each h ∈ X itholds l1h = l2h. But indeed (for t 6= 0)

l1h − l2htrick= l1(th)− l2(th)

t(6)= r2(th)− r1(th)

t= r2(th)

t− r1(th)

tK

t→00,

whence it follows that l1h − l2h = 0. ⊲

Corollary 2.1.3. F-derivative is defined uniquely.⊳ It follows from Theorems 2.1.1. and 2.1.2. ⊲

Theorem 2.1.4. If f is differentiable at x, then f is continuous at x.⊳ By the condition, f (x+h) = f (x)+ lh+r(h), where l ∈L (X,Y ) and r is small. Weneed to verify that if h → 0 then lh + r(h)→ 0. Since l is a continuous linear mapping,lh → 0 as h → 0. Now, for h 6= 0

‖r(h)‖ = ‖r(h)‖‖h‖︸︷︷︸→0

‖h‖ K‖h‖→0

0

(for h = 0 we have r(0) = 0), which just means that r(h)→ 0 if h → 0. ⊲

NB G-differentiability does NOT imply continuity (see Example 5).

2.2 Computation Rule and directional differentiability.

For practical computation of derivatives it is convenient to use the followingComputation Rule. Let f : X → Y be G-differentiable at a given point x , and let h ∈ Xbe given. Put for t ∈ R

ϕ(t) := f (x + th) ,


so that ϕ : R→ Y . Then ϕ is differentiable at 0, and

f ′(x)h = ϕ(0) = ∂

∂ t

∣∣∣∣t=0

f (x + th) .

(As to ϕ(0) see Example 1).)

f

x x+h

ϕ(0)=f(x) ϕ(1)=f(x+1h)

Y

t0 1

ϕ

⊳ Let f ′(x) = l. We have ϕ(t) = f (x + th) = f (x)+ l(th)+ r(th), so (for t 6= 0)

ϕ(t)− ϕ(0)t

=ϕ(0)= f (x)

l(th) + r(th)

t=

l∈Linlh + r(th)

t︸︷︷︸→0

→ lh

as t → 0, which does mean that ϕ(0) = lh. ⊲

The Computation Rule suggests the following definition. We say that a mappingf : X → Y is differentiable at a point x in a direction h if the function

ϕ : R→ Y, t 7→ f (x + th)

is differentiable at 0. In such a case we call the vector ϕ(0) ∈ Y the differential of f at xby the increment h, and we denote this differential by Dh f (x). Thus

Dh f (x) := ϕ(0) = ∂

∂ t

∣∣∣∣0

f (x + th) = limt→0

f (x + th) − f (x)

t. (1)

Corollary 2.2.1. If f : X → Y is G-differentiable at x then f is differentiable at x ineach direction, and

∀h ∈ X... Dh f (x) = f ′(x)h .

Vice versa, if f is differentiable at x in each direction and the mapping

l : h 7→ Dh f (x), X → Y

is LINEAR and CONTINUOUS, then f is G-differentiable at x, and f ′(x) = l.Remark. The mapping l in Corollary 2.2.1. is ever HOMOGENUOUS. More precisely, if fis differentiable at x in a direction h, then for any real number c it is differentiable at x inthe direction ch, and

Dch f (x) = cDh f (x) .

(This follows at once from the last expression for Dh f (x) in (1).) But this mapping l canbe non-linear (that is, non-additive), as the following example shows.

2.3. RULES OF DIFFERENTIATION 25

Example. The function f : R2 → R given (in the polar coordi-nates) by the formula

f (, θ) = sin 3θ,

the graph of which is shown on the picture (from: M. Krupka,Matematicka Analyza III, Opava 1999), is differentiable at EACH

point in EACH direction, but is NOT G-differentiable at the origin.(Verify!)

The next lemma is an extension of (1).

Lemma 2.2.2. (on f (x + th)). Let f : X → Y . Put for given x, h ∈ X

ϕ(t) := f (x + th) (t ∈ R).

Thenϕ(t) = Dh f (x + th) .

(If one side is defined then the other one is also defined, and they are equal.)

⊳ Dh f (x + th)(1)= lim

τ→0

f ((x + th)+ τh)− f (x + th)

τ

= limτ→0

ϕ(t + τ )− ϕ(t)τ

= ϕ(t). ⊲

2.3 Rules of differentiation

First of all, differentiation is a linear operation:Linearity of differentiation.(a) If f ∈ Dif(x) then for each c ∈ R we have also c f ∈Dif(x), and

(c f )′(x) = c f ′(x) .

(b) If f1, f2 ∈ Dif(x), then f1 + f2 ∈ Dif(x), and

( f1 + f2)′(x) = f ′1(x)+ f ′2(x) .

⊳ (a) We have (c f )(x + h) = c( f (x + h)) = c( f (x) + f ′(x)h + r(h)) = (c f )(x) +(c f ′(x))h + (cr)(h), so we need to verify that cr is small. But indeed (for h 6= 0)

‖(cr)(h)‖‖h‖ = ‖c(r(h))‖‖h‖ = |c| ‖r(h)‖‖h‖︸︷︷︸

r is smallK0

K‖h‖→0

0.

(b) We have analogously (in obvious notations)

( f1 + f2)(x + h) = ( f1 + f2)(x)+ ( f ′1(x)+ f ′2(x))h + (r1 + r2)(h),

so we just need to verify that r1 + r2 is small. But indeed


‖(r1 + r2)(h)‖‖h‖ = ‖r1(h)+ r2(h)‖

‖h‖ ≤ ‖r1(h)‖ + ‖r2(h)‖‖h‖

= ‖r1(h)‖‖h‖ +

‖r2(h)‖‖h‖

r1,r2 are smallK

‖h‖→00. ⊲

Product Rule. Let X,Y1, . . . ,Ym be normed spaces, and let fi : X → Yi , i = 1, . . . ,m.We denote by ( f1, . . . , fm ) the product mapping X → Y1 × · · · × Ym defined by theformula

Y1f1ր

X...

fmց

Ym

X( f1,..., fm)

K Y1 × . . .× Ym

( f1, . . . , fm)(x) := ( f1(x), . . . , fm(x)).

The mapping ( f1, . . . , fm) is differentiable (resp., G-dif-ferentiable, differentiable in a direction h) at x ∈ X iff eachmapping fi is differentiable (resp., G-differentiable, differen-tiable in h) at x, and

( f1, . . . , fm)′(x) = ( f ′1(x), . . . , f ′m(x))

Dh( f1, . . . , fm )(x) = (Dh f1(x), . . . ,Dh fm(x))(component-wisedifferentiation).

⊳ Consider, e.g., the case of F-differentiation. We have

( f1, . . . , fm)(x + h) = ( f1(x + h), . . . , fm (x + h))= ( f1(x)+ f ′1(x)h + r1(h), . . . , fm(x)+ f ′m(x)h + rm(h))= ( f1(x), . . . , fm(x))+ ( f ′1(x)h, . . . , f ′m (x)h)+ (r1(h), . . . , rm(h))= ( f1, . . . , fm )(x)+ ( f ′1(x), . . . , f ′m(x))h + (r1, . . . , rm)(h).

Now, ( f ′1(x), . . . , f ′m(x)) ∈ L (X,Y1 × · · · × Ym) iff each fi ∈ L (X,Yi ) (by thedefinitions of product vector space and product topology), and (r1, . . . , rm) is small iffeach ri is small. Indeed

(r1, . . . , rm)(h)

‖h‖ =(

r1(h)

‖h‖ , . . . ,rm(h)

‖h‖

)K

‖h‖→00⇔ ∀i ... ri (h)

‖h‖ K‖h‖→0

0,

since convergence in a product space is just convergence of each component. ⊲

Chain Rule. Let f : X → Y be differentiable at a point x ∈ X, and let g : Y → Z be

XfK Y

gK Z

x 7→ y

Xf ′(x)

K Yg′(x)

K Z

differentiable at the point y := f (x). Then the composition g ◦ fis differentiable at x, and the derivative of g ◦ f is equal to thecomposition of derivatives of f and g:

(g ◦ f )′(x) = g′(y) ◦ f ′(x) .

⊳ 1◦ Put f ′(x) =: l, g′(y) =: m, g(y) =: z. We have, by the conditions,

∀1x ∈ X... f (x +1x) = y + l 1x + r f (1x), (1)

∀1y ∈ Y... g(y +1y) = z + m1y + rg(1y), (2)

where ∥∥r f (1x)∥∥

‖1x‖ K‖1x‖→0

0, (3)∥∥rg(1y)

∥∥‖1y‖ K

‖1y‖→00. (4)


We need to verify that

∀1x ∈ X... (g ◦ f )(x +1x) = z + (m ◦ l)1x + r(1x), (5)

where ‖r(1x)‖‖1x‖ K

‖1x‖→00. (6)

2◦ Butr(1x) = mr f (1x)+ rg(1y), (7)

where1y := l 1x + r f (1x). (8)

Indeed

(g ◦ f )(x +1x) = g( f (x +1x))(1)= g(y + l 1x + r f (1x)︸︷︷︸

(7)=1y

)

(2)= z + m(l 1x + r f (1x))+ rg(1y)(7)= z + (m ◦ l)1x + r(1x).

3◦ Now,

‖r(1x)‖‖1x‖

(7)=∥∥mr f (1x)+ rg(1y)

∥∥‖1x‖ ≤

∥∥mr f (1x)∥∥+

∥∥rg(1y)∥∥

‖1x‖BI≤ ‖m‖

∥∥r f (1x)∥∥

‖1x‖︸︷︷︸(3)

K‖x‖→0

0

+∥∥rg(1y)

∥∥‖1y‖

trick

trick‖1y‖‖1x‖ .

So all will be proved if we show that(a) rg(1y)/‖1y‖ → 0 as ‖x‖ → 0;(b) ‖1y‖/‖1x‖ is bounded for sufficiently small ‖1x‖.

4◦ Proof of (a): r f is equal to 0 at 0 and is continuous (since f and l are). So ‖1y‖ → 0if ‖1x‖ → 0, and (a) is true by (4).5◦ Proof of (b): we have

‖1y‖‖1x‖

(8)=∥∥l 1x + r f (1x)

∥∥‖1x‖ ≤ ‖l 1x‖ +

∥∥r f (1x)∥∥

‖1x‖BI≤ ‖l‖ ‖1x‖ +

∥∥r f (1x)∥∥

‖1x‖ = ‖l‖ +∥∥r f (1x)

∥∥‖1x‖︸︷︷︸(3)

K‖1‖x→0

0

≤ c

for some c > 0 and sufficiently small ‖1x‖. ⊲

Important special cases.

1) If f = l ∈L (X,Y ) then (g ◦ l)′(x) = g′(lx) ◦ l .

2) If g = l ∈ L (Y, Z) then (l ◦ f )′(x) = l ◦ f ′(x) (we can “transfer” l◦ throughthe brackets).

3) If X = R, then (g ◦ f )(t) = g′( f (t)) · f (t) (recall that f (t) is an element of Y ).

4) In particular, if X = R and f = l then (g ◦ l)′(t) = g′(lt) · (l · 1) , and


5) if X = R and g = l then (l ◦ f )(t) = l · f (t) .

⊳ All this follows from the facts that l ′ ≡ l and that f (t) = f ′(t) · 1. ⊲

Below we refer to Special Cases 1), 2), 4), 5) as to l-Rule.NB For G-differentiability Chain Rule is NOT valid, as the following example shows.

g fo 1

Example. Let f : R→ R2, t 7→ (t, t2), and let g be the function from5) in (2.1) (where we used the letter f for it). Then f is G- (and evenF-)differentiable at 0, g is G-differentiable at f (0) = (0, 0), but g ◦ fis NOT G-differentiable at 0.

Lemma 2.3.1. (on evaluation). Let X,Y, Z be normed spaces, and let a mapping f : X →L (Y, Z) be differentiable at a point x . Let k be a fixed vector in Y , and let g : X → Zbe defined by the formula

g(x) := f (x) · k (the VALUE of f (x) at k).

Then g is differentiable at x, and

∀h ∈ X... g′(x)h = ( f ′(x)h)k.

⊳ Obviously, g = evk ◦ f (recall that evk : l 7→ l · k, see Chapter 1), hence, by l-Rule(evk ∈L (L (Y, Z), Z)),

g′(x)h = (evk ◦ f ′(x))h = evk( f ′(x)h) = ( f ′(x)h)k. ⊲

2.4 Partial derivatives

Here we consider two related things: differentiation in a (vector) subspace and partialdifferentiation.

Differentiation in a subspaceLet f : X → Y be a mapping between normed spaces and let X1 be a vector subspace

in X (the notation X1 ⋐ X). We say that f is F- (resp., G-)differentiable at a point x ∈ Xin the subspace X1 if f admits in x + X1 the representation

∀h1 ∈ X1... f (x + h1) = f (x)+ l1h1 + r1(h1),

where l1 ∈L (X1,Y ) and r1 : X1→ Y is F- (resp., G-) small. In such a case we write

f ∈ Dif X1(x) (resp., f ∈ G- Dif X1(x))

and we call l1 the derivative of f at x in the subspace X1:

l1 =: f ′X1(x).

Example. A mapping f : X → Y is differentiable at a point x in a (non-zero) directionh iff f is differentiable at x in the one-dimensional subspace Rh (= lin{h} ≡ span{h}),generated by h, and in such a case

f ′Rh(x) · th = tDh f (x) (t ∈ R).

2.4. PARTIAL DERIVATIVES 29

(Verify!)

Theorem 2.4.1. If f : X → Y is F- (resp., G-)differentiable at x, then f is F- (resp.,G-)differentiable at x in each subspace X1 ⋐ X, and the derivative in X1 is just therestriction of the “total” derivative:

f ′X1(x) = f ′(x)|X1 .

⊳ This follows at once from the obvious facts, that the restriction of a continuous linearmapping onto a vector subspace is also a continuous linear mapping, and the restrictionof a small mapping is also small. ⊲

Partial differentiationLet X1, . . . , Xn and Y be normed spaces. We say that a mapping f : X1×. . .×Xn →

Y is F- (resp., G-)differentiable at a point x = (x1, . . . , xn) in the i -th coordinate if themapping

f (x1, . . . , xi−1, ·, xi+1, . . . , xn) : X i → Y, xi 7→ f (x1, . . . , xi−1, xi , xi+1, . . . , xn)

(that is, the mapping with all other coordinates FIXED) is F- (resp., G-)differentiable atthe point xi . We denote the corresponding derivative by

∂ f (x)

∂X i(∈L (X i ,Y ))

and call it the partial derivative in X i at the point x .

Theorem 2.4.2. A mapping f : X1 × · · · × Xn → Y is differentiable (F- or G-) at x inthe i -th coordinates iff f is differentiable (in the same sense) at x in the subspace

0× · · · × 0× X i × 0× · · · × 0,

and

∀hi ∈ X i...

∂ f (x)

∂X i· hi = f ′0×···×X i×···×0(x) · (0, . . . , 0, hi , 0, . . . , 0).

⊳ This follows immediately from the definitions. ⊲

Theorem 2.4.3. (on total and partial derivatives). If a mapping f : X1 × · · · × Xn → Yis G-differentiable at a point x = (x1, . . . , xn) then f is G-differentiable at x in eachcoordinate, and

∀h = (h1, . . . , hn) ∈ X1 × · · · × Xn... f ′(x) · h =

n∑

i=1

∂ f (x)

∂X i· hi ,

or, more short,

f ′(x) = ∂ f (x)

∂X1⊕ · · · ⊕ ∂ f (x)

∂Xn≡

n⊕

i=1

∂ f (x)

∂X i.

Here l1 ⊕ · · · ⊕ ln , for li ∈ L (X i ,Y ), denotes the direct sum of the mappings li ,defined by the formula

l1 ⊕ · · · ⊕ ln : X1 × · · · × Xn → Y, (h1, . . . , hn) 7→ l1 · h1 + · · · + ln · hn .


⊳ f ′(x) · h = f ′(x) · (h1, . . . , hn) = f ′(x) ·n∑

i=1

(0, . . . , 0, hi , 0, . . . , 0)

Theorem 2.4.1.=n∑

i=1

f ′(x)(0, . . . , 0, hi , 0, . . . , 0)

·n∑

i=1

f ′0×...×X i×...×0(x) · (0, . . . , 0, hi , 0, . . . , 0)

Theorem 2.4.2.=n∑

i=1

∂ f (x)

∂X i· hi . ⊲

2.5 Finite-dimensional case

We ever denote by e1, . . . , en the standard basis in Rn:

ei := (0 . . . , 0, 1i-th place

, 0, . . . 0) ∈ Rn .

For a mapping f : Rn = R × . . .× R→ Y the partial derivative in the i -th coordinateapplied to the “vector” 1 ∈ R (that is, the “usual” partial derivative in the i-th coordinate)is traditionally denoted by

∂ f

∂xi(∈L (R,Y ) ≈ Y ).

By Theorem 2.4.2. (with hi = 1), and by the Example in 2.4 (with h = ei and t = 1),

∂ f (x)

∂xi= Dei f (x) . (1)

(Emphasize that ∂ f (x)/∂xi is an element of Y .)

Jacobi matrix

Theorem 2.5.1. (on representation). Let a mapping f = ( f1, . . . , fm ) : Rn → Rn be G-differentiable at a point x ∈ Rn . Then f ′(x) is represented as a linear mapping Rn → Rm

by the matrix of partials derivatives

J f (x) :=

∂ f1(x)∂x1

· · · ∂ f1(x)∂xn

.... . .

...∂ fm(x)∂x1

· · · ∂ fm(x)∂xn

.

This matrix is called Jacobi matrix of f at x .⊳ By linear algebra, we need to verify that the i -th column of the matrix represents thevector f ′(x) · ei . But indeed

f ′(x) · eiProd. Rule= ( f ′1(x), . . . , f ′m(x)) · ei = ( f ′1(x) · ei , . . . , f ′m(x) · ei ),

and

f ′j (x) · ei = Dei f j (x)(1)= ∂ f j (x)

∂xi. ⊲

2.5. FINITE-DIMENSIONAL CASE 31

Corollary 2.5.2. In conditions of Chain Rule, for mappings between finite-dimensionalspaces, Jacobi matrix of the composition is equal to the matrix product of Jacobi matricesof the composed mappings:

Jg◦ f (x) = Jg( f (x))J f (x).

⊳ This follows from Chain Rule and the fact that the matrix of a composition of two linearmappings is equal to the product of the matrices of these mappings. ⊲

Example. How to compute the derivative of the function f (x) = x x (x > 0)? Represent fas a composition: f = g ◦△, where△ : R→ R2, t 7→ (t, t), g : R2 → R, (x, y) 7→ x y .By l-Rule,

f (t) = g′(t, t) · (1, 1)Theorem 2.5.1.=

(∂g(t, t)

∂x,∂g(t, t)

∂y

)(11

)= ∂g(t, t)

∂x+ ∂g(t, t)

∂y

= yx y−1∣∣∣x=y=t

+ (ln x)x y∣∣x=y=t = (ln t + 1)t t .

GradientIn the special case of SCALAR functions f : Rn → R Jacobi matrix is the row

(∂ f (x)

∂x1· · · ∂ f (x)

∂xn

).

The corresponding vector in Rn is called the gradient of f at x and is denoted by grad f (x):

grad f (x) :=(∂ f (x)

∂x1, . . . ,

∂ f (x)

∂xn

)(∈ Rn).

In this situation Theorem 2.5.1. (Theorem on representation) says:

f ′(x) · h = grad f (x) · h , (2)

where the point to the right means scalar product. Indeed

(∂ f (x)

∂x1· · · ∂ f (x)

∂xn

)

h1...

hn

=

n∑

i=1

∂ f (x)

∂xihi .

For any UNIT vector Eν ∈ Rn (‖Eν‖2 = 1), the differential of a mapping f : Rn → R ata point x by the increment Eν is called the derivative at x in the direction Eν and is denotedtraditionally by ∂ f (x)/∂ν:

∂ f (x)

∂ν:= DEν f (x).

Theorem 2.5.3. If a function f : Rn → R is G-differentiable at a point x , then for anyunit vector Eν ∈ Rn it holds

∂ f (x)

∂ν= grad f (x) · Eν

(where the point means the scalar product).


⊳ ∂ f (x)/∂ν = DEν f (x) = f ′(x) · Eν (2)= grad f (x) · Eν. ⊲

x

grad f(x)

Corollary 2.5.4. The gradient vector of f at x yields the directionof the greatest growth of f at x and is orthogonal to the level lineof f passing through x.⊳ The scalar product grad f (x) · Eν is maximal for the unit vectorEν that has the SAME direction as grad f (x) and is equal to 0 for Eνorthogonal to the gradient. ⊲

2.6 Mean Value Theorem

In classic differential calculus, the following result plays an important role:

Theorem 2.6.1. (Lagrange). If a function f : R→ R is continuous in the closed interval[0, 1] and is differentiable in the open interval (0, 1), then there exists θ ∈ (0, 1) such that

f (1)− f (0) = f (θ).

This result is NOT true for functions with vector values, as the following exampleshows.

t=0t=1

Example. Let f : R→ R2, t 7→ (cos 2π t, sin 2π t). Then f (1)−f (0) = (0, 0), but

f (t)Prod. Rule= (−2π sin 2π t, 2π cos 2π t),

hence∥∥ f (t)

∥∥2 = 2π , which is never zero.

But the following ESTIMATE of the increment is true:

Theorem 2.6.2. (Mean Value Theorem, (MVT)). Let a function ϕ : R→ Y (where Y isa normed space) be continuous on [0, 1] and differentiable in (0, 1). Then

‖ϕ(1)− ϕ(0)‖ ≤ sup0<t<1

‖ϕ(t)‖ .

⊳ 1◦ Put y := ϕ(1)− ϕ(0). By Theorem 1.10.5. (Lemma from FA (see Chapter 1)) thereexists l ∈L (Y,R) such that

‖l‖ = 1, ly = ‖y‖ . (1)

2◦ Consider the composition

Rϕ→ Y

l→ R.

It is differentiable in (0, 1) by Chain Rule.

3◦ We have ‖y‖ (1)= ly = l(ϕ(1)− ϕ(0)) = (l ◦ ϕ)(1)− (l ◦ ϕ)(0). By Theorem 2.6.1. forsome θ ∈ (0, 1) it holds

(l ◦ ϕ)(1)− (l ◦ ϕ)(0) = (l ◦ ϕ)(θ) l-Rule= l · ϕ(θ) BI≤ ‖l‖︸︷︷︸=1

‖ϕ(θ)‖ ≤ sup0<t<1

‖ϕ(t)‖ . ⊲

2.7. CONTINUOUS DIFFERENTIABILITY 33

ϕ(1)

ϕ(0) sup||ϕ(t)||0<t<1

.

CAR-INTERPRETATION. If Y = R2 and we consider t ∈ Ras time, then ϕ describes a motion of a “car” in the plane, andϕ(t) is the velocity of this car. The inequality in Theorem 2.6.2.(MVT) means that our car in one hour will be INSIDE the circlewith the center at the original point, the radius of which is equalto the maximal value of the velocity of the car during this hour.

ϕ(t)

ϕ(1)-ϕ(0)

ϕ(0)

sup||ϕ(t)||0<t<1

ϕ(1)

.

.

Remark. In fact the following much more strong result is true:By the conditions of MVT, the increment ϕ(1) − ϕ(0) lies inthe closed convex hull of the set {ϕ(t) | 0 < t < 1} (shadowedon the picture).

Corollary 2.6.3. Let X,Y be normed spaces, let x, h ∈ X andlet a mapping f : X → Y has the following properties:a) the restiction of f onto the closed interval [x, x + h] (:={x + th | t ∈ [0, 1]}) is continuous andb) f is differentiable in the direction h in the open interval(x, x + h) (:= {x + th | t ∈ (0, 1)}). Then

‖ f (x + h)− f (x)‖ ≤ supy∈(x,x+h)

‖Dh f (y)‖ .

⊳ Put ϕ(t) := f (x + th), t ∈ R. By Lemma 2.2.2. (on f (x + th)), it holds ϕ(t) =Dh f (x + th). So our assertion follows from MVT (Theorem 2.6.2.). ⊲

Corollary 2.6.4. In the situation of Corollary 2.6.3., let f has the following properties:a) the restiction of f onto the interval [x, x + h] is comtinuous andb) f is G-differentiable in (x, x + h). Then

‖ f (x + h)− f (x)‖ ≤ ‖h‖ supy∈(x,x+h)

∥∥ f ′(y)∥∥ .

⊳ This follows from Corollary 2.6.3. and the fact that

‖Dh f (y)‖ =∥∥ f ′(y) · h

∥∥ BI≤∥∥ f ′(y)

∥∥ ‖h‖ . ⊲

2.7 Continuous differentiability

Let X,Y be normed spaces, let x ∈ X , and let f : X → Y be G-differentiable in an openneighbourhood U of x . We say that f is continuously G-differentiable at x and we write

f ∈ C1G(x),

if the derivative mapping

f ′ : U →L (X,Y ), y 7→ f ′(y)

is continuous at x . Thus

f ∈ C1G(x) :⇔

∥∥ f ′(x + h)− f ′(x)∥∥ K‖h‖→0

0. (1)

Theorem 2.7.1. (on Continuous Derivative). If f : X → Y is continuously G-differenti-able at a point x ∈ X, then f is F-differentiable at x.


⊳ 1◦ We have

f (x + h) = f (x)+ f ′(x)h + r(h), f ′(x) ∈L (X,Y ), r ∈ G-small.

Put for t ∈ Rψ(t) := r(th) = f (x + th)− f (x)− t f ′(x)h. (2)

Thenr(h) = ψ(1)− ψ(0). (3)

2◦We want to apply MVT, so compute the derivative ofψ . By Lemma 2.2.2. (on f (x+th)),

ψ(t) = Dh f (x + th)︸︷︷︸= f ′(x+th)h

− f ′(x)h = ( f ′(x + th)− f ′(x))h. (4)

3◦ Now,

‖r(h)‖‖h‖

(3)= ψ(1)− ψ(0)‖h‖

MVT≤ 1

‖h‖ sup0<t<1

∥∥ψ(t)∥∥

(4)= 1

‖h‖ sup0<t<1

∥∥( f ′(x + th)− f ′(x))h∥∥

︸︷︷︸BI≤‖( f ′(x+th)− f ′(x))‖‖h‖

≤ sup0<t<1

∥∥( f ′(x + th)− f ′(x))∥∥ (1)

K‖h‖→0

0,

since ‖th‖ = |t| ‖h‖ K‖h‖→0

0 uniformly in 0 < t < 1. Thus r is F-small, so f is

F-differentiable at x . ⊲

Class C1

Let X,Y be normed spaces and let U be an open set in X . We say that f : X → Y isof class C1 in U an we write

f ∈ C1(U)

if f is differentiable at each point of U (that is, in U ), and the derivative mapping

f ′ : U →L (X,Y ), x 7→ f ′(x)

is continuous.

Theorem 2.7.2. (C1-Theorem on Continuous Derivative). Let f : X → Y be continuouslyG-differentiable at each point of an open set U in X. Then f is of class C1 in U.⊳ This is an immediate corollary of Theorem 2.7.1. (on continuous derivative) ⊲

2.8 Continuous partial derivatives

At first we prove one result on continuous differentiability in n fixed directions.

Continuous differentials

Lemma 2.8.1. Let X,Y be normed spaces, and let h1, . . . , hn be fixed vectors in X. Let amapping f : X → Y be differentiable in the directions h1, . . . , hn in some neighbourhoodU of a point x , and let all the mappings

Dhi f : x 7→ Dhi f (x), U → Y (i = 1, . . . , n)

2.8. CONTINUOUS PARTIAL DERIVATIVES 35

be continuous at x. Then f is G-differentiable at x in the vector subspace H in X,generated (spanned) by the vectors hi (H = lin{h1, . . . , hn} ≡ span{h1, . . . , hn}).⊳ 1◦ If f IS G-differentiable at x in H , then

f ′H (x) · hi = Dhi f (x) (i = 1, . . . , n),

so we must have ∀c1, . . . , cn ∈ R

f ′(x)·(c1h1+. . .+cnhn) = c1 f ′(x)h1+. . .+cn f ′(x)hn = c1Dh1 f (x)+. . .+cnDhn f (x).(1)

(Since H is finite-dimensional, the so defined linear mapping f ′H (x) is CONTINUOUS.) Wehave to verify that for this f ′H (x) it holds

∀h ∈ H... f (x + h) = f (x)+ f ′(x)h + r(h),

r(th)

tK

t→00,

that is, that

∀h ∈ H...

f (x + th)− f (x)− t f ′H (x)ht

Kt→0

0.

This means by (1) that we need to verify that ∀c1, . . . , cn ∈ R...

f (x + t (c1h1 + . . .+ cnhn))− f (x)− t (c1Dh1 f (x)+ . . .+ cnDhn f (x))

tK

t→00.

By homogeneity of Dh f (x) in h, it holds ci Dhi f (x) = Dci hi f (x) so without loss ofgenerality we can assume that c1 = . . . = cn = 1 (take ci hi as NEW hi ). Further, byinduction argument, it is sufficient to consider the case n = 2. Thus we need to verify that

f (x + t (h1 + h2))− f (x)

t− (Dh1 f (x)+ Dh2 f (x)) K

t→00. (2)

2◦ Adding and subtracting f (x + th1) in the numerator, we can write the left-hand sideof (2) as the sum 1 + 2 , where

1 = f (x + th1)− f (x)t − Dh1 f (x),

2 = f (x + th1 + th2)− f (x + th1)− tDh2 f (x)t .

So it is sufficient to verify that 1 → 0 and 2 → 0 as t → 0.

3◦ 1 → 0 as t → 0 by the definition of Dh .x+th +th

x+th +θh1 2

1 2

x+th1x

4◦ Putting (see the picture)

(3)ϕ(θ) := f (x + th1 + θ th2)− θ tDh2 f (x) (θ ∈ R),

we can write 2 in the form

2 = ϕ(1)− ϕ(0)t

. (4)

5◦ By Lemma 2.2.2. (on f (x + th)), we obtain from (3)

ϕ(θ) = Dth2 f (x + th1 + θ th2)− tDh2 f (x)

homog.of Dh= t (Dh2 f (x + th1 + θ th2)− Dh2 f (x)). (5)


6◦ At last, ∥∥∥ 2∥∥∥ (4)= 1

|t| ‖ϕ(1)− ϕ(0)‖MVT≤ 1

|t| sup0<θ<1

‖ϕ(θ)‖

(5)= sup0<θ<1

∥∥Dh2 f (x + th1 + θ th2)− Dh2 f (x)∥∥ K

t→00,

since ‖th1 + θ th2‖ ≤ |t|(‖h1‖ + θ ‖h2‖) ≤0<θ<1

|t|(‖h1‖ + ‖h2‖) Kt→0

0 uniformly in θ

and Dh2 f is continuous at x . Thus, 2 → 0 as t → 0. ⊲

Continuous partial derivatives

Theorem 2.8.2. Let X1, . . . , Xn and Y be normed spaces and let a mapping f : X1 ×. . .× Xn → Y have all the partial derivatives ∂ f/∂X i in an open set U ⊂ X1× . . .× Xn

and these partial derivatives be continuous in U. Then f is of class C1 in U.⊳ For simplicity we consider only the case X1 = · · · = Xn = Y = R (that is, f : Rn →R), which is the most important for us. In this case continuity of the partial derivatives∂ f/∂X i means just continuity of the partial derivatives ∂ f/∂xi = Dei f . By Lemma onContinuous Differentials we conclude that f is G-differentiable in U . By Theorem onRepresentation,

f ′(x) =(∂ f (x)

∂x1, . . . ,

∂ f (x)

∂xn

).

Since each component of this vector continuously depends on x , we conclude, that f ′

continuously depends on x . Hence by Theorem 2.7.2. (C1-Theorem on Continuous Deri-vative), f is of class C1 in U . ⊲

Corollary 2.8.3. Let f = ( f1, . . . , fm) : Rn → Rm , and let all the partial derivatives∂ f j/∂xi (i = 1, . . . , n; j = 1, . . . ,m) be continuous in U ⊂ Rn . Then f is of class C1

in U.⊳ By the Product Rule, ∂ f/∂xi = (∂ f1/∂xi , . . . , ∂ fm/∂xi), so ∂ f/∂xi are continuous ifall ∂ f j/∂xi are. ⊲

Chapter 3

Inverse Function Theorem

3.1 Lipschitz functions

Let X,Y be normed spaces, and let A ⊂ X . We say that a mapping f : X → Y is Lipschitzon A with a constant k > 0, and we write

f ∈ LipA k

if

∀x1, x2 ∈ A... ‖ f (x1)− f (x2)‖ ≤ k ‖x1 − x2‖ .

We say that f is Lipschitz on A and we write f ∈ LipA if f is Lipschitz with some constantk. If A = X or if it is clear what A we mind, we omit A and write simply f ∈ Lip.

Examples.

1. | · | : R→ R, x 7→ |x |, is Lipschitz with the constant 1.

2. For any normed space X , the norm ‖·‖ is Lipschitz with the constant 1.

3. If f ∈L (X,Y ), then f ∈ Lip ‖ f ‖.⊳ ‖ f x1 − f x2‖ = ‖ f (x1 − x2)‖ ≤ ‖ f ‖ ‖x1 − x2‖. ⊲

x

x

4. The function x 7→ √|x |,R→ R, is NOT Lipschitz.

Theorem 3.1.1. If f is Lipschitz on an open set U then f is continuousin U.⊳ ‖ f (x + h)− f (x)‖ ≤ k ‖h‖ K

‖h‖→00. ⊲

x

x

1

x2

Theorem 3.1.2. If f ∈ C1G(x), then (for any ε > 0), f is Lipschitz in

some neighbourhood of x with the constant∥∥ f ′(x)

∥∥+ ε.⊳ By the definition of C1

G(x), there exists δ > 0 such that for anyy ∈ Bδ(x) it holds

∥∥ f ′(y)− f ′(x)∥∥ ≤ ε and hence

∥∥ f ′(y)∥∥ =

∥∥ f ′(y)− f ′(x)+ f ′(x)∥∥ ≤

∥∥ f ′(y)− f ′(x)∥∥

︸︷︷︸≤ε

+∥∥ f ′(x)

∥∥ ≤ ε+∥∥ f ′(x)

∥∥=: k.

(1)

37

38 CHAPTER 3. INVERSE FUNCTION THEOREM

Then ∀x1, x2 ∈ Bδ(x) it holds

‖ f (x1)− f (x2)‖MVT≤

≤k by (1)︷︸︸︷sup

y∈[x1,x2]

∥∥ f ′(y)∥∥ ‖x1 − x2‖ ≤ k ‖x1 − x2‖ . ⊲

3.2 Banach spaces

We say that a normed space X is a Banach space (in honour of a Polish mathematicianStefan Banach), and we write

X ∈ BS

if X is complete as a metric space (with the metric (x, y) := ‖x − y‖), that is, if anyCauchy sequence in X converges.

Note that in a normed space

{xn} ∈ Cauchy ⇔ ‖xm − xn‖ Km,n→∞ 0.

Examples.

1. R,Rn,C([0, 1]), ℓ2 are Banach spaces.

2. The vector subspace k in ℓ2 which consists from all FINITE sequences, that is, sequencesof the form x = (x1, . . . , xn, 0, 0, . . .) (n depends on x), equipped with the norm form ℓ2,is NOT a Banach space.

3.3 Contraction Lemma

It is the name of the following

Theorem 3.3.1. Let X ∈ BS, let A be a CLOSED subset in X, and let f be a mapping fromA into itself, f : A . If f ∈ LipA k with k < 1 (strictly!), then the operator f has oneand just one FIXED POINT x , that is, a point such that

x1x2 x0x

v

f

id

Ax2x1

x0

k a2

x 3 x4

kaa k a

3

(1)f (x) = x .

(Note that we can rewrite (1) as f (x) = id(x).) In thiscase we write

x ∈ Fix f .

⊳ 0◦ The idea of the proof is clear from the picture: thebroken line leads to x .1◦ Take an ARBITRARY point x0 ∈ A and put

x1 = f (x0), x2 = f (x1), . . . , xn+1 = f (xn), . . . .

We have

‖x1 − x0‖ =: a

‖x2 − x1‖ = ‖ f (x1)− f (x0)‖ ≤ k ‖x1 − x0‖ = ka, (2)

‖x3 − x2‖ = ‖ f (x2)− f (x1)‖ ≤ k ‖x2 − x1‖(2)≤ k2a,

3.4. ISOMORPHISMS 39

...

‖xn+1 − xn‖ ≤ kna. (3)

2◦ {xn} ∈Cauchy.

⊳⊳ ‖xm − xn‖ trick=suppose

m>n

‖ xm − xm−1︸︷︷︸+ xm−1 − xm−2︸︷︷︸+ · · · + xn+1 − xn︸︷︷︸ ‖

≤ ‖xm − xm−1‖︸︷︷︸≤akm−1 by (3)

+ · · · + ‖xn+1 − xn‖︸︷︷︸≤akn by (3)

≤ a(kn + · · · + km−1)

∑kn<∞

Km,n→∞ 0. ⊲⊲

3◦ Put x := lim xn (the limit exists since X ∈ BS). Then x ∈ A, since A is closed. Further,since f is continuous (by Theorem 3.1.1.) it holds

f (x) = limn→∞ f (xn)︸︷︷︸

=xn+1

= limn→∞ xn+1

obv.= limn→∞ xn = x,

hence x is a fixed point for f .4◦ This fixed point is unique. ⊳⊳ If x1 and x2 are both fixed points for f , then

‖ f (x1)︸︷︷︸=x1

− f (x2)︸︷︷︸=x2

‖ ≤ k︸︷︷︸<1

‖x1 − x2‖ ⇒ ‖x1 − x2‖ = 0⇒ x1 = x2. ⊲⊲ ⊲

3.4 Isomorphisms

Let X,Y ∈ NS, and let l ∈L (X,Y ). We say that l is an isomorphism and we write

l ∈ Iso(X,Y ) (or simply l ∈ Iso)

if l is a bijection, and if the inverse mapping l−1 (which is automatically linear, verify!) isalso continuous.

We say that X and Y are isomorphic (as normed spaces) and we write

X ≈ Y

if there exists an isomorphism from X onto Y .

Examples.

1. If ‖·‖1 and ‖·‖2 are two equivalent norms on a vector space X , then id : (X, ‖·‖1)→(X, ‖·‖2) is an isomorphism.

2. Rn × Rm ≈ Rn+m .

3. L (R, X) ≈ X .

4. L (Rn,R) ≈ Rn .

NB ∃X ∈ NS...L (X,R) 6≈ X .

5. L (ℓ2,R) ≈ ℓ2.

6. If l ∈L (R,R), then l ∈ Iso⇔ l 6= 0.

7. If l ∈L (Rn,Rn), then l ∈ Iso⇔ det l 6= 0 (as is known from linear algebra).


3.5 Inverse Function Theorem

Here is a generalization of the classic Inverse Function Theorem:

Theorem 3.5.1. Let X,Y ∈ BS, f : X → Y, x ∈ X, y := f (x). If f ∈ C1G(x)

and f ′(x) ∈ Iso(X,Y ), then there exists a neighbourhood U of x such that f is aHOMEOMORPHISM of U onto f (U). The inverse mapping f −1 is differentiable at y, and

( f −1)′(y) = ( f ′(x))−1. (1)

⊳ 1◦ Reduction to the case f (0) = 0, f ′(0) = id. Without loss of generality we canassume that X = Y, x = y = 0, f ′(0) = id. ⊳⊳ Put f (h) := f (x + h) − f (x).It is clear that f : X → Y satisfies the condition f (0) = 0 and has at 0 the samedifferentiability property as f has at x . Thus, without loss of generality x = 0, y = 0.

Now put l := f ′(0) and f := l−1 ◦ f (recall that l−1 ∈ L (Y, X), since l ∈ Iso).By Chain Rule, f ′(0) = l−1 ◦ f ′(0)︸︷︷︸

=l

= id. If the theorem is true for f , then it is true for

f = l ◦ f (since f −1 = ( f )−1◦l−1). Thus, without loss of generality X = Y, f ′(0) = id.

X KL

f

gY ⊲⊲

So, the decomposition f (x + h) = f (x)+ f ′(x)h + r(h), ‖r(h)‖‖h‖ K‖h‖→0

0, reduces

tof (h) = h + r(h), or f = id+r, (2)

where r satisfies the conditions

r(0)as always= 0, r ′(0)

as always= 0, r ′f ∈C1

G(0)∈ Cont(0),‖r(h)‖‖h‖ K

‖h‖→00. (3)

2◦ Reduction to the fixed point problem. Now note that to find the inverse function to fmeans to solve the equation

f (x) = y

with respect to y. But for a FIXED y,

f (x) = y ⇔ x ∈ Fix(id− f︸︷︷︸(2)= −r

+y). (4)

⊳⊳ (id− f + y)(x) = x ⇔ f (x) = y. ⊲⊲

g=−r+y

idf

y

x

−r=id−f

}

}

So (in view of Contraction Lemma) our goal is to find a set, where the mapping

(5)g := −r + y

is Lipschitz.3◦ By Theorem 3.1.2. applied to r , there exists ε > 0such that r ∈ LipBε

12 . Put

U := ( f |Bε )−1

(Bε/2) (⊂ Bε).

(Here −1 denotes the pre-image, not the inverse mapping!)

3.5. INVERSE FUNCTION THEOREM 41

y−r

X

X

r

x

id

yU

fε/2

ε

ε

4◦ f ∈ Bij(U,Bε/2). ⊳⊳ Let y ∈ Bε/2. Consider themapping

y − r : x 7→ y − r(x), X → X.

This mapping is Lipschitz on Bε with the constant 12 ,

since r is Lipschitz. Moreover, y− r maps Bε into itself;indeed,

‖x‖ ≤ ε⇒ ‖y − r(x)‖ ≤ ‖y‖︸︷︷︸≤ε/2

+ ‖r(x)‖︸︷︷︸≤ 1

2 ‖x‖︸︷︷︸≤ε

≤ ε.

By Contraction Lemma, there exists one and just one fixed point of y− r . But by (4)this means that there exists one and just one point x ∈ Bε such that f (x) = y, that is, bythe definition of U , there exists one and just one point x ∈ U such that f (x) = 0. ⊲⊲

5◦ Let now f −1 denotes the (existing!) inverse mapping to f : U → Bε/2. For conveni-ence introduce the following notation: for x ∈ U, y ∈ Bε/2,

x ↔ y :⇔ f (x) = y ⇔ x = f −1(y).

6◦ f −1 ∈ Lip 2. ⊳⊳ Let x1 ↔ y1, x2 ↔ y2. It holds

‖x1 − x2‖ id= f−r= ‖( f (x1)︸︷︷︸=y1

−r(x1))− ( f (x2)︸︷︷︸=y2

−r(x2))‖ ≤ ‖y1 − y2‖ + ‖r(x2)− r(x1)‖︸︷︷︸(3)≤ 1

2 ‖x2 − x1‖≤ ‖y1 − y2‖ + 1

2 ‖x1 − x2‖ .We conclude that ‖x1 − x2‖ ≤ 2 ‖y1 − y2‖, that is,

∥∥ f −1(y1)− f −1(y2)∥∥ ≤ 2 ‖y1 − y2‖.

⊲⊲

7◦ f ∈ Homeo(U,Bε/2). ⊳⊳ f = id+r3◦∈ Cont; f −1 6◦∈ Cont. ⊲⊲

8◦ ( f −1)′(0) = id (= ( f ′(0))−1). ⊳⊳ We need to verify that the mapping

s := f −1 − id

is small (recall that f −1(0) = 0, since f (0) = 0), that is, that∥∥∥ f −1(k)− k

∥∥∥‖k‖ K

‖k‖→00,

or‖h − k‖‖k‖ K

‖k‖→00 if h ↔ k.

But indeed‖h − k‖‖k‖ = ‖h − k‖

‖h‖︸︷︷︸r= f−id= ‖r(h)‖/‖h‖ K

‖h‖→00

‖h‖‖k‖︸︷︷︸6◦≤2

K‖k‖→0

0,

since ‖k‖ → 0⇒ ‖h‖︸︷︷︸6◦≤2‖k‖

→ 0. ⊲⊲ ⊲


Corollary 3.5.2. Let f = ( f1, . . . , fn) : Rn → Rn be continuously G-differentiableat x (this will be so, e.g., if all the partial derivatives ∂ f j/∂xi are continuous in someneighbourhood of x), and let Jacobi matrix J f (x) have non-zero determinant. Then thereexists a neighbourhood U of x such that f is a homeomorphism of U onto f (U), theinverse mapping f −1 is differentiable at y := f (x), and

J f −1(y) = (J f (x))−1.

⊳ Rn is a Banach space, and f ′(x) is an isomorphism iff det J f (x) 6= 0. ⊲

3.6 Implicite Function Theorem

An important corollary of the Inverse Function Theorem (3.5.1.) is:

Theorem 3.6.1. Let X,Y, Z ∈ BS, F : X × Y → Z. Put

M := F−1(0).

Let m := (x, y) ∈ M, that is, F(x, y) = 0, and let F ∈ C1G(m), ∂F/∂Y (m) ∈ Iso(Y, Z)

(so that Y and Z are isomorphic). Then there exists a neighbourhood U of x in X, amapping f : U → Y and a neighbourhood Nof m in X × Y such that

gr f = M ∩ N. (1)

This mapping f is differentiable at x , and

f ′(x) = −(∂F(m)

∂Y

)−1

◦ ∂F(m)

∂X. (2)

N

gr fm

Y

X

y

vx

UM

v

In other words, since the condition gr f ⊂ M means that F(x, f (x)) = 0 (for x ∈ U ),the theorem asserts that the equation

F(x, y) = 0

can be solved with respect to y:

y = f (x),

the resulting (“implicite”) function f being differentia-ble at x , and its derivative at x can be expressed in termsof partial derivatives of F at m.

Before the proof consider a model example.

Example. Let X = Y = Z = R, F(x, y) = x2 + y2 − 1, m = (0, 1). Here M is theunit circle with the center at 0. We have ∂F/∂x |(0,1) = 0, ∂F/∂y|(0,1) = 2 6= 0. The setM ∩ N (see the picture below) is the graph of the function

3.6. IMPLICITE FUNCTION THEOREM 43

F=1

y

z

x

G

G -1

F=1

x

3/41/21/40

−3/4

−1/4−1/2

−1

z

y

x

−1−3/4−1/2

−1/40

1/41/2

3/4F=1

3/4

3/41/2

1/21/4

1/4

0

0-1/4

-1/2-3/4

m

Ngr f

M

the viewfrom above

the front view

f (x) =√

1− x2 (x ∈ U = (− 12 ,

12 )),

and

f ′(0) = −∂F∂x

∣∣∣∣(0,1)

∂F∂y

∣∣∣∣(0,1)

= 0.

For the proof we need some lemmas.

Lemma 3.6.2. Let X1, . . . , Xn be Banach spaces. Then their product X1 × . . . × Xn isalso a Banach space.⊳ Completeness of all X i implies completeness of the product, since convergence in theproduct is just convergence in each component. ⊲

Lemma 3.6.3. Let f = ( f1, . . . , fm ) : X1 × . . . × Xn → Y1 × . . . × Ym , and letx ∈ (x1, . . . , xn) ∈ X1 × . . .× Xn . Then

∂ f j (x)

∂X i= (π j ◦ f ◦ ιi )′(x),

where ιi and π j are resp. the following imbeddings and projections:

ιi : X i → X1 × . . .× Xn, x 7→ (x1, . . . , xi−1, xi , xi+1, . . . , xn),

π j : Y1 × . . .× Ym → Y j , (y1, . . . , ym) 7→ y j .


⊳ This follows from the facts that

f j = π j ◦ f

and f j ◦ ιi is just the mapping f j with all the arguments but i -th one FIXED to be equalthe corresponding components of x . ⊲

Lemma 3.6.4. Let X,Y, Z be normed spaces, Y and Z being isomorphic, and let

A : X × Y → X × Z

be the linear mapping represented by the matrix

(id 0a b

):

A ∼(

id 0a b

),

where id = idX , 0 ∈L (Y, Z), a ∈L (X, Z), b ∈ Iso(Y, Z). Then A ∈ Iso(X × Y, X ×Z), and

A−1 ∼(

id 0−b−1 ◦ a b−1

).

Here

A ∼(

a11 a12a21 a22

)(ai j ∈ L(X j ,Yi )),

means of course that

Ah ∈ Y1 × Y2 is represented by

(a11 a12a21 a22

)(h1h2

):=(

a11h1 + a12h2a21h1 + a22h2

),

if

h ∈ X1 × X2 is represented by

(h1h2

)(hi ∈ X i ).

⊳ The direct computation yields(

idX 0a b

)◦(

idX 0b−1 ◦ a b−1

)=(

id ◦ id−0 ◦ b−1 ◦ a id ◦0+ 0 ◦ b−1

a ◦ id−b ◦ b−1 ◦ a a ◦ 0+ b ◦ b−1

)

=(

idX 00 idY

)∼ idX×Y ,

and analogously(

idX 0−b−1 ◦ a b−1

)◦(

idX 0a b

)=(

idX 00 idY

)∼ idX×Y . ⊲

Proof of Theorem 3.6.1.⊳ 0◦ The idea is to extend F to a mapping G, to which we can apply the Inverse FunctionTheorem (see the picture in Example!).1◦ Put

G := (π1, F) : X × Y → X × Z , (x, y) 7→ (x, F(x, y)).

Note that G does NOT change the first coordinate! (On the picture vertical lines remainvertical!)

3.6. IMPLICITE FUNCTION THEOREM 45

2◦ We have

G′ Prod. Rule= (π ′1, F ′)F∈C1

G (m)∈ C1G(m)

(the projection π1 : X × Y → X is of course of class C1 as a continuous linear mapping),and

G′(m)Theorem 2.5.1.on represent.∼

∂π1(m)∂X

∂π1(m)∂Y

∂F(m)∂X

∂F(m)∂Y

=

(idX 0∂F(m)∂X

∂F(m)∂Y

).

3◦ By Lemma 3.6.4., G′ ∈ Iso, and

(G′(m))−1 ∼(

idX 0

−(∂F(m)∂Y

)−1◦ ∂F(m)

∂X

(∂F(m)∂Y

)−1

).

4◦ By Lemma 3.6.2., both X × Y and X × Z are Banach spaces, and we can apply theInverse Function Theorem. We conclude that there exists a neighbourhood N of m in

N

U

G

G -1

m (x,0)v

UUxW

W

NG(N)

X × Y such that G is a homeomorphism of N onto G(N ), G−1 is differentiable at

F(m) = (x, 0),

and

(G−1)′((x, 0)) = (G′(m))−1.

Note that G−1 does NOT change the first coor-dinate, since G does not.

5◦ By properties of product topology, there exist a neighbourhood U of x in X and aneighbourhood W of 0 in Z , such that U ×W ⊂ G(N ). Put

N := G−1(U ×W ).

6◦ At last putf := π2 ◦ G−1 ◦ ι1,

where π2 is the projection X × Y → X, (x, y) 7→ x , and ι1 is the imbedding X →X × Z , x 7→ (x, 0).

X × YG−1

L X × Zπ2 ↓ ↑ ι1

Yf

L X

7◦ By Change Rule, f is differentiable at x , and

f ′(x) = (π2 ◦ G−1 ◦ ι1)′(x) Lemma 3.6.3.= ∂(G−1)2((x, 0))

∂X.

8◦ The latter partial derivative is just the (21)-element of the matrix representing

(G−1)′((x, 0))4◦= (G′(m))−1.

9◦ By 3◦, this element is equal to

−(∂F(m)

∂Y

)−1

◦ ∂F(m)

∂X.


Thus, (2) is proved.10◦And, by the very construction, gr f = M∩N . [Formal verification: Let (x, y) ∈ M∩N(so that x ∈ U ). Then

(x, y) ∈ M ⇒ F(x, y) = 0⇒ G(x, y) = (x, 0)⇒ (x, y) = G−1(x, 0)⇒ y = (π2 ◦ G−1 ◦ ι1︸︷︷︸

= f

)(x)⇒ (x, y) ∈ gr f.

Thus N ∩ M ⊂ gr f . Inverting the argument, we can analogously obtain the inverseinclusion.] ⊲

NB Vice versa, Inverse Function Theorem can be deduced from Implicite Function The-orem. [HINT: the equation y = f (x) can be written in the form F(x, y) = 0 withF(x, y) := y − f (x).]

Chapter 4

Higher derivatives

4.1 Multilinear mappings

Let a mapping f : X → Y is differentiable (everywhere). Its derivative is a mapping fromX into L (X,Y ):

f ′ : X →L (X,Y ), x 7→ f ′(x).

It is natural to define f ′′(x) as ( f ′)′(x), so

f ′′(x) ∈L (X,L (X,Y )).

It is also natural to consider the mapping

(h1, h2) 7→ ( f ′′(x) · h1︸︷︷︸∈L (X,Y )

) · h2, X × X → Y.

This mapping is BILINEAR, that is, linear in h2 for fixed h1 (evidently) and linear in h1 forfixed h2 (since f ′′(x) is a linear mapping from X into L (X,Y )).

Analogously higher derivatives lead to MULTILINEAR mappings. Let X1, . . . , Xn andY be vector spaces. We say that a mapping u : X1 × . . . × Xn → Y is mutilinear (orn-linear), and we write

u ∈ L(X1, . . . , Xn; Y ),if u is linear in each separate variable for fixed others, that is, if

∀i ∈ {1, . . . , n} ... u(x1, . . . , xi−1, αxi + βyi , xi+1, . . . , xn)

= αu(x1, . . . , xi , . . . , xn)+ βu(x1, . . . , yi , . . . , xn) (α, β ∈ R).

(For 2-linear mappings we say bilinear.)For multilinear mappings one uses one of the following notations:

u(x1, . . . , xn) ≡ u · x1 . . . xn ≡ ux1 . . . xn ≡ 〈u | x1, . . . , xn〉.

Examples.

1. The usual multiplication R× R→ R, (x, y) 7→ xy is bilinear.

2. The multiplication R× R× R→ R, (x, y, z) 7→ xyz is 3-linear.

3. The scalar product Rn×Rn → R, (x, y) 7→∑ni=1 xi yi , where x = (x1, . . . , xn), y =

(y1, . . . , yn), is bilinear.

47

48 CHAPTER 4. HIGHER DERIVATIVES

4. The vector product R3 ×R3 → R3, (Ex, Ey) 7→ Ex × Ey is bilinear.

5. The COMPOSITION

comp : L(X,Y )× L(Y, Z)→ L(X, Z), (l,m) 7→ m ◦ l

and the EVALUATION

ev : X × L(X,Y )→ Y, (x, l) 7→ lx

are bilinear.

6. The DETERMINANT mapping

det : R3 ×R3 ×R3 → R, (Ex, Ey, Ez) 7→

∣∣∣∣∣∣

x1 x2 x3y1 y2 y3z1 z2 z3

∣∣∣∣∣∣

(Ex = (x1, x2, x3), . . .) is 3-linear.It is easy to verify(!) that L(X1, . . . , Xn; Y ) is a VECTOR SPACE.

Operator normNow let X i and Y be normed spaces. Then the vector subspace of L(X1, . . . , Xn; Y )

consisting from all CONTINUOUS n-linear mappings we denote by

L (X1, . . . , Xn; Y ).Put for each u ∈ L(X1, . . . , Xn; Y )

‖u‖ := sup‖x1‖≤1...

‖xn‖≤1

‖ux1 . . . xn‖ (operator norm).

Examples.1. ‖multiplication‖ = 1;

2. ‖scalar product‖ = 1;

3. ‖vector product‖ = 1;

4. ‖comp‖ ≤ 1;

5. ‖ev‖ ≤ 1;

6. ‖det‖ = 1.

Basic inequality. Let u ∈ L(X1, . . . , Xn; Y ). Then for any (x1, . . . , xn) ∈ X1× . . .× Xn

‖ux1 . . . xn‖ ≤ ‖u‖ ‖x1‖ . . . ‖xn‖ (basic inequality (BI)).

⊳ If xi = 0 for some i then both sides are 0. Let none of xi is 0. Then

‖ux1 . . . xn‖ =u is

multilinear

∣∣∣∣∣∣u x1

‖x1‖︸︷︷︸∈B

X11

. . .xn

‖xn‖︸︷︷︸∈BXn

1︸︷︷︸≤

def of‖u‖

‖u‖

∣∣∣∣∣∣ ‖x1‖ . . . ‖xn‖ ≤ ‖u‖ ‖x1‖ . . . ‖xn‖ . ⊲

Normed space L (X1, . . . , Xn; Y )Theorem 4.1.1. Let u ∈ L(X1, . . . , Xn; Y ). Then the following conditions are equivalent:

4.1. MULTILINEAR MAPPINGS 49

a) u ∈L (X1, . . . , Xn; Y ), that is, u is continuous;b) u is continuous at 0;c) ‖u‖ <∞.

⊳ 0◦ For short consider the case n = 2 : u ∈ L(X; Y ; Z).1◦ (a)⇒(b): obviously.2◦ (b)⇒(a): Let (x, y) be an arbitrary point in X×Y . We need to show that u is continuousat (x, y). We have

||u(x + h, y + k)︸︷︷︸u is

multilin.=u(x,y)+u(x,k)+u(h,y)+u(h,k)

−u(x, y)|| = ‖uxk + uhy + uhk‖ ≤ ‖uxk‖︸︷︷︸1

+‖uhy‖︸︷︷︸2

+‖uhk‖︸︷︷︸3

.

So it is sufficient to verify that 1 , 2 , 3 → 0 as ‖h‖ , ‖k‖ → 0. If k = 0 then 1 = 0,if k 6= 0 then

1 =∥∥∥∥u(√‖k‖x

)( k√‖k‖

)∥∥∥∥ .

If ‖k‖ → 0 then√‖k‖ → 0 and hence

√‖k‖x → 0; further∥∥k/√‖k‖

∥∥ = ‖k‖/√‖k‖ =√‖k‖ → 0. Thus 1 → 0 as ‖k‖ → 0, by (b).

Quite analogously 2 → 0 as ‖h‖ → 0. At last 3 → 0 as ‖h‖ → 0, ‖k‖ → 0, by(b).3◦ (b)⇒(c): By (b), there exists δ > 0 such that

‖x‖ ≤ δ, ‖y‖ ≤ δ ⇒ ‖uxy‖ ≤ 1. (1)

Then‖u‖ = sup

‖x‖≤1‖y‖≤1

‖uxy‖︸︷︷︸=δ−2‖u(δx)(δy)‖︸︷︷︸

(1)≤ 1 if ‖x‖≤1,‖y‖≤1

≤ δ−2 <∞.

4◦ (c)⇒(b): If ‖x‖ , ‖y‖ → 0 then ‖uxy‖ BI≤ ‖u‖︸︷︷︸(c)<∞

‖x‖ ‖y‖ → 0. ⊲

Note that all the multilinear mappings from Examples 1)–6) are continuous (since allthey have norms ≤ 1). As to mappings from 1)–4) and 6), their continuity follows alsofrom

Theorem 4.1.2. In finite-dimensional case all the multilinear mappings are continuous.⊳ Analogously to the case of linear mappings. ⊲

Theorem 4.1.3. The operator norm is really a norm in L (X1, . . . , Xn; Y ).We EVER consider L (X1, . . . , Xn; Y ) as a normed space with the operator norm!⊳ By Theorem 1, the operator norm is FINITE on the whole space L (X1, . . . , Xn; Y ), andit is easy to verify that all 3 axioms of a norm are fulfilled. ⊲

Canonical isomorphisms

Theorem 4.1.4. For any natural k and n, k < n, it holds

L (X1, . . . , Xn; Y ) ≈L (X1, . . . , Xk ;L (Xk+1, . . . , Xn; Y ))


(the isomorphism of normed spaces), and the CANONICAL ISOMORPHISM

L (X1, . . . , Xn; Y )→L (X1, . . . , Xk ;L (Xk+1, . . . , Xn; Y )), u 7→ u,

u(x1, . . . , xk) := u(x1, . . . , xk, ·, . . . , ·︸︷︷︸n−k “free′′arguments

)

conserves the norm:‖u‖ = ‖u‖ .

⊳ 0◦ For short consider the case n = 2, k = 1. We need to verify that

L (X1, X2; Y ) ≈L (X1,L (X2,Y )).

1◦ ALGEBRAICAL ISOMORPHISM:

L(X1, X2; Y )alg.≈ L(X1,L(X2,Y )). (2)

Put for u ∈ L(X1, X2; Y )

u(x1) := u(x1, ·) (∈ L(X2,Y )),

and for v ∈ L(X1.L(X2,Y ))

v(x1, x2) := (v · x1) · x2 (∈ Y ).

It is easy to see that the mapping

u : X1 → L(X2,Y )

is linear, that is, u ∈ L(X1,L(X2,Y )), and the mapping

v : X1 × X2 → Y

is bilinear, that is, v ∈ L(X1, X2; Y ), and that the mappings

u 7→ u and v 7→ v

are linear and mutually inverse (≈u= u,

≈v= v). Hence u 7→ u is a linear bijection of

L(X1, X2; Y ) onto L(X1,L(X2; Y )), that is, (1) is true.

2◦ TOPOLOGICAL ISOMORPHISM: If u ∈ L (X1, X2,Y ) then ∀x1 ∈ X1... u(x1) =

u(x1, ·) ∈L (X2,Y ) (since u is continuous). Now the (linear) mapping

u : X1 →L (X2,Y )

is continuous since it has a finite norm equal to the norm of u:

‖u‖ = sup‖x‖≤1

‖ux1‖ = sup‖x1‖≤1

sup‖x2‖≤1︸︷︷︸

obv= sup‖x1‖≤1, ‖x2‖≤1

|| (ux1) · x2︸︷︷︸=u(x1,x2)

|| = ‖u‖ .

Thus, u ∈L (X1,L (X2,Y )).

4.1. MULTILINEAR MAPPINGS 51

Quite analogously it can be verified that if v ∈L (X1,L (X2,Y )) then v ∈L (X1, X2; Y ).We conclude that u 7→ u is a linear bijection of L (X1, X2; Y ) onto L (X1,L (X2,Y )).Since ‖u‖ = ‖u‖, both u 7→ u and v 7→ v have the norm 1 and hence are continuous.Thus they are isomorphisms of our normed spaces. ⊲

Remark. For X1 = X2 = Rn and Y = R, (1) is in fact the well-known (from linearalgebra) isomorphism between bilinear forms and linear operators in Rn .

Differentiation of multilinear mappings

Theorem 4.1.5. (Quasi-Leibniz Theorem (QL)). Any mapping u ∈L (X1, . . . , Xn; Y ) isdifferentiable, and its derivative is given by the formula

u′(x1, . . . , xn) · (h1, . . . hn) =n∑

i=1

u(x1, . . . , xi−1, hi , xi+1, . . . , xn),

or, more shortly

u′(x1, . . . , xn) =n⊕

i=1

u(x1, . . . , xi−1, ·, xi+1, . . . , xn). (3)

(The definition of the DIRECT SUM ⊕ni=1li ≡ l1 ⊕ . . .⊕ ln see in Chapter 2.)

⊳ This follows at once from Theorem on continuous partial derivatives. Indeed, u is linear

and continuous in each its argument, hence ∀i ∈ {1, . . . , n} ...∂u

∂X i(x1, . . . , xn) := u(x1, . . . , xi−1, ·, xi+1, . . . , xn) (∈L (X i ,Y )).

Now, each partial derivative ∂u/∂X i : X1 × . . .× Xn →L (X i ,Y ) is continuous as thecomposition of two continuous mappings:

∂u

∂X i= ui ◦ πi ,

where

πi : X1 × . . .× Xn → X1 × . . .× X i × . . .× Xn, (x1, . . . , xn) 7→ (x1, . . . , xi , . . . , xn),

ui : X1 × . . .× X i × . . .× Xn →L (X i ,Y ),

(x1, . . . , xi , . . . , xn) 7→ u(x1, . . . , xi−1, ·, xi+1, . . . , xn);πi is continuous as any projection, and ui is continuous by Theorem on canonical isomor-phism (since u is). ⊲

Remark. We can rewrite (3) so:

u′ = ⊕ ◦ (u1 ◦ π1, . . . , un ◦ πn),

where ⊕ denote the following mapping:

⊕ : L (X,Y )× . . .×L (Xn,Y )→L (X1 × . . .× Xn,Y ), (l1, . . . , l2) 7→n⊕

i=1

li .


It is evident that this mapping⊕ is linear, and it is easy to verify (!) that it is continuous.

Corollary 4.1.6. (Leibniz Theorem). Let f : X → Y and g : X → Z be differentiable ata point x , and let u ∈L (Y, Z;W ). Then the composition u ◦ ( f, g) is differentiable at x,and

(u ◦ ( f, g))′(x) = u(

f ′(x)h, g(x))+ u

(f (x), g′(x)h

).

X( f,g)

K Y × ZuK W.

⊳ This follows at once from Chain Rule and Quasi-Leibniz Theorem (QL):

(u ◦ ( f, g))′(x)hChainrule= (u′( ( f, g)(x)︸︷︷︸

=( f (x),g(x))

) ◦ ( f, g)′(x)︸︷︷︸=( f ′(x),g′(x))

) · h

= u′( f (x), g(x)) ◦ ( f ′(x)h, g′(x)h)QL= u( f ′(x)h, g(x))+ u( f (x), g′(x)h). ⊲

Examples.

1. If u is the usual multiplication R×R→ R, and if X = R, we obtain the classic Leibnizrule: ( f g)′ = f ′g + f g′.2. For the mapping q : Rn → R, x 7→ x2

1 + . . . + x21 = x · x ≡ x2, we have (here

f = g = id, u = scalar product)

g′(x) · h = x · h + h · x = 2x · h

(the first point denoting the application of a linear mapping, the other points denoting thescalar product!), so

g′(x) = 2x,

if we identify a vector x with the linear function h 7→ x · h. (Compare with the usual rule(x2)′ = 2x .)

4.2 Higher derivatives

Let a mapping f : X → Y be differentiable everywhere (or in an open set U ⊂ X). Thenwe can consider the derivative map

f ′ : X ′ →L (X,Y ), x 7→ f ′(x).

We say that f is two times differentiable at a point x , and we write

f ∈ Dif2(x),

if f ′ is differentiable at x ; we define the second derivative f ′′(x) of f at x as the derivativeof f ′ at x :

f ′′(x) := ( f ′)′(x) (∈L (X,L (X,Y ))).

By induction, we put

f (n+1)(x) :=(

f (n))′(x),

and we use in evident sense the notations

Difn(x), Difn(U), Difn .


Besides we put

Dif1 := Dif, Dif∞ :=∞⋂

n=1

Difn .

Thus we have

f : X → Yf ′ : X →L (X,Y )f ′′ : X →L (X,L (X,Y ))

...

f (n) : X →L (X,L (X, . . . ,L (X,Y ) . . .))...

The space L (X,L (X, . . . ,L (X,Y ) . . .)) of values of the n-th derivative f (n) isisomorphic (by repeatedly applied Theorem on isomorphism, see 4.1) to the space ofn-LINEAR mappings from X × . . .× X (n times) into Y :

L (X,L (X, . . . ,L (X,Y ) . . .)) ≈L (X, . . . , X︸︷︷︸n times

; Y ) =: L (n X; Y ).

The multilinear mapping, corresponding to f (n)(x), is given by the rule

f (n)(x)(h1, . . . , hn) := (. . . (( f n(x) · h1) · h2) . . .) · hn .

Usually we IDENTIFY f (n)(x) and f (n)(x), drop the wave and write

f (n)(x)h1, . . . , hn .

Example. For q : Rn → R, x 7→ x2 := x · x (see 4.1) we have q ′′ ≡ (2 scalar product),

that is, ∀x ∈ Rn... q ′′(x)h1h2 = 2h1 · h2. (Prove!)

4.3 Rules of differentiation

They are in essence the same as for the first derivative.

Linearity. If f, g ∈ Difn(x) then ∀α, β ∈ R... α f + βg ∈ Difn(α), and

(α f + βg)(n)(x) = α f (n)(x)+ βg(n)(x).

⊳ By induction. ⊲

Product Rule. Let f = ( f1, . . . , fn) : X → Y1× . . .×Ym . Then f ∈ Difn(x) iff eachfi ∈ Difn(x), and

f (n)(x) =(

f (n)1 (x), . . . , f (n)m (x)).

⊳ By induction. ⊲

Chain Rule. If f ∈ Difn(x) and g ∈ Difn( f (x)), then g ◦ f ∈ Difn(x). (The expliciteformula for (g ◦ f )(n)(x) is very cumbersome, and we drop it.)


⊳ For simplicity we consider only the case n = 2. By Chain Rule for the first derivative,

(g ◦ f )′ = comp( f ′, g′ ◦ f ),

wherecomp : L (X,Y ) ◦L (Y, Z)→L (X, Z), (l,m) 7→ m ◦ l.

Since f and g are 2 times differentiable at x and at f (x), resp., the mappings f ′ and g′

are differentiable at x and f (x), resp. The mapping comp is differentiable (everywhere)by Quasi-Leibniz Theorem. So (g ◦ f )′ is differentiable at x by Chain and Product Rulesfor the first derivative. But this means that g ◦ f is 2 times differentiable at x . ⊲

Computation Rule. Let f : X → Y be n times differentiable at x. Then for anyh1, . . . , hn ∈ X

f (n)(x)h1, . . . , hn =∂n

∂ t1 . . . ∂ tn

∣∣∣∣t1=...=tn=0

f (x + t1h1 + . . .+ tnhn)

:= ∂

∂ t1

∣∣∣∣t1=0

(∂

∂ t2

∣∣∣∣t2=0

(. . .

(∂

∂ tn

∣∣∣∣tn=0

f (x + t1h1 + . . .+ tnhn)

). . .

)).

For short we shall write the last expression as

∂

∂ t1

∣∣∣∣0. . .

∂

∂ tn

∣∣∣∣0

f (x + t1h1 + . . .+ tnhn).

⊳ For simplicity consider the case n = 2. It holds

∂

∂ t1

∣∣∣∣0

∂

∂ t2

∣∣∣∣0

f (x + t1h1 + t2h2)

︸︷︷︸=

C.R. forthe 1. der.

f ′(x+t1h1)h2trick= evh2 ·( f ′(x+t1h1))

= ∂

∂ t1

∣∣∣∣0

(evh2 · f ′(x + t1h1)

) l−Rule= evh2 ·∂

∂ t1

∣∣∣∣0

f ′(x + t1h1)

︸︷︷︸C.R. for

the 1. der.= ( f ′)′(x)·h1

= evh2 ·( f ′′(x)h1) =(

f ′′(x)h1)· h2 = f ′′(x)h1h2. ⊲

(Recall that evh denotes the (continuous linear) mapping of evaluation at a given pointh, see Chapter 1.)

l-Rules.

a) Let XfK Y

lK Z , f ∈ Dif p(x), l ∈L (Y, Z). Then

(l ◦ f )(p)(x)h1 . . . h p = l ·(

f (p)(x)h1 . . . h p

),

or, shortly,(l ◦ f )(p)(x) = l ◦ ( f (p)(x)),

where we consider the p-th derivative f (p)(x) as a p-linear mapping.In particular, if X = R then

(l ◦ f )(p)(x) = l · f (p)(x),

where we consider f (p)(x) as an element of Y .

4.4. HIGHER PARTIAL DERIVATIVES 55

b) Let XlK Y

fK Z , l ∈L (X,Y ), f ∈ Dif(p)(lx). Then

( f ◦ l)(x)h1 . . . h p = f (p)(lx)(lh1) . . . (lh p).

⊳ For short consider p = 2.a)

(l ◦ f )′′(x)h1h2Comp. Rule= ∂

∂ t1

∣∣∣∣0

∂

∂ t2

∣∣∣∣0(l ◦ f )(x + t1h1 + t2h2)

l−rule for 1. der.= ∂

∂ t1

∣∣∣∣0

(l · ∂∂ t2

∣∣∣∣0

f (x + t1h1 + t2h2)

)

l−rule for 1. der.= l ·(∂

∂ t1

∣∣∣∣0

∂

∂ t2

∣∣∣∣0

f (x + t1h1 + t2h2)

)

Comp. Rule= l · f ′′(x)h1h2.

b)

( f ◦ l)′′(x)h1h2Comp. Rule= ∂

∂ t1

∣∣∣∣0

∂

∂ t2

∣∣∣∣0( f ◦ l)(x + t1h1 + t2h2)︸︷︷︸l is linear= f (lx+t1lh1+t2lh2)

Comp. Rule= f ′′(lx)(lh1)(lh2). ⊲

4.4 Higher partial derivatives

Let f : X1 × . . . × Xn → Y . Of course we define partial derivatives of higher ordersinductively:

∂ p f (x)

∂X i1 . . . ∂X ip

:= ∂

∂X i1

(∂

∂X i2

(. . .

(∂ f

∂X ip

). . .

))(x),

or, shortly,

∂ p

∂X i1 . . . ∂X ip

:= ∂

∂X i1◦ · · · ◦ ∂

∂X ip

(i1, . . . , i p ∈ {1, . . . , n}).

So∂ p f (x)

∂X i1 . . . ∂X ip

∈L (X i1 ,L (X i2 , . . . ,L (X ip ,Y ) . . .))

≈Th. onisom.

L (X i1 , . . . , X ip ; Y ),

and we have identity ∂ p f (x)/∂X i1 . . . ∂X ip with the corresponding p-linear mapping:

∂ p f (x)

∂X i1 . . . ∂X ip

h1 . . . h p ≡(. . .

((∂ p f (x)

∂X i1h1

)h2

). . .

)h p

(hk ∈ X ik

).

As in the case of the first order, if each X i = R (that is, X1 × . . .× Xn = Rn), we put

∂ f (x)

∂X i1 . . . ∂X ip

:=(. . .

((∂ p f (x)

∂X i1 . . . ∂X ip

· 1)· 1). . .

)· 1 (∈ Y ) .


Lemma 4.4.1. Let a mapping f : X1× . . .× Xn → Y be p times differentiable at x. Thenfor any i1, . . . , i p ∈ {1, . . . , n} and for any hk ∈ X ik , k = 1, . . . , p, it holds

∂ p f (x)

∂X i1 . . . ∂X ip

h1 . . . h p = f (p)(x)h1 . . . h p,

wherehk := (0, . . . , 0, hk

ik, 0, . . . , 0),

that is, the partial derivative applied to the vectors h1, . . . , h p is just the “total” derivativeapplied to their images hath1, . . . , h p by the canonical imbeddings of X ik into the productspace X1 × . . .× Xn .⊳ 1◦ We use below the following result: If f : X → L (Y, Z) is differentiable at x thenfor any fixed h ∈ Y the mapping g : x 7→ f (x)h, X → Z is also differentiable at x , and

∀k ∈ X... g′(x)k =

(f ′(x)k

)h.

(See Chapter 2.)2◦ For short let p = 2. We have

f ′′(x)(0, . . . , hi1, . . . , 0)(0, . . . , h

i2, . . . , 0)

Comp.Rule= ∂

∂ t1

∣∣∣∣0

∂

∂ t1

∣∣∣∣0

f (x + t1(0, . . . , h, . . . , 0)+ t2(0, . . . , k, . . . , 0))

︸︷︷︸Comp.Rule= (∂ f (x+t1(0,...,h,...,0))/∂X i2 )k

0◦=(∂

∂ t1

∣∣∣∣0

∂ f (x + t1(0, . . . , h, . . . , 0))

∂X i2

)k

Comp.Rule=

((∂

∂X i1

∂

∂X i2

)(x)h

)k

def= ∂2 f (x)

∂X i1 · ∂X i2hk. ⊲

Theorem 4.4.2. (on representation). Let a mapping f : X1 × . . .× Xn → Y be p-timesdifferentiable at x. Then its p-th derivative at x can be represented by the matrix of thepartial derivatives:

f (p)(x) ∼(

∂ p f (x)

∂X i1 . . . ∂X ip

)

i1,...,ip∈{1,...,n}

in sense that

∀h1, . . . , h p ∈ X1 × . . .× Xn, hk = (hk1, . . . , hk

n)...

f (p)(x)h1 . . . h p =n∑

i1,...,ip=1

∂ p f (x)

∂X i1 . . . ∂X ip

h1i1 . . . h

pip.

⊳ In notations of the previous lemma,

f (p)(x)h1 . . . h p = (h11 + h1

n) . . . (hp1 + h p

n )

4.5. CLASS C p 57

f (p)(x) ismultilinear=

n∑

i1,...,ip=1

f (p)(x)h1i1+ ˆh1

ip︸︷︷︸4.4.1.= (∂ p( f (x))/∂(X i1 ...X i p ))h

1i1...h p

ip

. ⊲

Corollary 4.4.3. Let f : Rn → Y be p-times differentiable at x. Then f (p)(x) can berepresented by the following matrix with elements in Y :

f (p)(x) ∼(

∂ p f (x)

∂X i1 . . . ∂X ip

)

i1,...,ip∈{1,...,n},

in the sense that for any h1, . . . h p ∈ Rn (hk = (hk1, . . . , hk

n))

f (p)(x)h1 . . . h p =n∑

i1,...,ip=1

h1i1 . . . h

pip︸︷︷︸

∈R

∂ p f (x)

∂X i1 . . . ∂X ip︸︷︷︸∈Y

(∈ Y ).

Here all hkik

are real numbers, and h1i1. . . h p

ipis just the usual product of real numbers.

Remark. For the case p = 2 and Y = R we obtain as the representative of f ′′(x) a“usual” n × n-matrix (

∂2 f (x)

∂xi∂x j

)

i, j∈{1,...,n}.

This matrix is called the Hesse matrix of f at x (and its determinant is called the Hessianof f at x).

Example. For the mapping q : Rn → R, x 7→ x2 ≡ x · x we have

∀x ∈ Rn ... f ′′(x) ∼

2 0. . .

0 2

= 2 · 1

where 1 denotes the unit matrix. It corresponds of course with the fact we know thatf ′′ ≡ 2 id.

4.5 Class C p

We say that a mapping f : X → Y is p-times continuously differentiable (resp., p timescontinuously differentiable in an open set U ⊂ X or at a point x ∈ X) or that f is of classC p (resp., is of class C p in U or at x), and we write

f ∈ C p ( resp., f ∈ C p(U) or f ∈ C p(x)),

if f is p times differentiable everywhere (resp., in U or in a neighbourhood of x) and thederivative f (p) is continuous (resp., continuous in U or at x).

Thus,

f ∈ C p :⇔ f (p) ∈ Cont .


We put also

C0 := Cont, C∞ :=∞⋂

p=0

C p.

The mappings of class C p we call also, for short, C p-mappings.

Example. Any continuous linear mapping is of class C∞. (Indeed, the first derivative isa constant mapping, and all the other derivatives are zeros.)

Lemma 4.5.1.

C0 ⊃ C1 ⊃ C2 ⊃ . . . ⊃ C p ⊃ . . . ⊃ C∞ = Dif∞ .

⊳ 1◦ Dif p+1 ⊂ C p . ⊳⊳ f ∈ Dif p+1 ⇒ f (p) ∈ Dif ⇒ f (p) ∈ Cont⇒ f ∈ C p . ⊲⊲

2◦ C p+1 ⊂ C p. ⊳⊳ f ∈ C p+1 ⇒ f ∈ Dif p+1 1◦⇒ f ∈ C p. ⊲⊲

3◦ C∞ ⊂ C p ⊳⊳ Obviously. ⊲⊲

4◦ C∞ ⊂ Dif∞ ⊳⊳ Obviously. ⊲⊲

5◦ Dif∞ ⊂ C∞ ⊳⊳ f ∈ Dif∞ ⇒ ∀p ∈ N... f ∈ Dif p+1 1◦⇒ ∀p ∈ N

... f ∈ C p ⇒ f ∈C∞. ⊲⊲ ⊲

Lemma 4.5.2. For any k ∈ {0, 1, . . . , p}

f ∈ C p ⇔ f (k) ∈ C p−k .

⊳(

f (k))(p−k) = f (p). ⊲

Theorem 4.5.3. Any continuous multilinear mapping is of class C∞.⊳ Use induction. For linear mappings the assertion is true, by Example above. Let ourassertion is true for k-linear mappings with k ≤ n − 1, and let

u ∈L (X1, . . . , Xn; Y ).

By Remark after Quasi-Leibniz Theorem,

u′ = ⊕ ◦ (u1 ◦ π1, . . . , un ◦ πn),

where ui are continuous (n − 1)-linear mappings, and ⊕ and πi are continuous linearmappings. All these mappings are of class C∞, by the inductive assumption and hence areinfinitely differentiable. By Product and Chain Rules, u′ is also infinitely differentiable.Hence u is infinitely differentiable and therefore (by Lemma 4.5.1.) is of class C∞. ⊲

Remark. In fact the n-th derivative of a continuous n-linear mapping is a CONSTANT

mapping, and hence all the subsequent derivatives are ZEROS:

u(n) = const, u(n+1) = 0, u(u+2) = 0, . . .

Viz., if u ∈L (X1, . . . , Xn; Y ) then ∀x ∈ X1 × . . .× Xn...

u(n)(x)h1 . . . hn =

=∑

σ∈Sn

uhσ(1)1 . . . hσ(n)n

(hk = (hk

1, . . . , hkn) ∈ X1 × . . .× Xn

)(1)

4.6. SYMMETRY OF HIGHER DERIVATIVES 59

where Sn denotes the group of all permutations of the set {1, . . . , n}. [This can be provedby using Computation Rule. (Prove!)] E.g., for bilinear u

u′′(x)hk = uh1k2 + uk1h2. (2)

Note that in the case of MULTIPLICATION (u : R2 → R, (x, y) 7→ xy), Equation (1)follows at once from Representation Theorem:

u′′(x) ∼(

0 11 0

), hence u′′(x)hk = (h1 h2)

(0 11 0

)(k1

k2

)= h1k2 + k1h2.

Note also that it follows from (1) that u(n)(x)h1 . . . hn does NOT change by any permu-tation of the vectors h1, . . . , hn (for bilinear case it is quite obvious, see(2)). This is noaccident! See the next section.

Product

Theorem 4.5.4. ( f1, . . . , fn) ∈ C p ⇔ f1, . . . , fm ∈ C p.

⊳ This follows at once from Product rule for higher derivatives and from the topological

fact that(

f (p)1 , . . . , f (p)

m

)is continuous iff each f (p)

i is. ⊲

Composition

Theorem 4.5.5. f, g ∈ C p ⇒ g ◦ f ∈ C p.

⊳ By induction. For p = 0 all is O.K. (the composition of continuous mappings iscontinuous). Let our assertion is true for p − 1. We have, by Chain Rule,

(g ◦ f )′ = comp ◦( f ′, g′ ◦ f ).

The mapping comp : (l,m) 7→ m ◦ l is a continuous bilinear mapping (see 4.1) and henceis of class C∞ (by Theorem 4.5.3.). A fortiori it is of class C p−1, by Lemma 4.5.1. Thederivatives f ′ and g′ are both of class C p−1, by Lemma 4.5.2., and f is of class C p−1

by Lemma 4.5.1. Hence g′ ◦ f is of class C p−1 by the induction assumption. Then, byTheorem 4.5.4. (on product), ( f ′, g′ ◦ f ) is of class C p−1. So, once again by the inductiveassumption, the mapping comp ◦( f ′, g′ ◦ f ) is of class C p−1. Thus, (g ◦ f )′ ∈ C p−1,which means, by Lemma 4.5.2., that g ◦ f ∈ C p. ⊲

Case Rn → Rn

Criterion. A mapping ( f1, . . . , fm) : Rn → Rm is of class C p iff all the partial derivativesof the order ≤ p of each function f j exist and are continuous.⊳ Analogously to the case p = 1. ⊲

4.6 Symmetry of higher derivatives

Here we prove that for C p-mappings the derivative f (p)(x) is a symmetrical multilinearmapping.

A mapping f : Xn → Y (for arbitrary sets X and Y ) is called symmetrical if its valuedoes not change by any permutation of its arguments:

f ∈ Sym :⇔ ∀x1, . . . , xn ∈ X ∀σ ∈ Sn... f (xσ(1), . . . , xσ(n)) = f (x1, . . . , xn).


(Recall that Sn denotes the group of permutations of the set {1, . . . , n}.) The set of allsymmetrical n-linear mappings from Xn → Y (for vector spaces X and Y ) we denote by

Lsym(n X; Y ).

(Respectively, for continuous case we use the letter L .)

Lemma 4.6.1. Let for a mapping ϕ : R2 → R the second partial derivative ∂2ϕ/∂y∂x iscontinuous at the origin. Then

∂2ϕ(0, 0)

∂y∂x= lim

t↓0

12ϕ(0; (t, 0), (0, t))

t2 .

Here12ϕ denotes the SECOND DIFFERENCE ofϕ. Recall that the first difference1ϕ(x; h)of ϕ at x by h is defined so:

1ϕ(x; h) ≡ 1hϕ(x) := ϕ(x + h)− ϕ(x). (1)

Higher differences 1nϕ(x; h1, . . . , hn) of ϕ at x by h1, . . . , hn are defined inductively.E.g.,

12ϕ(x; h1, h2) := 1h2(1h1ϕ)(x) = 1h1ϕ(x + h2)−1h1ϕ(x)1= ϕ(x + h1 + h2)− ϕ(x + h2)− ϕ(x + h1)+ ϕ(x). (2)

x

x+h1

x+h +h1 2

x+h2

Note that (as it is clear from (2)) the second difference isSYMMETRICAL in the increments:

(3)12ϕ(x; h1, h2) = 12ϕ(x; h2, h1).

⊳ t−212ϕ(0; (t, 0), (0, t)) = t−21(t,0)(1(0,t)ϕ)(0, 0)

trick: g(x):=1(0,t)ϕ(x,0)=ϕ(x,t)−ϕ(x,0)= t−2(g(t)− g(0))

Lagr. Th.;for someθ∈(0,t)= t−2g′(θ)t

g′(x)=∂ϕ(x,t)∂x −∂ϕ(x,0)∂x= t−1

(∂ϕ(θ, t)

∂x− ∂ϕ(θ, 0)

∂x

)

Lagr. Th.;for someτ∈(0,t)= t−1 ∂

∂y

∂ϕ

∂x(θ, τ )t

= ∂2ϕ(θ, τ )

∂y∂xK

t↓0

∂2ϕ(0, 0)

∂y∂x,

since ∂2ϕ/∂y∂x is continuous at (0, 0) and (θ, τ )→ (0, 0) as t ↓ 0. ⊲

4.6. SYMMETRY OF HIGHER DERIVATIVES 61

Corollary 4.6.2. Let for a function ϕ : R2 → R the partial derivatives ∂2ϕ/∂y∂x and∂2ϕ/∂x∂y are both continuous at (0, 0). Then they are equal there:

∂2ϕ

∂y∂x

∣∣∣∣(0,0)= ∂2ϕ

∂x∂y

∣∣∣∣(0,0)

.

⊳ This follows from Lemma 4.6.1., by symmetry of the second difference in the incre-ments. ⊲

Lemma 4.6.3. Let for a mapping ϕ : R2 → Y (where Y is a normed space) the partialderivatives ∂2ϕ/∂x1∂x2 and ∂2ϕ/∂x2∂x1 are continuous at a point x . Then

∂2ϕ

∂x1∂x2

∣∣∣∣x= ∂2ϕ

∂x2∂x1

∣∣∣∣x.

⊳ 1◦ Without loss of generality we can assume that x = 0, since our function ϕ has thesame differentiability properties at x , as the function

ϕ : h 7→ ϕ(x + h),R2 → Y

at 0. (This follows at once from Chain Rule, since the mapping h 7→ x + h has at eachpoint the derivative equal to id.)2◦ Put

y := ∂2ϕ

∂x1∂x2

∣∣∣∣0− ∂2ϕ

∂x2∂x1(∈ Y ).

Our aim is to show that y = 0.3◦ By Lemma from Functional Analysis (see Chapter 1), there exist l ∈ L (Y,R) suchthat ‖l‖ = 1 and ly = ‖y‖. Then

‖y‖ = ly = l

(∂2ϕ

∂x1∂x2

∣∣∣∣0− ∂2ϕ

∂x2∂x1

∣∣∣∣0

)l−Rule= ∂2(l ◦ ϕ)

∂x1∂x2

∣∣∣∣0− ∂2(l ◦ ϕ)

∂x2∂x1

∣∣∣∣0

4.6.2.= 0,

since the second partial derivative

∂2(l ◦ ϕ)∂xi∂x j

=l−Rule for

X=R

l ◦ ∂2ϕ

∂xi∂x j(i, j = 1, 2)

is continuous at 0 together with ∂2ϕ/∂xi∂x j . ⊲

Lemma 4.6.4. Let ϕ : Rn → Y be of class C p . Then for any σ ∈ Sn and any i1, . . . , i p ∈{1, . . . , n} it holds

∂ pϕ

∂xiσ(1) . . . xiσ(p)

= ∂ pϕ

∂xi1 . . . xip

.

In other words, partial derivatives do not depend on the order in which we differentiate.⊳ 1◦ It is sufficient to prove this for p = 2, since then we can TRANSPOSE any twoNEIGHBOUR partial differentiations, and by such transpositions we can obtain any permu-tation.2◦ for p = 2 our assertion follows from Lemma 4.6.3., since all partial derivatives at thesecond order of a C2-mapping are continuous. ⊲

Theorem 4.6.5. Let f : X → Y be of class C p . Then for each x ∈ X the p-th derivativef (p)(x) is symmetrical.


⊳ f (p)(x)hσ(1) . . . hσ(p)

Comp.Rule= ∂ p

∂ tσ(1) . . . ∂ tσ(n)

∣∣∣∣0

f (x + tσ(1)hσ(1) + . . .+ tσ(p)hσ(p)︸︷︷︸obv= t1h1+...+tph p

)

4.6.4.= ∂ p

∂ t1 . . . ∂ tn

∣∣∣∣0

f (x + t1h1 + . . .+ tph p)

Comp.Rule= f (p)(x)h1 . . . h p. ⊲

(We can apply Lemma 4.6.4., since the mapping (t1, . . . , tn) 7→ f (x + t1h1 + . . . +tnhn),Rn → Y is of class C p as the composition of the C∞-mapping (t1, . . . tn) 7→x + t1h1 + . . .+ tnhn (a constant plus a (continuous) linear mapping) and f ∈ C p .)

Corollary 4.6.6. Let f : X1 × . . .× Xn → Y be of class C p . Then partial derivatives

∂ p f

∂X i1 . . . X ip

(i1, . . . , in ∈ {1, . . . , n})

do NOT depend on the order in which we differentiate.This means, e.g., that ∂2 f (x)/∂X1∂X2 and ∂2 f (x)/∂X2∂X1 define one and the same

bilinear mapping X1 × X2→ Y :

∀h1 ∈ X1, h2 ∈ X2...∂2 f (x)

∂X1∂X2h1h2 =

∂2 f (x)

∂X2∂X1h2h1.

⊳ This follows from Theorem in view of Lemma 4.4.1. ⊲

Remark. This corollary justifies notations of the type

∂3

∂X21∂X2

.

4.7 Polynomials

Let X,Y be vector spaces. We say that a mapping p : X → Y is a homogeneouspolynomial of degree n, and we write

p ∈ Pn(X,Y ),

if there exists an n-linear mapping

u ∈ L(n X; Y ) (= L(X, . . . , X︸︷︷︸n

; Y ))

such thatp = u ◦ △,

where △ ≡ △n is the diagonal mapping, defined so:

0 x

(x,x)

∆

△ : X → Xn := X × . . .× X︸︷︷︸n

, x 7→ (x, . . . , x).

In other words,

p(x) = u(x, . . . , x).

4.7. POLYNOMIALS 63

In the case where X,Y are NORMED spaces, we say that p is a continuous homoge-neous polynomial of degree n, and we write

p ∈ Pn(X,Y ),

if there exists u ∈L (n X; Y ) with the same property.If p = u ◦ △ we say that u generates p, or that p is generated by u.We put also

P0(X,Y ) := {all the CONSTANT mappings X→ Y}

(the polynomials of degree 0).In what follows we consider only homogeneous polynomials, and we will drop

“homogeneous”.

Examples.

1. Each linear mapping is a polynomial of degree 1, that is, P1 = L.

2. The power mapping x 7→ xn,R→ R is a continuous polynomial of degree n.

3. The maping (x, y) 7→ x3 + 4xy2,R2 → R is a continuous polynomial of degree 3.⊳ This polynomial is generated, e.g., by the following two 3-linear mappings (R2)3 → R:

((x1, y1), (x2, y2), (x3, y3)) 7→ x1x2x3 + 43 (x1 y2y3 + x2y3y1 + x3y1y2),

((x1, y1), (x2, y2), (x3, y3)) 7→ x1x2x3 + 4x1y2y3,

the former being symmetrical, and the latter being not. ⊲

4. Each n-linear mapping is a continuous polynomial of degree n.⊳ For n = 2, e.g., a bilinear mapping u : X1 × X2 → Y is generated by the followingbilinear mapping U : (X1 × X2)

2→ Y :

U : ((x1, y1), (x2, y2)) 7→ 12 (u(x1, y2)+ u(x2, y1)). ⊲

5. The function q : Rn → R, x 7→ x2 ≡ x · x = x21 + . . .+ x2

n is a continuous polynomialof degree 2 (generated by the scalar product).

6. More generally, for any linear operator A : Rn → Rn , with the matrix (ai j ), themapping (quadratic form)

Rn → R, x 7→ (Ax)scal.prod.· x =

n∑

i, j=1

ai j xi x j (x = (x1, . . . , xn))

is a polynomial of degree 2. (Prove!)

7. The function C([0, 1])→ R, x 7→∫ 1

0 x2(t) dt is a polynomial of degree 2. (Prove!)

Symmetrization

For any mapping f : Xn → Y , where Y is a VECTOR SPACE (Xn :=n︷︸︸︷

X × . . .× X),we define its SYMMETRIZATION sym f by the formula

∀x1, . . . , xn ∈ X... (sym f )(x1, . . . , xn) := 1

n!

∑

σ∈Sn

f (xσ(1), . . . , xσ(n)) .


Example. Let f : R2→ R, (x, y) 7→ x − 2y. Then

(sym f )(x, y) = 12 ( f (x, y)+ f (y, x)) = 1

2 ((x − 2y)+ (y − 2x)) = − x + y2 .

Lemma 4.7.1. Let u ∈ L(n; Y ). Thena) sym u ∈ Lsym(

n X; Y );b) u ∈ Lsym(

n X; Y )⇔ sym u = u.⊳ a) For any τ ∈ Sn it holds

(sym u)(xτ (1), . . . , xτ (n)

) def=we replace 1by σ(1) etc

1

n!

∑

σ∈Sn

u(xτ (σ (1))︸︷︷︸=(τ◦σ)(1)

, . . . , xτ (σ (n))︸︷︷︸=(τ◦σ)(n)

)

:=τ◦σ=if σ runs over Sn,

τ◦σ also runs over Sn

1

n!

∑

∈Sn

u(x(1), . . . , x(n))

def= (sym u)(x1, . . . , xn),

which means that sym u is symmetrical.b) ”⇒”: if u is symmetrical then

(sym u)h1 . . . hndef= 1

n!

∑

σ∈Sn

uhσ(1) . . . hσ(n)︸︷︷︸=uh1...hn

= 1

n!uh1 . . . hn = uh1 . . . hn,

hence sym u = u.”⇐”: if sym u = u then u is symmetrical by a). ⊲

Lemma 4.7.2. If a polynomial is generated by u then it is also generated by sym u.⊳ Let p = u ◦ △. Then

((sym u) ◦ △) = (sym u)(x, . . . , x) = 1

n!

∑

σ∈Sn

u(x, . . . , x) = u(x, . . . , x) = p(x),

that is, p = (sym u) ◦ △. ⊲

Lemma 4.7.3. Each polynomial is generated by an UNIQUE symmetrical multilinearmapping.

In other words, if u1 ◦ △ = u2 ◦ △ and u1, u2 ∈ Sym, then u1 = u2.⊳ For CONTINUOUS mappings this follows at once from Theorem on differention ofpolynomials it the next subsection, which says that if p = u ◦ △ and u ∈ Lsym, thenu = 1/n! · p(n)(0). For ”algebraical case” we give below a scetch of the proof (you mayomit it).

For any given h1, . . . , hn put

π(t1, . . . , tn) : = u(t1h1 + . . .+ tnhn, . . . , t1h1 + . . . , tnhn)

= tn1 uh1 . . . h1 + . . .+ tn

n uhn . . . hn .

Then ∂nπ/∂ t1 . . . ∂ tn|0 is equal to the coefficient by t1t2 . . . tn . This coefficient is equal,by symmetry of u, to n!uh1 . . . hn . Hence uh1 . . . hn is uniquely defined by π . But π isuniquely defined by p, since π(t1, . . . , tn) = p(t1h1 + . . .+ tnhn). ⊲

Corollary 4.7.4. If a polynomial is generated by two multilinear mappings then they haveone and the same symmetrization:

u1 ◦ △ = u2 ◦ △ ⇒ sym u1 = sym u2.

4.7. POLYNOMIALS 65

⊳ It follows from Lemmas 4.7.1.–4.7.3. ⊲

Differentiation of polynomialsThe following theorem is a generalization of the fact that

(xn)(k) = n(n − 1) . . . (n − k + 1)xn−k .

Theorem 4.7.5. Let p ∈ Pn(X,Y ), p = u ◦ △, u ∈Lsym(n X,Y ). Then

a) p is of class C∞;b) for any k ∈ {1, . . . , n} it holds

p(k) ∈ Pn−k(X,Lsym(k X; Y )),

viz., for k < np(k) = n(n − 1) . . . (n − k + 1)uk ◦ △n−k, (1)

whereuk(x1, . . . , xn−k) := u(x1, . . . , xn−k, ·, . . . , ·︸︷︷︸

k

), (2)

that is,

p(k)(x)h1 . . . hk = n(n − 1) . . . (n − k + 1)u(x, . . . , x︸︷︷︸n−k

, h1, . . . , hk), (3)

and for k = np(x) ≡ n!u.

c) For any natural k > np(k) = 0.

⊳ 1◦ p ∈ C∞ by Theorem on composition of C p-mappings, since p = u ◦ △, andboth u (as a continuous multilinear mapping) and△ (as a continuous linear mapping) areC∞-mappings.2◦ We have

p′(x)h = (u ◦ △)′(x)hChainRule= u′(△x) · △h = u′(x, . . . , x) ◦ (h, . . . , h)

Quasi−Leibniz= u(h, x, . . . , x)+ . . .+ u(x, . . . , x, h)

u∈Sym= nu(x, . . . , x, h),

which means that for k = 1 the formula (1) is true. Let us fix this:

(u ◦ △n)′ = nu1 ◦ △n−1. (4)

3◦ Let (1) is true for 1, . . . , k. Then

p(k) = n(n − 1) . . . (n − k)uk ◦ △n−k .

Hence

p(k+1) = (pk)′ = n(n − 1) . . . (n − k + 1)(uk ◦ △n−k)′

(4),applied touk◦△n−k= n(n − 1) . . . (n − k + 1)(n − k) (uk)1︸︷︷︸

(2)= uk+1

◦△(n−k)−1︸︷︷︸=△n−(k+1)

,


which means that (1) is true for k + 1, and, by induction, b) is true.4◦ Since p(n) = const, all the subsequent derivatives are zero. ⊲

Remark. You can obtain (1) using Computation Rule. (Do it!)

Corollary 4.7.6. Let pbe a continuous polynomial of degree n, generated by a (continuous)symmetrical n-linear mapping u. Then p(0) = 0 and

p(k)(0) ={

0, if k 6= n,n!u, if k = n.

4.8 Taylor Formula

At first a lemma:

Lemma 4.8.1. Let a mapping r : X → Y be k times differentiable at 0, and let

r(0) = 0, r ′(0) = 0, r ′′(0) = 0, . . . , r (k)(0) = 0.

Then

r(h) = o(‖h‖k

)(:⇔ r(h)

‖h‖kK

‖h‖→0, ‖h‖6=00

).

⊳ By induction. 1◦ For k = 1 we have

r(0+ h︸︷︷︸=h

) = r(0)︸︷︷︸=0

+ r ′(0)h︸︷︷︸=0

+r(h)

(that is, r is equal to its rest term), so r(h) = o(‖h‖) by the definition of differentiability.2◦ Let for k − 1 our assertion is true. Then

‖r(h)‖‖h‖k

= ‖r(h)− r(0)‖‖h‖k

MVT= 1

‖h‖ksup

0<t<1

∥∥r ′(th)∥∥ ‖h‖

trick= sup0<t<1

tk−1

∥∥r ′(th)∥∥

‖th‖k−1 K‖h‖→0

0,

by the induction assumption; indeed if ‖h‖ → 0 then ‖th‖ = |t| ‖h‖ → 0 uniformlyin t ∈ (0, 1), and hence

∥∥r ′(th)∥∥ /‖th‖k−1 → 0, since r ′ satisfies the conditions of the

lemma for k − 1. ⊲

Taylor Formula

Theorem 4.8.2. Let f : X → Y be of class Ck in U ⊂ X. Then for x ∈ U

f (x + h) = f (x)+ f ′(x)+ 12 f ′′(x)h2 + . . .+ 1

k! f (k)(x)hk + r(h), (1)

wheref (s)(x)hs := f (s)(x) h . . . h︸︷︷︸

s times

,

andr(h) = o(‖h‖k).

4.8. TAYLOR FORMULA 67

⊳ 0◦ We can rewrite (1) in the form

x

f

f(2)

r = f −k∑

i=1

1

i !pi ,

where

f (h) := f (x + h)− f (x),

and pi denotes the polynomial generated by f (i)(x) (whichis a continuous symmetrical i -linear mapping, since f is ofclass

C p in U ). Thus the graph of f (see the picture) is also the graph of f when consideredwith respect to the translated (on the picture dotted) axes. By Lemma 4.8.1., all we needis to verify that r(0) = 0, r ′(0) = 0, . . . , r (k)(0) = 0.1◦ r(0) = 0, since f (0) = 0 and each polynomial is equal to 0 at 0.2◦ For any j = 1, . . . , k

r ( j )(0)(2)= f ( j )(0)︸︷︷︸

obv= f ( j)(x)

−k∑

i=1

1

i !p( j )

i (0)︸︷︷︸4.7.6.=

{j ! f ( j)(x) if i = j0 if i 6= j

= 0. ⊲

Case Rn → R

Corollary 4.8.3. Let a function Rn → R have continuous partial derivatives up to orderk in an open set U ⊂ Rn . Then for any x ∈ U

f (x + h) = f (x)+n∑

i=1

∂ f (x)

∂xihi +

1

2!

n∑

i, j=1

∂2 f (x)

∂xi∂x jhi h j + . . .

+ 1

k!

n∑

i1,...,ik=1

∂k f (x)

∂xi1 . . . ∂xikhi1 . . . hik + r(h),

(h = (h1, . . . , hn) ∈ Rn)

where r(h) = o(‖h‖k).Here ‖·‖ is ANY norm in Rn . (If ‖·‖1 is another norm in Rn , then r(h) = o(‖h‖k1)

also, since any two norms in Rn are equivalent.)

Chapter 5

Extreme Problems

I. PROBLEMS WITHOUT CONSTRAINTS

5.1 Generalized theorem of Fermat

Definition. Let X be a topological space, and let f be a functional on X, f : X → R. Wesay that f has a local minimum at a point x ∈ X , and we write

x ∈ Locmin f,

if f attains at x its minimal value in some neighbourhood of x . Thus,

x ∈ Locmin f :⇔ ∃U ∈ Nbx ∀x ∈ U... f (x) ≥ f (x).

Theorem 5.1.1. Let X be a normed space, and let a functional f : X → R has a localminimum at a point x .

a) If for some h ∈ X there exists Dh f (x), then Dh f (x) = 0.b) If f is G-differentiable at x , then f ′(x) = 0.

⊳ a) If x ∈ Locmin f then 0 ∈ Locminϕ, where ϕ : R → R, t 7→ f (x + th). By theclassic Fermat’s theorem, ϕ(0) = 0; but Dh f (x) = ϕ(0).

b) It follows from a), since f ′(x)h = Dh f (x). ⊲

Example. If X = Rn and f has the partial derivatives of the first order at the pointx ∈ Locmin f , then all these partial derivatives are equal to 0:

∂ f (x)

∂x1= . . . = ∂ f (x)

∂xn= 0.

5.2 Necessary and sufficient conditions of locmin

Theorem 5.2.1. (on necessary conditions and sufficient conditions of the second order).Let X ∈ NS, f : X → R, f ∈ C2(x).

69

70 CHAPTER 5. EXTREME PROBLEMS

a) (Necessary conditions) If x ∈ Locmin f then f ′(x) = 0, and

∀h ∈ X... f ′′(x)h2 ≥ 0. (1)

b) (Sufficient conditions) If f ′(x) = 0 and ∃α > 0 such that

∀h ∈ X... f ′′(x)h2 ≥ α ‖h‖2 , (2)

then x ∈ Locmin f .

⊳ By Taylor formula,

f (x + h) = f (x)+ f ′(x)h + 1

2f ′′(x)h2 + r(h), (3)

wherer(h) = o(‖h‖2). (4)

a) Let x ∈ Locmin f . Then f ′(x) Theorem 5.1.1.= 0. Further let h ∈ X . It holds f (x + th) ≥f (x) for all sufficiently small t ∈ R. Hence,

12 t2 f ′′(x)h2+r(th) = 1

2 f ′′(x)(th)2+r(th)(3)= f (x + th) − f (x)︸︷︷︸

≥0

− f ′(x)︸︷︷︸0

(th) ≥ 0 (5)

for all sufficiently small t . Butr(th) = o(t2), (6)

since (without loss of generality h 6= 0)

|r(th)|t2 = ‖h‖2 |r(th)|‖th‖2︸︷︷︸

(4)→t→0

0

Kt→0

0.

So (5) is possible, only if f ′′(x)h2 ≥ 0.b) Let f ′(x) = 0, and let (2) be fulfilled. Then

f (x + h)− f (x)(3)= 1

2 f ′′(x)h2︸︷︷︸(2)≥ α‖h‖2

+r(h) ≥ α2 ‖h‖2 + o(‖h‖2) ≥ 0

for all sufficiently small ‖h‖. Hence, x ∈ Locmin f . ⊲

Conditions (1) and (2) in (5.2.1.) are, respecively, the condition of non-negativityand the condition of strict positivity of the second derivative f ′′(x) in the sense of thefollowing definition:Definition. Let X ∈ NS, u ∈ L(X, X;R) (bilinear functional).

a) u is said to be non-negative if the corresponding polynomial is non-negative, thatis, if

∀h ∈ X... u(h, h) ≥ 0, (7)

and positive if the corresponding polynomial is positive at any non-zero vector, thatis, if

∀h ∈ X\0 ... u(h, h) > 0. (8)

5.2. NECESSARY AND SUFFICIENT CONDITIONS OF LOCMIN 71

b) u is said to be strictly positive if

∃α > 0 ∀h ∈ X... u(h, h) ≥ α ‖h‖2 . (9)

It is evident thatstrict positivity⇒ positivity⇒ non-negativity

In general the inverse implications are not true:

(Counter-) examples.1. The functional R2 × R2 → R, ((x1, y1), (x2, y2)) 7→ x1x2 is non-negative, but is notpositive.2. The functional ℓ2 × ℓ2 → R, (x, y) 7→ ∑∞

i=1(1/i !)xi yi is (evidently) positive, but isnot strictly positive (verify!).

Finite-dimensional caseIn FINITE-DIMENSIONAL case positivity is equivalent to strict positivity: If u ∈

L(Rn,Rn;R) is positive, then u is strictly positive.⊳ Denote by S the unit sphere in Rn (defined by the equation ‖x‖ = 1), put p := u ◦ △,and consider the restriction p|S . It is clear that this restriction is continuous (since infinite dimensional case any bilinear functional is continuous). Further, S is compact, beingclosed (S = ‖·‖−1 (1)) and bounded. Hence, p|S attains its minimal value, say α. Since uis positive, we have α > 0. Thus,

‖x‖ = 1⇒ u(x, x) ≥ α > 0. (10)

So for any h 6= 0

u(h, h) = u

(‖h‖ h

‖h‖ , ‖h‖h

‖h‖

)= ‖h‖2 u

(h

‖h‖ ,h

‖h‖

)

︸︷︷︸(10)≥ α

≥ α ‖h‖2 ,

which means that u is strictly positive. ⊲

Further, in FINITE-DIMENSIONAL case the positivity condition (2) takes the form

∀h = (h1, . . . , hn) ∈ Rn\0 ...n∑

i, j=1

∂2 f (x)

∂xi∂x jhi h j > 0. (11)

This condition is none more then the condition of positive definiteness of Hesse matrix ofthe function f at the point x . Thus, by 1.9, for f : Rn → R, strict positivity of f ′′(x) isequivalent to positive-definiteness of Hesse matrix of f at x . The latter may be establishedwith the aid of SILVESTER CRITERION from algebra:

SILVESTER CRITERION. A squaire matrix A is positive definite iff all its principalminors det Ak (k = 1, . . . , n) are positive.

A a a a ... a11 12 13 1n1

A a a a ... a21 22 23 2n2

A a a a ... a31 32 33 3n3

A=A a a a ... an1 n2 n3 nnn

... ... ... ... ......

72 CHAPTER 5 EXTREME PROBLEMS

Remark. For the case of local MAXIMUM we have to replace all “>” in (1) and (2) by “<”.The strict NEGATIVITY of f ′′(x) is equivalent to NEGATIVE definiteness of the Hesse matrixA; the latter is equivalent to the conditions: det A1 < 0, det A2 > 0, det A3 < 0, det A4 >

0, . . .⊳ Apply Silvester criterion to −A. ⊲

II. PROBLEMS WITH CONSTRAINTS

5.3 Setting of the problem

Definition. Consider the following EXTREME PROBLEM WITH CONSTRAINTS (for definite-ness, the case of minimum): for a given function f : X → R (X ∈ NS) and a given setA ⊂ X (the constraints), to find all points in A, where the RESTRICTION f |A has its localminimum:

Locmin( f |A) =? (1)

Of course, we equippe A with the induced topology, so that

a ∈ Locmin f |A ⇔ ∃U ∈ Nba(X)∀x ∈ U ∩ A... f (x) ≥ f (a) (a ∈ A).

If A = X , we obtain a problem without constraints.Definition. By smooth (extreme) problem we shall mean a problem (1) with A given byan equation

A = g−1(0), (2)

where g : X → Y is a (sufficiently) smooth (e.g., of class C1) mapping from our normedspace X into some another normed space Y . In other words,

A = {x ∈ X | g(x) = 0}. (3)

Example. The problem with the constraints A ⊂ R2 given as follows: A = {(x, y}| x = 1}is a smooth problem with Y = R and g(x) = x − 1.

5.4 General (non-smooth) problems: necessary conditionof locmin

y

xA

grad faυ

At first consider a motivating example. Let f : R2 → R, A = {(x, y)|x ≥ 0}, f ∈ Dif,and let a = (x, y) ∈ Locmin f |A. Then

f ′(a) = 0 if a ∈ int A (that is, if x > 0),∂ f (a)∂x ≥ 0, ∂ f (a)

∂y = 0 if a ∈ fr A (that is, if x = 0).

This follows from a general theorem to be proved below,but it is clear by itself: in the first case we have in fact, locally(that is, in some neighbourhood of a), a problem withoutconstraints, so Fermat theorem is applicable.

In the second case (x = 0) our function f cannot have a strictly negative derivative in xat a, since it would mean that f STRICTLY decreases at a in x-direction, which contradicts

5.4. GENERAL PROBLEMS: NECESSARY CONDITION OF LOCMIN 73

to local minimality at a. The conditions ∂ f (a)/∂x ≥ 0, ∂ f (a)/∂y = 0 mean that thegradient (∂ f (a)/∂x, ∂ f (a)/∂y) of f and the unit outer NORMAL vector Ev = (−1, 0) toA at a have opposite directions. In general, as we shall see, the vector opposite to thegradient at a point of local minimum, must lie in the NORMAL CONE to A at this point.

Tangent vectorsDefinition. Let X be a normed space, A ⊂ X, a ∈ A. We say that a vector h ∈ X istangent to A at a, and we write h ∈ Ta A, if there exist a sequence {an} of points in A anda sequence {Tn} of positive real number, such that an converges to a and T−1

n (an − a)converges to h:

h ∈ Ta A :⇔ ∃{Tn} ⊂ (0,+∞)∃{an} ⊂ A : an → a,an − a

Tn→ h.

a

Aa

h

t (a −a)2 2-1

t (a −a)1 1-1

2

a 1

It is obvious that always 0 ∈ Ta A (take an = a) and that if h ∈ Ta A then th ∈ Ta A forany t > 0 (take Tn = t−1 Tn). This means that Ta A isa CONE with the vertex at 0 (the vertex belonging to thecone).

Examples.

1. For a motion f : R“time′′→ R3 the velocity f ′(t) at a

moment t is tangent to the trajectory at the point f (t).

x

f

f’(x)

2. For a differentiable function f : R→ R any vector ofthe graph of the derivative at a point x (considered as anelement of L (R,R)) is tangent to the graph of f at thepoint (x, f (x)).

3. If A is open then ANY vector is tangent to A at eachpoint:

∀a ∈ A... Ta A = X (verify!).

In particular

∀x ∈ X... Tx X = X.

4. T0{0} = {0}.5. If Y ⋐ X (this notation means that Y is a vector subspace in X), then

∀y ∈ Y... Ty Y = Y.

Lemma 5.4.1. Let f : X → Y be differentiable at a point a, and let

an → a,an − a

TnK

n→∞ h (an, h ∈ X,Tn > 0). (1)

Thenf (an)− f (a)

TnK

n→∞ f ′(a)h. (2)


⊳f (an)− f (a)

Tn

f ∈Dif(a)= ( f (a)+ f ′(a)(an − a)+ r(an − a))− f (a)

Tntrick= f (a)

an − a

Tn︸︷︷︸(1)Kh

︸︷︷︸f ′(a)∈Cont

K f ′(a)h

+ r(an − a)

‖an − a‖︸︷︷︸

‖an−a‖(1)K0

r∈SmallK0

‖an − a‖Tn︸︷︷︸(1)K‖h‖

︸︷︷︸→0

→ f ′(a)h. ⊲

Normal vectorsNORMAL vectors to a set A ⊂ X are not “in reality” vectors in X , they are COvectors,

that is, elements of the spaceX∗ :=L (X,R)

(which is called the DUAL space to X ; recall that R∗ ≈ R, (Rn)∗ ≈ Rn).Definition. Let X ∈ NS, A ⊂ X, a ∈ A. We say that an element h∗ ∈ X∗ is normal to Aat a, and we write

h∗ ∈ Na A,

if h∗ (as a linear function on X) is NON-POSITIVE on the tangent cone to A at a:

h∗ ∈ Na A :⇔ ∀h ∈ Ta A... h∗ · h ≤ 0. (3)

(Recall that we write lh ≡ l · x ≡ l(h) for linear l.)In the case X = Rn you can IDENTIFY a linear function

l(x1, . . . , xn) = l1x1 + . . .+ ln xn

with the vector (l1, . . . , ln) (in the SAME Rn!) and think about l · h as about the SCALAR

PRODUCT.Once again, it is clear that Na A is a cone, containing 0 as the vertex.

Examples.

1. For a (smooth) curve in R3, the normal cone at a point is the normal plane to the curveat this point.

K -

K+

0

2. Let K+ and K− be the positive and the negative quadrants in R2,resp. Then

N0 K+ = K−, N0 K− = K+.

3. For an OPEN set A the normal cone at any point is trivial:

∀a ∈ A... Na A = {0} (verify!).

In particular ∀x ∈ X... Nx X = {0}.

4. N0{0} = X .

5.5. SMOOTH PROBLEMS: SUFFICIENT CONDITIONS 75

Necessary condition of locmin

Theorem 5.4.2. Let X ∈ NS, f : X → R, a ∈ A ⊂ X, f ∈ Dif(a). If a ∈ Locmin f |Athen

− f ′(a) ∈ Na A.

⊳ Let us suppose that − f ′(a) 6∈ Na A. Then, by definition, ∃h ∈ Ta A:

− f ′(a)h > 0. (4)

By definition of a tangent vector, ∃an → a (an ∈ A) ∃Tn > 0:

an → a,an − a

Tn→ h.

So

0

for all sufficientlygreat n since

an→a and a∈Locmin f |A(recall that Tn>0)

≤ f (an)− f (a)

Tn

Lm.5.4.1.K f ′(a)h

(4)< 0, a contradiction! ⊲

Remark. That we require in the theorem f ∈ Dif(a), not merely f ∈ DifG(a), is essential,as the following counter-example shows:

y

z

x

0

A

gr f (0)

A

Example. Let A be the circle in R2 shown on the picture, andlet f : R2 → R be defined by the rule

f (x, y) ={

0 if (x, y) ∈ Ax if not.

Then f ∈ DifG(0), with f ′(0) = (1, 0), and 0 ∈ Locmin f |A,but

− f ′(0) = (−1, 0) 6∈ N0 A = {(x, y)| x = 0}, (= y-axis).

The point is that the set A is not “star-like”.NB In this example T0 gr f 6= gr f ′(0). ⊳⊳ gr f = (A×0)∪(gr f ′(0)\ A) (see the picture);T0 gr f = gr f ′(0) ∪ x-axis ⊲⊲

Remark. For f ∈ Dif, the generalized Fermat theorem follows from Theorem 5.4.2. andExample 3 from prewious set of examples.

5.5 Smooth problems: sufficient conditions

IDEA OF LAGRANGE. The idea of Lagrange was to reduce the problem with constraints inquestion:

Locmin f |A =? (A = g−1(0))

to a problem WITHOUT constraints for some new function 8 (instead of f ). The mostssimple way to construct 8 : X → R, starting from f and g, is to consider some LINEAR

(continuous) function λ : Y → R and to put

8 := f + λ ◦ g. (1)

Such a function 8 is called LAGRANGE FUNCTION, and the functional λ ∈ Y ∗ in (1) iscalled LAGRANGE MULTIPLIERS (plural!).


If Y = R, then λ ∈ R∗ ≈ R is just a number (Lagrange multiplier), if Y = Rn , thenλ ∈ (Rn)∗ ≈ Rn is a vector (λ1, . . . , λn) (Lagrange multipliers).

Theorem 5.5.1. (sufficient conditions for a smooth problem). Consider a smooth problem

Locmin f |A =?, A = g−1(0).

If for some λ ∈ Y ∗ Lagrange function 8 = f + λ ◦ g has a local minimum at a pointa ∈ A then

a ∈ Locmin f |A.

⊳ a ∈ Locmin( f + λ ◦ g)obv.⇒ a ∈ Locmin ( f + λ ◦ g)|A︸︷︷︸

obv.= f |A+ (λ ◦ g)|A︸︷︷︸g|A=0= 0

⇒ a ∈ Locmin f |A. ⊲

Remark. The condition “∃λ ∈ Y ∗ : a ∈ Locmin( f + λ ◦ g)” is NOT necessary for“a ∈ Locmin f |A”, as the following counter-example shows:

Example. X = R2, Y = R, f (x, y) = x, g(x, y) = x + x2; here A = {x = 0} ∪ {x =−1}, 0 ∈ Locmin f |A, but ∀λ ∈ R

... 0 6∈ Locmin( f + λg). (Verify!)

5.6 Smooth problems: tangent cone to A = g−1(0)

gr f(x,f(x))

x

gr f’(x)Y Y

X X

Theorem 5.6.1. (on the tangent cone to a graph).Let X,Y ∈ NS, f : X → Y , and let f ∈ Dif(x).Then

(1)T(x, f (x)) gr f = gr f ′(x).

Here gr f dentotes the graph of f :

gr f := {(x, f (x))| x ∈ X} ⊂ X × Y.

⊳ 1◦ gr f ′(x) ⊂ T(x, f (x)) gr f . ⊳⊳ Let (h, k) ∈ gr f ′(x), that is, k = f ′(x)h. Take ANY

sequence Tn ↓ 0, and put xn := x + Tn h, yn := f (xn). Then (xn, yn) ∈ gr f , and

(xn, yn) = (x + Tn h, f (x + Tn h))f ∈Cont(x),Tn h→0

Kn→∞ (x, f (x)),

(xn, yn)− (x, f (x))

Tn= (x + Tn h, f (x + Tn h))− (x, f (x))

Tn

obv=(

h,f (x + Tn)− f (x)

Tn h

)f ∈Dif(x)

Kn→∞ (h, f ′(x)h) = (h, k);

hence, (h, k) ∈ T(x, f (x)) gr f. ⊲⊲

2◦ T(x, f (x)) gr f ⊂ gr f ′(x).⊳⊳ Let (h, k) ∈ T(x, f (x)) gr f , that is, ∃{xn} ⊂ X, ∃{Tn} ⊂ (0,+∞):

(xn, f (xn)) Kn→∞ (x, f (x)), (2)

(xn, f (xn))− (x, f (x))

TnK

n→∞ (h, k). (3)

5.6. SMOOTH PROBLEMS: TANGENT CONE TO A = G−1(0) 77

Relation (2) means thatxn → x, f (xn)→ f (x). (4)

Relation (3) means that

xn − x

Tn→ h,

f (xn)− f (x)

Tn→ k. (5)

By Lemma 5.4.1., it follows from (4) and (5) that

f (xn)− f (x)

Tn→ f ′(x)h. (6)

Comparing (6) with the second relation in (5), we conclude (by the uniqueness of limit ina Hausdorff space) that k = f ′(x)h. But this means that (h, k) ∈ gr f ′(x) ⊲⊲ ⊲

Remark. In Step 1◦ we used just G-differentiability of f at x , but in Step 2◦ we haveused F-differentiability essentially, and this condition of F-differentiability is essentialfor validity of the theorem, as the following counter-example shows:

Example. Let A be a CIRCLE, shown on the picture, and let f : R2 → R be defined bythe rule

yA

x0

f (x, y) ={

0 if (x, y) 6∈ Ax if (x, y) ∈ A

Then f ∈ DifG(0), f ′(0) = 0, gr f ′(0) = R2 × 0, but T0 gr f =(R2 × 0)∪ (R(1, 0, 1)). (Verify! Compare Ex. 2.9! That example also issuited!)

Theorem 5.6.2. (on the tangent cone to g−1(0)). Let X,Y ∈ BS, g : X → Y, A =g−1(0), a ∈ A (that is, g(a) = 0), g ∈ C1

G(a), g′(a) ∈ Sur (that is, g′(a) is SURJECTIVE:g′(a)X = Y ), and let the kernel

K := ker g′(a) :=(g′(a))−1︸︷︷︸

pre−image, ratherthan the

inverse mapping!

(0) = {k| g′(a)k = 0} (7)

SPLITS the space X in the sense that there exists a vector subspace L in X such that:

(i) K , L ∈ BS (when equipped by the induced norm);(ii) X = K ⊕ L (that is, X = K + L and K ∩ L = {0});

(iii) X ≈ K × L (that is, more precisely, the mapping (k, l) 7→ k + l, K × L → X) isa (linear) homeomorphism). Then

Ta A = ker g′(a). (8)

Note, that K as the pre-image of a closed set is CLOSED, so the condition K ∈ BSis fulfilled automatically (a closed set in a complete metric space is also complete, whenequipped by the induced metric).

Note also, that in FINITE-DIMENSIONAL case (X = Rn) ANY vector subspace splitsthe whole space 1, so you can forget about this condition if you wish deal just withfinite-dimensional situation.

1⊳ Choose an orthonormal base in K , and extend it to an orthonormal basis in Rn ; the subspace, generatedby the “new” basis vectors, will be the desirable L . ⊲


⊳ 1◦ Without loss of generality (wlog) we can assume that a = 0. Otherwise we considera new mapping g : X → Y , defined by the rule g(h) = g(a + h). It is clear thatg′(0) = g′(a), and that A := g−1(0) = g−1(0)− a = A − a, that is, A is the translationof A byf the vector−a, so that T0 A = Ta A.

L

V

U

K

A

0

2◦ By (i)-(iii), we can assume that g is a mapping K × L → Y .Denote by g′K and g′L the corresponding partial derivatives:

(9)g′K (0)k =(k∈K )

g′(0) · (k, 0), g′L(0)l =(l∈L)

g′(0) · (0, l).

3◦ g′K (0) = 0. ⊳⊳ K = ker g′(0). ⊲⊲

4◦ g′L(0) ∈ Sur. ⊳⊳ Since g′(0) is surjective

∀y ∈ Y ∃(k, l) ∈ K × L : y = g′(0) · (k, l) (9)= g′K (0)k︸︷︷︸3◦=0

+g′L(0)l = g′L(0)l.

This means that g′L(0) is surjective. ⊲⊲

5◦ g′L(0) ∈ Inj (that is, is INJECTIVE). ⊳⊳ Let l ∈ L and g′L(0)l = 0. Then

g′(0) · (0, l) (9)= g′K (0)0︸︷︷︸0

+ g′L(0)l︸︷︷︸0

= 0,

which means that

(0, l) ∈ ker g′(0) = K × 0 (we identify K and K × 0!).

If follows (by (ii)) that l = 0. ⊲⊲

6◦ By 4◦ and 5◦, g′L(0) ∈ Bij (is BIJECTIVE). Hence, g′L(0) ∈ Iso(L,Y ), in finite dimensi-onal case automatically (any linear map is continuous!), and in general case by so-calledOpenness Principle from functional analysis.7◦ By Implicit Function Theorem, ∃U ∈ Nb0(K ) ∃V ∈ Nb0(L) ∃ϕ : U → V :

1. grϕ = A ∩ (U × V ),2. ϕ ∈ Dif(0),3. ϕ′(0) =−(g′L(0))−1

inverse map!◦ g′K (0)︸︷︷︸

3◦=0

= 0.

It follows from 3), that grϕ′(0) = K × 0 = ker g′(0).8◦ By Theorem 5.6.1., grϕ′(0) = T0 A. ⊲

Theorem 5.6.2. says in particular that the tangent cone to A is a VECTOR SUBSPACE

in X . In such a case any normal vector is ORTHOGONAL to each tangent vector:

Lemma 5.6.3. (on orthogonality). Let X ∈ NS, A ⊂ X, a ∈ A. If Ta A is a vectorsubspace in X, then

∀h ∈ Ta A ∀h∗ ∈ Na A... h∗ · h = 0.

If h∗ · h = 0 then we say that h∗ and h are orthogonal (in finite-dimensional case itis usual orthogonality).⊳ By the definition of a normal vector, h∗ · h ≤ 0. But we have also −h ∈ Ta A (sinceTa A is a vector subspace), so it holds also h∗ · (−h) ≤ 0, that is, h∗ · h ≥ 0. Hence,h∗ · h = 0. ⊲

Corollary 5.6.4. In conditions of Lemma 5.6.3.,

∀h∗ ∈ Na A... Ta A ⊂ ker h∗.

5.7. SMOOTH PROBLEMS: NECESSARY CONDITION 79

5.7 Smooth problems: necessary condition

Theorem 5.7.1. (on Lagrange multipliers). Let X,Y ∈ BS, and let (see the diagram)A := g−1(0), a ∈ A, f ∈ Dif(a), g ∈ C1

G(a), g′(a) ∈ Sur. If

a ∈ Locmin f |A, (1)

then∃λ ∈ Y ∗(=L (Y,R)) : ( f + λ ◦ g︸︷︷︸

=:8

)′(a) = 0. (2)

X fK R

gց λր

Y

Thus, the theorem says, that there exists Lagrange multipliers λ such that the corre-sponding Lagrange function satisfies at the point a Fermat condition.

Before the proof consider a model example:

Example. Let X = R2, Y = R, f (x, y) = x2 + y2, g(x, y) = x − 1. Here A ={(x, y)|x = 1}︸︷︷︸=:{x=1}

, and ∀b ∈ A... Tb A = {x = 0}, Nb A = {y = 0}.

d

y

levellines of f

A

level lines of g

grad f(a)=(2,0)

q x

N Aa

T Aa

Now, λ ∈ R∗ ≈ R may be here identified with a real numbers, so our Lagrange functionhas the form

8(x, y) = x2 + y2 + λ(x − 1).

Condition (2) gives (for a =: (x, y))

(3)8′(a) = (2x + λ, 2y) = (0, 0).

Condition a ∈ g−1(0) gives

(4)x − 1 = 0.

It follows from (3) and (4) that

x = 1, y = 0 (that is, a = (1, 0)), λ = −2.

Thus the unique candidate for a point of local minimum is a = (1, 0), and it is easyto verify that really a ∈ Locmin f |A.

The necessary condition− f ′(a) ∈ Na A means here (since Ta A is a vector subspaceof R2) that grad f |a⊥Ta A. So grad f |a is orthogonal at a both to the level line of f (asthe gradient of f ) and to the level line of g (which is just A). It follows (by the formula∂ϕ/∂ν = gradϕ · Eν), that both f and g have zero derivative in the direction of the commontangent line to these level lines, that is, zero derivative in y: ∂ f/∂y|a = ∂g/∂y|a = 0.Further, ∂g/∂x |a = 1 6= 0, so far some λ ∈ R it holds ∂ f/∂x |a = λ∂g/∂x |a (namely, forλ = 2, for we have ∂ f/∂x |a = 2). So, for this λ, both f and λg have ONE AND THE SAME

partial derivatives at a and hence one and the same derivative at a. Hence their differencef − λg has ZERO derivative at a.

We see that our desired Lagrange multiplier is

λ = −λ.


Roughly speaking, by adding λg to f we “rotate” the graph of f around the HORIZONTAL

line, passing through the point (a, f (a)) and parallel to the mentioned common tangentline, until we obtain the HORIZONTAL tangent plane to the graph.

THE PROOF. ⊳ 1◦ To avoid appealing functional analysis, we restrict ourselves by theFINITE-DIMENSIONAL case (X = Rn,Y = Rm).2◦ By Theorem 5.2.1.− f ′(a) ∈ Na A.3◦ By Theorem 5.6.2., Ta A = ker g′(a).

4◦ ker g′(a) ⊂ ker f ′(a). ⊳⊳ ker g′(a) 3◦= Ta A2◦;5.6.4.⊂ ker(− f ′(a)) obv= ker f ′(a). ⊲⊲

5◦ ALGEBRAICAL LEMMA (on passing through). Let X,Y, Z be vector spaces, and letϕ ∈ L(X, Z), γ ∈ L(X,Y ). Let γ be SURJECTIVE. Then the following two conditions areequivalent:(a) ker γ ⊂ kerϕ;(b) ∃λ ∈ L(Y, Z) : ϕ = λ ◦ γ (ϕ can be “passed through γ ”).

X ϕK Z

γց λր

Y

⊳⊳ (b)⇒(a): Let x ∈ ker γ , that is, γ x = 0. Then ϕx(b)= λ( γ x︸︷︷︸

0

) = 0, that is,

x ∈ kerϕ.(a)⇒(b): Take any element y ∈ Y . Since γ ∈ Sur, ∃x ∈ X : γ x = y. Put

γ y := ϕx .

This definition is correct, that is, doesn’t depend on the choice of x . Indeed, if we haveanother x ′ with the property γ x ′ = y then

γ (x ′ − x) = γ x ′ − γ x = 0 ⇛ x ′ − x ∈ ker γ(a)⇛ x ′ − x ∈ kerϕ ⇛ ϕ(x ′ − x) = 0

⇛ ϕx ′ = ϕx .

By the very construction, ϕ = λ ◦ γ . ⊲⊲

6◦ By 4◦, we can apply 5◦ to the diagram

X f ′(a)K R

g′(a)ց λր

Y

and conclude that ∃λ ∈ L(Y,R)(=L (Y,R), since Y = Rm) : f ′(a) = λ ◦ g′(a).7◦ Put λ = −λ. Then

( f + λ ◦ g)′(a)ChaineRule= f ′(a)+ λ ◦ g′(a) = f ′(a)− λ ◦ g′(a) 6◦= 0. ⊲

5.8 Problems with equations and inequalities

As an application consider a classic extreme problem with equations and inequalities tofind local minimums of a given function Rn → R on the set

A = {x ∈ Rn| g1(x) = 0, . . . , gk(x) = 0; gk+1(x) ≥ 0, . . . , gl(x) ≥ 0}.

5.8. PROBLEMS WITH EQUATIONS AND INEQUALITIES 81

(All the function are supposed to be sufficiently smooth.)Description of a method. At a point a ∈ Locmin f |A we have for i = k+ 1, . . . , l eithergi (a) = 0 or gi(a) > 0.

According to which of these two possibilities is realized, there 2l−k possibilities. Amethod of solution the extreme problem is to consider one by another all the possibilitiesand apply to each of them Theorem on Lagrange multipliers (TLM) with an appropriateg.

We illustrate this method on the following simple example:

Example. LetA = {x | g1(x) = 0, g2(x) ≥ 0}.

PutA1 := g−1

1 (0), A2 := g−12 (0), B2 := g−1

2 ((0,+∞)).The sets A1 and A2 are closed (as the pre-images of the closed set {0}), and the set B2 isopen (as the pre-image of the open set (0,+∞)). It is clear that

A = (A1 ∩ A2) ∪ (A1 ∩ B2),

the two intersections being disjoint. Let a ∈ Locmin f |A.There are two possibilities: 1) a ∈ A1 ∩ A2; 2) a ∈ A1 ∩ B2.In the first case

a ∈ Locmin f |Aa ∈ A1 ∩ A2

}A1∩A2⊂A⇒ a ∈ Locmin f |A1 ∩ A2︸︷︷︸

obv= (g1,g2)−1(0)

.

Thus we can apply TLM with g = (g1, g2) : Rn → R2.In the second case

a ∈ Locmin f |Aa ∈ A1 ∩ B2

}A1∩B2⊂A⇒ a ∈ Locmin f |A1∩B2 ⇒ a ∈ Locmin f | A1︸︷︷︸

g−11 (0)

.

(Proof of the last implication: since B2 ∈ Op, there exists (U ∈ Nba(Rn) : U ⊂ B2 ⇛

(A1 ∩ B2) ∩U = A1 ∩U.)Thus we can apply TLM with g = g1 : Rn → R.

Chapter 6

Riemann integral in Rn

6.1 Partitions and cubes

t0 t1 t2 t3 t4

a b

A partition of a (bounded closed) interval I = [a, b] is a finite sequence p = (t0, t1, . . . , tk),such that

a = t0 ≤ t1 ≤ . . . ≤ tk = b.

In such a case we writep ∈ Part I.

We say that the intervals Ji = [ti−1, ti ] are the intervals of the partition p, and we write

Ji ∈ Intv p.

A cube Q in Rn is a product I1 × . . .× In of n intervals Ii = [ai , bi ] (maybe ai = bi forsome i ), we write

Q ∈ Cube Rn.

The volume of a cube is defined as the product of the lengths of its edges:

vol Q := (b1 − a1) . . . (bn − an).

For example, any point x ∈ Rn considered as a one-point set {x} is a cube of zero volume.

S

A partition P of a cube Q = I1 × . . . × In is a sequence (p1, . . . , pn), where pi is apartition of the interval Ii :

P ∈ Part(I1 × . . .× In) :⇔ P = (p1, . . . pn), pi ∈ Part Ii .

A cube S of a partition P is a product J1× . . .× Jn , where each Ji isan interval of the partition pi :

S ∈ Cube P :⇔ S = J1 × . . .× Jn, Ji ∈ Intv pi .

Let P = (p1, . . . , pn) and P ′ = (p′1, . . . , p′n) be two partitions of a cube Q. We say thatP ′ is a refinement of P and we write

P ′ ≻ P

if for such i the sequence pi is a subsequence of p′i .

83

84 CHAPTER 6. RIEMANN INTEGRAL IN RN

6.2 Riemann integral

Let f : M → R,M ⊂ Rn . In this chapter we ALWAYS suppose that f is bounded, that is,its image f (M) is a bounded subset of R:

f ∈ BdM :⇔ f (M) ∈ Bd(R).

If f ∈ BdM then we can EXTEND f to a bounded function on the WHOLE Rn by putting

f (x) = 0 for x ∈ Rn\M.

So without loss of generality (wlog) we can (and we shall) assume that our functions aredefined on the whole space.

For a given cube Q in Rn and a given partition P of Q we define the lower sum LP fand the upper sum UP f of f , corresponding to P , by the formulas

LP f :=∑

S∈Cube P

(infS

f ) vol S, UP f :=∑

S∈Cube P

(supS

f ) vol S.

(Here, e.g., infS f denotes the infimum of f on S, that is, inf( f (S)).) By boundeness off , both the infS f and supS f are ever FINITE.

The lower integral of f over Q is defined as the SUPREMUM of all lower sums, andthe upper integral as the INFIMUM of all upper sums:

L∫

Qf := sup {LP f |P ∈ Part Q}︸︷︷︸

=:L

, U∫

Qf := inf {UP f |P ∈ Part Q}︸︷︷︸

=:U

.

L U

L fQ

U fQ

R

As we shall see in a minute (Lemma 6.3.3.), the set L lies TO THE LEFT of the set U, soboth integrals are finite, and the lower one is less (by “less” wemean “≤”, for “<” we say “strictly less”):

L∫

Qf ≤ U

∫

Qf.

We say that f is integrable over Q in the sense of Riemann if the lower sum is EQUAL

to the upper one:

L U

fQ

f ∈ (R) IntQ :⇔ L∫

Qf = U

∫

Qf.

In such a case this common value is called the Riemann integralof f over Q and is denoted by

(R)∫

Qf or (R)

∫

Qf (x1, . . . , xn) dx1 . . . dxn.

As a rule we shall drop (R) and “in the sense of Riemann.”

Examples.

1. f = const = c;∫

Q c = c vol Q. ⊳ For any partition P of Q

LP c =∑

S∈Cube P

c vol S = c∑

S∈Cube P

obv.= c vol Q,

and analogously for UP c. ⊲

6.3. CRITERION OF EXISTENCE OF RIEMANN INTEGRAL 85

2. f =

1α

1-1 0∫

[−1,1] f = 1 (does not depend on the value α at 0). ⊳ Exercise. ⊲

3. The Dirichlet function fDir : R→ R, defined by the rule

fDir(x) ={

1 if x 6∈ Q0 if x ∈ Q

is NOT integrable over, say, [0, 1]. ⊳ ∀P ∈ Part[0, 1]... LP fDir = 0,UP fDir = 1, so

L∫

[0,1] fDir = 0,U∫

[0,1] fDir = 1. ⊲

4. Any function CONTINUOUS on a cube (that is, in each point of this cube) IS integrableover this cube. This follows from the LEBESGUE THEOREM below. For n = 1 we obtainthe classic integral of one-dimensional analysis.

The Dirichlet function from example 3 is an example of so called indicator functions:Definition. The indicator (or characteristic) function of a subset M of a set X is definedby the rule

χM (x) :={

1 if x ∈ M0 if x 6∈ M

6.3 Criterion of existence of Riemann integral

Let f ∈ Bd(Rn), Q ∈ Cube Rn .

Lemma 6.3.1. ∀P ∈ Part Q... Lp f ≤ UP f .

⊳ ∀S ∈ Cube P... infS f ≤ supS f . ⊲

Lemma 6.3.2. If P, P ′ ∈ Part Q and P ′ ≻ P then

LP f ≤ LP ′ f ≤ UP ′ f ≤ UP f.

S’

S

j

⊳ The middle inequality is true by Lemma 6.3.1. Let us prove theleft one. Any cube S of the partition P is built from some cubesS′1, . . . , S′k of the partition P ′ (k depends on S), and so

vol S = vol S′1 + . . .+ vol S′k;hence

(infS

f ) vol S′ = (infS

f )︸︷︷︸≤infS′1 f

vol S′1 + . . .+ (infS

f )︸︷︷︸≤infS′k f

vol S′k

If we sum these inequalities over all S ∈ Cube P , we obtain LP f ≤ LP ′ f . The rightinequality may be proved analogically. ⊲


L U

L fP

L fP’’

U fP’

U fP’’

Lemma 6.3.3. ∀P, P ′ ∈ Part Q... LP f ≤ UP ′ f .

⊳ Take a partition P ′′ of Q such that P ′′ ≻ P and P ′′ ≻ P ′. Then

LP f6.3.2.≤ LP ′′ f

6.3.1.≤ UP ′′ f6.3.2.≤ UP ′ f. ⊲

Criterion of integrability. A bounded function f : Rn → R is integrable over a cube Qin Rn if and only if

L U

c

(1)∀ε > 0 ∃P ∈ Part Q : (0 ≤)UP f − LP f︸︷︷︸=:△P f

≤ ε.

⊳ 1◦ Let f ∈ IntQ , that is sup L = inf U = c, where

L := {LP f |P ∈ Part Q}, U := {UP f |P ∈ Part Q}.

L fP’ c U fP’’

L fP U fP

²ε/2 ²ε/2

Let ε > 0 be given. By the definitions of supremum and infimum

(2)∃P ′ ∈ Part Q : c − LP ′ f ≤ ε

2,

(3)∃P ′′ ∈ Part Q : UP ′′ f − c ≤ ε

2.

Let P be a refinement both of P ′ and P ′′. Then

UP f − LP f6.3.2.≤ UP ′′ f − LP ′ f

6.3.1.,6.3.2.≤ ε

2+ ε

2= ε. O.K.

2◦ Vice verse, let (1) be true. Then

}L f U fP P

UL

†ε

inf U︸︷︷︸≤UP f

− sup L︸︷︷︸≥LP f

≤ UP f − LP f ≤ ε.

Since ε was arbitrary we conclude that inf U− sup L ≤ 0, that is,

inf U ≤ sup L .

But by Lemma 6.3.3.,inf U ≥ sup L .

Hence,inf U = sup L,

which means that f ∈ IntQ . ⊲

Remark that the difference UP f −LP f which appears in Criterium, may be writtenin the form

1P f = UP f − LP f =∑

S∈Cube P

(supS

f − infS

f ) vol S.

This justifies the followingDefinition. Let f ∈ BdRn ,M ⊂ Rn . We define the oscillation of the function f on theset M so:

�M f := (supM

f )− (infM

f ).

If M = Rn we omit M in the notation.

6.4. NULL SETS 87

Example. � sin = 2.Thus

1P f =∑

S∈Cube P

(�S f ) vol S.

Lemma 6.3.4. (on monotony). If P ′ ≻ P then1P ′ f ≤ 1P f .⊳ It follows at once from Lemma 6.3.2. ⊲

Exercises.1

0 11/2

A 1.∫

[0,1]2 χA = 12 (not depending on taking A WITH the boundary or

WITHOUT).

2.∫

[a,b] id(=∫ b

a x dx)= 1

2 (b2 − a2) (do not use Newton-Leibniz formula!).

3. If f, g ∈ IntQ then f + g ∈ IntQ , and∫

Q( f + g) = ∫Q f + ∫Q g.[Hint: infS f + infS g ≤ infS( f + g), supS f + supS g ≥ supS( f + g).]

Below we omit for short “if. . . then. . . ”.

4.∫

Q c f = c∫

Q f .5. f ≤ g ⇒

∫Q f ≤

∫Q g.

6.∣∣∣∫

Q f∣∣∣ ≤

∫Q | f |. [Hint: �S | f | ≤ �S f .]

7. f = g on Q\F, #F < ∞ (F is FINITE) ⇒∫

Q f =∫

Q g (changing a function on afinite set does not change the integral).

8. ∀P ∈ Part Q...∫

Q f =∑S∈Cube P

∫S f .

6.4 Null sets

We say that a set N ⊂ Rn is a set of Lebesgue measure zero or a null set if for any ε > 0there exists (AT MOST) COUNTABLE family {Qi } of cubes in Rn , which covers N and issuch that the sum of the volumes of the cubes is less than ε:

N ∈ Null :⇔ ∀ε > 0 ∃{Qi }i∈N : Qi ∈ Cube Rn,⋃

i∈NQi ⊃ N,

∑

i∈Nvol Qi ≤ ε.

(We can, without loss of generality, assume that the family is just countable, since addingto our family any countable number of one-point set does not change the sum of volumes.)

In the integration theory null sets are “negligible” in a sense, as we shall see.

Remarks.

1. Emphasize that Qi may have zero volume.2. A cube Q has POSITIVE volume iff it has the non-empty interior:

vol Q > 0⇔ ◦Q 6= ∅.

3. We obtain an EQUIVALENT definition if we replace the condition ∪Qi ⊃ N by

⋃

i∈N

◦Qi ⊃ N.


(Prove as an EXERCISE. [Hint: for fixed l1, . . . , ln (the lengths of the edges of a cube)the function t 7→ (l1 + t) . . . (ln + t),R→ R is continuous and strictly increasingat 0.])

Examples.1. Any point (that is, a set {x}) is null. ⊳ {x} ∈ Cube Rn, vol{x} = 0. ⊲

2. Any finite set is null.

3. Any countable set is null. ⊳ Numerate the points of our set into a sequence {xi} andtake Qi := {xi }. ⊲

4. Any straight line in R2 is null. ⊳ EXERCISE. ⊲

5. A set in Rn which has an INTERIOR point is NOT null:◦M 6= ∅ ⇒ M 6∈ Null. In particular

no cube with positive volume is null; in fact, vol Q > 0⇔ Q 6∈ Null (prove!). (But thereexist NOT-NULL sets (even in R) with the EMPTY INTERIOR, cf. Exam. 6.7 2.)

Lemma 6.4.1. Any subset of a null set is null.⊳ Obviously. ⊲

Lemma 6.4.2. The union of a countable family of null sets is a null set.⊳ Let Ni ∈ Null for each i ∈ N, and let ε > 0 be given. Let us write ε = ε1 + ε2 + . . .,where each εi > 0. For each i there exists a countable family {Qi j } j∈N of cubes, such that

⋃

j

Qi j ⊃ Ni ,∑

j

vol Qi j ≤ εi .

Then the family {Qi j }i, j∈N (which is countable!) covers⋃

i Ni and satisfies the inequality∑

i, j

vol Qi j =∑

i

(∑

j

vol Qi j

︸︷︷︸≤εi

) ≤∑

i

εi = ε. ⊲

Lemma 6.4.3. If a null set N in Rn is COMPACT then for any ε > 0 there exists a FINITE

family Q1, . . . , Qk of cubes such that ∪ki=1 Qi ⊃ N,

∑ki=1 vol Qi ≤ ε.

⊳ By Remark 3, there exists a countable family {Qi } of cubes such that⋃

i∈N

◦Qi ⊃ N,

∑

i∈Nvol Qi ≤ ε.

By compactness of N we can choose a finite subcovering, and this finite family is whatwe need. ⊲

Remark 4. The compactness condition in Lemma 6.4.3. is essential (see Exercise 2 below).Exercises

1. The Cantor set, the intersection of the sequence

0

0

0

1

1

1

1/3� 2/3

1/9 2/9 2/3 7/9 8/9

n=1

n=2

n=3

is a (compact) null set.

2. Let M be the set of rational numbers between 0 and 1, M := Q ∩ [0, 1]. Then Mis NULL as a countable set. Prove that there exists no FINITE family I1, . . . , Ik ofintervals, such that ∪Ii ⊃ M and

∑length Ii < 1. [Hint: use induction in k.]

6.5. OSCILLATION 89

6.5 Oscillation

B (x)x

M

δ

Let X be a normed space, let M ⊂ X , and let f : X → R be a bounded function. Theoscillation of f on M at a point x ∈ X (usually x ∈ M) is defined bythe formula

ωM f (x) := limδ↓0

�Bδ(x)∩M f,

where � is the “global” oscillation, defined in Section 6.3:

�M f = supM

f − infM

f.

This limit exists since supBδ(x)∩M f ↓ and infBδ(x)∩M f ↑ as δ ↓ 0. If M = X we omitthe index M .

Examples.

1. f = 1 1/2 ; ω f (0) = 1, ω(−∞,0) f (0) = 0, ω[−1,0] f (0) = 12 .

2.

f (x) ={

sin 1|x | if x 6= 0,

0 if x = 0;ω f (0) = 2.

Remark. The valueωM f (x) does not change if we replace the norm in X by any equivalentnorm.

Lemma 6.5.1. Let X be a normed space, M ⊂ X, x ∈ M, and let f : X → R be abounded function. Then f is continuous at x if and only if the oscillation of f at x is equalto zero:

f |M ∈ Cont(x)⇔ ωM f (x) = 0.

⊳ ”⇒”: Let f |M is continuous at x . Consider arbitrary ε > 0. By supposed continuitythere exists δ > 0 such that

∀y ∈ Bδ(x) ∩ M... | f (y)− f (x)| ≤ ε

2 .

Thensup

Bδ(x)∩Mf ≤ f (x)+ ε

2 , infBδ(x)∩M

f ≥ f (x)− ε2 ,

whence it follows that�Bδ(x)∩M f ≤ ε.

HenceωM f (x) ≤ ε.


Since ε was arbitrary we conclude that ωM f (x) = 0.

y x}f(x)f(y)

B (x)δ

”⇐”: Let ωM f (x) = 0, and let ε > 0 be given. Then there exists δ > 0 such that

(1)�Bδ(x)∩M f ≤ ε.Hence for each y ∈ Bδ(x) ∩ M it holds

| f (x)− f (y)| obv≤ supBδ(x)∩M

f − infBδ(x)∩M

f(1)≤ ε,

which means that f |M is continuous at x . ⊲

Lemma 6.5.2. Let X be a normed space, M ⊂ X, and let f : X → R be a boundedfunction. Then for any ε > 0 the set

Aε := {x ∈ M|ωM f (x) < ε} (strict inequality!)

is OPEN IN M.⊳ Let x ∈ Aε. We need to show that there exist δ > 0 such that

B (y)

B (y)

M

xy

δ

γ

(2)◦Bδ(x) ∩ M ⊂ Aε.

But indeed (since ωM f (x) < ε) there exist δ > 0 and (0 <)ε′ < ε

such that

(3)�Bδ(x)∩M f ≤ ε′.

Let y ∈ ◦Bδ(x) ∩ M . Obviously there exists γ, 0 < γ ≤ δ, suchthat Bγ (y) ⊂ Bδ(x). Then Bγ (y) ∩ M ⊂ Bδ(x) ∩ M , and hence

�Bγ (y)∩M f ≤ �Bδ(x)∩M f(3)≤ ε′,

which implies thatωM f (y) ≤ ε′ < ε.

This means that y ∈ Aε, and (2) is true. ⊲

Let us return in Rn , equipped, say, by the Euclidean norm (‖·‖2).

Lemma 6.5.3. Let Q be a cube in Rn , and let f : Rn → R be a bounded function suchthat

∃ε > 0 ∀x ∈ Q... ωQ f (x) < ε (strict inequality!). (4)

Then there exists a partition P of Q such that∑

S∈Cube P

(�S f ) vol S < ε vol Q.

⊳ 1◦ Consider arbitrary point x ∈ Q. By (4),

∃δx > 0 : �Bδx (x)∩Q f < ε. (5)

Qx

QB (x)δx

x

Let Qx be a cube with the center at x such that◦Qx 6= ∅ and Qx ⊂ Bδx (x). (Such a cube

exists, since ‖·‖∞ ∼ ‖·‖2.) The cubes { ◦Qx} form an open coveringof Q. By compactness of Q, we can choose a finite subcovering,say

{ ◦Qx1, . . . ,

◦Qxk} (x1, . . . , xk ∈ Q).

6.6. LEBESGUE THEOREM 91

Thus,◦Qx1∪ . . . ∪ ◦Qxk

⊃ Q. Therefore if we put Qi := Qxi ∩ Q, it holds

Q1 ∪ . . . ∪ Qk = Q.

2◦ Obviously each set Qi is a cube, and there exists a partition P of Q such that each cubeS of P is cointained in some Qi , that is,

S ⊂ Qxi ∩ Q ⊂ Bδxi(xi ) ∩ Q.

Q4

Q2

Q1

Q3Then

�S f ≤ �Bδxi(xi )∩Q f

(5)< ε.

Hence,∑

S∈Cube P

(�S f )︸︷︷︸<ε

vol S < ε∑

S∈Cube P

vol S = ε vol Q. ⊲

6.6 Lebesgue theorem

The following result is fundamental.

Theorem 6.6.1. (Lebesgue). Let Q be a cube in Rn , and let f : Rn → R be a boundedfunction. Denote by discontQ f the set of all points where f |Q is not continuous:

discontQ f := {x ∈ Q| f |Q 6∈ Cont(x)}.

Then f is integrable over Q if and only if discontQ f is a null set:

f ∈ IntQ ⇔ discontQ f ∈ Null .

⊳ For short put A := discontQ f , and put for each ε > 0

Aε := {x ∈ Q| ωQ f (x) ≥ ε}.

(This set is COMPLEMENTARY in Q to the set Aε from Lemma 6.5.2. (with M = Q).) Wehave

ALm 6.5.2.= {x ∈ Q| ωQ f (x) > 0} obv=

∞⋃

k=1

{x ∈ Q| ωQ f (x) ≥ 1k },

that is,A =

⋃

k∈NA1/k. (1)

”⇒” 1◦ Let f ∈ IntQ . We need verify that A ∈ Nul. In view of (1) it is sufficient (byLemma 6.4.2.) to show that for each δ > 0

Aδ ∈ Nul . (2)

Let ε be an arbitrary positive number. By (Corollary of) Criterium of integrability (Section6.3) there exists a partition P of Q such that

∑

S∈Cube P

(�S f ) vol S ≤ ε. (3)


2◦ Denote by N the union of the boundaries of all cubes of P:

N :=⋃

S∈Cube P

fr S.

Obviously, N ∈ Null; hence there exists a countable (end even finite, since N is compact;see Lemma 6.4.3.) family {Qi } of cubes in Rn such that

⋃Qi ⊃ N,

∑vol Qi ≤ ε. (4)

3◦ Now denote by S the set of all cubes S of P such that at least one INTERIOR point of Sbelongs to Aδ:

S := {S ∈ Cube P|◦S ∩ Aδ 6= ∅}. (5)

It is clear that

∀S ∈ S... �S f ≥ δ (6)

(since for some x ∈ ◦S it holds ω f (x) ≥ δ). Further,

∑S∈S vol S

(6)≤

trick

∑S∈S δ

−1(�S f ) vol S = δ−1 ∑S∈S(�S f ) vol S

≤obv

δ−1∑S∈Cube P (�S f ) vol S

(3)≤ εδ−1.

(7)

4◦ The cubes from {Qi } and from S altogether cover Aδ , since each point of Aδ either liesin N or is interior for some cube S, and

∑vol Qi +

∑

S∈Svol S

(4),(7)≤ ε(1+ δ−1).

But here δ is fixed, and ε is arbitrary. We conclude that (2) is true.”⇐” 5◦ Let A ∈ Null. We prove that f ∈ IntQ , using the same Criterion. Let ε be

an arbitrary positive number. The set Aε = {x ∈ Q| ωQ f (x) ≤ ε} is null (since Aε ⊂ A)and is compact (Aε is bounded since Aε ⊂ Q, and Aε is closed since (Aε)c ∩ Q = {x ∈Q| ωQ f (x) < ε} is open in Q by Lemma 6.5.2., hence Aε is closed in Q and thereforeis closed (since Q is closed!)). By Lemma 6.4.3., there exists a finite number of cubesQ1, . . . , Qk such that

6.6. LEBESGUE THEOREM 93

Q1 Q2 Q3

Q4

A ε

(8)k⋃

i=1

◦Qi ⊃ Aε,

(9)k∑

i=1

vol Qi ≤ ε.

6◦ Put

Black := (∪ki=1 Qi ) ∩ Q

(shadowed on the second picture), and

White := Q\(∪ki=1

◦Qi ) (= cl(Q\Black)).

It is clear that

(10)∀x ∈White... ωQ f (x) < ε

(since White ⊂ (Aε)c).

White

P’

S’

PS’

Black

7◦ Obviously there exists a partition P ′ of Q suchthat each cube S′ of P ′ lies either in White or inBlack.

Put

W := {S′ ∈ Cube P ′| S′ ⊂White},B := {S′ ∈ Cube P ′| S′ ⊂ Black}.

It is clear from (9) that

(11)∑

S ′∈Bvol S′ ≤ ε.

8◦ For each S′ ∈ W there exists by Lemma 6.5.3.(in view of 10)) a partition PS ′ such that

(12)1PS′ f < ε vol S′.

9◦ Finally, there exists a partition P of Q such that

P ≻ P ′

and

∀S′ ∈W... P|S ′ ≻ PS ′ .

(Here P|S ′ denotes, naturally, the “restriction” ofthe partition P to S′.) Let us show that P is what weneed.10◦ For this end put

(0 ≤)M := �Q f

(M is finite, since f ∈ Bd). It is clear that

∀S ∈ Cube P... �S f ≤ M. (13)


11◦ Now,

1P f =∑

S∈Cube P

(�S f ) vol S =∑

S ′∈W

∑

S∈Cube PS⊂S ′

(�S f ) vol S

︸︷︷︸1

+∑

S ′∈B

∑

S∈Cube PS⊂S ′

(�S f ) vol S.

︸︷︷︸2

We have

1 = 1P|S′ f6.3.4.≤ 1PS′ f

(12)≤ ε vol S′, (14)

2(13)≤

∑

S ′∈B

∑

S∈Cube PS⊂S ′

M vol S = M∑

S ′∈B

∑

S∈Cube PS⊂S ′

vol Sobv= M

∑

S ′∈Bvol S′

(11)≤ Mε (15)

Thus

1P f =∑

S ′∈W1 + 2

(14),(15)≤

∑

S ′∈Wε vol S′+Mε

obv= ε∑

S ′∈Wvol S′

︸︷︷︸obv≤ vol Q

+Mε ≤ ε(vol Q+M).

But ε was arbitrary small. So by Criterium, f ∈ IntQ . ⊲

Exercises.1. Let

f (x) =

1q if x ∈ Q and x = p

q , p, q beingmutually prime integers,

0 ifx 6∈ Q.

Prove that discont f = Q (dense in R!). So f is integrable over any (bounded) interval.

3. Let f : Rn → Rm, f = ( f1, . . . , fm ), Q ∈ Cube(Rn), and let each componentfunction fi is integrable over Q. Let further g : Rm → R be a CONTINUOUS function.Prove that the composition g ◦ f

Rn fK Rm g

K R

is integrable over Q. (In particular the product f1 f2 of two integrable functions is inte-grable.) [Hint: discont(g ◦ f ) ⊂ ∪m

i=1 Discont fi .]

6.7 Jordan measurable sets

Now we define the integral over ARBITRARY (bounded) set. Let f be a bounded functionon Rn , and let M be a bounded set in Rn . We say that f is integrable over M and wewrite f ∈ IntM , if the product χM f (recall that χM denotes the indicator function of M)is integrable over some cube Q ⊃ M , and in such a case we put

∫

Mf :=

∫

QχM f.

6.7. JORDAN MEASURABLE SETS 95

(The result does not depend on the choice of Q, and sometimes we shall drop Q.)Further we say, that M is Jordan measurable if the constant function 1 is integrable

over M , and we define the volume of M as the corresponding integral:

M ∈ JMeas :⇔ ∃∫

M1 =

∫χM =: vol M.

This definition evidently agrees with our original definition of volume for cubes.

Theorem 6.7.1. A bounded set M in Rn is Jordan measurable iff its boundary is a nullset:

M ∈ JMeas⇔ fr M ∈ Null .

⊳ This follows at once from Lebesgue Theorem, since DiscontχM = fr M . ⊲

Remark. A null set (and even countable!) may be non-Jordan-measurable (Example 1below); an open set may be non-Jordan-measurable (Example 2). [All the null sets and allthe open ones are LEBESGUE MEASURABLE.]Example 1. The set Q ∩ [0, 1] is not Jordan measurable. (Cf. Example 1.2 3.)Example 2. We construct a bounded open set in R by the following procedure. Write

12 = ε1 + ε2 + . . . (εi > 0)

(e.g. εi = 2−i−1).Step 1. Take the interval of the length ε1 with the center common with the center of

the interval [0, 1]:

0 1

ε1

Step 2. Take 2 open intervals, each of the length 12ε2 with the centers common, resp.,

with the centers of 2 intervals complementary in [0, 1] to the open interval constructed inStep 1:

0 1

ε1ε212- ε2

12-

Step 3. Take 4 open intervals, each of length 14ε3, with the centers common, resp.,

with the centers of 4 intervals complementary in [0, 1] to the open intervals constructedin Steps 1 and 2:

0 1

ε1ε212− ε2

12−ε3

14− ε3

14− ε3

14− ε3

14−

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

The union M of all constructed by this procedure open intervals is a (bounded) openset, which is not Jordan measurable (but is LEBESGUE MEASURABLE, with LEBESGUE

MEASURE 1/2).

Exercises


1. Prove the assertion of Example 2. [Hint: prove at first that fr M = [0, 1]\M; thenprove by induction in k, that there exists no finite covering of fr M by intervals withthe sum of the lengths < 1/2 (cf. Lemma 6.4.3.).]

2. Any COMPACT null set is Jordan measurable (and its volume is equal to 0.)3. Prove that

vol M = 0⇒ M ∈ Null

and that if M ∈ JMeas then

vol M = 0⇐ M ∈ Null

4. f ∈ Bd, vol M = 0⇒∫

M f = 0.5.

vol M = 0⇔ ∀ε > 0 ∃k ∈ N ∃Q1, . . . , Qk ∈ Cube :⋃k

i=1 Qi ⊃ M,∑ki=1 vol Qi ≤ ε

⇔ ∀ε > 0 ∃k ∈ N ∃Q1, . . . , Qk ∈ Cube :⋃ ◦

Qi ⊃ M,∑ki=1 vol Qi ≤ ε

6. vol M = 0⇒ vol M = 0. (Remark that M ∈ Null 6⇒ M ∈ Null!)

6.8 Fubini Theorem

This theorem says about possibility to reduce calculation of the integral over a product tocalculation of integrals over the factors.

Theorem 6.8.1. (Fubini). Let A be a cube in Rn , let B be a cube in Rm , and let f :A × B → R be a (bounded) integrable function. Put for each x ∈ A

l(x) := L∫

Bf (x, ·), u(x) := U

∫

Bf (x, ·).

Then both the functions l and u are integrable over A, and∫

A×Bf =

∫

Al =

∫

Au.

(Recall that the “x-section” f (x, ·) of f is the function A→ R, y 7→ f (x, y).)⊳ 1◦ Obviously, any partition P of A× B may be written as a pair P = (PA, PB), wherePA ∈ Part A, PB ∈ Part B . We have

S ∈ Cube P ⇔ S = SA × SB, SA ∈ Cube PA, SB ∈ Cube PB .

2◦ For any P = (PA, PB) ∈ Part(A × B)

LP f =∑

S∈Cube P

(inf

Sf

)vol S =

∑

SA∈Cube PASB∈Cube PB

(inf

SA×SBf

)vol(SA × SB)︸︷︷︸=vol SA vol SB

=∑

SA∈Cube PA

∑

SB∈Cube PB

(inf

SA×SBf

)vol SB

︸︷︷︸1

vol SA.

6.8. FUBINI THEOREM 97

3◦ 1 ≤ infSA l. ⊳⊳ ∀x ∈ SA... infSA×SB f ≤ inf{x}×SB f = infSB f (x, ·). Hence

∀x ∈ SA... 1 ≤

∑

SB∈Cube PB

(infSB

f (x, ·)) vol SB = LPB f (x, ·) ≤ L∫

Bf (x, ·) = l(x).

We conclude that 1 ≤ infSA l. ⊲⊲

4◦ LP f ≤ LPA l.

⊳⊳ LP f2◦=

∑

SA∈Cube PA

1 vol SA3◦≤

∑

SA∈Cube PA

(infSA

l) volSA = LPA l. ⊲⊲

5◦ UP f ≥ UPA u.⊳⊳ Analogously. ⊲⊲

6◦ LPA l ≤ UPA u. ⊳⊳ L PA ll≤u≤ LPA u ≤ UPA u. ⊲⊲

7◦ By 3◦–5◦,LP f ≤ LPA l ≤ UPA u ≤ UP f.

It follows that

supP∈Part(A×B)

LP f

︸︷︷︸=∫

A×B f

≤ supPA∈Part(A)

LPA l

︸︷︷︸=L

∫A l

≤ infPA∈Part(A)

UPA l︸︷︷︸

=U∫

A l

≤ infP∈Part(A×B)

UP f︸︷︷︸=U

∫A×B f

.

Therefore

L∫

Al = U

∫

Al =

∫

A×Bf,

which means that∫

A l =∫

A×B f .8◦ The other equation may be proved analogously. ⊲

0 0

Dirichlet function

Example. Let

f (x, y) ={

1 if x = 12 , y 6∈ Q,

0 otherwise.

Then l = 0, u = χ{1/2}, and∫

[0,1]2f =

{∫[0,1] l =

∫[0,1] 0∫

[0,1] u =∫

[0,1] χ{1/2}

}= 0.

(Remark that f is integrable, though it has a NON-integrable section f (1/2, ·).)Notations. It is convenient to use the following “classic” notations:∫

f ≡∫

f (x, y) dx dy,∫

f (x, ·) ≡∫

f (x, y) dy,∫

f (·, y) ≡∫

f (x, y) dx

(and analogously for L∫,U∫

). E.g. we use these notations in the corollaries below.

Corollary 6.8.2. (change of the order of integrations).

(∫

A×Bf (x, y) dx dy =

)∫

A(L∫

Bf (x, y) dy) dx =

∫

B(L∫

Af (x, y) dx) dy

(and any from two “L” or both of them may be replaced by “U”).


Corollary 6.8.3. (reduction of a double integral to a repeated one). Let, in the conditionsof Fubini Theorem, for each x ∈ A the function f (x, ·) is integrable. Then

∫

A×Bf (x, y) dxdy =

∫

A

(∫

Bf (x, y) dy

)dx .

The condition of Corollary 6.8.3. is fulfilled, e.g., for continuous functions (since anysection of a continuous function is also continuous).

In particular, for f ∈ Cont we have (by induction)∫

[a1,b1]×...×[an ,bn]f ≡

∫ b1

a1

. . .

∫ bn

an

f (x1, . . . , xn) dx1 . . . dxn

=∫ bn

an

(. . .

(∫ b1

a1

f (x1, . . . , xn) dx1

). . .

)dxn.

Chapter 7

Partition of unity. Change ofvariables

7.1 Smooth indicators

supp f

f

For a function f : Rn → R its support supp f is defined as the closure of the set, wheref is not equal to 0:

supp f := cl{x ∈ Rn | f (x) 6= 0}.

We say that a C∞-function f is a smooth indicator of a set A ⊂ Rn if f |A = 1.

K

G

supp f

Theorem 7.1.1. For any open set G in Rn and any compact set K ⊂ Gthere exists a smooth indicator f of K with the support in G:

f |K = 1, supp f ⊂ G.

⊳ 1◦ Theorem is true for n = 1, K = [a, b], g = (c, d).( ([ [

c a b d⊳⊳ Step 1.

f1(x) :={

e− tg2 x if − π/2 < x < π/2,0 if not. 0-π/2 π/2

f1

It is easy to verify that f1 ∈ C∞ and supp f1 = [−π/2, π/2].

Step 2. ∀a < b ∃ f2 ∈ C∞ : f2 ≥ 0, supp f2 = [a, b]. a b

f2

⊳⊳ Put f2 := f1 ◦ l, where l =π/2

-π/2a b

⊲⊲

a b

fa,b1Step 3. ∀a < b ∃ fa,b ∈ C∞ : fa,b ≥ 0, fa,b|(−∞,a] = 0,

fa,b|[b,+∞) = 1.

⊳⊳⊳ fa,b(x) :=(∫ x

af2

)/

(∫ b

af2

). ⊲⊲⊲

99

100 CHAPTER 7. PARTITION OF UNITY. CHANGE OF VARIABLES

Step 4. Choose c and d such that c < c < a, b < d < d and put f := fc,d − fb,d .

⊳⊳⊳ c a b d

f

c d ⊲⊲⊲ ⊲⊲

2◦ Theorem is true for K = Q1,G =oQ2, where Q1, Q2 ∈ Cube(Rn).

⊳⊳ Let Q1 = [a1, b1] × . . . × [an, bn], Q2 = [c1, d1] × . . . × [cn, dn]. By 1◦, for eachi = 1, . . . , n there exists a smooth indicator fi

of [ai , bi ] with supp fi ⊂ (ci , di ). Put

f := f1 ⊗ . . .⊗ fn,

that is,

f (x1, . . . , xn) := f1(x1) · . . . · fn(xn).

It is clear that f is what we need. ⊲⊲

x K

Q’’xQ’x

3◦ General case. For any x ∈ K there exist cubes Q′x , Q′′xsuch that

x ∈oQ ′x , Q′x ⊂

oQ ′′x , Q′′x ⊂ G.

The cubesoQ′x cover K . By compactness of K , we can

choose a finite subcovering, sayoQ ′1, . . . ,

oQ ′k , with the

corresponding “outer” cubes Q′′1, . . . , Q′′k . By 2◦, for eachi = 1, . . . , k there exists a smooth indicator fi of Q′i with

supp fi ⊂oQ ′′i . Put

f :=k∑

i=1

fi .

It is clear that f ∈ C∞, f ≥ 0, f |K ≥ 1, supp f ⊂ G. At last, put

f := f0,1 ◦ f ,

where f0,1 = 0 1

1

(see Step 3 of 1◦). It is obvious that f is what we need. ⊲

7.2 Partition of unity

Let A ⊂ Rn , and let O be an open covering of A (the notation: O ∈ OC(A)). A family8of C∞-functions Rn → R is called a partition of unity for A submitted to O (the notation:8 ∈ PU(A, O)), if

1) ∀ϕ ∈ 8 ... 0 ≤ ϕ ≤ 1;2) ∀x ∈ A ∃U ∈ Nbx such that only FINITE number of functions from 8 are not

identically zero on U (the condition of local finiteness);

3) ∀x ∈ A...∑ϕ∈8 ϕ(x) = 1 (this sum is FINITE, by 2));

4) ∀ϕ ∈ 8 ∃U ∈ O : suppϕ ⊂ U (8 is submitted to O).

Remark. For any COMPACT set K ⊂ A there exists only FINITE number of functionsϕ ∈ 8such that ϕ|K 6= 0.

7.2. PARTITION OF UNITY 101

⊳ This follows from 2) and from compactness of K . ⊲

Theorem 7.2.1. For any A ⊂ Rn and any open covering O of A there exists a partitionof unity for A submitted to O.⊳ Case 1. A is compact. Without loss of generality we can assume that O is FINITE.

Q’’x

Q’x

U x

x

A

1◦ ∀x ∈ A ∃Ux ∈ O ∃Q′x , Q′′x ∈ Cube(Rn) : x ∈oQ ′x , Qx ⊂

oQ

′′x , Q′′x ⊂ Ux . The cubes

oQ ′

x cover A. By compactness of A

we can choose a finite subcovering, sayoQ ′1, . . . ,

oQ ′k , with the

corresponding outer cubes Q′′1, . . . Q′′k .

By Theorem on smooth indicators applied to Q′i andoQ ′′i , for

each i = 1, . . . , k there exists a smooth indicator fi for Q′i with

supp fi ⊂oQ ′′i .

2◦ Put onoQ ′1 ∪ . . .∪

oQ ′k =: G

9i := fi

f1 + . . .+ fk(i = 1, . . . , k). (1)

(Obviously, f1 + . . .+ fk ≥ 1 on G, hence this definition is correct.)3◦ Once again by Theorem on smooth indicators, applied this time to A and G, there existsa smooth indicator f0 of A with supp f0 ⊂ G. Put

ϕi :={

f0ψi on G,0 on Gc.

It is clear that ϕ1, . . . , ϕk are just what we need. Indeed,

suppϕi ⊂ supp fi ⊂oQ ′′i

1◦⊂ U for some U ∈ O,

and (k∑

i=1

ϕi

)∣∣∣∣∣A

=k∑

i=1

f0|A︸︷︷︸=1

(9i |A) =

(k∑

i=1

9i

)∣∣∣∣∣A

(1),A⊂G= 1.

Case 2. A = ∪∞i=1 Ai , Ai ∈ Comp, Ai ⊂◦Ai+1. Note that in this case A is open (since

A = ∪ ◦Ai ), each set Ki := Ai\◦Ai−1 is compact (verify!), each set Gi := ◦Ai+1/Ai−2 is

open, and

A

Gi

K i

Ki ⊂ Gi (see the picture). Put

Oi := {U ∩ Gi |U ∈ O}.Then Oi is an open covering of Ki , and by Case 1,

there exists a FINITE partition of unity8i for Ki submittedto Oi . Now put

ψ :=∞∑

i=1

∑

ϕ∈8i

ϕ.

This definition is correct since each point of A lies in some Gi , and for each i all thefunctions from8 j with j ≥ i + 3 have the supports OUTSIDE Gi , so on each Gi our ψ isthe sum of a FINITE number of functions ϕ.


At last put for each ϕ ∈ ∪∞i=18i

ϕ′ := ϕ

ψ.

The family of all such ϕ′ is what we need.Case 3. A is open. Put for i = 1, 2, . . .

AA

fr A

i

Bi

Ui Ui := {x ∈ Rn| dist(x, frA) < 1i },

Ai = A ∩U ci ∩ Bi .

Here Bi denotes the ball of radius i with the centerat 0, and dist(x,Y ) denotes the distance from a pointx to a set Y , that is defined by the formula

dist(x,Y ) := infy∈Y‖x − y‖ .

For any fixed set Y the function

Y := dist(·,Y )

is continuous (verify!).We claim that

∀i ... Ai ∈ Comp, Ai ⊂◦Ai+1, and A =

∞⋃

i=1

Ai .

⊳⊳ We use the following simple fact from topology: The difference of s set and an OPEN

neighbourhood of its frontier is closed. (Verify!)By this fact A ∩ U c

i ∈ Cl. Further, Bi is bounded and closed. Hence Ai is boundedand closed, that is, Ai ∈ Comp. (Note that Ui is open since Ui = −1

frA((− 1i ,+ 1

i )) andfrA is continuous function.) Other relations are obvious. ⊲⊲

Hence we can apply Case 2.General case. Put G := ∪U∈OU . By Case 3, there exists a partition of unity for G

submitted to O. It is of course also a partition of unity for A. ⊲

7.3 Partition of integral

Now we show that having a partition of unity8 for A we can represent the integral∫

A fas a sum of integrals

∫A ϕ f over ϕ ∈ 8.

Lemma 7.3.1. If A, B are Jordan measurable then

A ∪ B, A ∩ B, A\B, B\A (1)

are also Jordan measurable.⊳ Recall that a set is Jordan measurable iff its frontier is a null set. Thus frA and frB arenull sets and hence their union also is a null set. But the frontier of each from 4 sets in (1)lies in frA ∪ frB (verify!) and hence is null. ⊲

7.3. PARTITION OF INTEGRAL 103

Lemma 7.3.2. If A is a (bounded) Jordan measurable set then for any ε > 0 there existsa COMPACT Jordan measurable subset K of A such that

vol(A\K ) ≤ ε.

(Note that A\K is Jordan measurable by Lemma 7.3.1.)⊳ frA is a compact null set, hence by Lemma 1.4.3 there exists a finite number of cubesQ1, . . . , Qk such that

QKA i

k⋃

i=1

◦Qi ⊃ frA,

k∑

i=1

vol Qi ≤ ε.

Put

K := A\k⋃

i=1

◦Qi .

This set is bounded (obviously) and closed (as the difference of a set and an OPEN

neigbourhood of its frontier, see the end of the previous section). Hence K is compact.

By Lemma (7.3.1.), K is Jordan measurable (each◦Qi is obviously Jordan measurable).

At last, A\A ⊂ ∪Qi , hence

vol(A\K ) ≤∑

vol Qi ≤ ε. ⊲

Theorem 7.3.3. Let A be a (bounded) Jordan measurable set, and let f be a (bounded)function integrable over A. Let O be an open covering of A by Jordan measurable sets,and let 8 be a partition of unity for A submitted to O. Then

∫

Af =

∑

ϕ∈8

∫

Aϕ f, (2)

where the series converges ABSOLUTELY.(Recall that a series

∑ϕ∈8 aϕ (aϕ ∈ R) converges absolutely to s if for any ε > 0

there exists a FINITE set 80 ⊂ 8 such that for each FINITE set 8′, satisfying the condition80 ⊂ 8′ ⊂ 8, it holds ∣∣∣∣∣∣

s −∑

φ∈8′aφ

∣∣∣∣∣∣≤ ε.

In such a case the series∑φ∈8 |aφ| also (absolutely) converges.)

⊳ Consider an arbitrary ε > 0. By Lemma 7.3.2., there exists a compact Jordan measurableset K ⊂ A such that

vol(A\K ) ≤ ε. (3)

By Remark to the definition of a partition of unity, the set 80 of all functions ϕ from 8

such that ϕ|K 6= 0, is FINITE. For any finite 8′ such that 80 ⊂ 8′ ⊂ 8 it holds∣∣∣∣∣∣

∫

A−∑

ϕ∈8′

∫

Aϕ f

∣∣∣∣∣∣

sum isfinite=

∣∣∣∣∣∣

∫

A

f −

∑

ϕ∈8′ϕ f

∣∣∣∣∣∣≤∫

A

∣∣∣∣∣∣f

1−

∑

ϕ∈8′ϕ

∣∣∣∣∣∣


∑ϕ∈8′ ϕ|A=1≤ sup

A| f |

︸︷︷︸=:M

∫

A

∑

ϕ∈8ϕ −

∑

ϕ∈8′ϕ

= M

∫

A

∑

ϕ∈8\8′ϕ

=ϕ∈8\8′⇒ϕ|K=0

M∫

A\K

∑

ϕ∈8\8′ϕ

︸︷︷︸≤1

≤ M∫

A\K1 = M vol(A\K )

(3)≥ Mε.

Since ε was arbitrary we conclude that (2) is true. ⊲

Remark. 1. Since∑ϕ∈8 ϕ|A = 1, we can rewrite (2) so:

∫

A

∑

ϕ∈8ϕ f =

∑

ϕ∈8

∫

Aϕ f,

that is, we can change∫

and∑

with places.Remark. One can use (2) to EXTEND the definition of

∫A f to non-Jordan-measurable

or/and non-bounded sets A and non-bounded functions f . But we shall not need such anextension below.

7.4 Change of variables

The following result is a generalization of the known rule of classic analysis concerninga change of a variable in an integral.

By a change of variables in Rn we mean a (C1-)diffeomorphism g of an open setG ⊂ Rn onto an open set H ⊂ Rn , that is, a C1-bijection G → H such that the inversemapping g−1 : H → G is also of class C1.

Since any C1-mapping is continuous, both g and g−1 are continuous, thus g is ahomeomorphism.

Remember: any (C1-)diffeomorphism is a homeomorphism.Since g1 ◦ g = id and g ◦ g−1 = id, it follows by Chain Rule that for any x ∈ G and

for y := g(x)

(g−1)′(y) ◦ g′(x) = id, g′(x) ◦ (g−1)′(y) = id .

Hence (remember!)

∀x ∈ G... g′(x) ∈ Iso(Rn) (⇔ det g′(x) 6= 0).

Theorem 7.4.1. Let g : G → H (G, H ⊂ Rn) be a (C1-)diffeomorphism. Then for anyintegrable function f : H → R it holds (below we prefer write g(G) instead of H )

GgK H

fK R

∫

g(G)f =

∫

G| f ◦ g| det g′|. (1)

⊳ I. PRELIMINARIES. 1◦ This theorem is true for integrals in the extended sense mentionedin last Remark. But we shall prove this theorem only for our ”old” notion of the integral,and by this reason we shall suppose that our sets G and H are bounded and G is Jordanmeasurable. (It follows from (1) with f = 1 that H must then also be Jordan measurable.

7.4. CHANGE OF VARIABLES 105

2◦ We say that a Jordan measurable set A ⊂ G is nice for a diffeomorphism g and wewrite

A ∈ Nice(g),

if for any (bounded) integrable function f (on H )

∫

g(A)=∫

A( f ◦ g)| det g′|. (2)

It follows from (2) with f = 1 that g(A) is then also to be Jordan measurable. Thus ouraim is to prove that G is nice for g.

Ah k

Rf

h(A) k(h(A))II. CONDITIONAL PART.

1◦ A ∈ Nice(h), h(A) ∈ Nice(k) ⇒ A ∈Nice(k ◦ h).

⊳⊳ For any f integrable on k(h(A)) it holds∫

k(h(A))f =

∫

h(A)( f ◦ k)| det k ′| =

∫

A((( f ◦ k)| det k ′|) ◦ h)| det h′|

(ϕψ)◦γ=(ϕ◦γ )(ψ◦γ )=

∫

A( f ◦ k ◦ h)| det(k ′ ◦ h)|| det h′|

det(B◦C)=(det B)(det C)=

∫

A( f ◦ k ◦ h)| det ((k ′ ◦ h) ◦ h′)︸︷︷︸

Ch.rule= (k◦h)′

|. ⊲⊲

2◦ If an open Jordan measurable set A admits an open covering O by sets each of whichis a subset of A and is nice for a diffeomorphism g then A itself is nice for g.⊳⊳ For any set S ⊂ G put for short

S := g(S).

Since g is a homeomorphism, the sets U with U ∈ O form an open covering of A; notethat each U is Jordan measurable, for U is nice (see I, 2◦). Denote this covering by O.By Theorem 2.2 there exists a partition of unity 8 for A submitted to O. For any ϕ ∈ 8put ϕ := ϕ ◦ g−1 (so that ϕ = ϕ ◦ g). It is clear that the functions ϕ with ϕ ∈ 8, forma partition of unity for A submitted to O . Denote this partition by 8. We have for anyintegrable function f

U

A

g gsupp ϕ-1

U

A

supp ϕ=supp ϕ

∫

g(A)f

7.3.3.=∑

ϕ∈8

∫

Af ϕ

∃U∈Osupp ϕ⊂U=

∑

ϕ∈8

∫

Uf ϕ

U∈Nice(g)=∑

ϕ∈8

∫

U(( f ϕ) ◦ g)︸︷︷︸=( f ◦g) (ϕ ◦ g)︸︷︷︸

=ϕ

| det g′|

supp⊂U=∑

ϕ∈8

∫

A(( f ◦ g)| det g′|)ϕ 7.3.3.=

∫

A( f ◦ g)| det g′|. ⊲⊲


3◦ Let Q be a cube in g(G). If for any cube S ⊂ Q∫

S1 =

∫

g−1(S)| det g′| (3)

(that is, (2) is true for g−1(S) and f = 1), then both pre-images g−1(Q) and g−1(◦Q) are

nice for g.⊳⊳ a) The pre-image for any cube S ⊂ Q with vol S = 0 also has zero volume.⊳⊳⊳

0 =∫

S1(3)=∫

g−1(S)| det g′|︸︷︷︸≥m

(∗)> 0

≥ m∫

g−1(S)1

m>0⇒∫

g−1(S)1 = 0.

(∗): | det g′| is a continuous function on the compact g−1(S) which is nowhere 0. ⊲⊲⊲

b) For any cube S ⊂ Q and any bounded function f on G

∫

fr g−1(S)f = 0.

⊳⊳⊳ fr g−1(S)g∈Homeo= g−1(fr S). Since fr S is a finite union of cubes with zero volume

it follows by a) that fr g−1(S) is a finite union of zero volume sets and hence has itselfzero volume. But the integral of bounded function over a volume 0 set is equal to 0. ⊲⊲⊲

c) For any integrable function f on Q it holds

∀P ∈ Part(Q)... L P f =

∑

S∈Cube P

(infS

f ) vol S︸︷︷︸=∫S 1

(3)=∑

S∈Cube P

∫

g−1(S)(inf

Sf )

︸︷︷︸≤( f ◦g)|g−1(S)

| det g′|b)≤∫

g−1(Q)( f ◦ g)| det g′|.

It follows that∫

Q f ≤∫

g−1(Q)( f ◦ g)| det g′|. Analogously we conclude that the inverse

inequality is true (consider UP f ). Hence∫

Q f =∫

g−1(Q)( f ◦g)| det g′|. By b) we conclude

that also∫◦Q

f =∫

g−1(◦Q )( f ◦ g)| det g′|. ⊲⊲

III. ABSOLUTE PART. 1◦ For any permutation σ ∈ Sn , each Jordan measurable setA ⊂ Rn is nice for sσ , where

sσ (x1, . . . , xn) := (xσ(1), . . . , xσ(n)).

sσ

(It is clear that sσ is a linear bijection Rn → Rn

and hence is a diffeomorphism.)⊳⊳ The matrix of sσ has evidently the determi-

nant equal ±1, so

∫

A( f ◦ sσ ) | det s′σ |︸︷︷︸

=1

=∫

Af ◦ sσ =

∫

Af (xσ(1), . . . , xσ(n)) dx1 . . . dxn

7.4. CHANGE OF VARIABLES 107

change the order of integrationsby Fubini Theorem=

∫

sσ (A)f (y1, . . . , yn) dy1 . . . , dyn =

∫

sσ (A)f. ⊲⊲

2◦ Theorem is true for n = 1.⊳⊳ For any [α, β] ⊂ g(G)(⊂ R) it holds

∫

[α,β]1

obv= β − αNewton−Leibniz=

∫ g−1(β)

g−1(α)

g′consider two possible

cases: g′>0,g′<0=∫

g−1([α,β])|g′|.

Hence by II, 3◦ the pre-image of any open interval in g(G) is nice for g. Since thesepre-images (which are open intervals) cover G we conclude by II, 2◦ that G is nice for g.3◦ Now argue by induction. Let Theorem is true for n − 1.4◦ For any point x ∈ G there exists an open neigbourhood Ux such that

g|Ux = k ◦ h ◦ sσ , (4)

where σ ∈ Sn , and k and h are diffeomorphism, each of which DOES NOT CHANGE AT

LEAST ONE COORDINATE.⊳⊳ 5◦ We have, putting g =: (g1, . . . , gn),

det g′(x) =

∣∣∣∣∣∣∣∣∣

∂g1∂x1

. . .∂g1∂xn

.... . .

...∂gn∂x1

. . .∂gn∂xn

∣∣∣∣∣∣∣∣∣x

(∗)=(∂gn

∂x1Mn1 + . . .+

∂gn

∂xnMnn

)∣∣∣∣x.

(∗): decomposition of the determinant corresponding to the last row.Since det g′(x) 6= 0, we have

∂gn

∂xi

∣∣∣∣x6= 0, Mni |x 6= 0 for some i.

Take as σ the TRANSPOSITION of i and n. Then g ◦ sσ =: g satisfies the conditions

∂ g

∂xn

∣∣∣∣s−1σ (x)6= 0, Mnn

∣∣s−1σ (x) 6= 0.

If we decompose g into the composition of diffeomorphisms h and k as above, we obtainthe diserable decomposition (4), since g ◦ sσ = g ◦ sσ ◦ sσ︸︷︷︸

=id

= g.

x

h

k

g

(x ,...,x )1 n-1

nThus without loss of generality we can assume thati = n, so that

(5)∂gn

∂xn

∣∣∣∣x6= 0, Mnn |x 6= 0.

6◦ Put

(6)h(x) := (g1(x), . . . , gn−1(x), xn),

so that h DOES NOT CHANGE THE LAST COORDINATE.We have


det h′|x =

∣∣∣∣∣∣∣∣∣∣∣

∂g1∂x1

. . .∂g1∂xn−1

∂g1∂xn

... . . ....

...∂gn−1∂x1

. . .∂gn−1∂xn−1

∂gn−1∂xn

0 . . . 0 1

∣∣∣∣∣∣∣∣∣∣∣x

= Mnn |x(5)6= 0.

By Inverse Function Theorem, there exists an open neighbourhood Ux of x such that

h|Ux ∈ Diffeo .

7◦ Now putk := g ◦ h−1 (7)

(k is a diffeomorphism as a composition of two diffeomorphisms). This k DOES NOT

CHANGE THE FIRST n − 1 COORDINATES. Indeed, if x = (x1, . . . , xn) and h(x) =(y1, . . . , yn), then, by (6), y1 = g1(x), . . . , yn−1 = gn−1(x), yn = xn , so

k(y1, . . . , yn)(7)= g(x) = (g1(x), . . . , gn(x)) = (y1, . . . , yn−1, gn(x)). ⊲⊲

8◦ Ux ∈ Nice(h).

z z

h

y y

Q2

Q1

Q

⊳⊳ For short put x = (x1, . . . , xn−1︸︷︷︸=:y

, xn︸︷︷︸=:z

) = (y, z),

h(y, z) =: (a(y, z), z). For any cube Q = Q1 × Q2 inh(Ux ) it holds

∫

Q1

FubiniTheorem=

∫

Q2

dz∫

Q1

dy3◦=∫

Q2

dz∫

(a(·,z)−1)(Q1)

| det (a(·, z))′︸︷︷︸=∂h/∂y

|

det h′=∣∣∣∣∂h/∂y ∂h/∂z

0 1

∣∣∣∣=det ∂h/∂y

=∫

Q2

dz∫

(a(·,z)−1)(Q1)

| det h′|Fubini

Theorem=∫

h−1(Q)| det h′|.

By II, 3◦ we conclude that for any cube Q ⊂ h(Ux )we have h−1(◦Q) ∈ Nice(h). But such

the pre-images cover Ux , so, by II, 2◦, Ux is nice for h. ⊲⊲

9◦ h(Ux )Nice(k). ⊳⊳ Quite analogously. ⊲⊲

10◦ Ux ∈ Nice(g). ⊳⊳ This follows from (4), 1◦, 5◦, 6◦ and II, 1◦. ⊲⊲

11◦ G ∈ Nice(g) ⊳⊳ This follows II, 2◦, since the neighbourhoods Ux , x ∈ G, cover Gand are nice for g, by 7◦. ⊲⊲ ⊲

Corollary 7.4.2. Let g be a diffeomorphism of an open set G ⊂ Rn onto an open setH ⊂ Rn , and let A ⊂ G. If vol A = 0 then vol g(A) = 0.⊳ Exercise. ⊲

NB If g is merely a homeomorphism then vol A = 0 6⇒ vol g(A) = 0. (A counter-examplecan be constructed using two Cantor type sets, the usual one, with zero volume, and amodification, with a positive Lebesgue measure)

Chapter 8

Differential forms

8.1 Tensors

By tensor of rank k (or k-tensor), k = 1, 2, . . ., on a vector space X we mean a k-LINEAR

functionalu : X × . . .× X︸︷︷︸

k−times

→ R.

The set of all k-tensors on X we denote by Lk(X):

Lk(X) := L(X × . . .× X︸︷︷︸k

;R).

It is convenient to putL0(X) = R.

Notations. Our main special case is X = Rn , with points x = (x1, . . . , xn). We denoteby e1, . . . , en the canonical basis in Rn :

ei := (0, . . . , 0, 1[i], 0, . . . , 0),

and by πi the canonical projections in Rn:

πi x := xi .

It is clear that

πi e j = δi j :={

1 if i = j ,0 if not.

(1)

Examples.

1. L1(X) = L(X,R) =: X ′ (the dual vector space); 1-tensors are oft called covectors.

2. For a smooth function f : Rn → R

f (k)(x) ∈ Lksym(R

n) ⊂ Lk(Rn),

where Lksym(X) denotes the set of all SYMMETRIC k-linear functionals.

109

110 CHAPTER 8. DIFFERENTIAL FORMS

3. For any A ∈ L(Rn,Rn) we can define a 2-tensor u A by the formula

u A(h, k) := 〈Ah, k〉,

where 〈·, ·〉 denotes the scalar product in Rn . (The correspondence A 7→ u A is a bijectionL(Rn,Rn)→ L2(Rn)).

4. (Elasticity theory.) Let f (x) denotes the position of a point x of a (3-dimensional) bodyafter a deformation. The 2-tensor generated (in the sense of previous Example) by theoperator

12 ( f ′(x)+ f ′(x)T )− id

(here f ′(x) ∈ L(R3,R3); the symbol T denotes the transposed matrix; we identify linearoperators in Rn with their matrices ) is called deformation tensor.

F→ν→

The 2-tensor generated by the operator Eν 7→ EF , where Eν is the unit normal vector to someflat section of the body, and EF is the force that acts “on 1 cm2” ofthe surphace of that graf of our persected body for which Eν is OUTER

normal vector, is called the stress tensor. The known Hook law saysthat the stress tensor at a point linearly depend on the deformationtensor at this point. The corresponding matrix describes the elasticityproperties of our body at the point in question.

5. The determinat can be considered as a tensor:

det(h1, . . . , hn) :=

∣∣∣∣∣∣

h11 . . . h1n

. . . . . . . . .

hn1 . . . hnn

∣∣∣∣∣∣

(where hi = (hi1, . . . , hin ) ∈ Rn).The main operations over tensors are TENSOR PRODUCT and PULL-BACK.

Tensor productLet u ∈ Lp(X), v ∈ Lq(X), p, q ≥ 1. The tensor product u ⊗ v is defined by the formula

u ⊗ v(h1, . . . , h p+q ) := u(h1, . . . , h p)︸︷︷︸∈R

· v(h p+1, . . . , h p+q )︸︷︷︸∈R

.

It is clear that u ⊗ v ∈ Lp+q(X).For t ∈ L0(X) = R it is convenient to put

t ⊗ u := tu.

NB In general u ⊗ v 6= v ⊗ u.

Example. In Rn , πi ⊗ π j corresponds (in the sense of Example 3) above) to the operatorA with the matrix with just one non-zero element which is equal to 1:

0... 0

. . . 1 . . .

0... 0

[i]

[ j ]

⊳ (πi ⊗ π j )(h, k) = (πi h) · (π j k) = hi k j = 〈Ah, k〉. ⊲

8.1. TENSORS 111

Theorem 8.1.1. The operation⊗ is distributive and associative:

(u1 + u2)⊗ v = u1 ⊗ v + u2 ⊗ v, u ⊗ (v1 + v2) = u ⊗ v1 + u ⊗ v2,

t (u ⊗ v) = (tu)⊗ v = u ⊗ (tv), (u ⊗ v)⊗ w = u ⊗ (v ⊗ w).

⊳ Easy exercise. ⊲

Theorem 8.1.2. (on basis of Lk(Rn)) For any k = 1, 2, . . . the products πi1 ⊗ . . .⊗ πik(i j ∈ {1, . . . , n}) form a basis of the vector space Lk(Rn). Hence,

dim Lk(Rn) = nk .

⊳ 1◦ let u ∈ Lk(Rn). Then

u(h1, . . . , hk) = u

n∑

i1=1

h1i1 ei1 , . . . ,

n∑

ik=1

hkik eik

u∈Lk

=n∑

i1,...,ik=1

h1i1 . . . hkik︸︷︷︸=(πi1⊗...⊗πik )(h1,...,hk)

u(ei1 , . . . , eik )︸︷︷︸=:ai1 ...ik

=

n∑

i1,...,ik=1

ai1...tkπi1 ⊗ . . .⊗ πik

(h1, . . . , hk).

Hence,

u =n∑

i1,...,ik=1

πi1 ⊗ . . .⊗ πik ,

that is, our products span Lk(Rn).2◦ They are linearly independent. Indeed, let

u =n∑

i1,...,ik=1

ai1...ikπi1 ⊗ . . .⊗ πik = 0

Applying this to (ei ′1, . . . , ei ′k

), we obtain, by (1),

ai ′1...i′k= 0. ⊲

Pull-back

Xl

K Y

Lk(X)l∗

L Lk(Y )

Let X,Y be a vector spaces, and let l ∈ L(X,Y ). For any v ∈ Lk(Y ) we define thepull-back l∗v of v, putting

(l∗v)(h1, . . . , hk) := v(lh1, . . . , lhk).

It is clear that

l∗v ∈ Lk(X).

So we “pull” the tensor v “back” to X .


Rl∗y ′ ր տ y ′

Xl

K Y

X ′l∗

L Y ′

Example. For k = 1 we obtain the operator l∗ : Y ′ → X ′

which act so:

l∗y ′ = y ′ ◦ l.

⊳ (l∗y ′)h = y ′(lh) = (y ′ ◦ l)h. ⊲

This operator between the DUAL spaces is called the dual operator to l. Note that l∗

act in the OPPOSITE direction.

Theorem 8.1.3. Pull-back RESPECTS tensor product:

f ∗(u ⊗ v) = ( f ∗u)⊗ ( f ∗v).

⊳ Easy exercise. ⊲

8.2 Asymmetric tensors

A tensor u ∈ Lk(X) (k ≥ 2) is called antisymmetric if it has value 0 at any point(h1, . . . , hk) which has two equal components. The set of all antisymmetric tensors wedenote by 3k(X). Thus,

u ∈ 3k(X) :⇔ u(. . . , h, . . . , h, . . .) = 0.

It is convenient to put31(X) := L1(X)(= X ′),

30(X) := L0(X)(= R).

Remark. An equivalent description is such: a tensor is antisymmetric iff it changes thesign by any transposition of its arguments:

u ∈ 3k(X)⇔ u(. . . , h, . . . , k, . . .) = −u(. . . , k, . . . , h, . . .)

(all others arguments remain unchanged).⊳ “⇒”: u(. . . , h + k, . . . , h + k, . . .)︸︷︷︸

=0

= u(. . . , h, . . . , h, . . .)︸︷︷︸=0

+u(. . . , h, . . . , k, . . .)+

u(. . . , k, . . . , h, . . .)+ u(. . . , k, . . . , k, . . .)︸︷︷︸=0

,

hence u(. . . , h, . . . , k, . . .)+ u(. . . , k, . . . , h, . . .) = 0.“⇐”: u(. . . , h, . . . , h, . . .) = −u(. . . , h, . . . , h, . . .), hence u(. . . , h, . . . , h, . . .) = 0. ⊲

Examples.

1. det.

2. Let A ∈ L(Rn,Rn), and let u A be the corresponding 2-tensor. Then

u A ∈ 32(Rn)⇔ AT = −A.

(We identify A with the corresponding matrix.) ⊳ Exercise. ⊲ Operators A satisfying thecondition AT = −A, are also called antisymmetric. A typical example is rotation by 90◦ in

R2, e.g. counter-clock-wise, with the matrix

(0 −11 0

). The corresponding antisymmetric

tensor is just det.

8.2. ASYMMETRIC TENSORS 113

⊳ u A(h, k) = 〈Ah, k〉 = (k1, k2)

(0 −11 0

)(h1h2

)= h1k2 − h2k1 = det(h, k). ⊲

Thus, (0 −11 0

)←→ det .

Operator alt

From any tensor we can make an antisymmetric one. Viz., put for u ∈ Lk(X)

(alt u)(h1, . . . , hk) := 1

k!

∑

σ∈Sk

(sgn σ)u(hσ(1), . . . , hσ(k)).

(In the sum the signs alternate (+,−,+,−, . . .), whence the notation.)Recall that sgn σ denotes the sign of a permutation. It is clear that alt u is a k-tensor.

Examples.

1. alt〈·, ·〉 = 0. (Recall that 〈·, ·〉 denotes the scalar product, which is a symmetric 2-tensor.)More generally, alt sends ANY symmetric tensor to 0:

u ∈ Sym⇒ alt u = 0.

NB The inverse implication is not true! See Exercise 8.2.2. 3) below.

2. If u ↔ A (that is u = u A), then alt u ↔ 12 (A − AT ). (Verify!)

Theorem 8.2.1. The operator alt has the following properties:

a) alt ∈ L(Lk(X),3k(X)), that is, alt u is an antisymmetric tensor, and the mappingu 7→ alt u is linear;

b) u ∈ 3k ⇒ alt u = u, that is, 3k is INVARIANT under alt;c) alt2 = alt, that is, alt is an IDEMPOTENT operator; (alt2 u := alt(alt u));

d) alt u = 0⇒ ∀v... alt(u ⊗ v) = 0 (“bad sheep principle”: one bad sheep spoils all

the crew).

⊳ a) The sum defining (alt u)(. . . , h, . . . , h . . .) can be splitted onto pairs of the form

+u(. . . , h, . . . , h . . .)− u(. . . , h, . . . , h . . .),

where our two h appear on one and the same pair of places (different for differentpairs). Hence the sum is equal to 0.

b) ∀u ∈ 3k...

alt u(h1, . . . , hk) =1

k!

∑

σ∈Sk

(sgn σ) u(hσ(1), . . . , hσ(k))︸︷︷︸=(sgnσ)u(h1,...,hk)

= 1

k!(k!u(h1, . . . , hk)) = u(h1, . . . , hk).

c) By a), alt u ∈ 3k , hence

alt(alt u)b)= alt u.


d) (alt(u ⊗ v))h1 . . . h p+q

= 1

(p + q)!

∑

σ∈Sp+q

(sgn σ)(uhσ(1), . . . , hσ(p))(vhσ(p+1), . . . , hσ(p+q)).

If, for a fixed σ0 we consider all the σ such that

{σ(1), . . . , σ (p)} = {σ0(1), . . . , σ0(p)} (as NON-ORDED set!),

σ(p + 1) = σ0(p + 1), . . . , σ (p + q) = σ0(p + q),

then the sum over all such such σ is equal to 0, since alt u = 0. But the whole sumsplits onto such subsums. ⊲

Exercise 8.2.2. 1) alt ◦ sym = 0.2) sym ◦ alt = 0.3) Give an example of u ∈ L3(R3) such that u 6= 0, sym u = alt u = 0.NB Such an u is not the sum of its symmetric part sym u and antisymmetric part alt u.Only 2-tensors have this property.

Answer:

uei e j ek =

23 if (i, j, k) = (1, 2, 3),

− 13 if (i, j, k) = (3, 2, 1),0 otherwise.

Using the operation alt, we can make from tensor product an operation over antisy-mmetric tensors.

Exterior productFor u ∈ 3p(X) (p ≥ 1), v ∈ 3q(X) (q ≥ 1), the exerior product u ∧ v is defined by theformula

u ∧ v := (p + q)!

p! q!alt(u ⊗ v).

Remarks. 1) u ⊗ v itself is not in general antisymmetric (give an example!).2) The coefficient in this formula is chosen to obtain the coefficient 1 in the formula (1)below.

Example. In R2, π1 ∧ π2 = det. ⊳ (π1 ⊗ π2)(h, k) = π1h · π2k = h1k2; hence (π1 ∧π2)(h, k) = (1+1)!/(1! 1!) alt(π1⊗π2)(h, k) = 2( 1

2 (π1⊗π2)(h, k)− 12 (π1⊗π2)(k, h)) =

h1k2 − k1h2 = det(h, k). ⊲

Exercise 8.2.3. Prove that for u ∈ 3p, u ∈ 3q

(u ∧ v)h1 . . . h p+q =∑

σ∈Sp+q

σ(1)<σ(2)<...<σ(p)

σ (p+1)<...<σ(p+q)

(sgn σ)(uhσ(1) . . . hσ(p))(vhσ(p+1) . . . hσ(p+q)) (1)

Theorem 8.2.4. The operation ∧ has the following properties:

a) (u1 + u2) ∧ v = u1 ∧ v + u2 ∧ v, u ∧ (v1 + v2) = u ∧ v1 + u ∧ v2,t (u ∧ v) = (tu) ∧ v = u ∧ (tv) t ∈ R (distributivity);

b) u ∧ v = (−1)pqv ∧ u (u ∈ 3p, v ∈ 3q ) (“semi-commutativity”);

8.2. ASYMMETRIC TENSORS 115

c) (u ∧ v)∧w = u ∧ (v ∧w) = (p+ q + r)!/(p!q!r !) alt(u ⊗ v⊗w) =: u ∧ v ∧w(u ∈ 3p, v ∈ 3q , w ∈ 3r ) (associativity);

d) f ∗(u ∧ v) = ( f ∗u) ∧ ( f ∗v) (pull-back RESPECTS exterior product).

⊳ a) Obvious.b) Consider the permutation

σ0 =(

1 . . . p p + 1 . . . p + q1+ q . . . p + q 1 . . . q

)

It is clear that sgn σ0 = (−1)pq (1 is transposed q times, then 2 is transposed qtimes, . . . p is transposed q times).Each permutation σ ∈ Sp+q can be written an the form σ = σ ′ ◦ σ0. Then

σ(p+1) = σ ′(σ0(p+1)) = σ ′(1), . . . , σ (1) = σ ′(σ0(q+1)) = σ ′(q+1), . . . (2)

andsgn σ = (sgn σ ′)(sgn σ0) = (−1)pq sgn σ ′. (3)

Hence

(u ∧ v)(h1, . . . , h p+q ) =(p + q)!

p! q!︸︷︷︸=:c

alt(u ⊗ v)(h1, . . . , h p+q)

= c

(p + q)!

∑

σ∈Sp+q

(sgn σ)u(hσ(1),...,hσ(p))v(hσ(p+1), . . . , hσ(p+q))

σ=σ ′◦σ0(2),(3)= c

(p + q)!

∑

σ ′∈Sp+q

(−1)pq(sgn σ ′)(hσ ′(q+1), . . . , hσ ′(q+p))v(hσ ′(1), . . . , hσ ′(q))

= (−1)pqc alt(v ⊗ u)(h1, . . . , h p+q ) = (−1)pq(u ∧ v)(h1, . . . , h p+q).

c) To verify that

(u ∧ v) ∧w = (p + q + r)!

p! q! r !alt(u ⊗ v ⊗ w)

we need (after canceling constant factor) to verify that

alt((alt(u ⊗ v)) ⊗ v)︸︷︷︸[1]

= alt(u ⊗ v ⊗w)︸︷︷︸[2]

.

But [1]−[2]alt∈L= alt((alt(u⊗v))⊗w−u⊗v⊗w) distr.= alt((alt(u ⊗ v)− u ⊗ v)︸︷︷︸

[3]

⊗w)

bad Sheep Priciple= 0, since alt[3]alt∈L= alt2(u ⊗ v) − alt(u ⊗ v) alt2=alt= 0.

d) Easy exercise. ⊲

Corollary 8.2.5. For any antisymmetric tensor of ODD rank its exterior product by itselfis equal to 0.

⊳ If u ∈ 3k , k ∈ Odd, then u ∧ ub)= (−1)k

2

︸︷︷︸=−1

, whence 2(u ∧ u) = 0. ⊲


Basis of 3k(Rn)

Theorem 8.2.6. For any 1 ≤ k ≤ n the set {πi1 ∧ . . . ∧ πik | i1 < . . . < ik} is a basis ofthe vector space3k(Rn). hence

dim3k(Rn) =(

n

k

):= n!

k!(n − k)!.

⊳ By Theorem on basis of Lk(Rn), we can write any u ∈ 3k(Rn) as a linear combinationof πi1 ⊗ . . .⊗ πik :

u :=n∑

i1,...,ik=1

. . . πi1 ⊗ . . .⊗ πik (4)

(the first dots mean a number coefficient). It follows that

uTh 8.2.1.= alt u

(4)= alt∑

. . . πi1 ⊗ . . .⊗ πikalt∈L=

∑. . . alt(πi1 ⊗ . . .⊗ πik )

Th 8.2.4.=∑

. . . πi1 ∧ . . . ∧ . . . πikTh 8.2.4.,b),cor.=

∑

i1<...<ik

. . . πi1 ∧ πik .

Thus, our set spans3k(Rn). The linear independence can be proved just as in the caseof Lk(Rn). ⊲

Corollary 8.2.7. The space 3n(Rn) is 1-dimensional. Hence (since det ∈ 3n(Rn)) anyelement of 3n(Rn) has the form

c det (c ∈ R).

Corollary 8.2.8. In Rn ,π1 ∧ . . . ∧ πn = det .

⊳ By Corollary 8.2.7., π1 ∧ . . . ∧ πn = c det. Applying both sides to (e1, . . . , en) andtaking into account that πi e j = δi j , we conclude that c = 1. ⊲

Corollary 8.2.9. For k > n3k(Rn) = {0}.

⊳ This follows from the PROOF of Theorem 8.2.6. ⊲

Theorem on determinat

Theorem 8.2.10. Let A ∈ L(Rn,Rn). Then

A∗ det = (det A) det

Here det A denotes the determinant of the matrix of A in the canonical basis in Rn .NB Let X be an arbitrary n-dimensional vector space, and let A be a linear operator inX , A ∈ L(X, X). Then the determinant of the matrix of this operator in basis in X doesnot depend on the choice of the basis. ⊳ The matrix in a “new” basis has the formB M B−1, where M is the matrix in the “old” basis, and B is the “transition matrix”. Butdet(B M B−1) = det B det M(det B)−1 = det M . ⊲ So we can say about the determinantof an operator (in a finite-dimensional vector spaces).⊳ By Corollary 8.2.7., A∗ det = c det. Hence

8.3. DIFFERENTIAL FORMS 117

(A∗ det)e1 . . . en︸︷︷︸[1]

= c det e1 . . . en︸︷︷︸=1

= c

[1], def of A∗= det(Ae1) . . . (Aen)

def. of the matrixof an operator= det A

We conclude that c = det A. ⊲

Theorem 8.2.10. means that a linear operator A changes the volume by det A times. (Itfollows also from Theorem on Change!)

Corollary 8.2.11. If det A = 1, then A∗ det = det.In other words, an operator with the unit determinant does not change the volume.

Examples.

1. For any u1, . . . , un ∈ 31(Rn) (= L(Rn))

u1 ∧ . . .∧ un =

∣∣∣∣∣∣

u11 . . . u1n

. . . . . . . . .

un1 . . . unn

∣∣∣∣∣∣det,

where ui j are the coefficients of the linear function ui : Rn → R,

ui x = ui1x1 + . . .+ uin xn.

2. For any u1, . . . , uk ∈ 31(Rn), k ≤ n

(u1 ∧ . . .∧ uk)h1 . . . hk =

∣∣∣∣∣∣

u1h1 . . . u1hk

. . . . . . . . .

unh1 . . . unhn

∣∣∣∣∣∣.

In particular

(πi1 ∧ . . . ∧ πik )h1 . . . hk =

∣∣∣∣∣∣

h1i1 . . . h1ik. . . . . . . . .

hki1 . . . hkik

∣∣∣∣∣∣.

(Note that the determinant of the transposed matrix is the same.)

8.3 Differential forms

Let X be a normed space, and let U be an open set in X . A differential form ω of degree k(k = 0, 1, 2, . . .) (or k-form) on U is a smooth (that is, of class Cp for some p) mapping

ω : U → 3k(X),

that is, a “tensor field” on U , all the tensor being antisymmetric. As to smoothness, weconsiderω as a mapping into NORMED SPACE3k(X) (a vector subspace in Lk(X) equippedwith the induced norm). We denote the set of all k-forms on U by

�k(U).

Examples.

1. Any smooth function f : U → R is a 0-form.


2. For any smooth function f : U → R its derivative f ′ is a 1-form.

3. As a special case of the previous example, π ′i ∈ 31(Rn); for any point x ∈ Rn we haveπ ′i (x) = πi . Thus π ′i is a CONSTANT 1-form on Rn . It is denoted traditionally by

dxi .

So for any x ∈ Rn and any h = (h1, . . . , hn) ∈ Rn

(dxi )(x) · h = hi .

4. On Rn , a CONSTANT mapping ω(x) ≡ detk is k-form. We denote it also by detk , orsimply det.

The operations ∧ and ∗ for forms are “point-wise”.

Exterior productLet ω1 ∈ �p1(U), ω2 ∈ �p2(U). The exterior product ω1 ∧ ω2 is defined by the rule

∀x ∈ U... (ω1 ∧ ω2)(x) := (ω1(x)) ∧ (ω2(x)).

It is easy to verify that ω1 ∧ ω2 is a SMOOTH mapping U → 3p1+p2(X), so

ω1 ∧ ω1 ∈ �p1+p2(U).

Moreover we put for f ∈ �0(U)

f ∧ ω := f ω,

where∀x ∈ U( f ω)(x) := ( f (x))︸︷︷︸

∈R

(ω(x)).

Example. On Rn ,dx1 ∧ . . . ∧ dxn = det .

⊳ πi ∧ . . .∧ πn = det . ⊲

Theorem 8.3.1. Any k-form ω on Rn can be written in the form

ω =∑

i1<...<ik

fi1 ...ik dxi1 ∧ . . .∧ dxik ,

where fi1 ...ik are smooth (real-valued) functions.⊳ It follows at once from Theorem on basis on 3k(Rn). ⊲

Example. For f ∈ �0

f ′ = ∂ f

∂x1dx1 + . . .

∂ f

∂xndxn.

⊳ Apply both sides to h = (h1, . . . , hn). ⊲

8.3. DIFFERENTIAL FORMS 119

Pull-back

Xf

K Y

�k(X)f ∗

L �k(Y )

Let X,Y be normed spaces, let f be a smooth mapping from X into Y , and letω ∈ �k(Y ).We define the pull-back f ∗ω of ω by the rule

∀x ∈ X... ( f ∗ω)(x) := ( f ′(x))∗ω( f (x)),

where the star in the right-hand side means pull-back for tensors.Thus, the value of the pull-back of ω by f at x is the tensor pull-back by the

DERIVATIVE f ′(x) of the value of ω at f (x).More explicitly,

(( f ∗ω)(x))h1 . . . hk := ω( f (x))( f ′(x)h1) . . . ( f ′(x)hn).

Moreover we put for g ∈ �0(Y )f ∗g = g ◦ f.

Example. For a smooth mapping f : Rn → Rm , f = ( f1, . . . , fm), it holds

f ∗(dyi) =n∑

j=1

∂ fi

∂x jdx j (i = 1, . . . ,m)

((x = x1, . . . , xn) ∈ Rn , y = (y1, . . . , ym) ∈ Rm).

⊳ f ∗(dyi)(x) · h = (dyi) · ( f ′(x)h)

= dyi

∂ f1/∂x1 . . . ∂ f1/∂xn

. . . . . . . . .

∂ fm/∂x1 . . . ∂ fm/∂xn

h1...

hn

=n∑

j=1

∂ fi

∂x jh j︸︷︷︸

=(dx j )h

=

n∑

j=1

∂ fi

∂x jdx j

h. ⊲

Theorem 8.3.2. The pull-back operation over forms has the following properties

a) f ∗(ω1 + ω2) = f ∗ω1 + f ∗ω2 (linearity);b) f ∗(ω1 ∧ ω1) = ( f ∗ω1) ∧ ( f ∗ω2) (∗ respects ∧); in particular, for g ∈ �0

f ∗(gω) = (g ◦ f )( f ∗ω).

⊳ This follows at once from the definitions and the corresponding results for tensors. ⊲

Pull-back of determinant

Theorem 8.3.3. Let f : Rn → Rn (that is f : Rn → Rn) be smooth, and let g ∈ �0(Rn).

Rn f→ Rn g→ Rn .Then

f ∗(g det) = (g ◦ f )(det f ′) det

⊳ 1◦ f ∗(det) = (det f ′) det.

⊳⊳ ∀x ∈ Rn... ( f ∗ det)(x) = ( f ′(x))∗ det( f (x))︸︷︷︸

=det

Th. on det for tensors= (det f ′(x)) det ⊲⊲.

2◦ f ∗(g det)Th 8.3.2., b)= (g ◦ f ) f ∗(det)

1◦= (g ◦ f )(det f ′) det. ⊲

NB In the special case where f is a diffeomorphism, this Theorem describes the changeof a “weighted” volume by a change of variables—compare with Theorem on change ofvariables (where we write f instead of g and v.v.).


8.4 Exterion differentiation (operator d)

If we differentiate a form ω ∈ �k(X) (as a mapping X → 3k(X) between two normedspaces) then we obtain ω′(x) ∈ L(X,3k(X)) ⊂ L(X,Lk(X)) ∼ Lk+1(X). In generalω′(x) considered as an element of Lk+1(X) is not antisymmetric, that is, does not belongto 3k+1(X). So we appeal the operator alt.

Operator dLet X be a normed space, and let ω ∈ �k(X) we define the exterior derivative (or exteriordifferential) dω by the rule

∀x ∈ X... (dω)(x) := (k + 1) alt ω′(x) (∈ 3k+1(X)), (1)

where ω′(x) denotes the element of Lk+1(X) generated by ω′(x):

ω′(x) · h0h1 . . . hk := (ω′(x)h0)︸︷︷︸∈3k (X)

·h1 . . . hk . (2)

Is is easy to verify that if ω ∈ Cp then dω ∈ Cp−1, so is sufficiently smooth if ω is.Thus

dω ∈ �k+1(X).

Remark. The factor k + 1 is chosen to obtain the coefficient 1 in some formulas below.

Examples.

1. For f ∈ �0 we have d f = f ′.2. For any f ∈ �0

d2 f = 0,

whered2 f = d(d f ).

⊳ (d2 f )(x) = (d(d f ))(x)1)= (d( f ′))(x) (1)= (1+ 1) alt ˜( f ′)(x)︸︷︷︸

= f ′′(x)

= 2 alt f ′′(x) =f ′′(x)∈Sym

0. ⊲

3. In R2, d(xdy︸︷︷︸=:ω

) = det. ⊳ We have ω : (x, y) 7→ xπ2, so that ω ∈ L(R2,L(R2,R)),

hence ω′ ≡ ω. So

(dω)((x, y)) h︸︷︷︸=:(h1,h2)

k︸︷︷︸=:(k1,k2)

= (1+ 1) alt ˜ω′((x, y))︸︷︷︸=ω

hk = 2 12(ωhk − ωkh)

(2)= (ωh)︸︷︷︸h1π2

k − (ωk)︸︷︷︸k1π2

h = h1k2 − kh2 = det(h, k). ⊲

Exercise 8.4.1. Let A ∈ L(Rn,Rn) and ω(x) = 〈An, ·〉 (∈ L(Rn,R) = 31(Rn)). Provethat

dω ≡ u A−AT .

Remark. It can be shown that

dω(x) · h0 . . . hk =k∑

i=0

(−1)i (ω′(x)hi ) · h0 . . . hi . . . hk,

8.4. EXTERION DIFFERENTIATION (OPERATOR d) 121

where hi means that this term is to be canceled.

Theorem 8.4.2. The operator d has the following properties

a) d(ω1 + ω2) = dω1 + dω2 (linearity);b) d(ω1 ∧ ω2) = (dω1) ∧ ω2 + (−1)degω1ω1 ∧ dω2 (“semi-leibniz” rule);c) d2 = 0 (d2ω = d)dω) (“self-annihilation”);d) f ∗(dω) = d( f ∗ω) (pull-back respects d).

Here degω denotes the degree of ω.

⊳ a) Obvious.b) Let ω1 ∈ �r1 , ω2 ∈ �r2 . We have (for short we drop the argument x)

˜(ω1 ⊗ ω2)′h0 . . . hr1+r2Leibnitz rule= ((ω′1h0)⊗ ω2 + ω1 ⊗ (ω′2h0))h1 . . . hr1+r2

= ((ω′1h0)h1 . . . hr1)(ω2hr1+1 . . . hr1+r2)

+ (ω1h1 . . . hr1)((ω′2h0)hr1+1 . . . hr1+r2)

= (ω′1h0 . . . hr1)(ω2hr1+1 . . . hr1−r2)+ (ω1h1 . . . hr1)(ω′2h0hr1+1 . . . hr1+r2)

= (ω′1 ⊗ ω2)h0 . . . hr1+r2 + (ω2 ⊗ ω′1)h1 . . . hr1 h0hr1+1 . . . hr1+r2 .

To replace h0 to the first place we have to make r1 transpositions, which yieldsthe factor (−1)r1 . After alternating (applying of alt) we obtain what we need. Weomitting the details.

c) The idea is the same as in Example 2) above. Let ω ∈ �r . We have (omitting thedetails)

(d2ω)h0 . . . hr+1 = const(alt ˜(alt ω′)′)h0 . . . hr+1

= const∑

{h,k}⊂{h0 ...hr+1}

∑

...

((ω′′hk) . . .︸︷︷︸∗−(ω′′kh) . . .︸︷︷︸

∗)

∗— other r arguments in one and the same order

= const∑

...

((ω′′hk − ω′′kh)︸︷︷︸ω′′∈sym= 0

. . .) = 0

d) We consider the simplest case where ω ∈ �0, that is, ω is a function g:

Xf−→ Y

g−→ R.

We have

((d( f ∗g))(x))h = ( f ∗g)(x)h = (g ◦ f )′(x)h Chain rule= ( g′︸︷︷︸=dg

( f (x))) · ( f ′(x)h)

= (( f ∗(dg))(x))h,

whence d( f ∗g) = f ∗(dg). In more general case the idea is the same. ⊲

Remark. The semi-Leibnitz rule is NON-symmetric w.r. to ω1 and ω2. Only degω1 entersthe rule. The matter is that in our definition of ω′(x) we put the derivative to act ontothe FIRST argument. But we could choose anyone. So in essence dω is defined just UP TO

THE SIGN! “Physically”, ω and−ω are same, that is why when integrating over manifolds(Chapter 6) we need to choose one of two possible ORIENTATIONS of the manifold inquestion.Remark. It is instructive to see how semi-Leibnitz rule interacts with semi-commutativity:


d(ω2 ∧ ω1) = d((−1)r1r2ω1 ∧ ω2) = (−1)r1r2(dω1ω2 + (−1)r1ω1dω2)

= (−1)r1r2((−1)(r1+1)r2ω2 ∧ dω1 + (−1)r1(−1)r1(r2+1)dω2 ∧ ω1)

= dω2 ∧ ω1 + (−1)r2ω2dω1,

r1 “transforms” into r2, as it need to be!Exercise 8.4.3. Give the proof of d) with all details for the case ω ∈ �1.

d in Rn

Theorem 8.4.4. Let ω ∈ �r (Rn). If the canonical representation of ω is

ω =∑

i1<...<ir

fi1 ...ir dxi1 ∧ . . .∧ dxir

then dω is given by the formula

dω =∑

i1<...<ir

d fi1 ...ir ∧ dxi1 ∧ . . .∧ dxir

⊳ This follows at once from semi-Leibnitz rule for forms and from the self-annihilationproperty (d2); recall that f dxi1 ∧ . . . ∧ dxir = f ∧ dxi1 ∧ . . .∧ dxir . ⊲

Example. In R2, d(xdy) = dx ∧ dy = det.

Closed and exact formsIf dω = 0 (in an open set U ) one can says that ω is closed (in U ). If ω = dψ (in U ) forsome form ψ then one can says that ω is exact (in U ).

Each exact form is closed. ⊳ ω = dψ ⇒ dω = d2ψ = 0. ⊲

The inverse assertion is not in general true.

Example. The form

ω = −y

x2 + y2 dx + x

x2 + y2 dy

x

yh=(dx,dy)

rdθ

θ

on R2 \ {0} is closed (verify!), but is not exact. (This form arises naturally if one considerpolar coordinates r, θ (x = r cos θ , y = r sin θ ), and by thisreason is usually denoted by dθ , and though ω is not exact!(But ω IS exact on somewhat less sets,as it follows, e.g., fromPoicare lemma:))

Poincare lemma

Theorem 8.4.5. Let U be an open ball in Banach space. If a form ω is closed in U, thenit is exact in U.

We omit the proof.

Chapter 9

Stokes Formula for Chains

9.1 Chains

1

1

[0,1]2

By [0, 1]k we denote the unit cube [0, 1]× . . .× [0, 1] (k-times) in Rk . For k = 0 we put

[0, 1]0 := {0}, R0 := {0}.By a curved k-cube c in Rn we mean a continuous mapping

c : [0, 1]k → Rn.

For short we omit the word “curved” below. If im c ⊂ A ⊂ Rn we say that c is acube in A.

Examples.1. A 0-cube is just a point:

c

2. A 1-cube is a curve: 0 1

3. The embeding id : [0, 1]k → Rk is called the standard k-cube and is denoted by I k :

I k := idRk |[0,1]k .

ChainsA k-chain (in A) is a formal (finite) sum of k-cubes (in A) with integer coefficients, thatis, an expression of the form

2c1 + 3c2 − 5c3 + 100c4,

where ci are k-cubes. We identify a k-cube c with multiplication with the chain 1 · c.In natural way we define for k-chains multiplication by an integer number and

addition.

FacesFor a standard cube I n we define its faces I k

(i,α), i = 1, . . . , k, α = 0, 1, by the rule

I k(i,α) : [0, 1]k−1→ Rk, (x1, . . . , xi−1, α, xi , . . . , xk−1).

123

124 CHAPTER 9. STOKES FORMULA FOR CHAINS

Thus, we insert α into i th place and move all the “tail” to the right by one position.

cI k(i,α)

For any k-cube c we define its face c(i,α) by the rule

c(i,α) := c ◦ I k(i,α).

Operator ∂We define the boundary ∂c of a k-cube c as the following chain:

∂c :=k∑

i=1

∑

α=0,1

(−1)i+αc(i,α)

(an alternating sum of the faces). For chains we define the boundary “by linearity”:

∂∑

ai ci :=∑

ai∂ci (ai ∈ Z).

Examples.

1. ∂ I 1 = I 1(1,1) − I 1

(1,0); I 1(1,0) I 1

(1,1)

+−

2. ∂ I 2 = I 2(2,0) + I 2

(1,1) − I 2(2,1) − I 2(1, 0);

I 2(1,1)

I 2(2,0)

I 2(1,0)

I 2(2,1)

+

+

−

−

I 3(1,1)

I 3(2,0)

I 3(3,1)

3. ∂ I 3 three “visible” faces are positive (enter into sum with “+”)three “non-visible” faces are negative.

Theorem 9.1.1. ∂2 = 0 (that is, ∂(∂c) = 0 for any chain c).We do not need this result, so we omit the proof.

Exercise 9.1.2. Verify Theorem for I 3.

Closed and exact chainsA chain c is called closed if ∂c = 0, that is, if its boundary is the null chain; c is calledexact (in A) if c = ∂c′ for some chain (in A), that is, if c is the boundary of some chain(in A).

By the Theorem above, each exact chain is closed. But not each closed is exact:

c(0)=c(1)

Example.Consider the 1-cube c : [0, 1]→ R2, t 7→ (cos 2π t, sin 2π t)(the unit circle in R2 \ {0}. We have ∂c = 0 (verify), so that cis closed, but c is not exact in R2 \ {0}, since c is the boundaryof no chain in R2 \ {0}.

9.2 Integral over a chain

For any k-form ω in an open set U in Rk , such that U ⊃ [0, 1]k, k = 1, 2, . . ., we put

∫

I kω :=

∫

[0,1]kf (1)

9.3. STOKES FORMULA 125

where f is the function U → R, uniquely defined by the relation ω = f det. (Recall thatany k-form in Rk can be written (uniquelly) in such form; see Chapter 8.)

In more detailed record,∫

I kf dx1 ∧ . . . ∧ dxn :=

∫

[0,1]kf dx1 . . . dxn. (2)

NB This definition DOES DEPEND on the ORDER of the basis vectors, since det does depend!For k = 0 we put

∫

i0ω := ω(0), (here ω ∈ �0 is a FUNCTION).

[0, 1]k cK

GRn

c∗ω ↓ ↓ ω3k(Rk) 3k(Rn)

Now, let G be an open set in Rn , let ω be a k-form in G, and let c be a k-cube in G. Thenwe put

(3)∫

cω :=

∫

I kc∗ω

In particular, for a 0-cube c and a 0-form ω we have

∫

cω = ω(c(0)) (since c∗ω = ω ◦ c,if ω ∈ �0).

At last for a chain c =∑ ai ci we put

∫∑

ai ci

ω :=∑

ai

∫

ci

ω (4)

It is clear that so defined integral is LINEAR:∫

cαω = α

∫

cω,

∫

c(ω1 + ω2) =

∫

cω1 +

∫

cω2 (α ∈ R).

Lemma 9.2.1. (on integral over the boundary). Let c be a k-cube, and let ω be a k-form(both in G). Then ∫

∂cω =

∫

∂[0,1]kc∗ω.

⊳ Exercise. ⊲

9.3 Stokes formula

Similarly of properties of the operators d and ∂ (d2 = 0, ∂2 = 0), and of the notionsconcerning them (closeness, exactness) is not an accident. There is a deep relation betweend and ∂ , which is expressed by the following

Theorem 9.3.1. Let ω be a k− 1-form in G(∈ Op(Rn)), and let c be a k-chain in G. Then

∫

cdω =

∫

∂cω (Stokes formula) (1)

126 CHAPTER 9. STOKES FORMULA FOR CHAINS

⊳ 1◦ At first let us prove the folloving fact about pull-back by the standard faces:

(I k(i,α)

)∗dx j =

dx j if j < i ,0 if j = i ,

dx j−1 if j > i .(2)

⊳⊳

(I k(i,α)

)′(x) h︸︷︷︸=(h1,...,hk−1)

Exer.= (h1, . . . , hi−1, 0, hi , . . . , hk−1) whence

((I k(c,α))

∗dx j ) = dx j (Ik(i,α))

′(x)h = dx j · (h1, . . . , hi−1, 0, hi , . . . , hk−1)

=

h j if j < i ,0 if j = i ,

h j−1 if j > i .⊲⊲

2◦ Case c = I k , ω ∈ �k−1(Rk). Calculate the left-hand side in (1):

∫

I kdω

Th 8.3.1.=∫

I k

k∑

i=1

d fi dx1 . . . ∧ dxi ∧ . . .∧ dxk

Th 8.4.4.=∫

I k

k∑

i=1

d fi︸︷︷︸=D1 fi dx1+...+Dk fi dxk

∧dx1 . . . ∧ dxi ∧ . . . ∧ dxk

Th 8.2.4. and its Cor.=k∑

i=1

(−1)i−1∫

I kDi fi dx1 ∧ . . .∧ dxk︸︷︷︸

=det

(1)=k∑

i=1

(−1)i−1∫

[0,1]kDi fi

Fubini Th=k∑

i=1

(−1)i−1∫ 1

0. . .

∫ 1

0︸︷︷︸k−1

(∫ 1

0Di fi (x1, . . . , xk)dxi︸︷︷︸

Newt.-Leib. Th= fi (x1,...,1i,...,xk)− fi (x1,...,0

i,...,xk)=: 1

)dx1 . . . dxi . . . dxk

Fubini+trick!=k∑

i=1

(−1)i−1∫ 1

0. . .

∫ 1

0︸︷︷︸k

1 dx1 . . . dxk .

Calculate the right-hand side:

∫

∂ I kω

def of ∂=∫∑k

i=1∑α=0,1(−1)i+α I k

(i,α)

k∑

j=1

f j dx1 ∧ . . . ∧ dx j ∧ . . . dxk

(4)=k∑

i, j=1

∑

α=0,1

(−1)i+α∫

I k(i,α)

f j dx1 ∧ . . .∧ dx j ∧ . . . dxk

(3)=∑

i, j,α

(−1)i+α∫

I k−1

(I k(i,α)

)∗ (f j dx1 ∧ . . .∧ dx j ∧ . . . dxk

)

∗respects ∧=∑

i, j,α

(−1)i+α∫

I k−1

( (I k(i,α)

)∗f j

︸︷︷︸= f j◦I k

(i,α)

)

∧ (((I k(i,α))

∗dx1) ∧ . . . ∧ ((I k(i,α))

∗dx j−1) ∧ ((I k(i,α))

∗dx j+1) ∧ . . . ∧ ((I k(i,α))

∗dxk))︸︷︷︸(2)={

0 if i 6= jdx1 ∧ . . .∧ dxk−1 if i = j

9.3. STOKES FORMULA 127

=∑

i,α

(−1)i+α∫

I k−1

(fi ◦ I k

(i,α)

)dx1 ∧ . . . ∧ dxk−1︸︷︷︸

=det(1)=∑

i,α

(−1)i+α∫

[0,1]k−1fi ◦ I k

(i,α)

def of I k(i,α)+trick=

∑

i,α

(−1)i+α( ∫ 1

0. . .

∫ 1

0︸︷︷︸k−1

fi (x1, . . . , αi, . . . , xk−1)

)(∫ 1

0dxk

)

︸︷︷︸=1

Fubini Th=∑

i,α

(−1)i+α∫ 1

0. . .

∫ 1

0︸︷︷︸k

fi (x1, . . . , αi, . . . , xk−1)dx1 . . . dxk

y1 := x1, . . . , yi−1 := xi−1, yi := xk, yi+1 := xi , . . . , yk := xk−1

=∑

i,α

(−1)i+α∫ 1

0. . .

∫ 1

0fi (y1, . . . , α

i, . . . , yk−1)dy1 . . . dyk

obv.=∑

i,α

(−1)i+α∫ 1

0. . .

∫ 1

0( fi (y1, . . . , 1, . . . , yk−1)

− fi (y1, . . . , 0, . . . , yk−1))dy1 . . . dyk .

The result is the same.3◦ Case c ∈ k-Cube(G), ω ∈ �k(G).

∫

cdω

def=∫

I kc∗(dω)

∗ respects d=∫

I kd(c∗ω) 2◦=

∫

∂ I kc∗ω Lm 9.2.1.=

∫

∂cω. O.K.

4◦ General case c =∑αi ci , ci ∈ k-Cube(G), ω ∈ �k(G).∫

cdω

(4)=∑

ai

∫

ci

dω3◦=∑

αi

∫

∂ci

ω(4)=∫

∑ai∂ci︸︷︷︸

def of∂→=∂c

ω =∫

∂cω. O.K. ⊲

Chapter 10

Stokes Theorem for Manifolds

10.1 Manifolds in Rn

In this chapter by smooth mappings we mean C∞-mappings.We say that a subset M of Rn is a k-dimensional manifold (or simply k-manifold), and

we writeM ∈ Mfk(Rn),

if for each point x ∈ M the following condition is fulfilled:

∃U ∈ Op Nbx(Rn) ∃U ∈ Op(Rn) ∃8 ∈ Diffeo(U ,U) :

U ∩ M = 8(U ∩ Rn × 0︸︷︷︸∈Rn−k

). (1)

R n-k

R k

U~

8K

Rn

MU

x

(In other words, M is locally, up to a dif-feomorphism, a k-dimensional vector sub-space in Rn .) We call such a mapping 8 afull chart for M at x .

Examples.

1. Each single point set {x} is a 0-dimensi-onal manifold.

2. Each open set in Rn is an n-dimensional manifold.

8K

3. The unit circle in R2 with the center at 0is a 1-dimensional manifold. E.g. for the point(1, 0) a full chart is shown on the picture. (Givean analytic expression for8!)Exercise 10.1.1. Let G ∈ Op(Rn), and let g :G → Rp (p ≤ n) be a smooth mapping. Put

M := g−1(0).

If

∀x ∈ M... rank g′(x) = p

thenM ∈ Mfn−p(Rn).

129

130 CHAPTER 10. STOKES THEOREM FOR MANIFOLDS

Here rank denotes the rank of the corresponding matrix. Note that p is the maximalpossible value for the rank of n × p-matrix. [Hint: Let

g′(x) =

∂g1/∂x1 . . . ∂g1/∂xn

. . . . . . . . .

∂gp/∂x1 . . . ∂gp/∂xn

Wlog we can assume that it is the FIRST mirror that is not 0:∣∣∣∣∣∣

∂g1/∂x1 . . . ∂g1/∂x p

. . . . . . . . .

∂gp/∂x1 . . . ∂gp/∂x p

∣∣∣∣∣∣6= 0

Put9 := (id, g) : Rn → Rn−p ×Rp = Rn,

that is,9(x1, . . . , xn) := (x1, . . . , xn−p, g1(x), . . . , gp(x)).

Verify that 9 ′(x) ∈ Iso, and apply Inverse Function Theorem. The inverse mapping8 := 9−1 is a full chart for M .]

Charts

Un-kR

V~

x~

x0

V~

kRi π

kR

8K

UV

Mx

nR

Lemma 10.1.2. Let M ∈ Mfk(Rn), and let 8 :U → U be a full chart for M at x. Put

(2)V := U ∩ M, V := π(U ∩ (Rk × 0)),

(3)ϕ := 8 ◦ i, ψ := π ◦8−1,

where π and i are the canonical projection andinclusion, resp. (π(x1, . . . , xn) = (x1, . . . , xk), i(x1, . . . , xk) = (x1, . . . , xk, 0, . . . , 0)).Then ϕ : V → Rn is a smooth map and is a bijection of V onto V . Moreover

∀x ∈ V... ϕ′(x) ∈ Inj L(Rk,Rn) (⇔ rankϕ′(x) = k). (4)

We call ϕ a chart for M at x (generated by the full chart8), or a coordinate system onM at x . The element ϕ−1(x) (for any x ∈ V ) is called the representative of x in the chartϕ, and will be usually denoted by x .⊳ All but the assertion on the rank is obvious. As to this assertion, we have

ψ ◦ ϕ (3)= π ◦8−1 ◦8 ◦ i = π ◦ i = idRk ,

hence, by Chain Rule,ψ ′(x) ◦ ϕ′(x) = idRk .

This means that ϕ′(x) is injective, that is, has the maximal possible rank, k. ⊲

Example. If M = Rn , then idRn is a chart at all points at once.

Transition functionsLet we have two charts for M at x , ϕ1 and ϕ2 (see the diagram):

10.2. TANGENT SPACE 131

kRx~

V~

1

π1i1

V1

V2

ϕ12

ϕ21

Φ1

Φ1-1

Φ2

Φ2-1

kRx~

V~

π2i2

2

ϕ1 ϕ2

ϕ1 : V1 → V1, ϕ2 : V2 → V2.

Put V := V1 ∩ V2 and

ϕ12 := ϕ−12 ◦ ϕ1 : ϕ−1

1 (V )→ ϕ−12 (V ), (5)

ϕ21 := ϕ−11 ◦ ϕ2 : ϕ−1

2 (V )→ ϕ−11 (V ). (6)

So defined ϕ12 and ϕ21 are called the transition functions for these charts.In other words, a transition function sends the representative of x ∈ M in one chart

into the representative of x in the other one.

Lemma 10.1.3. The transition function ϕ12 and ϕ21 (see (5), (6)) are mutually inversediffeomorphism.⊳ It is clear from the diagram, that ϕ12 = π2 ◦8−1

2 ◦81 ◦ i1, ϕ21 = π1 ◦8−11 ◦82 ◦ i2.

Hence both ϕ12 and ϕ21 are smooth as compositions of smooth mappings. Now, it is clearfrom (5), (6), that ϕ12 and ϕ21 are mutually inverse. Hence they are mutually inversediffeomorphism. ⊲

10.2 Tangent space

Now we consider tangent vectors to a manifold in Rn .

n-kR

x~

U~

π1

Φ

Φ-1

ϕ

π 2

U

M

MTx

x

Theorem 10.2.1. Let M ∈ Mfk(Rn).Then for each point x ∈ M the tan-gent cone Tx M to M at x is a k-dimensional vector subspace in Rn;viz., for any char ϕ for M at x

Tx M = im ϕ′(x),

where x is the representative of x inthe chart ϕ (x = ϕ−1(x)).

⊳ Put g := π2 ◦8−1. Then obviously

M ∩U = g−1(0).

By theorem on tangent cone to g−1(0),

Tx M = Tx (M ∩U) = ker g′(x).

Thus we need to verify thatker g′(x) = imϕ′(x). (1)


But, indeed, we have (since 8−1(U ∪ M) ⊂ Rk × 0)

g ◦ ϕ = π2 ◦8−1 ◦ ϕ = 0Chain Rule⇒ g′(x) ◦ ϕ′(x) = 0⇒ imϕ′(x) ⊂ ker g′(x).

It remains to note that both imϕ′(x) and ker g′(x) have the same dimension k (since kerπ2has dimension k, and since both ϕ′(x) and (8−1)′(x) are injective). ⊲

Remark. It is convenient to imagine a tangent vector h from Tx M as an arrow with the

beginning at x and end at x + h (x x+h). Formally, starting from this point, wemean by a tangent vector to M at x a PAIR (x, h), where h ∈ Tx M , and we consider theset Tx M and Ty M as DISJOINT (and if they may coincide as vector subspaces in Rn .

Representatives of a tangent vectors

h~

ϕ’(x)

0 h

MTx

im ϕ’(x) =

0

Let ϕ be a chart for M at x , and let x be the representative of x in ϕ. The pre-image byϕ′(x) of a tangent vector h ∈ Tx M (this pre-image is UNIQUE: byLemma on charts, ϕ′(x) ∈ Inj) be called the representative of h in ϕand will be denoted by h).

Example. For Rn as a manifold in Rn we have

∀x ... Tx Rn = Rn,

and in the chart id

∀h ... h = h.

h~

ϕ12

ϕ21

M

h

ϕ1

x~1

ϕ2

1

x~2

h~

2

Lemma 10.2.2. Let h ∈ Tx M, let ϕ1ϕ2be two chats for M at x let x1, x2 be therepresentatives of x in ϕ1, ϕ2, resp.; andlet h1, h2 be the representatives of h. Then

(2)ϕ′12(x )h1 = h2

In other words, the derivative of tran-sition function sends the representative ofa tangent vector in one chart into the re-presentative in the other one.

⊳ ϕ1 = ϕ2 ◦ ϕ12 ⇒ ϕ′1(x1) = ϕ′2(x2) ◦ ϕ′12(x1)

⇒ ϕ′1(x1)h1︸︷︷︸=h

= ϕ′2(x2) · ϕ′12(x1)h1 ⇒ (2). ⊲

10.3 Mapping between manifolds. Vector fields and forms

A vector field v on a manifold M is a mapping from M into⋃

x∈M Tx M (all Tx M aremutually disjoint!), that sends a point x ∈ M into a tangent vector to M at x :

v : x 7→ v(x) ∈ Tx M.

A k-form ω on M is a mapping that sends x ∈ M into an antisymmetric tensor onTx M:

ω : x 7→ ω(x) ∈ 3k(Tx M).

10.3. MAPPING BETWEEN MANIFOLDS. VECTOR FIELDS AND FORMS 133

SMOOTHNESS of mappings between manifolds and of vector fields and forms on mani-fold we define as smoothness of their REPRESENTATIVES.

kR x~ϕ

nR xM

lRy~

ψ

mR

y

Nf

f~

The representative f OF A MAPPING f : M → N (M ∈ Mfk(Rn), N ∈ Mfl(Rm))in charts ϕ for M at x and ψ forN at y := f (x) is defined by theformula

f (ξ ) := f (ξ),

or, equivalently,

f := ψ−1 ◦ f ◦ ϕ : Rk → Rl

(for continuous f this last map-ping is defined in an appropriateneighbourhood of x).

In particular, the representative of a mapping

f : M → Rm

in a chart ϕ for M and the identity chart for Rm is

f = f ◦ ϕ = ϕ∗ f.

The representative v of a vector field v on M in a chart ϕ : U → U we define as themapping

v : U → Rk, x 7→ v(x);in other words,

ϕ′(x )v(x) = v(x).The representative ω of a k-form ω on M in a chart ϕ : U → U is defined by the rule

ω(x )h1 . . . hk = ω(x)h1 . . . hk,

or, equivalently,

ω(x )h1 . . . hk = ω(ϕ(x))(ϕ′(x )h1) . . . (ϕ′(x )hk),

that is,ω = ϕ∗ω.

SmoothnessWe say that a mapping f : M → N (M, N ∈ Mf) is differentiable at a point x ∈ M(resp., is smooth), if for any charts ϕ at x and ψ at y := f (x) the representative f of fin these charts is differentiable at x (resp., is smooth). We define the derivative f∗(x) off at x as the linear mapping from Tx M into Ty N :

f∗(x) ∈ L(Tx M,Ty N),

that acts by the rule

f∗(x)h := f ′(x )h.

In other words, f∗(x) is the linear mapping, represented by the derivative of the represen-tative of f at the representative of x .


(It is easy to verify (using Lemma 10.2.2., p.132), that if f has a differentiablerepresentation in some two charts at x and y, then its representative in any other twocharts at x and y will be also differentiable, and that our definition of the derivative doesnot depend on the choice of charts.)

Example. For any smooth mapping Rn → Rm its restriction to any manifold M in Rn isa smooth mapping M → Rm . (Verify!)

A vector field v on M is called smooth if the representative v of v in each chart forM is smooth. (It is easy to verify, that if v has smooth representatives in each chart froma family {ϕα}), ϕα : Uα → Uα such that

⋃Uα = M , then the representative of v in any

chart for M will be smooth, that is, v will be smooth.)A p-form ω on M is smooth (the record: ω ∈ �p(M)), if the representative ω of

ω in each chart for M is smooth. (Once again, it is easy to verify, that if ω has smoothrepresentatives for some family of chart “covering” M , then ω is smooth.)

Example. For any smooth form on Rn its restriction to M

ω|M (x) := ω(x)|Tx M×...×Tx M

is a smooth form on M .

Exterior derivativeWe define the exterior derivative dω of a form ω on M as the form, the representative ofwhich in any chart for M is the exterior derivative of the representative of ω:

dω := dω.

(it can be verified that this ”chart-wise“ definition is correct, that is, there exists just onesmooth form on M with this property.)

Exterior productLet ω1, ω2 be two forms on M . We define their exterior product point-wise:

∀x ∈ M... (ω1 ∧ ω2)(x) := ω1(x) ∧ ω2(x).

It is easy to verify that ω1 ∧ ω2 is also smooth, and in any chart for M

ω1 ∧ ω2 = ω1 ∧ ω2.

Pull-backLet M, N be manifolds, let f : M → N be a smooth mapping, and let ω be a k-form onN . We define the pull-back f ∗ω point-wise:

∀x ∈ M... ( f ∗ω)(x) := ( f∗(x))∗(ω( f (x)))

(compare with the definition for forms on vector spaces), that is,

∀h1, . . . , hk ∈ Tx M... ( f ∗ω)(x)h1, . . . , hk = ω( f (x))( f∗(x)h1) . . . ( f∗(x)hk).

Again, it can be verified that f ∗ω is smooth, and that in any charts for M and N

f ∗ω = ( f )∗ω.

Remarks. 1. The representative ω of a form ω on M in a chart ϕ for M is the pull-back:

ω = ϕ∗ω.2. Just as in the case of vector spaces, for manifolds also the operations ∗, d and∧ RESPECT

each other.

10.4. MANIFOLDS WITH A BOUNDARY 135

10.4 Manifolds with a boundary

A subset M of Rn is a k-dimensional manifold with a boundary, or simply a k-manifoldwith a boundary (the notation: M ∈ Mfk

∂(Rn)) if for each x ∈ M one of two condition (1)

and (2), is fulfilled, where (1) is the condition from 10.1, p.129, and (2) is the followingcondition:

∃U ∈ Op Nbx (Rn) ∃U ∈ Op(Rn) ∃8 ∈ Diffeo(U ,U) :

x := 8−1(x) ∈ Rk−1 × 0︸︷︷︸∈Rn−(k−1)

, (1)

andU ∩ M = 8(U ∩ (Rk−1 × R+ × 0︸︷︷︸

∈Rn−k

)). (2)

x

x

M

∂MU

Φ

U~

x~

k

x ,...,xk+1 n

x ,...,x1 k+1

(Here R+ := [0,+∞).) In other words, M is up to a diffeomorphism, a k-dimensionalhalf-space in Rn , the point in ques-tion lying on the boundary of thishalf-space.

Note that (1) and (2) cannot befulfilled simultaneously, since 8 ishomeomorphism, and a half-space(closed!) and the whole space arenot homeomorphic.

The set of all points x ∈ M , forwhich (2) is fulfilled, is called theboundary of M and is denoted by

∂M.

Example. If M ∈ Mfk(Rn) then ∂M = ∅ (though fr M 6= ∅ in general, e.g. for an openball).Exercise 10.4.1. Show that if M ∈ Mfk

∂(Rn), then ∂M ∈ Mfk−1(Rn) and M \ ∂M ∈

Mfk(Rn).All the notions introduced for manifold “without boundary” (full charts, charts, forms

vector fields, representatives etc.) can be naturally extended to the case of manifolds witha boundary. Little complications arises with differentiation (”usual“ and exterior) at pointsof the boundary ∂M , but ∂M has zero volume (by Corollary from Theorem ChangeVariables), so when dealing with INTEGRALS over manifolds (to be defined below), thevalues at boundary points are not essential, and we can just ignore these points. [Moreaccurately: we define a smooth form on a manifold with boundary as a form with smoothrepresentatives, and we DEFINE a smooth (representing) form on a closed half-space asa RESTRICTION to this half space of a smooth form on an OPEN set that contains thehalf-space. This solves all mentioned problems.]

10.5 Orientation

Let X be an n-dimensional vector space, and let

A := {a1, . . . , an} and B := {b1, . . . , bn}


be two bases for X . We define the sign of A with respect to B by the formula

sgnB A := sgn detB a1 . . . an,

where detB denotes the determinant with respect to basis B:

a j =n∑

i=1

ai j bi ⇒ detB a1 . . . an :=

∣∣∣∣∣∣

a11 . . . a1n

. . .

an1 . . . ann

∣∣∣∣∣∣.

(Note that the last determinant cannot be equal to 0, since a1, . . . , an are linearly indepen-dent.)

The relationA ∼ B :⇔ sgnB A = +1

is an equivalence relation (Verify!) An orientation of X is an equivalent class with respectto ∼. Obviously on X there are just 2 orientations. For a given basis, the orientation,containing B , we denote by

[B]

the other one by−[B].

We say that X is oriented if there is chosen one of 2 possible orientation on X . We denotedthis chosen class

or X

For oriented X we say that a basis A is positive if [A] = or X , and is negative[A] = − or X .

For Rn , the canonical orientation is [e1, . . . , en].

Examples.

1. ∀σ ∈ Sn... [eσ(1), . . . , eσ(n)] = (sgn σ)[e1, . . . , en] (Prove!)

2. [−e1, e2, . . . , en] = −[e1, . . . , en]. (Prove!)

Positive linear bijectionsLet X and Y be two n-dimensional ORIENTED vector spaces, or X = [A], or Y = [B], andlet l ∈ L(X,Y ) be a bijection. We put

sgn l := sgn det l

where det l denotes the determinant of the matrix of l with respect to the bases A and B:

det l := detB(la1) . . . (lan) ({a1, . . . , an} = A).

(It is easy to verify, that this definition of sgn l does not depend on the choice of positivebases A and B .) We say that l is positive if sgn l = +1 (resp., negative, if not).

Lemma 10.5.1. A positive linear bijection l between oriented finite-dimensional vectorspaces RESPECTS orientations, that is, sends each positive basis into a positive one.⊳ Let or X = [A], or Y . Let {c1, . . . , cn} be a positive basis in X , that is,

detA c1 . . . cn > 0. (1)

10.5. ORIENTATION 137

Then

detB(lc1) . . . (lcn) = (t∗ detB)c1 . . . cnTh on det= (det l)︸︷︷︸

sgn l=+1 >0

det Ac1 . . . cn︸︷︷︸(1) >0

> 0,

which means that the basis {lc1, . . . , lcn} is positive. ⊲

Oriented manifoldsWe say that a k-manifold M is oriented, and we write

M ∈ Or Mfk,

if for each x ∈ M the tangent space Tx M is oriented and if these orientations arecompatible in the sense that for any chart ϕ : V → V for M all all the mappings ϕ′(x),x ∈ V (which are linear bijections of Rk onto Tϕ(x) M) have one and the same sign (allare positive or all are negative), with respect to the canonical orientation of Rk .

It is obvious that if M is orientable (can be oriented), there are just 2 orientation onM . To fix an orientation on an orientable M , it is sufficient to claim any one chart ϕ aspositive, in the sense that for each point x from the domain of this chart ϕ′(x) is positive.

Examples.

1. For n ≥ 2 all the (n − 1) dimensional spheres in Rn are orientable.

2. The famous Mobius band is not oriented. (Exercise: define the Mobius band using TWO

full charts.)

Positive injectionsLet N,M ∈ Or Mfk , and let f be a (smooth) injection of M into N . We say that f ispositive (resp., negative), if for any x ∈ M the derivative f∗(x) : Tx M → T f (x) N ispositive (resp., negative). Thus, a positive mapping RESPECTS orientations.

x

-x

0

Example. The mapping f : Sk → Sk, x 7→ −x , where Sk denotesthe unit sphere in Rk+1, is positive if k is odd, and is negative if k is even.

Induced orientationLet M be an oriented k-manifold with a boundary.The induced orientationon ∂M is given by the rule: for each x ∈ ∂M

[h1, . . . , hk−1] ∈ or Tx (∂M) :⇔ [ν, h1, . . . , hk−1] = or Tx M,

where ν is the (uniquely defined) unit normal vector to M at x , such that −ν is a tangentvector to M at x (note that Tx M is here a (k-dimensional) HALF-space).

Examples.1. 2.

ν1

M(disc)

1

∂M

1ν

(ball)M

212

∂M

[ν, 1] = or M [1] = or ∂M [ν, 1, 2] = or M [1, 2] = or ∂M

Lemma 10.5.2. Let in the cube [0, 1]k its faces im I k(i,α) are equipped with with the

orientation induced by the canonical orientation in [0, 1]k . Then I k(i,α) is positive iff i + α

is even:sgn I k

(i,α) = (−1)1+α.


⊳ EXERCISE. [Hint: see Example 2 (p. 136).] ⊲

Corollary 10.5.3. Let c be a k-cube in Rn , and let im c be oriented by the condition thatc is positive. Let the faces im c(i,α) be equipped by the induced orientation. Then c(i,α) ispositive iff i + α is even:

sgn c(i,α) = (−1)i+α .

10.6 Integral of a form on an oriented manifold

Throughout this section M denotes an ORIENTED k-manifold with a boundary.

Integral over a chain on a manifoldLet c be a k-cube in M (that is, a k-cube in Rn , such that im c ⊂ M), and let ω ∈ �k(M).We put just as for cubes in Rn

∫

cω :=

∫

I kc∗ω.

Integrals over chains on M are defined once again ”by linearity“:

∫∑

ai ci

ω :=∑

ai

∫

ci

ω (the sums are FINITE).

Chart cubesWe say that a k-cube c in M is a chart cube if there exists a chart ϕ : V → V for M suchthat

V ⊃ [0, 1]k and c = ϕ|[0,1]k . (1)

Thus each chart cube is an injection, so its SIGN is determined (with respect to the canonicalorientation on [0, 1]k).

Let ω be a k-form on M . If there exists a chart k-cube c in M such that

suppω ⊂ im c (2)

then we put ∫

Mω := sgn c

∫

cω (3)

supp ω

c2 c1

[0,1] k c1-1(C)c2

-1(C)

supp c1*ωsupp c2

*ω

c2-1 c1o

IC:=im c1 im c2 M

kR

Theorem 10.6.1. This definition is correct,that is, does not depend on the choice of c: ifc1, c2 are two chart k-cubes in M such that

suppω ⊂ im c1 ∩ im c2,

then

sgn c1

∫

c1

ω = sgn c2

∫

c2

ω.

10.6. INTEGRAL OF A FORM ON AN ORIENTED MANIFOLD 139

⊳ sgn c1

∫

c1

ω = sgn c1

∫

I kc∗1ω = sgn c1

∫

I k(c2 ◦ (c−1

2 ◦ c1)︸︷︷︸defined only on c−1

1 (C),but suppω ⊂ C

)∗ω

= sgn c1

∫

I k(c−1

2 ◦ c1)∗(c∗2ω︸︷︷︸

as a k-formon Rk

= g det

)

Th on det= sgn c1

∫

I k(g ◦ c−1

2 c1) det(c−12 ◦ c1)

′︸︷︷︸

= sgn(c−12 ◦ c1)

′︸︷︷︸

Chain Rule anddef. of sgn= (sgn c2)(sgn c1)

| det(c−12 ◦c1)|

det

= sgn c2

∫

[0,1]k(g ◦ (c−1

2 ◦ c1))| det(c−12 ◦ c1)

′| =Th on change

of var’s

sgn c2

∫

[0,1]kg

c∗2ω=g det= sgn c2

∫

I kc∗2ω = sgn c2

∫

c2

ω. ⊲

Partitions of manifolds

M

U

Uc

c ϕ

ϕ

Lemma 10.6.2. Let M be compact. There exists afinite covering O of M by RELATIVELY open (thatis, open in M equipped with the topology inducedfrom Rn) sets U, each of which is contained in theimage of some chart cube.⊳ It follows obviously from the fact that for eachpoint of M it is fulfilled either (1) or (2) (see thepicture). ⊲

Lemma 10.6.3. For any compact M there exists a finite partition of unity 9 on M,submitted to the covering O from Lemma 10.6.2.

(The definition of a partition of unity for manifolds is the same as early, merelynow functions ϕ ∈ 8 are functions on M; but we know what is a smooth function on amanifold.)⊳ This follows from the following lemma ⊲

Lemma 10.6.4. For any M and any covering O of M by relatively open sets there existsa partition of unity on M, submitted to O.⊳ Each U ∈ O can be represented as U ′ ∩ M , where U ′ is an open set in Rn . The familyO′ of all such U ′ is an open covering of M in Rn . Let 8′ be a partition of unity for Msubmitted to O′ Then 8 := {ϕ|M : ϕ ∈ 8′} is what we need. (Note that ϕ|M is a smoothfunction on M (see 10.3).) ⊲

Integral over a manifoldFor a compact M and a k-form ω on M we put

∫

Mω :=

∑

ϕ∈8

∫

Mϕω


where ϕ is a FINITE partition of unity for M submitted to a coveringO of M by RELATIVELY

OPEN sets, each of which is contained in the image of some CHART cube, and the integralsin the sum are defined by the formula (3) on p. 138.

Such a covering O and such a partition 8 do exists by Lemmas 10.6.2. and 10.6.3.,and it can be shown that the so defined integral

∫M ω does not depend on the choice of O

and 8.

10.7 Stokes Theorem

Theorem 10.7.1. (Stokes) Let M be a compact oriented k-manifold with a boundary inRn , and let ω ∈ �k−1(M). Then

∫

Mdω =

∫

∂Mω

where ∂M is equipped by the induced orientation.

⊳ Case 1: There exists a chart k-cube c in M such that suppω ⊂ int(im c).(Of course suppω := cl {x ∈ M : ω(x) 6= 0}, the closure IN M , but since M is compact,it is the same as the closure in Rn .)

supp ωim c

∂M

Wlog we can assume that c is positive (if there exists a negative chart cube with thementioned property, then obviously there exists also a positivechart cube with the same property). We have

∫

Mdω =

∫

cdω =

∫

I kc∗dω =

∫

I kd(c∗ω)

Stokes Thfor chains=

∫

∂ I kc∗ω =

∫

∂cω = 0,

since ω = 0 on

im(∂c) :=⋃

i,α

im c(i,α).

(Note that Stokes Theorem for chains is applicable here, since c∗ω is, by the definition ofa smooth form on manifold, smooth in some open neighbourhood of [0, 1]k .)

But also∫∂M ω = 0, since ω = 0 on ∂M O.K.

10.8. CLASSICAL SPECIAL CASES 141

supp ωim c

Case 2: There exists a chart k-cube c in M , such that

(1)∂M ∩ im(∂c) = im c(k,0),

(2)suppω ⊂ rel int(im c).

Again wlog c is positive, so that

∫

Mdω

as in Case 1=∫

∂cω =

∑

i,α

(−1)1+α∫

c(i,α)ω

ω = 0 on all thefaces but c(k,0)= (−1)k

∫

c(k,0)ω

(3)= (−1)k sgn c(0,k)︸︷︷︸Lm 10.5.2.= (−1)k

∫

∂Mω =

∫

∂Mω. O.K.

General case: We have∫

Mdω

def=finite!∑

ϕ∈8

∫

Mϕdω

trick!=∑dϕ=0,

since∑ϕ=1

∑

ϕ∈8

∫

M(dϕ ∧ ω + ϕdω)

Cases 1,2=∑

ϕ∈8

∫

∂Mϕω

def=∫

∂Mω. ⊲

10.8 Classical special cases

In this section we discuss classical notions of divergence and rotor. For this end we needsome special 1- and 2-forms, named length element and area element, resp.

Length element

Mτ

Let M be an oriented 1-manifold (maybe with a boundary) in R3, let τ denote the unitpositive (that is, respecting the orientation) tangent vector to M .

The length element ds on M is the 1-form ON M , defined by therule

(1)ds(x)h := 〈τ, h〉 (h ∈ Tx M)

Theorem 10.8.1. Let M ∈ Or Mf1∂(R

3), let τ = (τ1, τ2, τ3) be the unit positive positivevector field on M, and let ds be the length element on M. Then

a) ds = τ1dx + τ2dy + τ3dz,b) τ1ds = dx, τ2ds = dy, τ3ds = dz.

Of course, here dx, dy, dz denote the RESTRICTIONS on M of the 1-forms dx, dy, dz onR3.

⊳ a) ds · h(1)= 〈τ, h〉 = τ1h1 + τ2h2 + τ3h3 = τ1dx · h + τ2dy · h + τ3dz · h =

(τ1dx + τ2dy + τ3dz)h.b) Let h ∈ Tx M . Then h = ατ for some α ∈ R. Hence

(τ1ds)h = τ1(ds · h) (1)= τ1︸︷︷︸dx ·τ〈τ, ατ 〉︸︷︷︸=α

= dx · ατ︸︷︷︸=h

= dx · h,

and analogously for dy, dz. ⊲


Definition. The length of a 1-dimensional compact oriented manifold M in R3 is∫

M ds.

Area element

h1

h2

ν

Let M be an oriented 2-manifold in R3 (maybe with a boundary), and let ν be the unitnormal vector to M , positive in the sense that

[h1, h2] ∈ or M ⇒ [ν, h1, h2] = or R3.

The area element dA on M is the 2-form on M , defined bythe rule

dA(x)hk := det(h, k, ν) (h, k ∈ Tx M).

Theorem 10.8.2. Let M ∈ Or Mf2∂(R

3), let ν = (ν1, ν2, ν3) be the positive unit normalvector field on M, and let dA be the areal element on M. Then

a) dA = ν1dy ∧ dz + ν2dz ∧ dx + ν3dx ∧ dy;b) ν1dA = dy ∧ dz, ν2dA = dz ∧ dx, ν3dA = dx ∧ dy.

⊳ 1◦ For h, k ∈ R3 define the vector product h × k by the rule

∀t ∈ R3 ... 〈h × k, t〉 := det(h, k, t). (2)

Applying (2) to t = ei , we conclude that

h × k =(∣∣∣∣

h2 h3k2 k3

∣∣∣∣ ,∣∣∣∣

h3 h1k3 k1

∣∣∣∣ ,∣∣∣∣

h1 h2k1 k2

∣∣∣∣). (3)

It follows at once from (2) thath × k 6 lin{h, k}. (4)

(Indeed if t is a linear combination of h and k, then det(h, k, t) = 0.) Hence

∀h, k ∈ Tx M... h × k = αν for some α ∈ R. (5)

2◦ Proof of a). ⊳⊳ ∀h.k ∈ Tx M... dA(x)hk = det(h, k, ν) = ν1

∣∣∣∣h2 h3k2 k3

∣∣∣∣︸︷︷︸=(dy∧dz)hk

+ . . . =

(ν1dy ∧ dz + . . .)hk. ⊲⊲

3◦ Proof of b). ⊳⊳ (ν1dA)hk = ν1(dAhk) = ν1 det(h, k, ν)(2)= ν1〈h × k, ν〉 (5)= αν1 =

α〈e1, ν〉 = 〈e1, αν〉 (5)= 〈e1, h × k〉 (3)=∣∣∣∣

h2 h3k2 k3

∣∣∣∣ = (dy ∧ dz)hk. ⊲⊲ ⊲

Definition. The area of 2-dimenstional compact oriented manifold M in R3 is∫

M dA.

Theorem on rotorLet M ∈ Or Mf2

∂(R3), and let F = (F1, F2, F3) be a vector field in R3. Put

ω := F1dx + F2dy + F3dz.

Then

dω = (D2 F3 − D3 F2)dy ∧ dz + (D3 F1 − D1 F3)dz ∧ dx + (D1 F2 − D2 F1)dx ∧ dy.

10.8. CLASSICAL SPECIAL CASES 143

We define the rotor rot F of the vector field F as the vector vector formed by the coefficientsof this form:

rot F :=(∣∣∣∣

D2 D3F2 F3

∣∣∣∣ ,∣∣∣∣

D3 D1F3 F1

∣∣∣∣ ,∣∣∣∣

D1 D2F1 F2

∣∣∣∣).

Stokes theorem yields∫

M((D2 F3 − D3 F2︸︷︷︸

(rot F)1

) dy ∧ dz︸︷︷︸ν1dA

+ . . .) =∫

∂M(F1 dx︸︷︷︸=τ1ds

+ . . .),

νrot F

M

τ∂M

F

whence it follows that

(6)∫

M〈rot F, ν〉dA =

∫

∂M〈F, τ 〉ds.

Physical meaning of (6) is: the FLOW of the rotor of avector field through a surface is equal to the CIRCULATION ofthis vector field over the contour.

Remark. 1. Physically, rot F is not a vector, since the direction of so defined rot F dependson definition of d.

2. In English language literature one use more oft the termini “curl” instead of “rotor”.3. The so-called Green formula

∫∂M αdx + βdy = ∫

M (∂β/∂x − ∂α/∂y)dxdy is aspecial case of (6), corresponding to F3 = 0 and to a FLAT surface, parallel to x, y plane.

Theorem on divergenceLet now M ∈ Or Mf2

∂(R3) (equipped with the orientation of R3), and let again F be a

vector field in R3. Put

ω := F1dy ∧ dz + F2dz ∧ dx + F3dx ∧ dy.

Thendω = (D1 F1 + D2 F2 + D3 F3) det .

The quantity in the brackets is called the divergence of the field F and is denoted by div F :

div F := D1 F1 + D2 F2 + D3 F3.

Stokes formula yields∫

M(D1 F1 + . . .) det =

∫

∂

(F1 dy ∧ dz︸︷︷︸=ν1dA

+ . . .),

whence it follows that ∫

Mdiv F =

∫

∂M〈F, ν〉dA. (7)

F

ν The physical sense of (7) is: the flow of the vectorfield through a closed surface OUTSIDE is equal to theintegral of the divergence of the field over the regioninside this surface.

Chapter 11

Functions of complex variable

11.1 Analytic functions

We can identify the set C of complex numbers with the real plain R2. More precisely, ifwe put

α : C→ R2, z 7→ (Rez,Imz) =: z (we call z the representative of z),

β : R2 → C, (x, y) 7→ x + iy,

thenα ◦ β = id, β ◦ α = id .

By this identification, the norm in C, defined by the formula

‖z‖ := |z|,

coincides with Euclidean norm in R2. So we can identify C and R2 also as topologicalspaces.

Ifx = cos θ, y = sin θ,

y

x

(x,y)ρ

θ

we say that , θ are polar coordinates of (x, y) and of z = x + iy and we write

(x, y) ∼ {, θ}, and z ∼ {, θ}.E.g., 0 ∼ {0, 0}, 0 ∼ {0, 22π}, i ∼ {1, π/2}, i ∼

{1,−3π/2}.

Lemma 11.1.1. If z1 ∼ {1, θ1} , z2 ∼ {2, θ2}, then z1z2 ∼ {12, θ1 + θ2}.⊳ (1 cos θ1 + i1 sin θ1)(2 cos θ2 + i2 sin θ2) = 12(cos θ1 cos θ2 − sin θ1 sin θ2︸︷︷︸

=cos(θ1+θ2)

) +

i12(sin θ1 cos θ2 − cos θ1 sin θ2︸︷︷︸=sin(θ1+θ2)

). ⊲

Lemma 11.1.2. Let c = a + i ∼ {, θ}. Consider the operator of multiplication by c

A : C→ C, z 7→ cz,

145

146 CHAPTER 11. FUNCTIONS OF COMPLEX VARIABLE

and denote by A the corresponding operator in R2:

A := α ◦ A ◦ β.

CA

K Cβ ↑ ↓ αR2 A

K R2

Then A ∈ L(R2,R2), and the matrix of A is equal to(

a −bb a

)=

(cos θ − sin θsin θ cos θ

).

In other words, A is the composition of the TURN by the angle θ and blowing up withthe coefficient (≥ 0).

⊳ (a + ib)(x + iy) = (ax − by)+ i(ay + bx), and

(a −bb a

)(xy

)=(

ax − byay + bx

). ⊲

C-differentiabilityWe say that a function f : C→ C is C-differentiable at a point z ∈ C, and we write

f ∈ DifC(z),

if there exists the limit (in the norm)

limz→z(z 6=z)

f (z)− f (z)

z − z=: f ′(z) ∈ C,

called the (C-) derivative of f at z. If G is an open subset of C, and f is differentiable ateach point of G, then we say that f is C-differentiable in G, and we write

f ∈ DifC(G).

Examples.

1. z 7→ zn is C-differentiable everywhere if n = 0, 1, 2 . . ., and is C-differentiable in C\0if n = −1,−2, . . .; (zn)′ = nzn−1.

2. z 7→ z is nowhere C-differentiable. (Verify!)

Cf

K Cβ ↑ ↓ αR2 f

K R2

Notation: For any f : C→ C we denote by f the corresponding“real” mapping:

f := α ◦ f ◦ β,

that isf (x, y) := (Re( f (x + iy))︸︷︷︸

=:u(x,y)

,Im( f (x + iy))︸︷︷︸=:v(x,y)

)

For short we write f = u + iv. Thus

f = u + iv :⇔ f = (u, v).

Theorem 11.1.3. Let f = u + iv. The following conditions are equivalent:

a) f ∈ DifC(z), f ′(z) = a + ib ∼ {, θ};

11.2. COMPLEX FORMS 147

b) f ∈ Dif(z), f ′(z) =(∂u/∂x ∂u/∂y∂v/∂x ∂v/∂y

)∣∣∣∣z=(

a −bb a

)=

(cos θ − sin θsin θ cos θ

).

In other words, C-differentiability is a SPECIAL case of differentiability where thederivative is the composition of turn an a blowing up with non-negative coefficient. ⊳ Justas in the classic real case, f is C-differentiable at z iff it holds the decomposition

f (z + ζ ) = f (z)+ f ′(z)ζ + r(ζ ),r(ζ )

|ζ | K|ζ |→0

0.

In “R2-language” this means that

f (z + ζ ) = f (z)+ f ′(z )ζ + r (ζ ),r (ζ )

‖ζ‖ K‖ζ‖→0

0;

here f ′(z) denotes the linear operator R2 → R2, corresponding to the operator of mul-tiplication by f ′(z) (Lemma 11.1.2.). So our assertion follows from Lemma 11.1.2. ⊲

Cauchy-Riemann conditionsLet G ∈ Op(C). We say that a function f : G→ C, f = u+iv, satisfies Cauchy-Riemann(or d’Alambert-Euler) conditions, and we write

f ∈ CR(G),

if u, v ∈ C1(G) and∂u

∂x= ∂v

∂y,

∂u

∂y= −∂v

∂x.

Analytic functionsLet G ∈ Op(C). We say that a function f : G → C, is analytic, and we write

f ∈ An(G),

if f ∈ DifC(G) and f ′ : G → C is continuous.NB It is possible to show that if f is C-differentiable in G then f ′ IS AUTOMATICALLY

continuous. But it is a hard theorem.

Example. z 7→ zn is analytic in the whole C if n ∈ N, and is analytic in C\0 if n ∈ Z\N.

Theorem 11.1.4. A function f : G → C is analytic iff it satisfies Cauchy-Riemannconditions:

An(G) = CR(G).

⊳ This follows at once from Theorem 11.1.3. ⊲

11.2 Complex forms

For any M ∈ Mfk∂(R

n) and any p ∈ {0, 1, . . . , k} we define

�pC(M)

as the set of pairs (ω1, ω2) ∈ �p(M)×�p(M) written as ω1 + iω2; for short we put

ω + i0 =: ω.


We equipe this set by the following structure of vector space over C (below a + ib ∈ C):

(a + ib)(ω1 + iω2) := (aω1 + bω2)+ i(bω1 + aω2), (1)

(ω1 + iω2)+ (ω3 + iω4) := (ω1 + iω3)+ i(ω2 + iω4). (2)

Define the conjugate form by the rule

ω1 + iω2 := ω1 − iω2, (3)

and define the operations ∧, d, ∗,∫

M component-wise:

(ω1 + iω2) ∧ (ω3 + iω4) := (ω1 ∧ ω3 − ω2 ∧ ω4)+ i(ω2 ∧ ω3 + ω1ω4), (4)

d(ω1 + iω2) := dω1 + idω2, (5)

g∗(ω1 + iω2) := g∗ω1 + ig∗ω2 (g : M → N), (6)∫

M(ω1 + iω2) :=

∫

Mω1 + i

∫

Mω2. (7)

Examples.

1. Let x and y denote, in accord with a classical tradition, the projections (x, y) 7→ x and(x, y) 7→ y in R2. Then

z := x + iy ∈ �0C(R2), z= x − iy ∈ �0

C(R2),

dz = dx + idy ∈ �1C(R2), dz=dz = dx − idy ∈ �1

C(R2).

2. Each function f : G → C, f : u + iv, G ∈ Op(C), with sufficiently smooth u, v maybe considered as a complex 0-form:

f ∈ �0C(G).

(In the last relation we consider G as an open set in R2.) What means “sufficiently”,depends on the context. Sometimes just continuity is sufficient.NB All the result for “real” forms retain (as it is easy to verify) for complex ones, e.g. forC-forms as for real ones, it holds

d(ω1 ∧ ω2) = dω1 ∧ ω2 + (−1)degω1ω1 ∧ dω2, (8)

d2 = 0, (9)∫

Mdω =

∫

∂Mω (for compact oriented M). (10)

Theorem 11.2.1. Let G ∈ Op(R2), and let f, g ∈ �0C(G) (that is, f and g are functions

G → C). Then the following assertions are true:

a) f ∈ An(G)⇔ d( f dz) = 0;b) f ∈ An(G)⇒ d f = f ′dz;c) d f = gdz ⇒ f ∈ An(G), g = f ′.

This can be summarized so:

d( f dz) = 0⇔ f ∈ An⇔ d f = gdz⇒ g = f ′. (11)

11.3. INTEGRALS OF COMPLEX 1-FORMS 149

NB The equation d( f dz) = 0 is equivalent to d f ∧dz = 0 (since d2 = 0), and the equationd f = gdz may be formally written as

d f

dz= g.

Both equations say that d f and dz are proportional (“parallel”).⊳ Let f = u + iv, g = p + iq .

a) d( f dz)d2=0= d f ∧ dz = (du + idv) ∧ (dx − idy)(4)= (du ∧ dx − dv ∧ dy)+ i(dv ∧ dx + du ∧ dy)

du = (∂u/∂x)dx + (∂u/∂y)dy, dv = (∂v/∂x)dx + (∂v/∂y)dy

=(−(∂u

∂y+ ∂v∂x

)+ i

(∂u

∂x− ∂v∂y

))dx ∧ dy.

Hence

d( f dz) = 0⇔(∂u

∂y+ ∂v∂x= 0,

∂u

∂x+ ∂v∂y= 0

)def.⇔ f ∈ CR(G)

Th.11.1.4.⇔ f ∈ An(G).

b) f ∈ An(G)Th.11.1.4.⇒ f ∈ CR(G)⇒

d f = du + idv =(∂u

∂xdx + ∂u

∂ydy

)+ i

(∂v

∂xdx + ∂v

∂ydy

)

∂u/∂x = ∂v/∂y =: a, ∂v/∂x = −∂u/∂y =: b

= (adx − bdy)+ i(bdx + ady)(1)= (a + ib)(dx + idy)

Th.11.1.3.= f ′(z)dz.

c) d f = gdz ⇒((∂u/∂x)dx + (∂v/∂y)dy)+ i ((∂v/∂x)dx + (∂v/∂y)dy) = (p + iq)(dx + idy)

Compare Re and Im{(∂u/∂x)dx + (∂u/∂y)dy = pdx − qdy

(∂v/∂x)dx + (∂v/∂y)dy = qdy + pdy

Compare coeficients by dx and dy{

∂u/∂x = ∂v/∂y = p

−∂u/∂y = ∂v/∂x = qTh.11.1.3.⇒

{f ∈ CR(G)

f ′ = p + iq = g.⊲

11.3 Integrals of complex 1-forms

c(0)==c(1)

Theorem 11.3.1. Let G ∈ Op(C), let c : [0, 1] → G be a closedcurve (that is, a 1-cube with c(0) = c(1)), and let f ∈ �0

C(G) (that

is, f : G → C). Then∫

cd f = 0.

⊳∫

c d f = ∫I 1 c∗d f = ∫I 1 d (c∗ f )︸︷︷︸= f ◦c=:g

= ∫I 1 g′ = ∫[0,1] g′ = ∫ 10 g′(t)dt = g(1)− g(0) =

f (c(1))− f (c(0)) = 0. ⊲


Example. For any n ∈ Z \ {−1}, and for any closed curve in C \ {0}∫

czndz = 0.

⊳ For n 6= −1 we have zndz = d

(1

n + 1zn+1

)

︸︷︷︸∈An(C\0)

. ⊲

C

Theorem 11.3.2. Let C ∈ Comp Or Mf1(R2), and let f ∈ �0C(C). Then

∫

Cd f = 0

⊳

∫

Cd f

Stokes Th=∫

∂Cf∂M=∅= 0. ⊲

Let M be a compact 2-manifold in R2. Then M is orientable (verify!), and we alwayssuppose that M is equipped with the orientation from R2. Further

◦M = intR2 M = M \ ∂M

(verify!). We say that a function f : M → C is analytic on M , and we write

f ∈ An(M),

if it is analytic in◦M (we identify R2 and C!) and is continuous on M . Thus,

An(M) := An(◦M) ∩ C(M).

M

∂M

Theorem 11.3.3. (Cauchy) Let M ∈ Comp Mf2∂(R

2), and let f ∈An(M). Then

∫

∂Mf dz = 0

⊳

∫

∂Mf dz

Stokes Th.=∫

Md( f dz)︸︷︷︸

Th. 11.2.1. =0

= 0. ⊲

If f /∈ An(M), the integral can be non-zero, as the following example shows.

∂Bρ(0)

Example. Let M = B(0) (the disc of radius with the center at0). Then

∫

∂ B(0)

dz

z= 2π i.

(Note that (z 7→ 1/z) ∈ An(C \ 0).)⊳ It is clear that ∫

∂ B(0)

dz

z=∫

c

dz

z,

where c : [0, 1]→ R2, t 7→ ( cos 2π t︸︷︷︸=:c1

, sin 2π t︸︷︷︸=:c2

). But

11.4. CAUCHY FORMULA 151

∫

c

dz

z=∫

c

dx + idy

x + iy=∫

c

(x − iy)(dx + idy)

x2 + y2 =∫

cc∗((x − iy)(dx + idy)

x2 + y2

)

=∫ 1

0

(c1(t)− ic2(t))(c′1(t)dt + ic′2(t)dt

c21(t)+ c2

2(t))

= 2π i∫ 1

0(cos 2π t − i sin 2π t)(cos 2π t + i sin 2π t)︸︷︷︸

=1

= 2π i. ⊲

Lemma 11.3.4. Let B := B(a), f ∈ C(∂B). Then

∣∣∣∣∫

∂Bf dz

∣∣∣∣ ≤ 2πmax∂B| f |.

⊳ Let a =: a1 + ia2. Put c(t) := (a1 + cos 2π t︸︷︷︸=:c1

, a2 + sinπ t︸︷︷︸=:c2

). Then

∣∣∣∣∫

∂Bf dz

∣∣∣∣ =∣∣∣∣∫

cf dz

∣∣∣∣ =

∣∣∣∣∣∣∣

∫

I 1( f ◦ c) c∗dz︸︷︷︸

=(c′1+ic′2)dt

∣∣∣∣∣∣∣=∣∣∣∣∫ 1

0f (c(t))(c′1 + ic′2)dt

∣∣∣∣

Exam. 11.3.5.≤ max0≤t≤1

| f (c(t))(c′1 + ic′2|︸︷︷︸= | f (c(t))| |c′1 + ic′2|︸︷︷︸

=2π

≤ 2πmax∂B| f |. ⊲

Exercise 11.3.5. Prove, using Mean Value Theorem, that for any continuous functionf : [0, 1]→ C ∥∥∥∥

∫ 1

0f (t)dt

∥∥∥∥ ≤ max0≤t≤1

| f (t)|.

Exercise 11.3.6. More general, prove that if C ∈ Comp Or Mf1(R2), f ∈ �0C(C), then

∥∥∥∥∫

Cf dz

∥∥∥∥ ≤∫

C| f |ds.

11.4 Cauchy formula

Example 11.3.6. is a special typical case of the following

Theorem 11.4.1. Let M ∈ Comp Mf2∂(R

2), and let f ∈ An(M). Then for each point◦M(= M \ ∂M)

∫

∂M

f (z)dz

z − a= 2π i f (a) (Cauchy Formula).

In other words, the values of an analytic function inside a region are uniquely definedby the values on the boundary.

⊳ Let ε > 0 be such that the disc B := Bε(a) lies in◦M . Obviously f (z)/(z − a) ∈

An(M \ ◦M). Hence by Cauchy Theorem


Ba

M \ B 0 =∫

∂(M\◦B)

f (z)dz

z − a=∫

∂M

f (z)dz

z − a−∫

∂B

f (z)dz

z − a.

So it suffices to show that∫

∂B

f (z)dz

z − aK

ε↓02π i f (a). (1)

Since f ∈ DifC(a), we have

f (z) = f (a)+ f ′(a)(z − a)+ r(z − a),r(ζ )

|ζ | K|ζ |→0

0. (2)

Therefore,∫

∂B

f (z)dz

z − a= f (a)

∫

∂B

dz

z − a︸︷︷︸[1]

+∫

∂B

(f ′(a)+ r(z − a)

z − a

)

︸︷︷︸=:g(z)

dz.

︸︷︷︸[2]

(3)

Just as in Example 11.3.6., [1] = 2π i. As to [2], the function g is by (2) bounded (in thenorm) in some neighbourhood of a, hence, by Theorem 11.3.1., [2]→ 0 as ε ↓ 0, and (1)is proved. ⊲


2), f ∈ An(M). Then f ∈ C∞C(◦M), and

∀n ∈ N ∀a ∈ M... f (n)(a) = n!

2π i

∫

∂M

f (z)dz

(z − a)n+1 . (4)

Thus if f is just one time continuously C-differentiable, it is infinitely C-differen-tiable! ⊳ For n = 1 this this is Theorem 11.4.1.. Let it be true for n − 1, that is,

∀a ∈ M... f (n−1)(a) = n − 1

2π i

∫

∂M

f (z)dz

(z − a)n. (5)

Differentiation of (5) in a (which is possible as it can be shown) yields (4). ⊲

NB For ANY continuous function ϕ : ∂M → C, Cauchy formula

f (z) := 1

2π i

∫

∂M

ϕdζ

ζ − z(6)

defines an analytic function f :◦M → C, but in general this f , extended to ∂M as ϕ, is

NOT continuous! IF ϕ = f |∂M for some f ∈ An(M), then Cauchy formula does reproducethe original function f .

11.5 Representation by series

We say that a series∑∞

n=0 cn , c ∈ C, converges, and we write

∞∑

n=0

cn ;,

11.5. REPRESENTATION BY SERIES 153

if the sequence of partial sums sN :=∑Nn=0 cn converges (in the norm ‖z‖ = z) to some

c ∈ C , that is, if |sN − c| KN→∞

0, and in such case we write∑∞

n=0 cn = c.

Uniform convergenceConsider a FUNCTION SERIES

∑∞n=0 fn , fn : C→ C. If

∑∞n=0 fn ; uniformly in z ∈ A

we write ∑fn;

;A.

We say that a function series is majorized on A ⊂ C by a real series∑

tn , tn ≥ 0, if

∀z ∈ A ∀n ∈ N... | fn(z)| ≤ tn .

Lemma 11.5.1. Let∑∞

n=0 tn , be a CONVERGING real sequence, and let function series∑∞n=0 fn , fn : C→ C, is majorized on A ⊂ C by

∑tn . Then

∑fn;

;A.

⊳ ∀z ∈ A...

∣∣∣∣∣N∑

n=0

f (z)−M∑

n=0

∣∣∣∣∣if e.g. N > M=

∣∣∣∣∣N∑

n=M+1

f (z)

∣∣∣∣∣ ≤N∑

n=M+1

| f (z)| ≤N∑

n=M+1

tn KN,M→∞

0. ⊲

Member-wise integration and differentiation

Theorem 11.5.2. Let C ∈ Comp Or Mf1∂(R

2), and let fn ∈ �0C(C) (that is, fn = un+ ivn ,

where un, vn are (at least) continuous functions C → R). If fn Kn→∞ f UNIFORMLY on

C then ∫

Cfndz →

∫

Cf dz.

⊳ Taking into account the definition of∫

M ω, this follows from the corresponding theoremfor real-valued functions on [0, 1]. ⊲

Corollary 11.5.3. Let C ∈ Comp Or Mf1∂(R

2), f ∈ �0C(C) and

∑∞n=0 fn;

;C. Then

∫

C

( ∞∑

n=0

fn

)dz =

∞∑

n=0

∫

Cfndz.


2), and let fn ∈ An(M). If fn Kn→∞ f UNIFOR-

MLY on M, then also fn ∈ An(M), and

∀a ∈ ◦M ∀k ∈ N... f (k)n (a) K

n→∞ f (k)(a).

⊳ 1◦ f ∈ C(M) as an uniform limit of continuous function.

2◦ For each a ∈ ◦M we have obviously fn(z)/(z − a) Kn→∞ f (z)/(z − a) UNIFORMLY ON

∂M , hence

f (a) = lim fn(a)Cauchy formula= lim

1

2π i

∫

∂M

fn(z)dz

z − aTh 11.5.2.= 1

2π i

∫

∂M

f (z)dz

z − a.

So by NB from 11.4, f ∈ An(◦M). Thus, in view of 1◦, f ∈ An(M). 3◦ ∀a ∈ ◦M ∀k ∈ N

...

f (k)n (a)Th. 11.4.2.= k!

2π i

∫

∂M

fn(z)dz

(z − a)k+1

Th 11.5.2.K

k!

2π i

∫

∂M

f (z)dz

(z − a)k+1Th. 11.4.2.= f (k)(a). ⊲


Corollary 11.5.5. Let M ∈ Comp Mf2∂(R

2), fn ∈ An(M),∑∞

n=0 fn;;M f . Then also

f ∈ An(M), and

∀k ∈ N... f (k) =

∞∑

n=0

f k in◦M .

Convergence discBρ(a)

az^

Theorem 11.5.6. (Abel) Let a, z ∈ C, a 6= z. If a series∑∞

n=0 cn(z−a)n

converges for z = z then for each > 0 such that < |z−a| this seriesconverges UNIFORMLY on B(a).⊳ 1◦ Since

∑cn(z − a)n ;, the sequence |cn(z − a)n| converges to

zero and hence it is bounded, e.g. by α > 0.2◦ Now, let z ∈ B(a). Then |z − a|/|z − a| ≤ /|z − a| =: k < 1, hence

|cn(z − a)n| ≤ |cn(z − a)n|︸︷︷︸1◦≤α

(|z − a|/|z − a|︸︷︷︸≤k

)n ≤ αkn .

Thus, our series is majorized on B(r) by the converging real series∑αkn , and our

assertion follows from Lemma 11.5.1. ⊲

It follows from this theorem that the set of all points z, where a power series∑

cn(z−a)n converges, is either merely {a}, or it the whole C, or lies between

◦B(a) and B(a)

for some > 0. In other words, the CONVERGENCE REGION is, up to the boundary, a DISC

with the center at a. We call it the convergence disc.

Taylor formula

M

az

h

∂M

Lemma 11.5.7. Let M ∈ Comp Mf2∂(R

2), let z, a ∈ ◦M(= M \ ∂M),h := z − a, and let f ∈ An(M). Then for each n ∈ N

f (z) =n∑

k=0

1

k!f (k)(a)hk + hn+1

2π i

∫

∂M

f (ζ )dζ

(ζ − z)(ζ − a)n + 1︸︷︷︸=:rn (h)

.

⊳ f (z)Cauchy formula= 1

2π i

∫

∂M

f (ζ )dζ

ζ − ztrick= 1

2π i

∫

∂M

f (ζ )dζ

(ζ − a)

(1− z − a

ζ − a

) =

q := (z − a)/(ζ − a) = h(ζ − a), (1− qn+1)/(1− q) = 1+ q + . . .+ qn

= 1

2π i

∫

∂M

f (ζ )

ζ − a

(n∑

k=0

(h

ζ − a

)n

+ 1

1− h

ζ − a

(h

ζ − a

)n+1)

dζ

=n∑

k=0

hk 1

2π i

∫

∂M

f (ζ )dζ

(ζ − a)k+1︸︷︷︸

Th.11.4.2. = f (k)(a)/n

+hn+1

2π i

∫

∂M

f (ζ )dζ

(ζ − z)(ζ − a)n+1. ⊲

11.6. ELEMENTARY FUNCTIONS 155

Definition. Let G ∈ Op(R2), f ∈ An(G), a ∈ G. The series∑∞

n=0 1/n! · f (n)(a)(z−a)n

is called Taylor series for f at a and is denoted by tsa f :

tsa f :=∞∑

n=0

1

n!f (n)(a)(z − a)n.

Theorem 11.5.8. Let B be a disc in R2 of a positive radius with center at a.

a) Let f ∈ An(◦B). Then tsa f ; f in B.

b) Let∑∞

n=0 cn(z − a)n ; f in◦B. Then f ∈ An(

◦B), and tsa f =∑∞n=0 cn(z − a)n .

Thus an analytic function can be uniquely represented in each “disc of analyticity”by a power series w.r. to the center of the disc.

Bρ(a)a

z

B

⊳ a) Let z ∈ ◦B, h = z − a. Choose > 0 such that z ∈ ◦B(a) ⊂◦B.

Then f ∈ An(B(a)) and k := h/ < 1. Hence we have for the restterm rn in Taylor formula

|rn(h)| =∣∣∣∣∣hn + 1

2π i

∫

∂B(a)

f (ζ )dζ

(ζ − z)(ζ − a)n

∣∣∣∣∣

Lm 11.3.4.≤ 2π i|h|n+1

2πmaxζ∈∂B

∣∣∣∣f (ζ )

ζ − z

∣∣∣∣1

n+1 = const ·kn+1K

n→∞ 0,

that is, tsa f ; f at z.

b) If∑

cn(z − a)n ; in◦B, then, by Abel Theorem, this series converges uniformly

on each closed disc which is contained in◦B; hence, by Corollary 11.5.5., its sum f is

analytic in◦B, and ∀ f ∈ N

... f (k)(a) = ∑∞n=0

ddzk

∣∣∣∣z=a

(cn(z − a)n) = n!cn . It follows

that tsa f =∑ cn(z − a)n . ⊲

NB This theorem allows to EXTEND analytic functions, if the convergence disc of theTaylor series is greater than the original disc of analyticity. These extensions can lead toDIFFERENT values at one and the same “outer” point!

11.6 Elementary functions

We DEFINE exp, sin, etc by the SAME series representation as in real analysis:

ez := 1+ z + z2

2!+ z3

3!+ z4

4!+ z5

5!+ . . . , (1)

sin z := z − z3

3!+ z5

5!− . . . , (2)

cos z := 1− z2

2!+ z4

4!− . . . , (3)

sh z := z + z3

3!+ z5

5!+ . . . , (4)

ch z := 1+ z2

2!+ z4

4!+ . . . . (5)


Since all these series are majorized in each disc B by the real series∑∞

n=0 n/n! all they

converge in the whole C uniformly on each closed disc. So by Corollary 11.5.5., theirsums are analytic functions in the whole C.

It follows from (1)–(5) that

sin z = eiz − e−iz

2i= −i sh(iz), sh z = ez − e−z

2= −i sin(iz), (6)

cos z = eiz + e−iz

2i= ch(iz), ch z = ez + e−z

2= cos iz, (7)

andez = ch z + sh z = cos(iz)− i sin(iz). (8)

Putting z := iθ , θ ∈ R, we obtain

eiθ = cos θ + i sin θ (Euler formula). (9)

In particular

eiπ = −1. (10)

This formula connect three the most fundamental numbers in mathematics, e, π and i, andis one of most beautiful mathematical formulas.

Relation (9) implies that eiθ = cos θ + i sin θ , which means that

eiθ ∼ {, θ} ( ≥ 0, θ ∈ R). (11)

Further the following basic property of the exponent function remains true:

ez1+z2 = ez1ez2 . (12)

⊳ Multiplying the absolutely converging series for ez1 and ez2 , we obtain

ez1ez2 =∞∑

k=0

zk1

k!

∞∑

l=0

f zl2l! =

∞∑

n=0

n∑

k=0

1

k!(n − k)!︸︷︷︸1

n!

(n

k

)zk

1zn−k2

=∞∑

n=0

(z1 + z2)n

n!= ez1+z2 . ⊲

LogarithmAt last we define ln z as a complex number c such that ec = z. Such a number is definednon-uniquely. Indeed,

∀n ∈ Z... ec+2π in (12)= ec e2π i

︸︷︷︸=1

= ec = z. (13)

Sinceeln+iθ (12)= eln eiθ ,

we see that if z ∼ {, θ} then one of the values of ln z is ln + iθ . By (13), ln+ iθ+2π in,n ∈ Z ,are also values of ln z. It can be verified that there is no other ones. Thus

z ∼ {, θ} ⇒ ln z = ln + i(2πn + θ). (14)

The fact that ln is a MULTI-VALUED function is just another form of the fact that therepresentation z ∼ {, θ} is non-unique.

11.7. RESIDUES 157

11.7 Residues

B a

U

Let a ∈ C, U ∈ Op Nba(C), and f ∈ An(U \ {a}). The residue of f at a is defined by theformula

(1)resa f := 1

2π i

∫

∂Bf (z)dz,

where B is any disc with the center at a such that B ⊂ U and

a ∈ ◦B.

B\B1 2

This definition does not depend in the choice of B . Indeed if B1 and B2 are two different

such discs, e.g. B1 ⊂◦B2, then

0Cauchy Th=

∫

∂(B2\B1)

f (z)dz =∫

∂B2

f (z)dz −∫

∂B1

f (z)dz.

Examples.

1. If f is analytic in a neighbourbood of a, then

resaf (z)

z − a= f (a). (2)

⊳ This follows at once from Cauchy formula. ⊲

2. As a special case of previous example ( f = 1) we obtain

resa1

z − a= 1. (3)

3. For k = 2, 3, . . .

resa1

(z − a)k= 0. (4)

(See Example on page 150.)

a1 a2

∂M

Theorem 11.7.1. (on residues) Let M ∈ Comp Mf2∂(R

2), let

a1, . . . , an ∈◦M(= M\∂M), and let f ∈ An(M\{a1, . . . , an}).

Then∫

∂Mf dz =

n∑

k=1

resak f.

⊳ Let Bk be mutually disjoint discs with the centers at ak such that Bk ⊂◦M . Then

0Cauchy Th=

∫

∂(M\⋃ ◦Bk)

f dz =∫

∂Mf dz −

∑∫

∂Bk

f dz

︸︷︷︸2π i·resak f

(note that ∂Bk as a part of ∂(M \⋃ ◦Bk) has the orientation opposite to orientation of ∂Bk

itself). ⊲

This theorem reduces calculation of integrals to calculation of residues.


Calculation of residuesA point a ∈ C is a pole of degree n (or n-pole), n ∈ N, of a function f : U \ a → C,(U ∈ Op Nba(C)) if the function (z − a)n f admits an analytic extension to U , but thefunction (z − a)n−1 f does not.

Examples.

1. 0 is a 1-pole for 1/z, is a 2-pole for 1/z2, etc.

2. 1/sin z has 1-poles at z = 0,±π,±2π, . . ..

Theorem 11.7.2.

a) Let a be an n-pole for f . The there exist a closed disc B with the center at a such

that◦B 6= ∅ and f can be (uniquely) represented in

◦B\ a by some series

∞∑

k=−n

ck(z − a)k, (ck ∈ C),

called the Laurent series for f at a; that is,

f =∞∑

k=−n

ck(z − a)k in◦B\ a.

b) The residue of f at a is equal to the coefficient by (z − a)−1:

resa f = c−1.

c) This residue can be calculated by differentiation:

resa f = 1

(n − 1)!

dn−1

dzn−1

∣∣∣∣a((z − a)n f )

= 1

(n − 1)!limz→a

dn−1

dzn−1

((z − a)n f (z)

).

⊳ EXERCISE for you. [Hint: Apply Theorem 11.5.8. on Taylor series to (z−a)n f and thenuse Examples 2 and 3 on page 157.] ⊲

Chapter 12

Ordinary differential equations

12.1 Analytic setting

A differential equation is an equation of the form

F(

x, u(x), u′(x), u′′(x), . . . , u(p)(x))= 0, (1)

whereu : X → Y,

u′ : X →L (X,Y )

...

u(p) : X →L (X, . . . , X︸︷︷︸p

; Y ),

andF : X × Y ×L (X,Y )× . . .×L (X, . . . , X; Y )→ Z .

Here X,Y, Z are (say) normed spaces; U is the unknown function, and (1) is to be fulfilledat each point x of some open set U ⊂ X .

The order p of the highest derivative in (1) is called order of the equation.

ClassificationIf X = R we have an ordinary differential equation (ODE);If X = Rn we have a partial differential equation (PDE);If Y = R we have a scalar differential equation;If Y = Rn we have a vector differential equation;If Z = R we have ONE differential equation;If Z = Rn we have a system of differential equations.

ODE’sIn ODE’s the unknown function is a function of ONE independent variable, which physi-cally can be interpreted as TIME (and is denoted usually by t), and (1) can be interpretedas a LAW of some process, of some evolution.

159

160 CHAPTER 12. ORDINARY DIFFERENTIAL EQUATIONS

The derivatives in t are often denoted (after Newton) by dots above, e.g.,

x := dx

dt, x := d2x

dt2 .

Examples.

1. Inertial motion equation: x = 0, x : R→ R3; this eqaution describes a free motion ofa point in the space (no external forces).

2. Pendulum equation: x + x = 0, x : R → R; this equationdescribes the motion of a point, on which the force act, that isproportional to the displacement of the point, from an equilibriumposition and tends to return the point back.

3. Exponent equation: x = kx , x : R → R; it describes a process where the speed ofgrown of something is proportional to the present quantity of this something; e.g., fork > 0 it may be a chain reaction, for k < 0 it may be a nuclear decay.

Thus, a general ODE looks as follows (now we write t instead of x , and x instead ofu, respectively X instead of Y ):

F(

t, x, x, x, . . . , x (n))= 0. (2)

Since all the derivatives of x : R→ X are again functions R→ X , now we have

F : R× Xn+1 → Z .

A solution of (2) is an n times differentiable function ϕ : (a, b)→ X (where (a, b) is annon-empty interval in R, such that

∀t ∈ (a, b)... F

(t, ϕ(t), ϕ(t), . . . , ϕ(n)(t)

)= 0.

NB Not every equation, containig derivatives in t , is an ODE. E.g., the equation x(x = 0)is NOT.

Reduction to a first order equationWe shall consider ONLY ODE’s SOLVED w.r. to the highest derivative:

dnx

dtn= F

(t, x, x, x, . . . , x (n−1)

)= 0. (3)

Theorem 12.1.1. Equation (3) is equivalent to the following system of n ODE’s of the firstorder:

x1 = x2x2 = x3

...

xn−1 = xn

xn = F(t, x1, x2, . . . , xn)

(4)

⊳ If ϕ : (a, b)→ X is a solution of (3), then(ϕ, ϕ, ϕ, . . . , ϕ(n−1))

12.2. GEOMETRIC SETTING 161

is a solution of (4); v.v., if

(ϕ1, ϕ2, . . . , ϕn) : (a, b)→ Xn

is a solution of (4), then ϕ1 is a solution of (3). ⊲

By this theorem we can restrict ourselves by ODE’s of the first order

x = f (x, t) , x : R→ X, f : X ×R→ X.

12.2 Geometric setting

X

t

x

Geometrically, the unknown function x : R → X in an ODE x = f (x, t) is a CURVE

in X . The space X is called the phase space of the equation. Thegraph of a solution ϕ is called an integral curve of the equation.The space X ×R, where this graph lies, is called the extend phasespace.

Examples.

1. Inertial motion equation x = 0 (x : R→ R3) is equivalent to the system

x1 = x2

x2 = 0

}(x1, x2 : R→ R3),

which can be written as x = Ax , where x : R→ R6, and A ∈L (R6,R6), viz.

A =

0 0 0 0 0 00 0 0 0 0 00 0 0 0 0 01 0 0 0 0 00 1 0 0 0 00 0 1 0 0 0

.

Here the phase space is R6 (position-velocity).

2. Pendulum equation x + x = 0 is equivalent to the system

x1 = x2

x2 = −x1

}

which can be written as x = Ax , where x : R→ R2,

A =(

0 +1−1 0

)∈L (R2,R2)

(the operator of rotation by 90◦ �) The phase space is R2.

3. For the exponent equation x = kx the phase space is R.


Fields corresponding to ODE’s

X

t

A geometric interpretation of the equation x = f (x, t) is such. At each point of theextended phase space a DIRECTION is given (the derivative of acurve at a point “is” the tangent line to the graph). To find asolution of the equation means to find an integral curve of thisfield of directions, that is, to find a curve, such that at each pointthe tangent line coincides with the directions of the field at thispoint.

X

In the special case where f does NOT depend on the t (autonomic ODE’s) another inter-pretation is possible, in the phase space itself. Viz., at each pointx of the phase space a VECTOR f (x) is given (the derivative canbe considered as a vector in X (the velocity)). That is, we have aVECTOR FIELD. To find a solution of the equation x = f (x)meansto find a motion of a point in the phase space such that the velocityat each moment of time coincides with the value of the vector fieldat the point where we come at this moment.

Examples.

x

t

1. The simplest equation x = f (x), x : R→ R, f : R→ R; the field of directions doesnot depend on x . The solution is unique up to vertical translationof the graph. We know this already, of course:

x(t) =∫ t

x0

f (τ )dτ + const (for any t0 ∈ I ).

(we suppose that f is sufficiently “nice”.)

2. x + x = 0; (x1, x2) = (x2,−x1). The velocity field is drownon the picture. Obviously the solutions are motions along circleswith center at 0, e.g., (cos t,− sin t). Hence, the correspondingsolutions of the original equation x + x = 0 are cos t and sin t .

x

t

3. x = kx . The field of directions for k > 0 looks as on the picture. Up to a horizontaltranslation of the graph there are just 3 solutions. Of course, weknow them:

x = x0ekt .

For x0 > 0, = 0 or < 0 we obtain the 3 types of solutions.

Initial conditionsIn general the function f in the right-hand side of our equation

x = f (x, t) (1)

is defined only on an open subset9 of X ×R. We say that ϕ : R→ X is a solution of (1)in an interval (a, b) (−∞ ≤ a < b ≤ +∞), and we write

ϕ ∈ Sol(a,b)(1),

if ϕ ∈ Dif((a, b)), {(ϕ(t), t)|tt ∈ (a, b)} ⊂ 9 and ∀t ∈ (a, b)... ϕ(t) = f (ϕ, t).

We say that ϕ is a solution of (1), and we write

ϕ ∈ Sol(1),

12.3. BASIC THEOREM 163

if ϕ ∈ Sol(a,b)(1) for some a, b.Let (x0, t0) ∈ 9 . We say that ϕ is a solution of (1) (in (a, b)) with the initial condition

(x0, t0), and we write

ϕ ∈ Sol(1)t0,x0 (resp., ϕ ∈ Sol(a,b)(1)t0,x0)

if ϕ ∈ Sol(1) (resp., ϕ ∈ Sol(a,b)(1)) and

ϕ(t0) = x0.

This condition means physically that at given initial moment t0 our process has value x0,and means geometrically that the graph of ϕ, the integral curve, passes through the point(x0, t0).

12.3 Basic Theorem

Here we discuss the questions of existence and uniqueness of a solution of an equation

x = f (x, t), x : R→ X, (1)

with a given initial condition, and dependence of the solution on the initial condition.

Theorem 12.3.1. (Peano) Let X = Rn , 9 ∈ Op(Rn ×R). If f ∈ C(9, X), then

∀(x0, t0) ∈ 9 ∃ϕ ∈ Sol(1)t0,x0 .

In other words, if f is continuous then a solution of (1) always exists. (We omitthe proof of this theorem). But in general a solution with a given initial condition is NOT

unique, as the following example shows.

x

Example. The equation x = 3x2/3, x : R → R, has 2 solutionwith the initial condition (0, 0): ϕ1(t) = 0 and ϕ2(t) = t3. Thereason is that the right-hand side x2/3 decrease too quickly asx → 0:

x

3x 2/3

If f is of CLASS C1 in x , such a patology cannot occur, as we shall see.

Picard methodConsider the equation (1) for X = R, f ∈ C(R×R,R). The main idea of the method is:to find a solution of (1) means just to find a fixed point of some operator. More precisely:

ϕ ∈ Sol(1)t0,x0 ⇔ ϕ ∈ Fix A,

where A is defined by the formula

(Aϕ) := x0 +∫ t

t0f (ϕ(τ ), τ )dτ. (2)


⊳ (Aϕ)·(t) (2)= f (ϕ(t), t); if ϕ ∈ Sol(1)t0,x0 , that is, if ϕ = f (ϕ(t), t) and ϕ(t0) = x0,

then (Aϕ)· = ϕ and (Aϕ)(t0)(2)= x0 = ϕ(t0); hence Aϕ = ϕ, that is, ϕ ∈ Fix A. V.v., if

ϕ ∈ Fix A, that is, Aϕ = ϕ, then ϕ = (Aϕ)· (2)= f (ϕ(t), t) and ϕ(t0) = (Aϕ)(t0) (2)= x0. ⊲

Picard method is to construct a solution of (1) as a fixed point of A, that is, as a limitof sequence ϕ0, ϕ1 := Aϕ0, ϕ2 := Aϕ1, . . ., where ϕ0 is an initial approximation to thesolution.

Examples.

x

tt0

x0

1ϕ

0ϕ

1. x = f (t), x(t0) = x0. Field of directions does not depend on x . Put

ϕ0 : =x0.

Then already the FIRST approximation

ϕ1(t) = (Aϕ)(t) = x0 +∫ t

t0f (τ )dτ

yields the solution ϕ = f , ϕ1(t0) = x0.

x

t

x0

1ϕ

0ϕ

2ϕx0et

2. x = x , x(0) = x0. Again put ϕ0 : =x0. Then ϕ1(t) = x0 +∫ t

t0x0dτ = x0(1 + t),

ϕ2(t) = x0+∫ t

t0x0(1+ t)dτ = x0+ (1+ t + t2/2), . . . , ϕn(t) =

x0(1 + t + t2/2 + . . .+ tn/n!), so that ϕn(t) → x0et . And x0et

is indeed the solution.To justify Picard method we shall show that in an appropriate

space the Picard operator A is a contraction. We need for this endto define integral of a VECTOR function of real variable.

Integrals of vector functionsLet X be a Banach space (e.g., Rn), and let f ∈ C(R, X). The integral

∫ b

af (t)dt (∈ X)

is defined just as usually (by means of partial sums).

Lemma 12.3.2. ∥∥∥∥∫ b

af (t)dt

∥∥∥∥ ≤∣∣∣∣∫ b

a‖ f (t)‖dt

∣∣∣∣ .

(Here it may be a > b!)⊳ It follows from the corresponding inequality for partial sums:

∥∥∥∑

f (ti )1i

∥∥∥ ≤∑‖ f (ti )1i‖ =

∣∣∣∑‖ f (ti )‖1i

∣∣∣ . ⊲

Lemma 12.3.3.d

dt

∣∣∣t=t

∫ b

af (τ )dτ = f (t).

⊳ Just as usually. ⊲

Lemma 12.3.4. (Newton-Leibniz Theorem). Let ϕ ∈ C1(R, X). Then

∫ b

aϕ(t)dt = ϕ(b)− ϕ(a).

12.3. BASIC THEOREM 165

⊳ As usually. ⊲

Basic Theorem

Theorem 12.3.5. (on existence, uniqueness and continuous dependence on initial condi-tion). Let X ∈ BS, W ∈ Op(x × R), f ∈ C1(W, X), (x0, t0) ∈ W, and we consider theequation

x = f (x, t). (3)

Then there exist an open interval I with the center at t0 and an open ball B in X with thecenter at x0, such that

∀x ∈ B ∃!ϕx ∈ SolI (3)t0,x0

and for any CLOSED interval J ⊂ I the mapping

x 7→ ϕx |J , B → C(J, X)

is continuous.

b

aW

x0

t0

⊳ 1◦ At first we construct a subspace M of of a Banachspace, where a modification of Picard operator is a con-traction Take a > 0, b > 0 such that the cylinder

C := Bb(x0)× Ia(t0)

lies in W . (We denote by Ia(t) the closed ball Ba(t) in R.)Put

S := sup(x,t)∈C

‖ f (x, t)‖, (4)

L := sup(x,t)∈C

‖D1 f (x, t)‖ (5)

(where D1 f ≡ ∂ f/∂x : 9 →L (X, X)). These supremums are finite and attained sinceC is compact.

a

b

a’

b’

Now choose a′ > 0 and b′ > 0 so that the cone K′ := {(x, t) : |t − t0| ≤ a′, ‖x − x0‖ ≤S|t − t0|} and all its translations by the vectors(b, 0), b ∈ Bb′(x0), lies in the cylinder C:

(6)K := K′ + (Bb′(x0)× {0}) ⊂ C.

Consider the “small” cylinder

C′ := Bb′(x0)× Ia′(t0) ⊂ C.

We put

M := {v ∈ C(C, X)|∀(x, t) ∈ C′ ... ‖v(x, t)‖ ≤ S|t − t0|}.

In particular

∀v ∈M... v(·, t0) = 0. (7)

An element of M is shown on the following two pictures (note that on the left picture thephase space X is represented by a LINE, and on the right one by a PLAIN):


t

xXX

vv

a’

b’

t0

x0

t0

v(x, )

-(x ,0)0

0

The graph of v. The graph of v(x, ·) for fixed x .Emphasize that M depends on a′, b′, S.2◦ Now define a modified Picard operator A on M by the formula

(Av)(x, t) :=∫ t

t0f (x + v(x, τ ), τ )dτ ((x, t) ∈ C

′). (8)

This definition is correct since, by (6), the argument of f lies in 9 .

3◦ A maps M into itself. ⊳⊳ ∀(x, t) ∈ C′...

‖Av(x, t)‖ =∥∥∥∥∫ t

t0f (x + v(x, τ ), τ )dτ

∥∥∥∥Lm12.3.2.≤

∥∥∥∥∥

∫ t

t0‖ f (x + v(x, τ ), τ )‖︸︷︷︸

≤S

dτ

∥∥∥∥∥ ≤ S|t − t0|. ⊲⊲

4◦ A is a contraction for sufficiently small a′. Indeed A ∈ LipLa′ , where L is from (5).

⊳⊳ ∀v1, v2 ∈M...

‖Av1 − Av2‖ = sup(x,t)∈C′

‖ (Av1 − Av2)(x, t)︸︷︷︸(8)=∫ t

t0( f (x+v1(x,τ ),τ )− f (x+v2(x,τ )))dτ︸︷︷︸

1

‖ ≤ sup(x,t)∈C′

∣∣∣∣∣

∫ t

t0

∥∥∥ 1∥∥∥

︸︷︷︸MVT≤ L ‖v1(x, τ )− v2(x, τ )‖︸︷︷︸

obv.≤ ‖v1−v2‖

dτ

∣∣∣∣∣

Lm 12.3.2.≤ sup(x,t)∈C′

L |t − t0|︸︷︷︸≤a′

‖v1 − v2‖ ≤ La′‖v1 − v2‖. ⊲⊲

5◦ By Fixed point Theorem (= Contraction Lemma) there exists v ∈M such that Av = v.Put

u(x, t) := x + v(x, t) ((x, t) ∈ C′).

For any given x ∈ Bb′(x0) (an initial value) we have

u(x, ·) ∈ Sol(3)t0,x0 .

⊳⊳ddt u(x, t) = d

dt(x + v(x, t)︸︷︷︸

=Av(x,t)

)

(8)= d

dt

(x +

∫ t

t0f (x + v(x, τ ), τ )dτ

)Lm 12.3.3.= f (x + v(x, τ ), τ )

= f (u(x, τ ), τ ),

and u(x, t0) = x + v(x, t0)︸︷︷︸(7)=0

= x . ⊲⊲

12.4. METHODS OF SOLUTIONS 167

6◦ Our solution depends continuously on the initial value x since v is a continuousmapping.7◦ Uniqueness: Take b′ := 0, and consider the corresponding set M and operator A. (NowM consists from functions defined just on Ia′(t0).) Obviously,

ϕ ∈ Sol(3)t0,x0 ⇔ ϕ − x0 ∈ Fix A,

but the fixed point of A is unique (by Contraction Lemma). Hence (on the interval◦Ia′(t0))

the solution is unique. ⊲

12.4 Methods of solutions

As Liouville showed, in general it is impossible to solve a given ODE in explicite form (in“quadratures”), that is, in form of finite combination of elementary and algebraic functionsand of integrals of them. E.g., such a simple equations as dy/dx = y2−x cannot be solvedin quadratures.

There are general methods of APPROXIMATIVE solution of ODE’s, in particular methodsbased on Picard approximations.

Rather full theory of explicite solution is only for LINEAR ODE’s, which we considerin the next two sections.

Here we discuss special but important case where solutions can be calculated ratherexplicitly.

No dependence on x

x = f (t).

This equation describes a process, the speed of which does not depend on its state, but isfully determined “from outside”. The solution satisfying an initial condition x(t0) = x0 isgiven by the “classic” formula of analysis

x(t) = x0 +∫ t

t0f (τ )dτ.

No dependence on t

x = f (x) (X = R). (1)

This equation describes an “automatic” process where the behavior of the process isdefined entirely by its present state.

Theorem 12.4.1. Let f ∈ C1((a, b)), x0 ∈ (a, b), f (x0) 6= 0 (−∞ ≤ a < b ≤ +∞).Then for any t0 ∈ R the solution ϕ of equation (1) with initial condition (x0, t0) (whichdoes exist by Basic Theorem) satisfies the relation

t − t0 =∫ ϕ(t)

x0

dξ

f (ξ). (2)


x

b

a

t0

0x

t

In other words, our solution x = ϕ(t) can be found by solving of the equation

t − t0 =∫ x

x0

dξ

f (ξ)

w.r. to x .⊳ Let ϕ ∈ Sol(1)t0,x0 . Then ϕ(t0) = f (x0) 6= 0. By inverseFunction Theorem, the inverse function ϕ−1 =: ψ is definedlocally (in a neighbourhoodof x0).

We have ψ(x0) = t0, anddψ

dx

∣∣∣∣ξ

= 1

f (ξ).

Since f (x0) 6= 0, the function 1/ f (ξ) is continuous in a neighbourhood of x0. By Newton-Leibniz Theorem,

ψ(x)− ψ(x0) =∫ x

x0

(dξ)

f (ξ).

Putting here x = ϕ(t), we obtain (2). ⊲

Separable variables

x = g(x)

f (t)(X = R).

Here x and t enter “separately”. For better symmetry let us write y instead of x and xinstead of t:

dy

dx= g(y)

f (x). (3)

Theorem 12.4.2. Let f and g are of class C1 in some neighbourhoods of points x0 andy0, resp.; let f (x0) 6= 0, g(y0) 6= 0, and let y = F(x) be a solution of (3) with theinitial condition F(x0) = y0 (such solution does exist by Basic Theorem). Then F is givenimplicitly by the equation ∫ x

x0

dξ

f (ξ)=∫ y

y0

dη

g(η).

Thus the solution method is: to rewrite (3) formally as dx/ f (x) = dy/g(y) and tointegrate in the corresponding limits.⊳ Consider two new ODE’s

x = f (x), (4)

y = g(y). (5)

By Basic Theorem, there exist ϕ ∈ Sol(4)0,x0 , ψ ∈ Sol(5)0,y0 , defined on a (w.l.o.g.)COMMON open interval I:

ϕ(t) = f (ϕ(t)), ϕ(0) = x0, (6)

ψ(t) = f (ψ(t)), ψ(0) = x0. (7)

t

x y

ϕ ψϕ−1

u

We have ϕ(0) = f (x0) 6= 0, ψ(0) = g(y0) 6= 0. By Inverse Function Theorem, thereexist (locally) the inverse functions, ϕ−1. We claim that

u := ψ ◦ ϕ−1 ∈ Sol(3)x0,y0 .

Indeed

12.5. LINEAR EQUATIONS 169

u(x) =t :=ϕ−1(x)

ψ(t)(ϕ−1).(x) = ψ(t)

ϕ(t)(6),(7)=

y:=u(x)=ψ(t)g(y)

f (x),

u(x0) = ψ(ϕ−1(x0))(6)= ψ(0) (7)= y0.

But by Theorem 12.4.1.,

∫ ϕ(t)

x0

dξ

f (ξ)= t − t0 =

∫ ψ(t)

y0

dη

g(η).

Putting ϕ(t) = x , ψ(t) = y, we see that u = ψ ◦ ϕ−1 : x 7→ y, and∫ x

x0

dξ

f (ξ=∫ y

y0

dη

g(η),

which is what we need. ⊲

12.5 Linear equations

By a linear (homogenious) ODE we mean an equation

x = A(t)x (x : I → X, X ∈ NS), (1)

where A(t) for each t ∈ I is a continuous LINEAR operator in X , and the mapping

A : I →L (X, X)

is sufficiently smooth. Thus the right-hand side of (1) is linear (and continuous) in x . Inthe case X = Rn the equation (1) takes the form

x1 = a11(t)x1 + . . .+ a1n(t)xn...

xn = an1(t)x1 + . . .+ ann(t)xn

(2)

so usually one calls (1) a linear ODE with variable coefficients.

x

Example. Pendulum of variable length: x = −ω2(t)x , (x : R →R). This equation when written in the form (2) is

x1 = x2

x2 = −ω2(t)x1

}(3)

In the form (1) it looks as

x = A(t)x, where A(t) =(

0 1−ω2(t) 0

), x : R→ R2.

A very pleasant feature of linear equations is that they have solutions defined in thewhole interval I :

Theorem 12.5.1. Any solution of (1) can be extended to I .⊳ The idea of the proof is such. Since on any COMPACT subinterval J of I the norm ‖A‖is bounded (as a continuous function), we have ‖x‖ = ‖A(t)x‖︸︷︷︸

≤C

‖x‖ ≤ C‖x‖. on J . It


follows that any solution grows not faster than eCt (in the norm) on J , and hence cannotgo away to infinity on J . An accurate proof see e.g., in [9, p. 196]. ⊲

tt0

x0

I

NB For non-linear equations it can be that a solution does not admitan extension to whole I . E.g., for the equation x = x2 one suchsolution x(t) = −1/(t − 1) is shown on the picture.

Vector space of solutions

Theorem 12.5.2. The set S of all the solutions of (1) (defined in the whole I ) is a vectorspace. This space is isomorphic to the phase space X.

⊳ 1◦ If ϕ1, ϕ2 ∈ Sol(1) then ∀α1, α2 ∈ R... (α1ϕ1 + α1ϕ1)

· = α1ϕ1 + α1ϕ2(1)= α1 Aϕ1 +

α2 Aϕ2 = A is linear = A(α1ϕ1 + α2ϕ2), that is, α1ϕ1 + α2ϕ2 ∈ Sol(1). Thus, S is avector space.

ta b

In particular

0 ∈ Sol(1),

and

ϕ ∈ Sol(1)⇒ −ϕ ∈ Sol(1);the picture of integral curves is SYMMETRIC (see the picture).

2◦ Fix any t ∈ I and consider the mapping

δt S :→ X ϕ :7→ ϕ(t),

which sends each solution ϕ into its value at the moment t . Obviously, δt is linear. Theimage of δt is the whole X , since by Basic Theorem for any x ∈ X there exists a solution ϕwith ϕ(t) = x . The kernel of δt is {0}, since again by Basic Theorem, there exists just onesolution ϕ with ϕ(t) = 0, and this solution is evidently ϕ = 0. Thus δt is both surjectiveand injective. ⊲

Fundamental system of solutionsLet X = Rn . Then, by Theorem 12.5.2., S ≈ Rn . Any basis ϕ1, . . . , ϕn of S is calleda fundamental system of solutions for (1). Thus:

a) Each Equation (1) (in Rn) has a fundamental system of solutions.b) If ϕ1, . . . , ϕn is a fundamental system of solutions then any solution ϕ is a linear

combination of ϕ1, . . . , ϕn .c) Any n + 1 solutions are linearly depend.d) For any t1, t2 ∈ I the mapping

X X

gt2t1

tt2t1

gt2t1 := δt2 ◦ (δt1)

−1 : X → X

(the transformation of the phase space in the timefrom t1 up to t2 is a linear isomorphism.

12.5. LINEAR EQUATIONS 171

Example. For the pendulum equation x =(

0 1−1 0

)x (x : R→ R2) the system

{(cos t,− sin t), (sin t, cos t)}

is a fundamental system of solutions. (Verify! [Hint:

∣∣∣∣cos t − sin tsin t cos t

∣∣∣∣ = 1.])

Scalar linear equation of the n-th orderConsider a linear (in x) homogeneous equation of the n-th order with variable coefficients

x (n) = a1(t)x(n−1) + . . .+ an(t)x, (x : I → R, ai ∈ C(I,R)). (4)

We know from Theorem 12.1.1. that (4) is equivalent to an equation

Ex = A(t)Ex, (Ex : I → Rn, A ∈ C(I,L (Rn,Rn)).

In view of this equivalence, it follows from Theorem 12.5.2. that the following result istrue:

Theorem 12.5.3. The set S of all solutions of (4) (defined on the whole I ) is a vectorspace, which is isomorphic to Rn . This isomorphism is realized by the mapping

S→ Rn, ϕ 7→ (ϕ(t), ϕ(t), . . . , ϕ(n−1)(t)),

where t is an arbitrary fixed point in I .Any basis of this n-dimensional vector space is called fundamental system of solutions

for (4).

Example. For the pendulum equation x + x = 0 the functions cos t , sin t form a funda-mental system of solutions (see example on 171).

Finding solutions with given initial conditionsLet ϕ be a solution of (4). We say that ϕ satisfies an initial condition

(Ex0, t0) ∈ Rn × I

if(ϕ(t0), ϕ(t0), . . . , ϕ

(n−1)(t0) = Ex0,

that is,ϕ(t0) = x01ϕ(t0) = x02

...

ϕ(n−1)(t0) = x0n

(5)

Let we know a fundamental system of solutions ϕ1, . . . , ϕn for (4). If we need to findthe solution with initial conditions (5), we look for the solution in the form

ϕ = c1ϕ1 + . . .+ cnϕn (ci ∈ R).

Then (5) yieldsc1ϕ1(t0)+ . . .+ cnϕn(t0) = x01

...

c1ϕ(n−1)1 (t0)+ . . .+ cnϕ

(n−1)n (t0) = x01

(6)


Solving this linear algebraic system, we obtain the desired values c1, . . . , cn .Remark. The determinant of the system (6)

∣∣∣∣∣∣∣∣∣

ϕ1(t0) . . . ϕn(t0)ϕ1(t0) . . . ϕn(t0)...

...

ϕ(n−1)1 (t0) . . . ϕ

(n−1)n (t0)

∣∣∣∣∣∣∣∣∣

is called Wronskian of the system {ϕ1, . . . , ϕn}. Since the solution of (6) MUST exist forany Ex0 we conclude that Wronskian of any fundamental system of solutions of (4) isNO-ZERO for each t ∈ T .

Variation of constantsFor solving of NON-homogenious linear equations the following METHOD OF VARIATION

OF CONSTANTS is available:In order to solve an equation

x = A(t)x + h(t), x : I → Rn, A ∈ C(I,L (Rn,Rn), h ∈ C(I,Rn),

supposing we know a fundamental system ϕ1, . . . , ϕn of solutions of the correspondinghomogenious equation x = A(t)x , we look for the solutions in the form

ϕ(t) = c1(t)ϕ1(t)+ . . .+ cn(t)ϕ(t) (ϕ, ϕi : I → Rn)

(with VARIABLE “constants” ci !). Then we obtain these unknown functions

(c1, . . . , cn) := c : I → Rn

a “simplest” linear equation of the form

c = f (t) ( f : I → Rn),

which we know to solve.

Example. Consider the equation

x + x = f (t) x : I → R, f ∈ C(I,R), 0 ∈ I, (7)

with initial condition (0, a), a = (a1, a2) ∈ R2, that is

x(0) = a1, x(0) = a2. (8)

Reduction to a 1. order system yields

x1 = x2x2 = −x1 + f

},

x1(0) = a1x2(0) = a2

}. (9)

The corresponding homogenious system{

x1 = x2x2 = −x1

has a well known (Example on page 171) a fundamental system of solutions

{(cos t,− sin t), (sin t, cos t)}.

12.6. LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 173

So look for the solution in the form

(x1, x2) = c1(t)(cos t,− sin t)+ c2(t)(sin t, cos t).

The substitution into (9) gives after simplifications

c1 cos t + c2 sin t = 0−c1 sin t + c2 cos t = f

},

c1(0) = a1c2(0) = a2

},

whence we obtain

c1 = − f sin t, c1(0) = a1 ⇒ c1 = a1 −∫ t

0 f (τ ) sin τdτ,c2 = f cos t, c2(0) = a2 ⇒ c2 = a2 +

∫ t0 f (τ ) cos τdτ.

If follows that the answer is (x(t) = x1(t)!):

x(t)(8)=(

x(0)−∫ t

0f (τ ) sin τdτ

)cos t +

(x(0)+

∫ t

0f (τ ) cos τdτ

)sin t .

12.6 Linear equations with constant coefficients

Here we study an equation

x = Ax, x : R→ X, X ∈ NS, A ∈L (X, X). (1)

We suppose that the space L (X, X) (with the operator norm) is COMPLETE; for example,it is ever true for X = Rn .

In the simplest case X = R we have an equation

x = ax, x : R→ R, a ∈ R.

The solution is well-known

x = x0eat , x0 = x(0).

In the general case the result is just the same:

Theorem 12.6.1. Any solution x of (1) can be extended to the whole R and is given by theformula

x = eAt x0, x0 = x(0).

Here for any operator A ∈L (X, X) we put

eA := id+A + 12 A2 + 1

3! A3 + . . . ∈L (X, X), Ak := A ◦ . . . ◦ A︸︷︷︸k-times

This series converges in L (X, X), since it is majorized by the converging non-negativereal series

∑∞k=0 ‖A‖k/k! (indeed, ‖Ak‖ = ‖ A ◦ . . . ◦ A︸︷︷︸

k-times

‖ ≤ ‖A‖k).

⊳ The theorem can be proved essentially in the same way as in classic 1-dimensional case(using member wise differentiation of series). ⊲

Thus principally we know the solution of (1), but the problem is how to CALCULATE

eAt for concrete A. Even for X = R it is non-trivial problem.


Case of diagonal operatorsLet X = Rn . We shall identify an operator A ∈ L (Rn,Rn) with its matrix. If A isdiagonal, that is,

A =

λ1 0. . .

0 λn

,

then it is easy to calculate eAt :

eAt =

eλ1t 0. . .

0 eλn t

.

⊳ Ak =

λk

1 0. . .

0 λkn

, hence

∑ Ak tk

k!=∑

λk1tk

k! 0. . .

0λk

n tk

k!

=

eλ1t 0. . .

0 eλn t

. ⊲

Hence the solution of equation

x = Ax, x(0) = x0 = (x01, . . . , x0n)

isx = eAt x0 = (x01eλ1t , . . . , x0neλn t ).

Note that λk are just the EIGENVALUES of our diagonal operator. Thus, each componentof the solution has the form

c eλt

where λ is an eigenvalue of A.NB In OTHER bases the components of the solutions will be LINEAR COMBINATIONS of theexponents eλk t .

General caseIn occurs that in general case each component of the solution of an equation

x = Ax, x : R→ Rn, A ∈L (Rn,Rn), (2)

is a linear combination of n members of the form

tmReeλt or tm

Imeλt

where λ is an eigenvalue of A (λ ∈ C) and m is a natural number less than multiplicity ofλ.

Recall that the eigenvalues of a are roots of the characteristic equation

det(A − λE) = 0 E denotes the unit matrix

and that the multiplicity of an eigenvalue is just the multiplicity of the root.

12.6. LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 175

Case of one scalar equation of n-th orderx (n) = a1x (n−1) + . . .+ anx, x : R→ C, a j ∈ C. (3)

As we know (3) can be considered as a special case of (2). It follows that:Any solution of (3) has the form

x(t) =k∑

l=1

eλl t pl(t), (4)

where λ1, . . . , λk are the different roots of the characteristic equation

λn = a1λn−1 + . . .+ an, (5)

and pl is a polynomial of degree less than the multiplicity of the root λl .NB This result remains true for non-homogenious equations x (n) = a1x (n−1)+ . . .+anx+f (t), if f (t) has the form (4).

Examples.

1. x+ x = 0. The characteristic equation λ2+1 has the roots±i; we haveRee±it = cos t ,Ime±it = ± sin t . The functions cos t, sin t form a fundamental system solutions. Thegeneral solution is c1 cos t + c2 sin t .

2. x − x = 0. The characteristic equation λ2 − 1 has the roots ±1. The correspondingfunctions et and e−t form a fundamental system of solutions. The general solution isc1et + c2e−t . (In particular sh t and ch t are solutions.)

3. x = 0. The characteristic equation λ2 = 0 has one 2-multiple root 0. The correspondingfunctions from (4) are 1 and t . They form a fundamental system of solutions. The generalsolution is c1 + c2t (as well known, of course).

PRIKLADY A CVICENI1. Prirozena topologie Rn

Prıklady

1. Dokazte, ze ctverec M = {(x, y) ∈ Rn; |x | + |y| ≤ 1} je kompaktnı mnozina.

Resenı: Stacı ukazat, ze mnozina M je uzavrena a ohranicena. Uzavrenost lze dokazatprımo z definice uzavrene mnoziny; muzeme ale vyuzıt spojitosti zobrazenı f (x, y) =|x | + |y|. Platı M = f −1(M) = (−∞, 1], jedna se tedy o vzor uzavrene mnoziny prispojitem zobrazenı.Jelikoz pro kazde (x, y) ∈ M platı

√x2 + y2 ≤ |x | + |y| ≤ 1, je mnozina M ohranicena.

2. Dokazte, ze kanonicka projekce π i : Rn → R, π i (x) = x i , je spojita.

Resenı: Necht’U ⊂ R je otevrena mnozina. Dokazeme, ze (π i )−1(U) = V , kde

V = R× . . .×R︸︷︷︸i − 1 cinitelu

×U × R× . . .× R︸︷︷︸n − i cinitelu

.

Necht’x ∈ V . Platı π i (x) = x i ∈ U , a tedy x ∈ (π i )−1(U). Opacne, je-li x ∈ (π i )−1(U),pak π i (x) ∈ U . Jelikoz π i (x) = x i , je x i ∈ U , a tedy x ∈ V .

3. Dokazte, ze zobrazenı f : R→ R2 je spojite prave tehdy, kdyz je spojita kazda jehoslozka.

Resenı: ,,⇒“ Pro slozky f 1, f 2 zobrazenı f platı f 1 = π1 ◦ f , f 2 = π2 ◦ f (π1, π2

jsou kanonicke projekce). Je-li tedy spojite zobrazenı f , jsou spojite i jeho slozky (jakoztokompozice spojitych zobrazenı).,,⇐“ Necht’I 1, I 2 jsou otevrene intervaly. Pro dukaz spojitosti zobrazenı f stacı dokazat,ze mnozina f −1(I 1 × I 2) je otevrena (zduvodnete!). Oznacme V 1 = ( f 1)−1(I 1), V 2 =( f 2)−1(I 2) a V = V 1 ∩V 2. Jsou-li zobrazenı f 1 a f 2 spojita, je mnozina V (jako prunikdvou otevrenych mnozin) otevrena. Je-li y ∈ f (V ), existuje x ∈ V takove, ze f (x) = y.Tedy f 1(x) ∈ I 1, f 2(x) ∈ I 2 a y = ( f 1(x), f 2(x)) ∈ I 1 × I 2.

4. Ma funkce f : R2 \ {(0, 0)} → R definovana f (x, y) = (x4 − y4)/(x2 + y2) pro(x, y) 6= (0, 0) limitu v bode (0, 0)?

Resenı: Mame

f (x, y) = (x2 + y2)(x2 − y2)

x2 + y2 = x2 − y2.

Funkce f je tedy definovana na libovolnem okolı bodu (0, 0) mınus tento bod. Navıc

lim(x,y)→(0,0)

f (x, y) = lim(x,y)→(0,0)

(x2 − y2) = 0.

5. Vypoctete

lim(x,y)→(0,0)

tg(x − y)

x − y.

177

178 PRIKLADY A CVICENI

Resenı: Necht’g(x, y) = x − y

h(t) ={

1 pro t = 0;tg(t)/t jinde.

Jelikoz

limt→0

tg(t)

t= 1,

je funkce h spojita. Navıc lim(x,y)→(0,0)(x − y) = 0 a pro (x, y) 6= (0, 0) platı

h ◦ g(x, y) = tg(x − y)

x − y.

Je tedy

lim(x,y)→(0,0)

tg(x − y)

x − y= h(0) = 1.

6. Rozhodnete, je-li funkce f : R2→ R spojita v bode (0, 0), jestlize

f (x, y) ={(x2 − y2)/(x2 + 2y2) pro (x, y) 6= (0, 0);

0 pro (x, y) = (0, 0).

Resenı: Tato funkce je samozrejme spojita ve vsech bodech (x, y) takovych, ze x2+2y2 6=0, to jest, vsude mimo bod (0, 0). Abychom vyresili otazku v bode (0, 0), odhadnemeodpoved’a pote se pokusıme nas odhad overit. V tomto prıpade odhadneme, ze se jednao nespojitost. Pokusıme se tedy najıt takovou cestu, po nız, kdyz se budeme priblizovatk (0, 0), limita z f (x, y) bude ruzna od f (0, 0).Predpokladejme, ze (x, y) → (0, 0) po prımce y = x . Potom (x, y) = (t, t) a nauvazovane prımce platı

lim(x,y)→(0,0)

f (x, y) = limt→0

t2 − t2

t2 + 2t2 = 0 = f (0, 0).

Nas prvnı odhad cesty tedy nevysel, protoze jsme se po nı priblızili k hodnote f (0, 0).Pokusıme se priblızit k bodu (0, 0) po prımce y = 2x , to jest, (x, y) = (t, 2t). Na tetoprımce tedy platı

lim(x,y)→(0,0)

f (x, y) = limt→0

t2 − 4t2

t2 + 8t2 = limt→0

−3t2

9t2 = limt→0−1

3= −1

36= 0 = f (0, 0).

Tedy lim(x,y)→(0,0) f (x, y) neexistuje a funkce f nenı spojita v bode (0, 0).

7. Rozhodnete, je-li funkce f : R2→ R spojita v bode (0, 0), jestlize

f (x, y) ={(x3y − xy3)/(x2 + 2y2) pro (x, y) 6= (0, 0);

0 pro (x, y) = (0, 0).

PRIROZENA TOPOLOGIE Rn 179

Resenı: V tomto prıpade budeme ocekavat v bode (0, 0) spojitost. Abychom to overili,musıme ukazat, ze f (x, y) → 0 pro (x, y) → (0, 0). Nejlepe toho dosahneme tak, zenajdeme vyraz, jehoz absolutnı hodnota je vetsı nez | f (x, y)| a ktery zrejme konvergujek 0, kdyz z = (x, y)→ (0, 0). Vsimneme si, ze |x | ≤ ‖z‖ a |y| ≤ ‖z‖. Pak

| f (x, y)| =∣∣xy(x2 − y2)

∣∣x2 + y2 = |x ||y||x + y||x − y|

x2 + y2 ≤ |x ||y|(|x | + |y|)(|x | + |y|)x2 + y2

≤ ‖z‖‖z‖(‖z‖ + ‖z‖)(‖z‖ + ‖z‖)‖z‖2 = 4‖z‖2 = 4(x2 + y2).

Jelikoz pro (x, y)→ (0, 0) mame 4(x2 + y2)→ 0, dostavame, ze | f (x, y)| → 0.

8. Najdete

lim(x,y)→(∞,∞)

x + y

x2 + y2.

Resenı: Prejdeme k polarnım souradnicım. Tedy x = cosϕ, y = sinϕ a

lim(x,y)→(∞,∞)

x + y

x2 + y2 = limϕ∈[0,π/2]

sinϕ + cosϕ

2 sin2 ϕ + 2 cos2 ϕ

= limϕ∈[0,π/2]

sinϕ + cosϕ

= 0.

Cvicenı

1. Najdete vnitrek, vnejsek, hranici a uzaver mnoziny A pokuda) A = {(1/n, 1/n) ; n ∈ N} ; b) A = {(x, y) ∈ R2; x > 0, y = sin(x)}.

2. Rozhodnete, zda mnozina M ⊂ R2 je otevrena, uzavrena, ohranicena kompaktnı asouvisla, kde

a) M = {(x, y) ∈ R2; xy < 1, y ≥ 0, x ≥ 0};b) M = {(x, y) ∈ R2; x2 < y < x3, 1 < x < 2};c) M = {((−1)5, 2/k2) ∈ R2; k ∈ N};d) M = {(1,−k/(1− k)) ∈ R2; k ∈ N \ {1}};e) M = {(k/(3k + 2), (k2 + 1)/(2− k)) ∈ R2; k ∈ N \ {2}};f) M = {(x, y) ∈ R2; 0 ≤ y ≤

√2x − x2, 0 ≤ x ≤ 2}.

3. Uved’te prıklad mnozin A, B ⊂ R2 takovych, ze A = clA a clB = frB . Existujemnozina C ⊂ R2 takova, ze frC = intC?

4. Uved’te prıklad otevrene mnoziny A ⊂ R2 a spojiteho zobrazenı f : R2 → R2

takovych, ze f (A) bude uzavrena.

5. Dokazte, ze pro kazde dve mnoziny A, B ⊂ R2 platı int(A \ B) ⊂ intA \ intB , auved’te prıklad, ve kterem neplatı opacna inkluze.

6. Ukazte, ze funkce f : R2 \ {(0, 0)} → R, f (x, y) = (2xy)/(x2 + y2) je ohranicena.

7. Dokazte, ze kazde konstantnı zobrazenı f : Rn → Rm je spojite.

8. Dokazte, ze kazda konvergentnı posloupnost {xi ∈ Rn}∞i=1 je ohranicena.


9. Dokazte, ze topologie na R2 generovana systemem vsech otevrenych ctvercu {(x, y) ∈R2; max{|x − x0|, |y − y0|} < a, x0, y0 ∈ R, a ∈ R+} je ekvivalentnı s topologiıgenerovanou systemem {(x, y) ∈ R2; |x − x0| + |y − y0| < a, x0, y0 ∈ R, a ∈ R+}.

10. Bud’te f : R2 → R spojita funkce a A ⊂ R2 takove, ze clA = R2 a f |A = 1. Ukazte,ze potom f = 1.

11. Rozhodnete, zda mnozina M = {x ∈ Rn; x1 > x2 > . . . > xn} je otevrena.

12. Necht’ f : R2 → R je spojita funkce a A = {(x, y) ∈ R2; f (x, y) > 0}, B ={(x, y) ∈ R2; f (x, y) = 0} a C = {(x, y) ∈ R2; f (x, y) < 0}. Je nektera z mnozinA, B,C otevrena? Jak tomu bude s kompaktnostı?

13. Necht’ f, g : R2 → R jsou spojite funkce. Ukazte, ze potom mnozina A = {(x, y) ∈R2; f (x, y) = 1+ g(x, y)} je uzavrena.

14. Rozhodnete, zda pro funkci f : R2 → R, f (x, y) = sin(x) a pro libovolnou mnozinuA ∈ R2 platı f (intA) = int f (A).

15. Dokazte nebo vyvrat’te: Necht’ f : Rn → Rn je spojite zobrazenı. Pak pro kazdoumnozinu A ⊂ Rn platı f (clA) = cl f (A).

16. Povazujme prvky mnoziny R9 za ctvercove matice typu 3× 3. Dokazte, ze mnozinaM ⊂ R9 tvorena regularnımi maticemi je otevrena.

17. Najdete obraz definicnıho oboru a nacrtnete graf funkce f : Rn → Rm . Najdetemnozinu vsech bodu, ve kterych je uvedena funkce spojita.

a) f : R→ R2, f (x) = (sin x, cos x); b) f : R→ R2, f (x) = (sgnx, x);c) f : R→ R2, f (x) = (χ(x), sin x); d) f : R2 → R, f (x, y) = sgn(xy).

18. Rozhodnete, ktere z nasledujıcıch posloupnostı {xk}∞k=1 konvergujı a v takovemprıpade najdete jejich limity.

a) xk =(−1

k, 1

); b) xk =

(sin 2k

1+ k + k2 , e−k2+1);

c) xk =(

ln1

k,

1

k, 0, k2

); d) xk =

(sin k

k,

k

sin(1/k)

).

19. V prıpade, ze nasledujıcı limity existujı, najdete je. Pokud neexistujı, pokuste sezduvodnit proc.

a) lim(x,y)→(2,0)

ln xy

x2 + y2 ; b) lim(x,y)→(0,0)

(x2 + y2) sin1

x;

c) lim(x,y)→(2,0)

lnx2 − y2

x − y; d) lim

(x,y)→(0,2)sin xy

x;

e) lim(x,y)→(1,1)

1

(x − 1)3 + (y − 1)3; f) lim

(x,y)→(∞,k)

(1+ y

x

)x;

g) lim(x,y)→(2,3)

y − 3

x + y − 5; h) lim

(x,y)→(0,0)x

x + y;

i) lim(x,y)→(2,2)

x3 − y3

x4 − y4 ; j) lim(x,y)→(0,0)

x3y − xy3

(x − y)3;

PRIROZENA TOPOLOGIE Rn 181

k) lim(x,y)→(0,0)

sin(x2 + y2)

x2 + y2 ; l) lim(x,y)→(1,1)

1

(x − 1)2 + (y − 1)2;

m) lim(x,y,z)→(0,0,0)

x2 + y2 − z2

x2 + y2 + z2; n) lim

(x,y)→(0,0)2xy

xy + 2x − y;

o) lim(x,y)→(0,0)

2x3 + 5x3

x2 + y2 ; p) lim(x,y,z)→(0,0,0)

x2y2z2

x6 + y6 + z6 ;

q) lim(x,y)→(0,0)

x2 + y2√

x2 + y2 + 4− 2; r) lim

(x,y)→(0,0)x2 + y2

arctan(x/y);

s) lim(x,y)→(0,1)

√x2 + (y − 1)2 + 1− 1

x2 + (y − 1)2.

20. Najdete vsechny body, ve kterych jsou nasledujıcı funkce Rn → R spojite:

a) f (x, y) ={

xy/(x2 + y2) pro (x, y) 6= (0, 0);0 pro (x, y) = (0, 0);

b) f (x, y) ={(x2y + xy2)/(x2 + y2) pro (x, y) 6= (0, 0);

0 pro (x, y) = (0, 0);

c) f (x, y) ={

ln√

x2 + y2 pro (x, y) 6= (0, 0);0 pro (x, y) = (0, 0);

d) f (x, y) ={(x2y + 3x2)/(x2 + y2) pro (x, y) 6= (0, 0);

0 pro (x, y) = (0, 0);

e) f (x, y) ={

1/(1− x2 − y2) pro x2 + y2 6= 1;0 pro x2 + y2 = 1;

f) f (x, y) ={

x2/(x2 + y2) pro (x, y) 6= (0, 0);12 pro (x, y) = (0, 0);

g) f (x, y, z) ={

xyz/(x2 + y2 + z2) pro (x, y, z) 6= (0, 0, 0);0 pro (x, y, z) = (0, 0, 0);

h) f (x, y) ={

x2y/(x4 + y2) pro (x, y) 6= (0, 0);0 pro (x, y) = (0, 0).

21. Ukazte, ze jestlize f : R2 → R je spojita v (x0, y0), pak fx0 , definovana fx0(y) =f (x0, y), je spojita v bode y = y0 a fy0 , definovana fy0(x) = f (x, y0), je spojita v bodex = x0.

22. Spojitost fx0 v y = y0 a fy0 v x = x0 (viz. predchozı cvicenı), ale nezarucuje spojitostf v bode (x0, y0). Overte toto tvrzenı na prvnı funkci ze cvicenı 20.

23. Necht’pro (x, y) takova, ze x2 + y2 6= 0,

f (x, y) = sin(x2 + y2)

x2 + y2 .

Jak musı byt definovano f (0, 0), aby byla funkce f : R2→ R spojita v bode (0, 0)?


24. Necht’ f : (R \ {0})× R→ R,

f (x, y) = (x2 + y2) arctg

(x

y

).

Lze tuto funkci rozsırit na body (0, y), aby byla stale spojita pro ktera y a jakou hodnotuf (0, y) musı mıt.

25. Necht’ f : R2 \ {(x, y) ∈ R2; x = y} → R,

f (x, y) = e−1/|x−y|.

Lze tuto funkci spojite rozsırit i na prımku y = x? Jak?

26. Necht’ f : R2 → R,

f (x, y) =

sin xy

xpro x 6= 0;

y pro x = 0.

Ma tato funkce nejake body nespojitosti?

Vysledky

15. ) Neplatı pro A ⊂ R, A = Q a f (x) = 0. 17. a) Im f = [−1, 1] × [−1, 1],b) Im f = (−1, 1×R)∪(0, 0),, c) Im f = 0, 1×[−1, 1], d) Im f = −1, 0, 1. 18. a) (0, 1),b) (0, 0), c)∞, d) (0, 1). 19. a) −∞, b) 0, c) 2, d) 2, e) neexistuje, f) ek , g) neexistuje,h) neexistuje, k) 1, l) ∞, o) 0, 20. a) Nespojita v (0, 0), b) spojita vsude, c) nespojitav (0, 0), d) nespojita v (0, 0), e) nespojita na kruhu x2 + y2 = 1, f) nespojita v (0, 0)g) spojita vsude, h) nespojita v (0, 0). 23. ) f (0, 0) = 1. 24. ) Ano, f (0, y) = 0. 25. ) Ano,f (x, x) = 0. 26. ) Nema.

2. Derivace prvnıho radu

Prıklady

1. Rozhodnete, je-li funkce f : R → R3, f (x) = (cos x, sin x, x), diferencovatelnav bode π/2.

Resenı: Vsechny slozky funkce f jsou spojite a diferencovatelne v kazdem bode x ∈ R.Tedy f je spojita a diferencovatelna v π/2. Mame

f ′(x) = (− sin x, cos x, 1),

tedyf ′(π/2) = (−1, 0, 1).

2. Najdete parcialnı derivace zobrazenı f : R2 → R, f (x1, x2) = x21 + x2 cos x1 v bode

(π/2, 1).

Resenı: Platı D1 f (π/2, 1) = g′(π/2), kde g(x1) = f (x1, 1) = x21 + cos x1. Tedy

D1 f (π/2, 1) = π − 1. Podobne D2 f (π/2, 1) = h′(1), kde h(x2) = f (π/2, x2) = π/2.Tedy D2 f (π/2, 1) = 0.

3. Dokazte, ze funkce f : Rn → R, f (x) = x1 + 2x2 + . . . + nxn je diferencovatelna

v bode x0 =(

1, 12 , . . . , 1/n

)a najdete D f (x0).

Resenı: Pro k = 1, . . . , n platı Dk f (x0) = k. Dale, mame

limh→0

f (x0 + h)− f (x0)− (h1 + 2h2 + . . .+ nhn)

‖h‖ = limh→0

0

‖h‖ = 0.

Funkce f je tedy diferencovatelna v bode x0 a platı D f (x0)(h) = h1 + 2h2 + . . .+ nhn .Druha moznost: Pro kazde x ∈ Rn a k = 1, . . . , n platı Dk f (x0) = k. Funkce f tedy

ma spojite parcialnı derivace. To znamena, ze je diferencovatelna a platı D f (x0)(h) =h1 + 2h2 + . . .+ nhn .

Tretı moznost: Funkce f je linearnı. Je tedy diferencovatelna v kazdem bode a platıD f = f .

4. Dokazte, ze funkce f : R2 → R f (x1, x2) = x21 + x2 cos x1 je diferencovatelna v bode

(π/2, 1).

Resenı: Z prıkladu 2 plyne, ze existuje-li derivace funkce f v bode (π/2, 1), platı

D f (π/2, 1)(h1, h2) = (π − 1, 0)

(h1

h2

)= (π − 1)h1.

Stacı tedy provest nasledujıcı vypocet:

0 ≤ limh→0

∣∣(π/2+ h1)2 + (1+ h2) cos(π/2+ h1)− (π/2)2 − cos(π/2)− (π − 1)h1

∣∣√

h21 + h2

2

183


= limh→0

|h21 − (1+ h2) sin h1 + h1|√

h21 + h2

2

≤ limh→0

h21√

h21 + h2

2

+ limh→0

| sin h1 + h1|√h2

1 + h22

+ limh→0

|h2 sin h1|√h2

1 + h22

≤ limh1→0

h21

|h1|+ lim

h1→0

∣∣∣∣sin h1 + h1

h1

∣∣∣∣+ limh2→0|h2| · lim

h1→0

| sin h1||h1|

= 0+ 0+ 0 · 1 = 0.

Druha moznost: Jelikoz funkce D1 f (x1, x2) = 2x1− x2 sin x1 a D2 f (x1, x2) = cos x1jsou spojite, je funkce f spojite diferencovatelna, a tedy i diferencovatelna.

5. Rozhodnete, zda funkce f : R→ R dana predpisem

f (x1, x2) ={

x1 jestlize |x1| ≤ |x2|,x2 jestlize |x1| > |x2|,

je diferencovatelna v bode (0, 0).

Resenı: Platı f (x1, 0) = f (0, x2) = 0, mame tedy D1 f (0, 0) = D2 f (0, 0) = 0. V prı-pade, ze je funkce f diferencovatelna, tedy musı byt D f (0, 0) = 0. Jelikoz vsak

lim(x,y)→(0,0)

f (x1, x2)√x2

1 + x22

neexistuje (stacı polozit x1 = x2), nenı funkce f v bode (0, 0) diferencovatelna.

6. Najdete parcialnı derivace funkce g ◦ f v bode (1, 1, 1), jestlize

f (x1, x2, x3) =(

2x1x2x3

x21 + x2

2 − x23

), g(x1, x2) =

(x1x2

x1/x2

).

Resenı: Platı

(g ◦ f )(x1, x2, x3) = g(2x1x2x3, x21 + x2

2 − x23) =

(2x1x2x3(x2

1 + x22 − x2

3)

2x1x2x3/(x21 + x2

2 − x23 )

)

Lze tedy postupovat prımym vypoctem. My ale vyuzijeme vetu o derivaci slozenefunkce:

D1 f 1(x1, x2, x3) = 2x2x3 D2 f 1(x1, x2, x3) = 2x1x3 D3 f 1(x1, x2, x3) = 2x1x2

D1 f 2(x1, x2, x3) = 2x1 D2 f 2(x1, x2, x3) = 2x2 D3 f 2(x1, x2, x3) = −2x3

D1g1(x1, x2) = x2 D2g1(x1, x2) = x1

D1g2(x1, x2) = 1/x2 D2g2(x1, x2) = −x1/x22

mame tedy

f ′(1, 1, 1) =(

2 2 22 2 −2

), g′(2, 1) =

(1 21 −2

).

Jelikoz parcialnı derivace funkcı f a g jsou spojite, jsou tyto funkce diferencovatelne a

(g ◦ f )′(1, 1, 1) = g′( f (1, 1, 1)) · f ′(1, 1, 1) =(

1 21 −2

)·(

2 2 22 2 −2

)

=(

6 6 −2−2 −2 6

).

Je tedy

DERIVACE PRVNIHO RADU 185

D1(g ◦ f )1(1, 1, 1) = 6, D2(g ◦ f )1(1, 1, 1) = 6, D3(g ◦ f )1(1, 1, 1) = −2,D1(g ◦ f )2(1, 1, 1) = −2, D2(g ◦ f )2(1, 1, 1) = −2, D3(g ◦ f )2(1, 1, 1) = 6.

7. Vyjadrete pomocı parcialnıch derivacı diferencovatelnych zobrazenı g : R → R2 af : R2 → R funkci ( f ◦ g)′′. Do vysledku dosad’te zobrazenı

f (x1, x2) = x21 + x2

2 , g(x) =(

x sin x

x cos x

).

Resenı: Platı

( f ◦ g)′′ = (( f ◦ g)′)′ = ((D1 f ) ◦ g · (g1)′ + (D2 f ) ◦ g · (g2)′)′

= ((D11 f ) ◦ g · (g1)′ + (D12 f ) ◦ g · (g2)′)(g1)′ + (D1 f ) ◦ g · (g1)′′

+ ((D21 f ) ◦ g · (g1)′ + (D22 f ) ◦ g · (g2)′)(g2)′ + (D2 f ) ◦ g · (g2)′′

= (D11 f ) ◦ g · ((g1)′)2 + 2(D12 f ) ◦ g · (g1)′(g2)′ + (D22 f ) ◦ g · ((g2)′)2

+ (D1 f ) ◦ g · (g1)′′ + (D2 f ) ◦ g · (g2)′′.

Zkouska pro zadana zobrazenı: Platı f ◦ g(x) = x2, tedy ( f ◦ g)′′ = 2. Dosazenımvztahu:

D1 f (x1, x2) = 2x1, (D1 f )g(x) = 2x sin x ,D2 f (x1, x2) = 2x2, (D2 f )g(x) = 2x cos x ,D11 f (x1, x2) = D22 f (x1, x2) = 2, (D11 f )(g(x)) = (D11 f )(g(x)) = 2,D12 f (x1, x2) = D21 f (x1, x2) = 0, (D12 f )(g(x)) = (D21 f )(g(x)) = 0,(g1)′(x) = sin x + x cos x , (g1)′′(x) = 2 cos x − x sin x ,(g2)′(x) = cos x − x sin x , (g2)′′(x) = −2 sin x − x cos x .

Dostavame

(D11 f )(g(x))) · ((g1)′)2(x)+ 2(D12 f )(g(x)) · (g1)′(x)(g2)′(x)+ (D22 f )(g(x)) · ((g2)′)2(x)+ (D1 f )(g(x)) · (g1)′′(x)+ (D2 f )(g(x)) · (g2)′′(x)= 2(sin x + x cos x)2 + 2 · 0+ 2(cos x − sin x)2 + 2x sin x(2 cos x − x sin x)+ 2x cos x(−2 sin x − x cos x)= 2(sin2 x + 2x sin x cos x + x2 cos2 x + cos2 x − 2x sin x cos x + x2 sin2 x+ 2x sin x cos x − x2 sin2 x − 2x sin x cos x − x2 cos2 x) = 2.

Cvicenı

1. Najdete parcialnı derivace funkce f (x1, x2) = x21 + x2

2 v bode x0 = (1, 2).

2. Pomocı definice derivace dokazte diferencovatelnost funkce f (x1, x2) = x21−2x1+x2

2v bode (1, 0) a urcete D f (1, 0).

3. Uved’te prıklad funkce f : R2 → R, ktera ma obe parcialnı derivace v bode (0, 0)rovny 0 a pritom zde nenı diferencovatelna.

4. Najdete parcialnı derivace funkce f , jestlizea) f (x1, x2) = ex1 sin x2, b) f (x1, x2, x3) = ex1x2 sin(x2x3)+ x2

2 ln(x1x2x3),

c) f (x1, x2) =x1

x2+ x2

x1, d) f (x1, x2) = arctg

x1

x2,


e) f (x1, x2) =1√

x1 −√x2, f) f (x1, x2) = ln

(x1 −

√x2

1 + x22

),

g) f (x1, x2, x3) = x3x1/x21 , h) f (x1, x2) = (3x2

1 + x22 )

4x1+x2,

i) f (x1, x2, x3) = xx

x32

1 , j) f (x1, x2, x3) = (x1 − x2x3)x1x2 .

5. Najdete D2 f (1, x2), jestlize x2 > 0 a

f (x1, x2) = xx

xx

x21

11

1 + (ln x1)(arctg(arctg(arctg(sin(cos(x1x2))− ln(x1x2))))).

6. Necht’g : R→ R je spojita funkce. Najdete parcialnı derivace funkcıa) f (x1, x2) =

∫ x1x2

g(t) dt, b) f (x1, x2) =∫ x1x2

0 g(t) dt .

7. Je dana funkce f : R→ R, f (x) = x4 − 2x . Vypoctete D f (2)(x), D f (x)(2).

8. Uved’te prıklad funkcı f, g : R2→ R takovych, ze neexistujı D f (0, 0) a Dg(0, 0), aleexistuje D( f + g)(0, 0).

9. Uved’te prıklad fuknce f : R2→ R takove, zea) D f (x, y) neexistuje pro zadne (x, y) ∈ R2;b) D f (x, y) existuje pouze pro (x, y) ∈ R2 takove, ze x = 0.

10. Necht’ f : R2 → R je funkce splnujıcı podmınku 0 ≤ f (x, y) ≤ x2 + y2. Ukazte,ze potom existuje D f (0, 0). (Navod: Odhadnete D f (0, 0) a pote odhad overte z definicediferencialu.)

11. Je funkce f : R2 → R, f (x, y) = |x5| spojite diferencovatelna? (Navod: Uzijtestejneho postupu jako v predchozım cvicenı)

12. Rozhodnete, ktere z funkcı

a) f (x1, x2) =

x1x2

x21 + x2

2

jestlize x21 + x2

2 6= 0,

0 jestlize x21 + x2

2 = 0,

b) f (x1, x2) =

(x21 + x2

2) sin1√

x21 + x2

2

jestlize x21 + x2

2 6= 0,

0 jestlize x21 + x2

2 = 0,

c) f (x1, x2) = max2(|x1|, |x2|),jsou direfencovatelne v bode (0, 0).

13. Je dano spojite diferencovatelne zobrazenı f : Rn → Rn . Dokazte, ze mnozina vsechx ∈ Rn takovych, ze D f (x) je surjektivnı, je otevrena.

14. Necht’ f : Rn → Rm je linearnı zobrazenı. Dokazte, ze D f (x) = f . Na zaklade tohodokazte, ze pro libovolna diferencovatelna zobrazenı g, h : Rn → Rm a bod x ∈ Rn platı:

a) D(g + h)(x) = Dg(x)+ Dh(x),b) D(g)(x) = (Dg1(x), . . . ,Dgm(x)),


c) D(g − Dg(x))(x) = 0.

15. Pro diferencovatelne zobrazenı f : R2 → R2 platı

f (0, 0) =(

11

), f ′(0, 0) =

(1 −11 2

).

Rozhodnete, zda je funkce g ◦ f , kde g(x1, x2) = x21 − x2

2 , diferencovatelna. V kladnemprıpade urcete D(g ◦ f )(0, 0).

16. Necht’ f : R3 → R3,

f (x, y, z) =

cos2(y)sin(x + y)sin(x + z)

,

dale g : R3 → R, g(x, y, z) =√

x2 + y2 + z2 a h : R→ R2, h(x) =(

ex

x e

). Vypoctete

D1(h ◦ g ◦ f )1(0, π/2, 0).

17. Necht’ f : R3 → R2,

f (x, y, z) =(

xyzx2 + y2 − z2

).

Bud’g : R2 → R2 diferencovatelna a jejı diferencial je v (1, 1) je Dg(1, 1) =(

2 −11 1

).

Spoctete, existuje-li, D(g ◦ f )(1, 1, 1).

18. Vypoctete parcialnı derivace funkce

F(x, y, z) =(

f (x y, yz, zx )

g((xy)z, (yz)x , (zx)y)

),

kde f, g : R3 → R jsou diferencovatelne funkce.

19. Najdete parcialnı derivace funkcıa) F(x1, x2) = f (g(x1)h(x2), g(x1)+ h(x2)),b) F(x1, x2) = f (x1, g(x1), h(x1, x2)),c) F(x1, x2) = f (g(x1, x2), g(x2, x1)),d) F(x1, x2, x3) = f (g(x1 + x2), h(x1 + x3)),e) F(x1, x2) = f (x1, x2, x1),f) F(x1, x2, x3) = f (x x2

1 , x x32 , x x1

3 ),kde g, h jsou diferencovatene funkce R→ R prıpadne R2→ R prıpadne R3→ R.

20. Bud’ f : R2 → R2, f (x, y) =(

y2

−x

). Vypocıtejte D f (1,−1)

(−1

1

).

21. Urcete parcialnı derivace prvnıho radu slozene funkce F = f ◦ g, kde f : R2 → R,g : R2 → R2,

f (x1, x2) = x21 x2 − x2

2 x1


g(y1, y2) =(

y1 cos y2y2 sin y1

).

22. Najdete diferencial slozene funkce F = f ◦ g, kde f : R3 → R, g : R→ R3,

f (x1, x2, x3) = e2x1(x2 − x3),

g(x) =

x2 sin xcos x

.

23. Necht’ f : R2 → R je diferencovatelna funkce takova, ze f (3, 0) = −2 a D f (3, 0) =(−2, 2). Dale g : R2 → R2,

g(x, y) =(

xy2 + 2x2y − 1

).

Najdete tecnou rovinu ke grafu funkce f ◦ g v bode (1, 1, f ◦ g(1, 1)).

24. Necht’ f : R2 → R je diferencovatelna funkce takova, ze f (1, 2) = −2 a D f (1, 2) =(−1, 2). Dale g : R2 → R2,

g(x, y) =(

x2 + y2 + 1x2 − y2 + 2

).

Najdete tecnou rovinu ke grafu funkce f ◦ g v bode (0, 0).

25. Ukazte, ze pro funkci f : R2 \ {(x, y) ∈ R2; x2 6= ay2} → R

f (x, y) = y

y2 − a2x2 , a > 0

platı ze D1(D1 f ) = a2D2(D2 f ).

26. Necht’ f (x1, x2) = x1 − x1x22 , vypoctete smerovou derivaci DEu f (0, 0), DEu f (2, 1),

DEu f (1, 2), kdea) Eu = (1, 1), b) Eu = (−1, 0),c) Eu = (−a, 0) a > 0, d) Eu = (a, a) a > 0,e) Eu = (a,−a) a > 0.

27. Necht’ f (x1, x2, x3) = x22 sin(x1x2x3), vypoctete smerovou derivaci podle vektoru

Eu = (π, π, 1) v bode (1, 1, π).

Vysledky

1. ) Dx1 f (1, 2) = 2, Dx2 f (1, 2) = 4. 2. ) D f (1, 0) = 0. 3. ) f (x, y) = 0 pro x = 0nebo y = 0 a f (x, y) = 1 jinak. 4. a) Dx1(x1, x2) = ex1 sin x2, Dx2 f (x1, x2) =ex1 cos x2; b) Dx1(x1, x2, x3) = x2 ex1x2 sin(x2x3) + x2

2/x1 , x1x3 ex1x2 cos(x2x3) +2x2 ln(x1x2x3)+x2, Dx2(x1, x2, x3) = x2 ex1x2 cos(x2x3)+x2

2 ; c) Dx1 f (x1, x2) = 1/x2−x2/x2

1 , Dx2 f (x1, x2) = (1 − x1)/x2; d) Dx1 f (x1, x2) = x2/(x21 + x2

2), Dx2 f (x1, x2) =−x1/(x2

1 + x22); e) Dx1 f (x1, x2) = (2(

√x1 −√

2)2√

x1)−1, Dx2 f (x1, x2) = (2(

√x1 −


√2)2√

x2)−1; f) Dx1 f (x1, x2) = (1− x1/

√x2

1 + x22)/(x1 −

√x2

1 + x22), Dx2 f (x1, x2) =

−(x2/

√x2

1 + x22)/(x1 −

√x2

1 + x22 ); g) Dx1 f (x1, x2, x3) = x3x1/x2−1

1 /x2,

Dx2 f (x1, x2, x3) = x3x1/x21 ln x1(−1/x2

2), Dx3 f (x1, x2, x3) = x1/x21 ; i) Dx1 f (x1, x2, x3) =

x x32 x

xx32 −1

1 , Dx2 f (x1, x2, x3) = xx

x32

1 ln x1·x3x x3−12 , Dx3 f (x1, x2, x3) = x

xx32

1 ln x1·x x32 ln x2.

5. ) D2 f (1, x2) = 0. 6. a) Dx1 f (x1, x2) = g(x1), Dx2 f (x1, x2) = −g(x2); b) Dx1 f (x1, x2) =x2g(x1x2), Dx2 f (x1, x2) = x1g(x1x2). 7. ) D f (2)(x) = 30x , D f (x)(2) = 8x2 − 4.8. ) Naprıklad f (x, y) = −g(x, y) = 1 pro y = 0 a f (x, y) = −g(x, y) = 0 proostatnı. 9. a) Naprıklad f (x, y) = 1 pokud x, y ∈ Q a f (x, y) = 0 jinak; b) naprı-klad f (x, y) = x2χ(x). 12. a) Ne; b) ano; c) ano. 15. ) Ano, D(g ◦ f ) = (0,−6).

17. ) D(g ◦ f ) =(

03

03−4−1

). 20. ) D f (1,−1)

(−11

)=(−21

). 21. ) Dy1( f ◦ g)(y1, y2) =

2y2(y1 sin y1 cos y2− sin y1) cos y2+ y1y2(y1 cos2 y2− 2y2 sin y1 cos y2) cos y1, Dy2( f ◦g) = −2y1y2(y1 sin y1 cos y2− sin y1) sin y2+ y1 y2(y1 cos2 y2− 2y2 sin y1 cos y2) sin y122. ) D( f ◦ g)(x) = 5 e2x sin x . 23. ) z = 2x − 2y − 8. 24. ) z = 0. 27. ) Du f (1, 1, π) =(−π2,−π2, 1).

3. Veta o implicitnı a inverznı funkci

Prıklady

1. Necht’

f (x, y) =(

ex cos yex sin y

).

Dokazte, ze kazdy bod (x0, y0) ∈ R2 ma okolı V takove, ze pro kazde (x ′, y ′) ∈ f (V ) marovnice

f (x, y) =(

x ′

y ′

)

prave jedno resenı (x, y) ∈ V .

Resenı: Mame dokazat, ze existuje okolı V bodu (x0, y0), na nemz je zobrazenı f proste.Toto zobrazenı je spojite diferencovatelne, stacı tedy podle vety o inverznım zobrazenıoverit, ze det f ′(x0, y0) 6= 0. A ono

det f ′(x0, y0) =∣∣∣∣ex0 cos y0 − ex0 sin y0ex0 sin y0 ex0 cos y0

∣∣∣∣ = e2x0 6= 0.

2. Uvazujme rovnicix2 + 4y2 − 3z2 = 6

a bod a0 = (3, 0, 1).

a) Definuje implicitne tato rovnice promennou y jako funkci x a z na nejakem okolıbodu (x0, z0) = (3, 1)? Pokud ano, najdete jejı parcialnı derivace podle x a z.

b) Definuje implicitne tato rovnice promennou z jako funkci x a y na nejakem okolıbodu (x0, y0) = (3, 0)? Pokud ano, najdete jejı parcialnı derivace podle x a y.

Resenı: a) Jelikoz D2 F(3, 0, 1) = 0, veta o implicitnı funkci nam nerıka nic o tom, jestlije y definovano jako funkce x a z. Presto muzeme usoudit, ze tomu tak nenı. Vsimnemesi, ze jinak by takove y muselo splnovat

y(x, z) =√

6+ 3z2 − x2.

V bode (x0, z0) platı 6 + 3z20 = x2

0 . Jestlize se x malinko zvetsı, vyraz pod odmocninoubude zaporny a dana rovnice tedy nemuze definovat y na zadnem okolı bodu x0, z0.

b) Jelikoz D3 F(3, 0, 1) 6= 0, muzeme aplikovat vetu o implicitnı funkci a zjistıme, zedana rovnice definuje z jako funkci x a y na nejakem okolı U bodu (3, 0). Navıc, na tomtookolı mame

D1 f (x, y) = −D1 F(x, y, z)

D3 F(x, y, z)= − 2x

−6z= x

3z,

D2 f (x, y) = −D2 F(x, y, z)

D3 F(x, y, z)= − 8y

−6z= 4y

3z,

kde F = x2 + 4y2 − 3z2 − 6.

3. Rozhodnete, zda pro F(x, y, z)= 2x/z+2y/z−8 na nejakem okolı bodu (2, 2, 1) rovniceF(x, y, z) = 0 implicitne definuje nejakou funkci. Pokud ano, najdete jejı parcialnıderivace v bode (2, 2).

190

VETA O IMPLICITNI A INVERZNI FUNKCI 191

Resenı: Funkce F je na okolı bodu (2, 2, 1) spojite diferencovatelna, pro existenci impli-citnı funkce tedy stacı, aby D3 F(2, 2, 1) 6= 0.

D3 F(2, 2, 1) = −2x/z x

z2 ln 2− 2y/z y

z2 ln 2,

tedyD3 F(2, 2, 1) = −16 ln 2 6= 0.

Oznacme implicitnı funkci f . Platı f (2, 2) = 1 a pro kazde (x, y) z nejakeho okolı bodu(2, 2)

2x/ f (x,y) + 2y/ f (x,y) = 8.

Z techto vztahu jiz snadno parcialnı derivace zobrazenı f v bode (2, 2) vypocıtame.

4. Necht’

F(x, y, z, u) =(

x2y + xy2 + z2 − u2

ex+y − u

).

Ukazte, ze F(0, 0, 1, 1) = (0, 0) a ze F(x, y, z, u) = (0, 0) definuje (z, u) jako diferen-covatelne zobrazenı ( f 1, f 2) promennych x a y na nejakem okolı bodu (0, 0). Najdetejeho parcialnı derivace funkce f 1 podle x a y v bode (0, 0).

Resenı: Platnost F(0, 0, 1, 1) = (0, 0) je evidentnı. Dale

(D3 F1(x, y, z, u) D4 F1(x, y, z, u)D3 F2(x, y, z, u) D4 F2(x, y, z, u)

)=(

2z −2u0 1

)

a v bode (z0, u0) = (1, 1) tedy

(2z0 −2u00 1

)=(

2 −20 −1

).

Jelikoz

det

(2 −20 −1

)= −2 6= 0,

uvedena rovnice definuje (z, u) jako diferencovatelne zobrazenı f promennych x a y nanejakem okolı bodu (0, 0). Zavedeme-li si nynı zobrazenı G : R2 → R4, G(x, y) =(x, y, f (x, y)), vidıme, ze

(0, 0) = F(x, y, f (x, y)) = F ◦ G(x, y)

a tedy0 = F ′(x, y, f (x, y)) = F ′(G(x, y)) · G′(x, y).

Po nekolika upravach a vyuzitı toho, ze A−1 = (det A)−1 adj A, kde A je invertibilnımatice, nakonec zjistıme, ze

D1 f 1 = −det

(D1 F1(0, 0, 1, 1) D4 F1(0, 0, 1, 1)D1 F2(0, 0, 1, 1) D4 F2(0, 0, 1, 1)

)

det

(D3 F1(0, 0, 1, 1) D4 F1(0, 0, 1, 1)D3 F2(0, 0, 1, 1) D4 F2(0, 0, 1, 1)

) = − 2−2 = 1,


D2 f 1 = −det

(D2 F1(0, 0, 1, 1) D4 F1(0, 0, 1, 1)D2 F2(0, 0, 1, 1) D4 F2(0, 0, 1, 1)

)

det

(D3 F1(0, 0, 1, 1) D4 F1(0, 0, 1, 1)D3 F2(0, 0, 1, 1) D4 F2(0, 0, 1, 1)

) = − 2−2 = 1.

Cvicenı

1. Rozhodnete, zda existuje okolı U ⊂ R cısla 1, na nemz je funkce f (x) = x x prosta.

2. Rozhodnete, zda existujı otevrene mnoziny U, V ⊂ R2 takove, ze (2, π) ∈ U azobrazenı F : U → V ,

F(x, y) =(

x cos yx sin y

),

je bijekce. Pokud ano, najdete tato okolı a vypoctete F−1 : V → U .

3. Rozhodnete, zda existuje okolı U ⊂ R2 bodu (1, e), na nemz je funkce F : R2 → R2,

F(x, y) =(

x y

yx

),

proste.

4. Necht’U, V ⊂ R2, (0, 1) ∈ V , jsou otevrene mnoziny a f : U → V zobrazenı takove,ze pro kazde (x, y) ∈ V platı

f −1(x, y) =(

exy + xexy + y

).

Vypoctete D2 f 1(1, 2).

5. Necht’funkce f : R→ R je definovana predpisem

f (x) ={

x/2+ x2 sin(x/2) pro x 6= 0;0 pro x = 0.

Dokazte, ze f ′(0) 6= 0, ale na zadnem okolı bodu 0 neexistuje funkce inverznı.

6. Uved’te prıklad zobrazenı f : R2 → R2, ktere ma inverzi f −1 : R2 → R2 takovou, zeD f −1(0, 0) = 0.

7. Necht’ f : R2 → R je spojite diferencovatelna funkce. Dokazte, ze neexistuje funkcef −1.

8. Rozhodnete, zda existuje okolı U ⊂ R2 bodu (0, 0) takove, ze pro kazde (x, y) ∈ Uma rovnice

cos(xz)− sin(yz) = z

resenı.

9. Zjistete, zda existuje cıslo ε takove, ze pro kazde x ∈ (1− ε, 1+ ε)ma resenı soustava

xyz = 1

yzx = 1.


10. Rozhodnete, zda existuje okolı U bodu x0 = 0 takove, ze pro kazde x ∈ U masoustava

x + y + z = ez

x + y + z = e2yz

resenı. Poznamenejme, ze bod (x, y, z) = (0, 1, 0) je resenım teto soustavy.

11. Najdete takova x, y, z, u, v,w, ke kterym existuje okolı, na nemz je mozno z rovnic

u2 + v2 + w2 = 1u2

x2 +v2

y2 +w2

z2 = 1

vyjadrit u = f (x, y, z, w) a v = g(x, y, z, w)?

12. Rozhodnete, zda existuje okolı U bodu (0, 0) takove, ze pro kazde (x, y) ∈ U marovnice

cos(xz)− sin(yz) = z

resenı.

13. Ukazte, ze rovnicez3 + z(x2 + y2)+ 1 = 0

ma jednoznacne resenı z = f (x, y), pro vsechna x, y ∈ R a najdete parcialnı derivacefunkce f .

14. Rozhodnete, zda existuje funkce f definovana na nejakem okolı bodu −1, ktera natomto okolı splnuje

x2 − 2x f (x)+ 2( f (x))2 + 2x + 1 = 0

a ktera ma v bode−1 lokalnı extrem.

15. Necht’ F : R2 → R, F(x, y) = 8(x2 − y2) + (y2 + x2)2. Rozhodnete, zda rovniceF(x, y) = 0 na okolı bodu (1,

√3) definuje

a) x jako funkci promenne y;b) y jako funkci promenne x .

16. Rozhodnete, zda existuje okolı bodu 0, na kterem rovnice

tan√

x2 + y2 − x

y= 0

implicitne definuje x jako funkci y.

17. Necht’ F : R3 → R, F(x, y, z) = x + y + z − sin(xyz). Rozhodnete, zda rov-nice F(x, y, z) = 0 na okolı bodu (0, 0, 0) definuje z jako diferencovatelnou funkcipromennych x, y a najdete jejı parcialnı derivace (pokud existujı).

18. Existuje inverznı funkce k funkci f : R2→ R2,

f (x, y) =(

xy cos yxy sin x

)


na okolı bodu (0, π2/4)? Pokud ano, najdete jejı diferencial.

19. Existuje inverznı funkce k funkci f : R2 → R2,

f (x, y) =(

xy ex+y

xy exy

)

na okolı bodu (1, 1)? Pokud ano, najdete jejı diferencial.

20. Napiste rovnici tecny a normaly k plose

x ln y + y ln z + z ln x = 0

v bode (1, 1, 1).

21. Spojite diferencovatelna funkce F : R2 → R splnuje F(0, 0) = 0, D2 F(0, 0) 6= 0 apro kazde (x, y) ∈ R2 platı F(x, y) = F(y, x). Napiste rovnici tecny k mnozine

M = {(x, y) ∈ R2; F(x, y) = 0}

v bode (0, 0).

22. Necht’ funkce g : R2 → R je diferencovatelna v bode (0, 0) a necht’ g(0, 0) = 0,D1g(0, 0) = 2, D2g(0, 0) = 2. Oznacme gx(y) = g(x, y) a predpokladejme, ze pro kazdex existuje inverznı funkce g−1

x . Polozme h(x, y) = g−1x (y). Vypoctete parcialnı derivace

funkce h v bode (0, 0).

23. Necht’ funkce F : R2 → R je diferencovatelna v bode (0, 0). Oznacme fx (y) =f (x, y) a predpokladejme, ze pro kazde x existuje inverznı funkce f −1

x . Polozme g(x, y) =f −1x (y). Dale predpokladejme, ze pro kazde x existuje inverznı funkce g−1

x a polozmeh(x, y) = g−1

x (y). Dokazte, ze

D1 f (0, 0)D1g(0, 0)D1h(0, 0) = D2 f (0, 0)D2g(0, 0)D2h(0, 0) = −1.

24. Necht’(x0, y0, z0, u0) = (1, 4, 4,−5) a

F(x, y, z, u) =(

x + 2y − z + u−2x + y + 2z + 2u

)=(

00

).

Rozhodnete, zda na nejakem okolı bodu (x0, y0) = (1, 4) uvedena rovnice implicitnedefinuje (z, u) jako zobrazenı ( f 1, f 2)(x, y). Pokud ano, najdete jeho parcialnı derivace.

25. Uved’te prıklad spojite diferencovatelne funkce F : R2 → R takove, aby D2 F(0, 0) =0 a mnozina M = {(x, y) ∈ R2; F(x, y) = 0} byla grafem diferencovatelne funkce.

26. Necht’(x0, y0, z0, u0, v0) = (1, 1,−1, 3, 3) a

F(x, y, z, u, v) =(−x + 2y − 5z + 3u − 5v

x + 4y + 2z + 2u − 3v

)=(

00

).

Rozhodnete, zda na nejakem okolı bodu (x0, y0, z0) = (1, 1,−1) uvedena rovnice im-plicitne definuje (u, v) jako zobrazenı ( f 1, f 2)(x, y, z). Pokud ano, najdete parcialnıderivace f 1 podle x , y a z.


27. Dokazte vetu o inverznım zobrazenı pomocı vety o implicitnım zobrazenı.

28. Dokazte vetu o implicitnım zobrazenı pomocı vety o inverznım zobrazenı.

Vysledky

15. a) ano, b) ne. 16. ) ano. 17. ) ano, ∂ f/∂x(0, 0) = ∂ f/∂y(0, 0) = −1. 18. ) Neexis-tuje, protoze body (0, a), (0, b) se zobrazı na stejny bod. 19. ) Neexistuje, protoze body(1, a), (a, 1) se zobrazı na stejny bod.

4. Extremy funkcı vıce promennych

Prıklady

1. Najdete vsechny body, ve kterych funkce f : R2→ R,

f (x, y) = 3x + 12y − x3 − y3

nabyva lokalnı extrem.

Resenı: Nejdrıve najdeme f ′(x, y). Platı

f ′(x, y) = (3− 3x2, 12− 3y2).

Podezrele body zıskame tak, ze f ′ polozıme rovnu 0. Resenım zıskanych rovnic dostaneme3 − 3x2 = 0, tedy x2 = 1 a |x | = 1, 12 − 3y2 = 0, tedy y2 = 4 a |y| = 2. Dostavametak body (1,−2), (1, 2), (−1, 2), (−1,−2). Abychom je mohli klasifikovat, potrebujemeznat druhe parcialnı derivace:

D11 f (x, y) = −6x, D12 f (x, y) = D21 f (x, y) = 0, D22 f (x, y) = −6y.

Potom

det f ′′(x, y) = det

(−6x 00 −6y

)= 36xy.

V bode (1, 2) mame D11 f (1, 2) < 0 a det f ′′(1, 2) = 72. To znamena, ze v bode (1, 2)nabyva funkce f lokalnıho maxima a f (1, 2) = 18. Dale dostavame det f ′′(1, 2) =det f ′′(2, 1) = −72 < 0 a body (1,−2) a (−1, 2) jsou tedy inflexnı. V bode (−1,−2)mame D11 f (−1,−2) > 0 a det f ′′(−1,−2) = 72 > 0. To znamena, ze v bode (−1,−2)nabyva funkce f lokalnıho minima a f (−1,−2) = −18.

2. Najdete vsechny body, ve kterych funkcea) f (x, y) = x2 − 2xy + y2, b) f (x, y) = x3 − 3xy2 + y2,

nabyva lokalnı extrem.

Resenı: a) Zde platı

f ′(x, y) = (2x − 2y, −2x + 2y) = (0, 0),

jestlize x = y. Mame tedy spoustu bodu podezrelych z extremu. Dale platı

D11 f (x, y) = 2, D12 f (x, y) = D21 f (x, y) = −2, D22 f (x, y) = 2.

Tedy

det f ′′(x, y) = det

(2 −2−2 2

)= 0.

To znamena, ze toto kriterium nam nedava zadnou odpoved’. Pokud si ale uvedomıme, ze

f (x, y) = (x − y)2,

zjistıme, ze v kazdem podezrelem bode nabyva funkce f absolutnıho minima.

196

EXTREMY FUNKCI VICE PROMENNYCH 197

b) Zde platıf ′(x, y) = (3x2 − 3y2, −6xy + 2y) = (0, 0),

jestlize

x2 − y2 = (x − y)(x + y) = 0,

−6xy + 2y = 2y(−3x + 1) = 0.

Z prvnı rovnice tedy y = ±x a z druhe y = 0 nebo x = 13 . Podezrelymi body tedy jsou

(0, 0), ( 13 ,

13 ) a ( 1

3 ,− 13 ). Dale platı

D11 f (x, y) = 6x, D12 f (x, y) = D21 f (x, y) = −6y, D22 f (x, y) = −6x + 2.

Tedy

D11 f ( 13 ,

13 ) = D11 f ( 1

3 ,− 13 ) = 2, D11 f (0, 0) = 0, D12(

13 ,

13 ) = −2,

D12(13 ,− 1

3 ) = −2, D12 f (0, 0) = 0, D22 f ( 13 ,

13 ) = D22 f ( 1

3 ,− 13 ) = 0,

D22 f (0, 0) = 0

adet f ′′( 1

3 ,13 ) = det f ′′( 1

3 ,− 13 ) = −4 < 0, det f ′′(0, 0) = 0.

To znamena, ze body ( 13 ,

13 ), (

13 ,− 1

3 ) jsou inflexnı a v prıpade bodu (0, 0) nam totokriterium nedava zadnou odpoved’. Ovsem, na prımce y = 0, funkce f |y=0(x) = x3

nema v bode x = 0 zadny lokalnı extrem a bod (0, 0) je tedy take inflexnı.

3. Najdete lokalnı extremy funkce

f (x, y) = 27x2y + 14y3 − 69y − 54x .

Resenı: Platı

D1 f (x, y) = 54xy − 54, D2 f (x, y) = 27x2 + 42y2 − 69.

Resenım soustavy rovnic

D1 f (x, y) = 0

D2 f (x, y) = 0

jsou body(x1, y1) = (1, 1), (x2, y2) = (−1,−1),

(x3, y3) = (√

14/2, 3/√

14), (x4, y4) = (−√

14/2,−3/√

14).

Dale

D11 f (x, y) = 54y, D12 f (x, y) = D21 f (x, y) = 54x, D22 f (x, y) = 84y.

Prımym vypoctem nebo pomocı Sylvestrova kriteria lze zjistit, ze matice(

54y 54x54x 84y

)


je v bode (x1, y1) pozitivne definitnı, v bode (x2, y2) negativne definitnı a v bodech(x3, y3) a (x4, y4) indefinitnı. Funkce f ma tedy v bode (x1, y1) lokalnı minimum, v bode(x2, y2) lokalnı maximum a platı f (x1, y1) = −82 a f (x2, y2) = 82.

4. Najdete lokalnı extremy funkce

f (x, y) = x + y

na mnozine M dane rovnostı1

x2+ 1

y2= 1.

Resenı: Jelikoz pro kazdy bod (x, y) ∈ M a funkci

g(x, y) = 1

x2+ 1

y2− 1

platı Dg(x, y) 6= 0, lze na resenı ulohy pouzıt metodu Lagrangeovych multiplikatoru.Hledame tedy body (x, y) ∈ M a cıslo λ takove, ze pro Lagrangeovu funkci

L(x, y, λ) = x + y − λ(

1

x2 +1

y2 − 1

)

platı

D1 L(x, y, λ) = 1+ 2λ1

x3 = 0,

D2 L(x, y, λ) = 1+ 2λ1

y3 = 0.

Tato soustava ma nasledujıcı dve resenı:

(x1, y1, λ1) = (√

2,√

2,−√

2), (x2, y2, λ2) = (−√

2,−√

2,√

2).

Dale(

D11L(x, y, λ) D12L(x, y, λ)D21L(x, y, λ) D22L(x, y, λ)

)=

−6

λ

x4 0

0 −6λ

y4

.

Tato matice je v bode (x1, y1, λ1) pozitivne definitnı a v bode (x2, y2, λ2) negativnedefinitnı. Funkce f ma tedy na mnozine M v bode (x1, y1) lokalnı minimum a v bode(x2, y2) lokalnı maximum. Platı f (x1, y1) = 2

√2 a f (x2, y2) = −2

√2.

5. Najdete nejmensı a nejvetsı hodnotu funkce

f (x, y) = x2 + y2 − 2x + 4y

na mnozine M dane rovnostı x2 + y2 = 25.

Resenı: Jelikoz mnozina M je kompaktnı a funkce f spojita, existuje maximum a minimumfunkce f na mnozine M . Jelikoz pro kazdy bod (x, y) ∈ M a funkci

g(x, y) = x2 + y2 − 25


platı D1g(x, y),D2g(x, y) 6= (0, 0), lze na resenı ulohy pouzıt metodu Lagrangeovychmultiplikatoru. Jestlize (x0, y0) je bod extremu funkce f na mnozine M , existuje cısloλ ∈ R takove, ze pro Lagrangeovu funkci

L(x, y, λ) = f (x, y)− λg = x2 + y2 − 2x + 4y − λ(x2 + y2 − 25)

platıD1 L(x0, y0, λ) = 0,

D2 L(x0, y0, λ) = 0,

tedy

(1− λ)x0 = 1

(1− λ)y0 = −2

x20 + y2

0 = 25.

Vyresenım teto soustavy dostaneme (x0, y0) ∈ {(√

5,−2√

5), (−√

5, 2√

5)}. Jelikozf (√

5,−2√

5) = 25+ 6√

5 a f (−√

5, 2√

5) = 25+ 10√

5, je nejmensı hodnota funkcef na mnozine M rovna 25+ 6

√5 a nejvetsı 25+ 10

√5.

Druha moznost: Mnozinu M parametrizujeme rovnicemi x = 5 cos(t) a y = 5 sin(t),kde t ∈ [0, 2π]. Nynı zuzıme funkci f na mnozinu M . Tedy

g(t) = f |M = 25 cos2(t)+ 25 sin2(t)− 2 cos(t)+ 4 sin(t) = 25− 2 cos(t)+ 4 sin(t).

Nynı hledame extremy funkce g na [0, 2π], to je ale jednoduche, protoze se jedna o funkcijedne promenne. Proto polozme

g′(t) = 2 sin(t)+ 4 cos(t) = 0,

coz po uprave davasin(t)

cos(t)= y

x= −2.

Odtud y = −2x a dosazenım do rovnice pro mnozinu M mame x2+4x2 = 25. A konecne(x1, y1) = (

√5,−2

√5) a (x2, y2) = (−

√5, 2√

5). Zbytek prıkladu doresıme tak, jak jeuvedeno vyse.


f (x, y) = x2 + y2 − 2x + 4y

na mnozine M dane nerovnostı x2 + y2 ≤ 25.

Resenı: Mnozina M je kompaktnı a funkce f spojita; maximum a minimum funkce f namnozine M tedy existuje. Nejprve hledame extrem funkce f uvnitr mnoziny M . Necht’(x0, y0) je bod, v nemz funkce f nabyva na mnozine M maxima nebo minima. Je-li(x0, y0) ∈ intM , platı

D1 f (x0, y0) = 0,

D2 f (x0, y0) = 0,

tedy

2x0 − 2 = 0,

2y0 + 4 = 0


a (x0, y0) = (1,−2). Je-li (x0, y0) ∈ frM , je bodem extremu funkce f na mnozine frM .Podle predchozıho prıkladu tedy (x0, y0) ∈ {(

√5,−2

√5), (−

√5, 2√

5)}.Celkove, je-li bod (x0, y0) bodem extremu funkce f na mnozine M , platı

(x0, y0) ∈ {(1,−2), (√

5,−2√

5), (−√

5, 2√

5)}.

Jelikoz f (1,−2) = −5, f (√

5,−2√

5) = 25+ 6√

5 a f (−√

5, 2√

5) = 25 + 10√

5, jenejmensı hodnota funkce f na mnozine M rovna−5 a nejvetsı 25+ 10

√5.


f (x, y) = x2 + y2 − 2x + 4y

na mnozine M dane nerovnostmi x2 + y2 ≤ 25, y ≥ 0.

Resenı: Existence maxima a minima opet plyne z kompaktnosti mnoziny M a spojitostizobrazenı f . Oznacme (x0, y0) bod extremu funkce f na mnozine M . Nynı mohou nastattri moznosti. Je-li (x0, y0) ∈ intM , platı

D1 f (x0, y0) = 0,

D2 f (x0, y0) = 0.

Tyto podmınky ovsem zadny bod mnoziny M nesplnuje. Je-li (x0, y0) ∈ frM , y0 > 0,je (podle resenı predchozıho prıkladu) (x0, y0) = (−

√5, 2√

5). Je-li x0 ∈ [−5, 5], y0 =0, lze bod (x0, y0) opet najıt pomocı Lagrangeovy funkce, jednodussı ovsem je najıtpodezrele body funkce g(x) = f |y=0(x, y) na intervalu [−5, 5]: Krajnı body (−5, 0),(5, 0) jsou podezrele automaticky a protoze g′(x) = 2x−2, je stacionarnım (podezrelym)bodem bod (1, 0). Tedy celkove:

(x0, y0) ∈ {(−√

5, 2√

5), (−5, 0), (1, 0), (5, 0)}.

Jelikoz f (−√

5, 2√

5) = 25 + 10√

5, f (−5, 0) = 35, f (1, 0) = −1, f (5, 0) = 15 jenejmensı hodnota funkce f na mnozine M rovna−1 a nejvetsı 25+ 10

√5.

Cvicenı

1. Najdete lokalnı extremy funkce f : R2→ R, jestlizea) f (x, y) = 1+ 6y − y2 − xy − x2; b) f (x, y) = x2 − 2y2 − 3x + 5y − 1;c) f (x, y) = (y − x − 3)2; d) f (x, y) = x2y3(12− x − y);e) f (x, y) = x + y + 4 cos(x) cos(y); f) f (x, y) = 4x2 − xy + y2;g) f (x, y) = sin2(x)+ cos2(y); h) f (x, y) = x2 − xy − y2 + 5y − 1.i) f (x, y) = x2 + y4; j) f (x, y) = x2 + x2y2;k) f (x, y) = f (x, y) = x2 + y3; l) f (x, y) = x2y2

m) f (x, y) =√

x2 + y2.

2. Najdete lokalnı extremy funkce f : R2→ R (respektive R3 → R), jestlizea) f (x, y) = x2 − y2 − 2xy − 4x ;b) f (x, y) = (8x2 − 6xy + 3y2) e2x+3y;c) f (x, y, z) = 35− 6x + 2z + x2 − 2xy + 2y2 + 2yz + 3z2.


3. Najdete vsechny lokalnı extremy funkce f : R3 \ {(x, y, z); x = 0 nebo y =0 nebo z = 0} → R,

f (x, y, z) = xyz + 12

x+ 24

y+ 72

z.

4. Najdete vsechny lokalnı extremy funkce f : (−1,∞)× (−1,∞)→ R,

f (x, y) = y√

1+ x + x√

1+ y.

5. Najdete maximum funkce f : R+ ×R+ → R,

f (x, y) = xy − 4y − 8x

x2y2 .

6. Najdete lokalnı extremy funkce f na mnozine g(x, y) = 0 (resp. g(x, y, z) = 0),jestlize

a) f (x, y) = xy − x + y − 1, g(x, y) = x + y − 1;b) f (x, y, z) = xyz, g(x, y, z) = x2 + y2 + z2 − 3;

c) f (x, y, z) = xyz, g(x, y, z) =(

x + y + z − 5xy + yz + zx − 8

);

d) f (x, y) = 6− 4x − 3y, g(x, y) = x2 + y2 − 1;e) f (x, y) = xy − x + y − 1, g(x, y) = x + y − 1;f) f (x, y) = x2 + 2y2, g(x, y) = x2 − 2x + 2y2 + 4y;

g) f (x, y, z) = x2 + y2 + z2, g(x, y, z) =(

x + y − 3z + 7x − y + z − 3

);

h) f (x, y) = x + y, g(x, y) = 1/x2 + 1/y2.

7. Najdete maximum a minimum (pokud existuje) funkce f : R2 → R (respektiveR3 → R) na mnozine M , jestlize

a) f (x, y) = (3x2 + 2y2) e−x2−y2, M = {(x, y) ∈ R2; x2 + y2 ≤ 4};

b) f (x, y, z) = x2 − y2 + z3, M = {(x, y, z) ∈ R3; x2 + y2 + z2 ≤ 1};c) f (x, y) = x2 − 2y2 + 4xy − 6x, M = {(x, y) ∈ R2; x, y ≥ 0, y ≤ −x + 3};d) f (x, y) = y, M = {(x, y) ∈ R2; y ≤ 1, y3 ≤ x2};e) f (x, y) = exy, M = {(x, y) ∈ R2; x2 − y2 ≤ 1};f) f (x, y, z) = x − 2y + 2z, M = {(x, y, z) ∈ R3; x2 + y2 + z2 = 9};g) f (x, y, z) = xy + 2xz + 2yz, M = {(x, y, z) ∈ R3; xyz = 4};h) f (x, y) = x2 + y2, M = {(x, y) ∈ R2; |x + y| ≤ 1, |x − y| ≤ 1}.

8. Najdete lokalnı i globalnı extremy funkce f : R2 → R (respektive R3 → R) namnozine M , jestlize

a) f (x, y) =√

x2 + y2, M = {(x, y) ∈ R2; x2 + y2 ≤ 12};b) f (x, y) = xy, M = {(x, y) ∈ R2; x + y = 1};c) f (x, y, z) = x + 2y − z, M = {(x, y, z); x2 + y2 + z2 = 2, z = x2 + y2};d) f (x, y, z) = xy + xz, M = {(x, y, z) ∈ R3; x2 + y2 = 1, xz = 1};e) f (x, y) = 4xy M = {(x, y) ∈ R2 : x4 + y4 = 2};f) f (x, y) = 5x + 4y M = {(x, y) ∈ R2 : x2 − y2 = 9}.

9. Bud’ f : R2 → R, f (x, y) = cos(x) cos(y) cos(x + y). Najdete nejmensı a nejvetsıhodnotu funkce f na ctverci s vrcholy (0, 0), (0, π), (π, π), (π, 0).


10. Najdete supremum a infimum (pokud existuje) funkce f (x, y, z) = x+ y na mnozine

M = {(x, y, z) ∈ R3; x2 + z2 ≤ 1, y2 + z2 ≤ 2}.

11. Najdete maximum funkce f : R3 → R,

f (x, y, z) = 900z − 700x − 400y

na mnozine dane rovnostı z = x/3+ y/3− 1/x − 1/y + 5.

12. Necht’ f : R2 → R,f (x, y) = exy,

a M = {(x, y) ∈ R2; √|x | + √|y| ≤ 1}. Vypoctete f (M).

13. Dokazte, ze pro kazde x, y, z ∈ R, x, y, z ≥ 0, platı

3√

xyz ≤ x + y + z

3.

(Navod: Hledejte maximum funkce f (x, y, z) = 3√

xyz na mnozine 13 (x + y + z) = k.)

14. Naleznete vzdalenost elipsy x2 + 2y2 + 2(x + 1)y = 1 od bodu (0, 0).

15. Na parabole y2 = 4x naleznete bod, ktery je nejblıze prımce x − y + 4 = 0.

16. Naleznete kvadr nejvetsıho objemu, jestlize delka jeho uhloprıcky je rovna 2√

3.

17. Naleznete polomer r a vysku h kuzele nejvetsıho objemu, aby jeho plast’byl roven S.

18. Naleznete kvadr, ktery ma pri danem povrchu S maximalnı objem.

19. Najdete takova ctyri realna nezaporna cısla x, y, z, u se souctem s, aby jejich soucinbyl nejvetsı.

20. Urcete polomer r a vysku h valce, ktery ma pri danem povrchu S maximalnı objem.

21. Mezi vsemi trojuhelnıky o obvodu O naleznete ten s maximalnım obsahem.

22. Bud’ M ⊂ Rn kompaktnı mnozina a x bod nelezıcı v M . Dokazte, ze existuje body ∈ M , ktery ma ze vsech bodu mnoziny M od bodu x nejkratsı vzdalenost.

23. Bud’ M ⊂ Rn uzavrena mnozina a x bod nelezıcı v M . Dokazte, ze existuje body ∈ M , ktery ma ze vsech bodu mnoziny M od bodu x nejkratsı vzdalenost.

Vysledky

1. a) Maximum v (−2, 4), d) maximum v (4, 6), h) nema extrem; i) minimum v (0, 0);j) neostre minimum v (0, c), c ∈ R; k) nema extrem; l) neostre minimum na obouosach; m) minimum v (0, 0). 2. a) Nema extrem, b) minimum v (0, 0), c) minimumv (8, 5,−2). 3. ) Minimum v (1, 2, 6), maximum v (−1,−2,−6). 6. a) Maximumv (− 1

2 ,32 ); d) minimum v ( 4

5 ,35 ), λ = 5

2 , maximum v (− 45 ,− 3

5 ), λ = − 52 ; e) Nelze

pouzıt Lagran. metodu, dosazenım maximum v (− 12 ,

32 ); f) minimum v (0, 0), λ = 0,


maximum v (2,−2), λ = −2; g) minimum v (0,−1, 2), λ1 = 1, λ2 = −1; h) mini-mum v (

√2,√

2), λ =√

2, maximum v (−√

2,−√

2), λ = −√

2. 8. e) lok. maximum(λ = −1) v (1, 1), (−1,−1) a lok. minimum (λ = 1) v (1,−1), (−1, 1), f) lok. maximum(λ = −1/2) v (−5, 4) a lok. minimum (λ = 1/2) v (5,−4). 15. ) (1, 2). 16. ) Krychleo strane 2. 17. ) r =

√S/( 4√

3√π), h =

√2S/( 4√

3√π). 18. ) Krychle o strane

√S/6.

19. ) x = y = z = u = s/4. 20. ) r = h/2 = √S/(6π). 21. ) a, b, c = O/3.

5. Integralnı pocet na Rn

Prıklady

1. Vypoctete ∫

A(2x + 3y − 2) dx dy,

kde A = [−2, 3]× [1, 4].

Resenı: Funkce (2x + 3y − 2) je spojita na mnozine A, coz je meritelna mnozina, a f jetedy na A integrovatelna. Mnozinu A si muzeme vyjadrit nerovnostmi

−2 ≤ x ≤ 3,

1 ≤ y ≤ 4.

Podle Fubiniovy vety muzeme dany integral pocıtat dvema zpusoby. Vybereme si jeden znich.∫

A(2x + 3y − 2) dx dy =

∫ 3

−2

(∫ 4

1(2x + 3y − 2) dy

)dx =

=∫ 3

−2

[2xy + 3y2

2− 2y

]4

1dx =

∫ 3

−2(6x + 16,5) dx = 97,5.

2. Vypoctete ∫

Axy dx dy,

kde A je mnozina vsech bodu z R2 ohranicena parabolou y = 2x−x2 a prımkou y = −x.

Resenı: Bylo by vhodne namalovat si obrazek. Souradnice prusecıku P1, P2 uvedeneparaboly a uvedene prımky najdeme resenım systemu rovnic

y = 2x − x2,

y = −x .

Odtud dostavame P1 = (0, 0) a P2 = (3,−3). Mnozina A je tedy dana nerovnostmi

0 ≤ x ≤ 3,

−x ≤ y ≤ −x2 + 2x .

Jelikoz je funkce F spojita na meritelne mnozine A, platı

∫

Axy dy dx =

∫ 3

0

(∫ −x2+2x

−xxy dy

)dx =

∫ 3

0

[xy2

2

]−x2+2x

−xdx =

= 12

∫ 3

0(x5 − 4x4 + 3x3) dx = − 243

40 .

3. Vypoctete ∫

Ay dx dy dz,

204

INTEGRALNI POCET NA Rn 205

kde A = {(x, y, z) ∈ R3; x ≥ 0, y ≥ 0, z ≥ 0, 2x + 2y + z − 6 ≤ 0}.

Resenı: Opet by bylo vhodne namalovat si obrazek. Z podmınek z ≥ 0, 2x+2y+z−6≤0 pro z vyplyva 0 ≤ z ≤ 6− 2x − 2y. Mnozina A jsou tedy takove body z R3, pro ktereplatı

0 ≤ x ≤ 3,

0 ≤ y ≤ 3− x,

0 ≤ z ≤ 6− 2x − 2y.

Funkce f (x, y, z) = y je spojita na meritelne mnozine A a je na nı tedy integrovatelna.Platı ∫

Ay dx dy dz =

∫ 3

0

∫ 3−x

0

∫ 6−2x−2y

0y dz dy dx = 27

3 .

4. Vypoctete ∫

Ae−x2−y2

dx dy,

pricemz mnozina A je dana nerovnostmi 1 ≤ x2 + y2 ≤ 9, y ≥ 0.

Resenı: Pouzijeme transformaci do polarnıch souradnic. Mnozina A je pri teto transfor-maci obrazem mnoziny B dane nerovnostmi 1 ≤ r ≤ 3, 0 ≤ ϕ ≤ π , jak se muzemepresvedcit pouzitım transformacnıch vztahu v uvedenych nerovnostech. Toto zobrazenıje na mnozine urcene nerovnostmi r > 0, 0 < ϕ < 2π proste a spojite diferencovatelne.Funkce f (x, y) = e−x2−y2

je spojita a tedy integrovatelna na mnozine A. Proto podleFubiniovy vety platı

∫

Ae−x2−y2

dx dy =∫

Be−r2 cos2 ϕ−r2 sin2 ϕ |r | dϕ dr = π

2

(1

e− 1

e9

).

5. Vypoctete∫

A

√

1− x2

a2 −y2

b2 −z2

c2 dx dy dz,

kdyz A je dana nerovnostıx2

a2 −y2

b2 −z2

c2 ≤ 1.

Resenı: Pri vypoctu pouzijeme zobrazenı dane rovnicemi

x = ar cosϑ cosϕ,

y = br cosϑ sin ϕ,

z = cz sinϑ.

Jakobian tohoto zobrazenı je abcr2 cosϑ . Toto zobrazenı je proste a spojite diferencova-telne na mnozine dane nerovnostmi r > 0, 0 < ϕ < 2π, −π/2 < ϑ < π/2. Mno-zina A je pri uvedene transformaci obrazem mnoziny B dane nerovnostmi 0 ≤ r ≤ 1,


0 ≤ ϕ ≤ 2ϕ, −π/2 ≤ ϑ ≤ π/2. Funkce

f (x, y, z) =√

1− x2

a2 −y2

b2 −z2

c2

je spojita a tedy integrovatelna na mnozine A. Proto platı∫

A

√

1− x2

a2 −y2

b2 −z2

c2 dx dy dz =

=∫

B

√1− r2 cos2 ϕ cos2 ϑ − r2 sin2 ϕ cos2 ϑ − r2 sin2 ϑ |abcr2 cosϑ|×× dr dϕ dϑ =

=∫ 1

0

∫ 2π

0

∫ π/2

−π/2

√1− r2abcr2 cos2 ϑ dϑ dϕ dr = π2

4abc.

6. Najdete pomocı dvojneho integralu obsah casti A roviny ohranicene krivkami y =√x, y = 2

√x a x = 4.

Resenı: Cast A roviny ohranicena danymi krivkami je mnozina urcena nerovnostmi

0 ≤ x ≤ 4,√x ≤ y ≤ 2

√x

a proto dostavame

S =∫

Adx dy =

∫ 4

0

∫ 2√

x

√x

dx dy = 163 .

7. Najdete objem mnoziny A ohranicene plochami

z = x2 + y2, y = x2, y = 1, z = 0.

Resenı: Dana mnozina A je zdola ohranicena rovinou xy (neboli z = 0), zhora rotacnımparaboloidem z = x2 + y2 a z boku ,,parabolickym valcem“ y = x2 a rovinou y = 0.Mnozina A je tedy valcovite teleso, pro ktere platı 0 ≤ z ≤ x2+ y2 pro kazde (x, y) ∈ B ,kde B je dana nerovnostmi

−1 ≤ x ≤ 1,

x2 ≤ y ≤ 1.

Proto platı

V =∫

B

(x2 + y2

)dx dy =

∫ 1

−1

∫ 1

x2

(x2 + y2

)dy dx = 88

105 .

8. Najdete objem mnoziny A ohranicene plochami

x2 + y2 + z2 = a2, x2 + y2 + z2 = b2, x2 + y2 − z2 = 0,

pricemz z ≥ 0 a 0 ≤ a ≤ b.


Resenı: Dana mnozina A je teleso ohranicene dvema koncentrickymi kulovymi plochamise stredem v pocatku a ,,polovinou“ kuzelove plochy x2 + y2 = z2 s vrcholem ve streduobou kulovych ploch. Objem tohoto telesa vyhodne najdeme ve sferickem souradnicovemsystemu. Mnozina A je obrazem kvadru

a ≤ r ≤ b,

0 ≤ ϕ ≤ 2π,π

4≤ ϑ ≤ π

2

pri zobrazenı danem rovnicemi

x = r cosϑ cosϕ,

y = r cosϑ sinϕ,

z = z sinϑ.

Toto zobrazenı je proste a spojite diferencovatelne na mnozine urcene nerovnostmi

r > 0, 0 < ϕ < 2π, −π2< ϑ <

π

2

a jeho Jakobian je roven r2 cosϑ . Proto platı

V =∫

Adx dy dz =

∫ b

a

∫ 2π

0

∫ π/2

π/4r2 cosϑ dϑ dϕ dr = π

3

(2−√

2) (

b3 − a3).

9. Najdete objemovy element na kruznici S se stredem v bode (0, 0) a polomerem R.

Resenı: Rovnice kruznice s pozadovanymi parametry je x2 + y2 = R2. Vsımavy studentjiste snadno pozna, ze normalovy vektor v bode (x, y) lezıcı na kruznici je naprıkladu = (x, y) (stejne tak i v = (2x, 2y)), jednotkovy normalovy vektor v bode (x, y)je tedy roven n = (x/R, y/R) (tento krok bude jasnejsı uvedomıme-li si, ze ‖u‖ =√

x2 + y2 = R). Nynı, mame-li normalovy vektor v libovolnem bode kruznice (nekdy setake rıka jednotkove normalove pole na kruznici), zıskame objemovy element na kruznicikontrakcı objemoveho elementu na R2 tımto zmınenym vektorovym polem. Objemovyelement na R2 je dx ∧ dy (nekdy zkracene jen dx dy). Vzorec pro kontrakci vnejsıhosoucinu 1-forem nasleduje

iξ (ω ∧ η) = iξ (ω)η − iξ (η)ω.

Dosadıme-li, dostavame

dS = in( dx ∧ dy) = in( dx) dy − in( dy) dx = x

Rdy − y

Rdx .

10. Overte spravnost vzorce pro obvod kruznice (O = 2πR) pomocı integrace objemovehoelementu.

Resenı: Z predchozıho prıkladu vıme, ze objemovy element na kruznici S o polomeru R jedS = x/R dy − y/R dx . Obvod spocıtame pokud zintegrujeme tento objemovy elementpres tuto kruznici. Tedy

O =∫

SdS =

∫

S

x

Rdy − y

Rdx .


Parametrizujme kruznici S tak, ze polozıme x = R cos t , y = R sin t a t ∈ [0, 2π]. Nynıdosadıme do naseho integralu

O =∫ 2π

0

R cos t

Rd(R sin t)− R sin t

Rd(R cos t) =

∫ 2π

0R(cos2 t + sin2 t) dt

= R[t]2π0 = 2πR.

11. Vypocıtejte krivkovy integral prvnıho druhu

∫

C

√1+ x2 + y2 dC,

kde C je spirala dana parametrizacı x = ϕ cosϕ, y = ϕ sin ϕ pro ϕ ∈ [0, 2π].

Resenı: Pouzijeme vzorec pro vypocet integralu po krivce C s parametrizacı Ŵ : I →R2, (x, y) 7→ (ϕ(t), ψ(t)):

∫

Cf (x, y) dC =

∫

If ◦ Ŵ(t)‖Ŵ′(t)‖ dt

∫

If (ϕ(t), ψ(t))

√(ϕ′(t))2 + (ψ ′(t))2 dt .

Jelikoz je parametrizace dana v zadanı, lehce dostavame:∫

C

√1+ x2 + y2 dC =

∫

[0,2π]

√1+ ϕ2‖(cosϕ − ϕ sin ϕ, sinϕ + ϕ cosϕ)‖ dϕ =

=∫ 2π

0

√1+ ϕ2

√(cosϕ − ϕ sinϕ)2 + (sin ϕ + ϕ cosϕ)2 dϕ

=∫ 2π

0

√1+ ϕ2

×√

cos2 ϕ − 2ϕ cosϕ sinϕ + ϕ2 sin2 ϕ + sin2 ϕ + 2ϕ sin ϕ cosϕ + ϕ2 cos2 ϕ dϕ

=∫ 2π

0

√1+ ϕ2

√1+ ϕ2 dϕ =

∫ 2π

01+ ϕ2 dϕ =

[ϕ + ϕ

3

3

]2π

0= 2π + 8π3

3.

Mimo jine, jsme zıskali objemovy element na teto spirale v polarnıch souradnicıch:dC =

√1+ ϕ2 dϕ.

12. Vypocıtejte plosny integral druheho druhu

∫

Sx dy dz + y dx dz + (z2 − 1) dx dy,

kde S je cast valcove plochy x2 + y2 = 1 ohranicena rovinami z = 0 a z = 2.

Resenı: Valcovou plochu muzeme parametrizovat takto:

8(h, ϕ) =

x = cosϕy = sin ϕ

z = h

,

odtud dostavame, ze dx = − sin ϕ dϕ + 0 dh, dy = cosϕ dϕ + 0 dh, dz = 0 dϕ + dh.Definicnım oborem nası parametrizace je obdelnık I = [0, 2π]× [0, 1], proto


∫

Sx dy dz + y dx dz + (z2 − 1) dx dy

=∫

Icosϕ cosϕ dϕ dh − sin ϕ(− sinϕ) dϕ dh + (h2 − 1)(− sinϕ) dϕ cosϕ dϕ

=∫

Icos2 ϕ − sin2 ϕ dϕ dh =

∫ 1

0

∫ 2π

0cos 2ϕ dϕ dh =

∫ 1

0

1

2[sin 2ϕ]2π

0 dh = 0.

13. Vypoctete plosny integral prvnıho druhu∫

Sx dS,

kde S je je cast grafu z = x2 + y2 lezıcı nad pulkruhem x2 + y2 < 1, x ≥ 0.

Resenı: Vyuzijeme nasledujıcıho vzorce

∫

Sf (x, y, z) dS =

∫

Uf ◦ σ(u, v)

∥∥∥∥∂σ

∂u(u, v)× ∂σ

∂v(u, v)

∥∥∥∥ du dv,

v nemz σ : U → S je parametrizace plochy S. V nasem prıpade je σ : (r, ϕ) 7→(r cosϕ, r sin ϕ, r2) a U = [0, 1]× [−π/2, π/2]. Proto:∫

Sx dS =

∫

Ur cosϕ‖(−2r, 0, 1)‖ dr dϕ =

∫ π/2

−π/2

∫ 1

0r cosϕ

√4r2 + 1 dr dϕ =

=∫ π/2

−π/2

1

12cosϕ

[(1+ 4r2)2/3

]10 dϕ = 5

√5− 1

12

∫ π/2

−π/2cos(ϕ) dϕ

= 5√

5− 1

12

[sin ϕ

]π/2−π/2 =

5√

5− 1

6.

14. Vypocıtejte integral∫

S x dy dz + y dz dx + z dx dy, kde S je hranice krychle [0, 1]3.

Resenı: Je mozno plochu S rozdelit na sest castı (jednotlive steny krychle) a pocıtat integralpres tyto casti. My ale vyuzijeme Stokesovu vetu, tedy

∫

Sx dy dz + y dz dx + z dx dy =

∫

[0,1]3d(x dy dz + y dz dx + z dx dy)

=∫

[0,1]3dx dy dz + dy dz dx + dz dx dy

= 3∫

[0,1]3dx dy dz = 3[x]1

0[y]10[z]1

0 = 3.

Cvicenı

1. Znazornete mnoziny dane systemem nerovnostıa) y ≤ x, y ≥ x2; b) x2 + y2 − 1 ≥ 0, 4− x2 − y2 ≤ 0, |y| ≥ |x |,c) 1 ≤ |x | ≤ 2, 1 ≤ |y| ≤ 2; d) 0 ≤ x ≤ 2a,

√2ax − x2 ≤ y ≤ 2ax, a > 0.

2. Vypoctetea)∫ 1

0

∫ xx2 xy2 dy dx; b)

∫ π/20

∫ 1cos y x4 dx dy.


3. Vypoctetea)∫

A

(x2 + y2 − 2x − 2y + 4

)dx dy, kde A = [0, 2]× [0, 2];

b)∫

A x2y cos(xy2

)dx dy, kde A = [0, π/2]× [0, 2];

c)∫

A xy2√z dx dy dz, kde A = [−2, 1]× [1, 3]× [2, 4].

4. Zamente poradı integrovanıa)∫ 2

1

∫ 43 f (x, y) dx dy; b)

∫ 20

∫ 6−x2x f (x, y) dy dx;

c)∫ 1

0

∫ 1−y

−√

1−y2f (x, y) dx dy; d)

∫ 10

∫ 1−x0

∫ x+y0 f (x, y, z) dz dy dx .

5. Napiste Fubiniovu vetu pro dvojny integral∫

A f (x, y) dx dy, pokuda) A je lichobeznık s vrcholy (1, 1), (3, 1) (2, 2), (1, 2);b) A je mnozina ohranicena hyperbolou y2− x2 = 1 a kruznicı x2 + y2 = 9, pricemz

obsahuje bod (0, 0).

6. Vypoctetea)∫

A

(5x2 − 2xy

)dx dy, kdyz A je trojuhelnık s vrcholy (0, 0), (2, 0), (0, 1);

b)∫

A(x − y) dx dy, kdyz A je mnozina ohranicena prımkami y = 0, y = x,x + y = 2;

c)∫

A

√xy − y2 dx dy, kdyz A je dana nerovnostmi 0 ≤ y ≤ b, y ≤ x ≤ 10y;

d)∫

A(|x | + |y|) dx dy, kdyz A je dana nerovnostmi |x | + |y| ≤ 4;e)∫

A(12− 3x − 4y) dx dy, kdyz A je dana nerovnostmi x2 + 4y2 ≤ 4;f)∫

A x/3 dx dy, kdyz A je ohranicena krivkou x = 2+ sin y a prımkami x = 0,y = 0, y = 2π ;

g)∫

A sgn(x2 − y2 + 2

)dx dy, kdyz A je kruh x2 + y2 ≤ 2.

7. Vypoctetea)∫ 1

0

∫ 20

∫ 30 (x + y + z) dx dy dz; b)

∫ 10

∫ x0

∫ xy0 x3y2z dz dy dx;

c)∫ e−1

0

∫ e−x−10

∫ x+y+ e0 ln(x − y − z)/(z − e)(z + y − e) dz dy dx .

8. Vypoctete uvedeny integral, kde mnozina A je dana uvedenymi nerovnostmia)∫

A z2 dx dy dz, A :√

x2 + y2 ≤ z ≤√

2− x2 − y2, 0 ≤ y ≤√

1− x2,

0 ≤ x ≤ 1;b)∫

A z dx dy dz, A : x2/a2 + y2/b2 + z2/c2 ≤ 1, z ≥ 0, kde a, b, c > 0;c)∫

A

(x2 + y2 + z2

)dx dy dz, A : y2 + z2 ≤ x2, x2 + y2 + z2 ≤ R2, x ≥ 0.

9. Vypoctete integral∫

A

√1− x2 + y2

1+ x2 + y2 dx dy,

kde A ⊂ R2 je oblast ohranicena krivkami x = 0, y = 0 a x2 + y2 = 1.

10. Vypoctete uvedeny integral, kde A je mnozina ohranicena uvedenymi plochamia)∫

A(2x+3y−z) dx dy dz, A : z = 0, z = a, x = 0, y = 0, x+ y = b, a, b > 0;b)∫

A xyz dx dy dz, A : x2 + y2 + z2 = 1, x = 0, y = 0, z = 0, x, y, z > 0;c)∫

A y cos(z + x) dx dy dz, A : y = √x, y = 0, z = 0, x + z = π/2.

11. Vypoctetea)∫

A

(x2 + y2 + z2 + u2

)dx dy dz du, kde A = [0, 1]4;


b)∫

A u4 ey2dx dy dz du, kde A je mnozina dana nerovnostmmi 0 ≤ z ≤ u, 0 ≤ u ≤

1, 0 ≤ y ≤ zu, 0 ≤ x ≤ yzu;c)∫

A dx1 dx2 . . . dxn, kde A je mnozina dana nerovnostmi x1 ≥ 0, x2 ≥ 0, . . . , xn ≥0, x1 + x2 + . . .+ xn ≤ 1.

12. Vypoctete dvojne integraly na mnozine A transformacı do polarnıch souradnica)∫

A(1− 2x − 3y) dx dy, kde A je kruh x2 + y2 ≤ 2;b)∫

A ln(x2 + y2

)/(x2 + y2

)dx dy, kde A je mezikruzı 1 ≤ x2 + y2 ≤ e;

c)∫

A sin√

x2 + y2 dx dy, kde A je mezikruzı π2 ≤ x2 + y2 ≤ 4π2.

13. Vypocıtejte pomocı transformace do zobecnelych polarnıch souradnic x = ar cosϕ, y =br sin ϕ dvojny integral

∫

A

√

1− x2

a2 −y2

b2 dx dy, kde A je vnitrek elipsyx2

a2 +y2

b2 ≤ 1.

14. Vypoctete trojne integraly na mnozine A transformacı do valcovych souradnica)∫

A dx dy dz, A : x2 + y2 ≤ 1, x ≥ 0, 0 ≤ z ≤ 6;

b)∫

A z√

x2 + y2 dx dy dz, kde A je mnozina ohranicena rovinami y = 0, z = 0, z =a > 0 a valcovou plochou x2 + y2 = 2x ;

c)∫

A

(x2 + y2

)dx dy dz, kde A je mnozina ohranicena paraboloidem 2z = x2 + y2

a rovinou z = 2.

15. Vypoctete trojne integraly na mnozine A transformacı do sferickych souradnica)∫

A

√x2 + y2 + z2 dx dy dz, kde A je cast koule x2 + y2 + z2 ≤ 1, x, y, z ≥ 0;

b)∫

A

(x2 + y2

)dx dy dz, A : z ≥ 0, 4 ≥ x2 + y2 + z2 ≥ 9;

c)∫

A

√x2 + y2 + z2 dx dy dz, kde A je koule x2 + y2 + z2 ≤ z.

16. Vypoctete obsah oblasti ohranicene prımkami

2x − y = 0, 2x − y = 7, x − 4y + 7 = 0, x − 4y + 14 = 0.

17. Vypoctete obsah oblastı ohranicenych elipsou x2/a2 + y2/b2 = 1 a prımkou x/a +y/b = 1.

18. Vypoctete obsah oblasti ohranicene danymi krivkamia) y = (x − a)2/a, a > 0, x2 + y2 = a2;b) y = −2, y = x + 2, y = 2, y2 = x .

19. Transformacı do polarnıch souradnic najdete obsah casti roviny ohranicene kruznicemi(x − a)2 + y2 = a2 a x2 + (y − a)2 = a2.

20. Pomocı zobecnelych polarnıch souradnic

x = ar cosp ϕ,

y = br sinp ϕ,

kde a, b, p jsou vhodne zvolene konstanty, vypoctete obsah casti roviny ohranicene da-nymi krivkami

a)(x2/a2 + y2/b2

)2 = xy/c; b) (x/a + y/b)2 = x/a − y/b, y > 0;


c) x3 + y3 = axy; d)√

x +√y = √a, x + y = a > 0;

21. Trasformacı do zobecnelych sferickych souradnic x = ar cosϑ cosϕ, y = br cosϑ sin ϕ,z = cr sinϑ vypoctete integral

∫

V

√

1− x2

a2 −y2

b2 −z2

c2 dx dy dz,

kde V je elipsoid se stredem v bode (0, 0, 0) a poloosami a, b, c > 0.

22. Vypoctete integral ∫

Sx2y2z dx dy,

kde S = {(x, y, z) ∈ R3; x2 + y2 + z2 = a, z ≤ 0} a a > 0.

23. Najdete objemy teles ohranicenych danymi plochamia) rovinami: x − y + z = 6, x + y = 2, x = y, y = 0, z = 0;b) valcovymi plochami: z = 4− y2, y = x2 a rovinou z = 0;c) paraboloidy: z = 4− x2 − y2, 2z = 2+ x2 + y2;d) plochami: z = e−x2−y2

, z = 0, 1 = x2 + y2.

24. Vypocıtejte krivkove integraly druheho druhu:a)∫

C(x2 + y2) dx + (x2 − y2) dy, kde C je trojuhelnık s vrcholy (0, 0), (1, 0), (0, 1)

orientovany ve smeru danem poradım ve vyctu vrcholu;b)∫

C(x + y dx − x − y dy)/(x2 + y2), kde C je kruznice se stredem v pocatku apolomerem 1;

c)∫

C xy dx + (x + y) dy, kde C = {(x, y) ∈ R2; x = y ∈ [0, 1]};d)∫

C xy dx + (x + y) dy, kde C = {(x, y) ∈ R2; x = t2, y = t2, t ∈ [0, 1]};e)∫

C xy dx + (x + y) dy, kde C = {(x, y) ∈ R2; x = t, y = t2, t ∈ [0, 1]};f)∫

C sin y dx + sin x dy, kde C je usecka z bodu (0, π) do bodu (π, 0),g)∫

C ( dx − dy)/(x + y), kde C je hranice ctverce s vrcholy (1, 0), (0, 1), (−1, 0) a(0,−1) orientovana ve smeru danem poradım ve vyctu vrcholu.

h)∫

C 2xy dx + x2 dy, kde C je libovolna krivka vychazejıcı z bodu (0, 0) a koncıcıv bode (2, 1);

i)∫

C y f (xy) dx + x f (xy) dy, kde f je spojita funkce a C je krivka z bodu (1, 1) dobodu (2, 3);

j)∫

C(x4+4x+ y3) dx+ (6x2y2−5y4) dy, kde C je krivka z bodu (−2,−1) do bodu

(3, 0);k)∫

C(1/y) dx − (x/y2) dy, kde C je usecka z bodu (1, 1) do bodu (2, 3).

25. Vypocıtejte krivkove integraly prvnıho druhu:a)∫

C xy dC , kde je krivka C = {(x, y) ∈ R2 : x2/4+ y2 = 1, x, y ≥ 0};b)∫

C x2y dC , kde je krivka C = {(x, y) ∈ R2 : x2 + y2 = 4, y ≤ 0};c)∫

C x dC , kde je krivka C = {(t, t2) ∈ R2 : −1 ≤ t ≤ 3};d)∫

C x2 dC , kde je krivka C = {(t, ln(t)) ∈ R2 : 1 ≤ t ≤ 3};e)∫

C |y| dC , kde je krivka C = {(t2/2, t) ∈ R2 : t ≤ 1/2};f)∫

C

√x2 + y2 dC , kde je krivka C = {(x, y) ∈ R2 : x2 + y2 = x};

g)∫

C y dC , kde je krivka C = {(x, y) ∈ R2 : (x2 + y2)2 = x2 − y2, x, y ≥ 0};h)∫

C 1/(x − y) dC , kde C je usecka z bodu (0,−2) do bodu (4, 0);


i)∫

C(1+√

2y) dC , kde je krivka C = {(t − sin t, 1− cos t) ∈ R2 : 0 ≤ t ≤ 2π};j)∫

C 5− x2 − 2xy + 3y2 dC , kde je krivka C = {(x, y) ∈ R2 : x2 + y2 = 4};k)∫

C z dC , kde je krivka C = {(x, y, z) ∈ R3 : x2 + y2 + z2 = 1};l)∫

C x + y dC , kde C je ctvrtina kruznice x2 + y2 + z2 = R2, x = y lezıcı v prvnımoktantu (x, y, z ≥ 0);

26. Vypocıtejte plosne integraly prvnıho druhua)∫

S 15|xy| dS, kde S je cast sedlove plochy z = xy lezıcı uvnitr valce x2 + y2 ≤ 1;

b)∫

S dS, kde S je cast kuzele z =√

x2 + y2 lezıcı uvnitr valce x2 + y2 ≤ 2y;c)∫

S xy dS, kde S je cast valce x2 + y2 = 4, x, y ≥ 0 a 0 ≤ z ≤ 3;d)∫

S xyz dS, kde S je cast roviny z = 2x+3y+ z = 6 lezıcı uvnitr valce x2+ y2 ≤ 1;e)∫

S x dS, kde S je cast kulove plochy v prvnım kvadrantu se stredem v pocatku apolomerem R;

f)∫

S x2y2 dS, kde S je hornı pulka kulove plochy se stredem v pocatku a polomerem R;g)∫

S x2y2z dx dy, kde S = {(x, y, z) ∈ R3; x2 + y2 + z2 = 1, z ≤ 0};h)∫

S

√x2 + y2 dS, kde S je cast kuzelove plochy x2/a2 + y2 − z2/c2 = 0,

0 ≤ z ≤ b.

27. Vypocıtejte plosne integraly druheho druhu:a)∫

S x dy dz + y dz dx + z dx dy, kde S je povrch krychle [0, 1]3;b)∫

S x2 dy dzy2 dz dx + z2 dx dy, kde S je povrch krychle [0, 1]3;c)∫

S 6xz dx dy, kde S je cast roviny x + y + z = 1, kde x, y, z ≥ 1;d)∫

S y dy dz + y2 dz dx + yz dx dy, kde S je cast roviny x + z = 1 uvnitr valcex2 + y2 ≤ 1;

e)∫

S x dy dz + y2 dz dx + yz dx dy, kde S je kuzel x2 + y2 = (z − 1)2 a 0 ≤ z ≤ 1;f)∫

S y2 dz dx + z dx dy, kde S je cast sedlove plochy z = xy lezıcı nad ctvrt kruhemx2 + y2 ≤ 1 a x, y ≥ 0;

g)∫

S y2 dz dx+z dx dy, kde S je cast sedlove plochy z = x2−y2 lezıcı nad obdelnıkem0 ≤ x ≤ 1 a |y| ≤ 1;

h)∫

S z dx dy − (x + y) dz dx , kde S je ctvrt paraboloidu z = x2 + y2, x, y ≥ 0 a0 ≤ z ≤ 1;

i)∫

S dy dz+z2 dx dz, kde S je cast kuzele x2 = z2+y2 lezıcı uvnitr valce x2+y2 ≤ 1;

28. Najdete objemove elementy na nasledujıcıch krivkach respektive plochach.a) Kulova plocha v R3 se stredem v pocatku a polomerem R;b) Rovina v R3 dana rovnicı z = ax + by + c;c) Rotacnı paraboloid v R3 s rovnicı z = x2 + y2;d) Sroubovice v R3 dana parametrizacı x = cos t, y = sin t, z = t .

29. Nasledujıcı krivkove integraly se pokuste vypocıtat Stokesovou vetou.a)∫

C(xy + sin(x/y))/x dx − sin(x/y)/y dy, kde C je hranice trojuhelnıku s vrcholy(0, 0), (1, 1) a (0, 2) orientovana poradım ve vyctu vrcholu;

b)∫

C ey dx + y ex dy, kde C je hranice obdelnıku [0, 3]× [1, 2],c) Obvod jednotkoveho kruhu, tedy

∫S x dy − y dx , S je jednotkovy kruh.

Vysledky

2. a) 140 , 3. a) 32

3 , b)−π/16, c) 5(

4−√

2)

. 4. a)∫ 4

3

∫ 21 f (x, y) dx dy, b)

∫ 40

∫ y/20 f (x, y) dx dy+


∫ 64

∫ 6−y0 f (x, y) dx dy, c)

∫ 0−1

∫√1−x2

0 f (x, y) dx dy+∫ 1

0

∫ 1−x0 f (x, y) dx dy. 5. a) Napr.∫ 2

1

∫ −y+41 f (x, y) dx dy, 6. a) 3, b) 1, c) 6b3, d) 64

3 , e) 24π , f) 32π , g) 2π . 7. a) 18, b) 1

110 ,

8. a) π(√

8− 1)/15, b) πabc2/4, c) 2πR5/5. 9. ) π(π − 2)/8. 10. a) 5ab3/6− (a2b2)/4,b) 1

84 , c) π2/16 − 12 . 11. a) 4

3 , b) (2 e − 5)/32, c) 1/n!. 12. a) 2π , b) π/2, c) −6π2.13. ) 2πab/3 14. a) 3π , b) b3a2π/6. 15. a) π/8, b) 844

15 π , c) π/10. 16. ) 492 . 17. ) S1 =

ab(π−2a)/4, S2 = πab( 34+2a/π). 18. a) a2(π/4− 1

3 ), b) 403 . 19. ) πa2/2. 21. ) abcπ/4.

23. a) 163 , b) 128

√2/21, c) 3π . 24. a) 0, b) −2π , c) 4

3 ; d) 43 ; e) 17

12 , f) 0, g) −4.h) 4, i) F(5) − F(1), F-primitivnı funkce k f , j) 62, k) − 1

3 . 25. a) 14/9, b) −32/3,

c) (37√

37 − 5√

5)/12, d) (10√

10 − 2√

2)/3, e) (4√

2 − 2)/3, f) 2, g) 1 −√

2/2,h)√

5 ln(2), i) 8+ 2π√

2, j) 18π , k) ??, l) 12 R2, 26. a) 4(

√2+ 1); b) π

√2; c) 12; d) 0;

e) πR3/4, f) 2πR6/15, h) 23πa2√

a2 + b2. 27. a) 3, b) 3; c) 14 ; d) 0; e) π/3; f) 7

120 ;g) 0; h) π ; i) 0; 28. a) dS = x/R dy dz − y/R dz dx + z/R dx dy, b) dS = (−a dy dz +b dz dx + dx dy)/

√a2 + b2 + 1, c) dS = (−2x dy dz + 2y dz dx + dx dy)/

√4z + 1

d) (x dy − y dx + dz)/√

2. 29. a)−1, b) 2 e− e2/2− 32 , c) 2π .

6. Funkce komplexnı promenne

Prıklady

1. Z definice derivace dokazte ze pro f : C→ C, f (z) = z2, platı f ′(z) = 2z.

Resenı: Hodnota f ′(z0) je dana vztahem

f ′(z0) = lim1z→0

f (z0 +1z)− f (z0)

1z.

Dosad’me do tohoto vztahu nasi funkci. Dostavamef ′(z0) = lim1z→0((z0 +1z)2− (z0)

2)/1z = lim1z→0(z20 + 2z01z +1z2− z2

0)/1z =lim1z→0(2z01z +1z2)/1z = lim1z→0 2z01z/1z + lim1z→01z2/1z = 2z0 + 0.

2. Vypocıtejte integral

I =∫

Cz2 dz,

kde C je oblouk kruznice |z| = 1 ohraniceny body z1 = eαi a z2 = eβi (α < β).

Resenı: Podıvejme se nejprve blıze na funkci f (z) = z2, pomocı realne a imaginarnıcasti argumentu ma tato funkce tvar f (z) = z2 = (x + iy)2 = (x2 − y2) + (2xy)i =u(x, y)+ iv(x, y), kde u(x, y) = x2 + y2 a v(x, y) = 2xy.

Samotny integral budeme pocıtat podle nasledujıcıho vzorce∫

Cf (z) dz =

∫

C(u dx − v dy)+ i

∫

C(u dy + v dx)

V nasem prıpade tedy dostavame∫

Cz2 dz =

∫

C(x2 − y2 dx + 2xy dy)+ i

∫

C(x2 − y2 dy + 2xy dx)

Nynı parametrizujeme oblouk nasledovne: x = r cos t , y = r sin t odtud dx = −r sin t dta dy = r cos t dt . Coz dava

I =∫ β

α

(sin3 t − 3 cos2 t sin t) dt + i∫ β

α

(cos3 t − 3 sin3 t cos t) dt

=[

cos(3t)

3

]β

α

+ i

[sin(3t)

3

]β

α

= 13 (sin(3β)+ cos(3β)− sin(3α)− cos(3α)).

Pokud je C cela kruznice, potom I = 0, coz odpovıda faktu, ze integral z analytickefunkce po uzavrene krivce je roven 0.

3. Vypocıtejte integral ∫

C3z3 dz,

kde C je oblouk spiraly dane parametrizacı ϕ : [0, 8π] ∋ t 7→ t et i.

Resenı: Nejprirozenejsı by bylo pouzıt zadanou parametrizaci a ,,dosadit“ ji do integralu(presneji receno provest pull-back formy). Vsimeme si ale, ze funkce za integralem je

215


analyticka v cele C proto integral z nı nezavisı na integracnı ceste. Proto si muzeme zvolituplne jinou (jednodussı) krivku spojujıcı pocatecnı ϕ(0) = 0 a koncovy ϕ(8π) = 8 bod.Zvolme si tedy ,,lepsı“ parametrizaci, treba ψ : [0, 1] ∋ t 7→ 8t . Dostaneme

∫

C3z3 dz =

∫ 1

03t3 d(8t) =

∫ 1

024t3 dt = 6

[t4]1

0= 6.

4. Vypocıtejte integral ∫

C

1

zdz,

kde C je ctverec dany vrcholy 1+ i, −1+ i, −1− i a 1− i s orientacı danou poradım vevyctu vrcholu.

Resenı: V prıpade, ze by se jednalo o analytickou funkci integral by vysel roven 0. My aletakove stestı nemame. Tato funkce ma v bode 0 singularitu. Integral po jakekoli zavrenekrivce obepınajıcı singularnı bod je roven 2π i-nasobku residua funkce v tomto bode. Tedy

I =∫

C

1

zdz = 2π i res0

1

z.

Stacı tedy spocıtat vyse uvedene residuum. Vyuzijeme nasledujıcıho vztahu

resa f = 1

(n − 1)!

dn−1

dzn−1

∣∣∣∣a

((z − a)n f (z)

),

kde a je polem n-teho radu funkce f . V nasem prıpade f (z) = 1/z je 0 polem prvnıhoradu. O tom se lze presvedcit tım, ze limita limz→0(z − 0) f (z) existuje a je konecna.Dosad’me tedy do vzorce pro reziduum

resa f = 1

(n − 1)!

dn−1

dzn−1

∣∣∣∣a

((z − a)n f (z)

)= 1

d1−1

dz1−1

∣∣∣∣0

((z − 0)

1

z

)= 1.

Proto ∫

C

1

zdz = 2π i.

5. Vypoctete integral ∫ ∞

−∞

1

x2 + 1dx

kde x ∈ R.

Resenı: Rozsırıme-li funkci, kterou mame integrovat na cele C, tedy formalne definujemefunkci g : C→ C, g(z) = 1/(z2+1), lze nas integral vypocıtat tak, ze budeme integrovatfunkci g po ,,uzavrene“krivce C := R∪{∞} ⊂ C. Uvnitr C (naprıklad v oblastiImz > 0)lezı jediny singularnı bod g a to z = i. Tento bod je jednonasobnym polem funkce g.Muzeme tedy pouzıt Cauchyho vetu o reziduıch a dostaneme

∫ ∞

−∞

1

x2 + 1dx = 2π i resi g(z) = 2π i lim

z→i

z − i

(z − i)(z + i)= π.

Jiste si snadno overıte, ze stejny vysledek dostaneme pokud budeme uvazovat oblastImz < 0.

FUNKCE KOMPLEXNI PROMENNE 217

Pokud se rozhodneme pocıtat nas integral ,,klasicky“ potom dostanme∫ ∞

−∞

1

x2 + 1dx = lim

c→∞

∫ c

−c

1

x2 + 1dx = lim

c→∞[arctan(x)]c−c =

π

2+ π

2= π.

Cvicenı

1. Z definice derivace dokazte nasledujıcı vztahya) ((a + bi)z)′ = (a + bi); b) (zn)′ = nzn−1, kde n ∈ N;

2. Rozhodnete, zda je funkce f : C → C analyticka v C prıpadne na nejake otevrenemnozine, pokud

a) f (z) = z; b) f (z) = |z|;c) f (z) = z ez; d) f (z) = z;e) f (z) = z2 + (3+ 4i)z/(z − (1+ i)).

3. Najdete rezidua nasledujıcıch funkcı ve vsech jejich singularnıch bodech.a) f (z) = 1/(z − z3); b) f (z) = z/(1+ z)3;c) f (z) = 1/(z2 + i)3; d) f (z) = 1/( ez + 1);e) f (z) = 1/ sin(x).

4. Pomocı vety o derivaci (integraci) rady clen po clenu najdete derivaci (primitivnı funkci)k nasledujıcım funkcım

a) sin(z); b) cos(z); c) ez .

5. Vypoctete integraly:a)∫

C(5z − 2)/(z(z− 1)) dz, kde C je kruznice |z| = 2 orientovana v kladnem smeru;b)∫

C 1/(z3(z + 4)) dz, kde C je kruznice |z + 2| = 3 orientovana v kladnem smeru;c)∫

C 1/(z3(z + 4)) dz, kde C je kruznice |z| = 2 orientovana v kladnem smeru;d)∫

C(3z3 + 2)/((z − 1)(z2 + 9)) dz, kde C je kruznice |z − 2| = 2 orientovanav kladnem smeru;

e)∫

C tan z dz, kde C je kruznice |z| = 2 orientovana v kladnem smeru;f)∫

C(3z3+ 2)/((z− 1)(z2+ 9)) dz, kde C je kruznice |z| = 4 orientovana v kladnemsmeru;

g)∫

C 1/z(z + 1)(z − 2) dz, kde C je kruznice |z − 3| = 2 orientovana v kladnemsmeru.

6. Vypocıtejte integral z funkce f po jednotkove kladne orientovane kruznici C se stredemv 0 jestlize

a) f (z) = z−2e−z; b) f (z) = z e1/z.

7. Pomocı reziduı komplexnıch funkcı vypocıtejte nasledujıcı integraly z realnych funkcı.a)∫∞−∞ x2/((x2 + 9)(x2 + 4)) dx; b)

∫∞−∞ 1/(x2 + 1) dx;

c)∫∞−∞ 1/(x2 + 2x + 2)) dx; d)

∫∞−∞ x/((x2 + 1)(x2 + 2x + 2)) dx;

e)∫∞−∞ x2/(x2 + 1)2 dx .

Vysledky2. a) ano na C, b) ne na C, ano na C \ {0}, c) ano na C, d) ne nikde, e) ne na C, ano naC \ {1 + i}. 4. a) cos(z), b) − sin(z), c) ez . 5. a) 10π i, b) 0, c) π i/32, d) π i, e) −4π ,f) 6π i. 6. a) −2π i, b) π i. 7. a) π/100, b) π , c) π , d) −π/5, e) π/2.

7. Obycejne diferencialnı rovnice

Ve vetsine nasledujıcıch prıkladu a cvicenı jsou y a z funkce jedne realne promenneoznacovane jako x .

Prıklady

1. Overte, ze funkce y(x) = x√

1− x2 je resenım diferencialnı rovnice

y ′ = x − 2x3

y

na intervalu (0, 1).

Resenı: Nejdrıve si spocıtejme y ′, tedy y ′ =(

x√

1− x2)′= (1− 2x2)/

√1− x2. Nynı

dosad’me do prave strany diferencialnı rovnice funkci y:

x − 2x3

y= x − 2x3

x√

1− x2= 1− 2x2

√1− x2

= y ′

2. V diferencialnı rovnici

y ′ = y2

xy − x2

proved’te transformaci y(x) = z(x) · x.

Resenı: Nejprve si z transformace vypocteme y ′. Dostavame y ′ = z′x + z (nezapo-menme, ze z je funkcı promenne x). Transformaci i vypocıtanou derivaci y ′ dosad’me dodiferencialnı rovnice

z′x + z = z2

z − 1.

3. Metodou separace promennych vyreste diferencialnı rovnici

1

xy ′ = 1

x2 + 1.

Resenı: Za predpokladu x 6= 0 upravıme rovnici na tvar

y ′ = x

x2 + 1.

Nynı jiz rovnice ma ,,separovane“ promenne. Nekdy se takove rovnice zapisujı ve tvaru

dy

dx= x

x2 + 1a nebo dy = x

x2 + 1dx

218

OBYCEJNE DIFERENCIALNI ROVNICE 219

Nynı ,,zintegrujeme“ obe strany (presneji receno aplikujeme vetu o diferencialnıch rovni-cıch se separovanymi promennymi tj. rovnicıch ve tvaru y ′F(y) = G(x)). Dostavame

y =∫

x

x2 + 1dx + C = 1

2 ln(x2 + 1)+ C.

Vsechna resenı nası rovnice lze na R \ {0} zapsat ve tvaru y = 12 ln(x2 + 1)+ C , kde

C ∈ R je libovolne cıslo.

4. Najdete vsechna resenı diferencialnı rovnice

y ′ = y2

xy − x2 .

Resenı: Delıme-li citatele i jmenovatele x , dostaneme rovnici ve tvaru

y ′ = (y/x)2

(y/x)− 1.

Tedy nase rovnice je homogennı a jevı se jako vyhodne pouzıt transformaci y = zx , tedyy ′ = z′x + z, coz ji prevadı na tvar

z′x + z = z2

z − 1.

Je videt, ze jde o rovnici se separovatelnymi promennymi, tedy po jejich separaci (apredpokladu z, x 6= 0) dostaneme

z − 1

zdz = dx

x.

Zintegrujeme-li poslednı rovnici obdrzıme z − ln |z| = ln |x | + ln C , kde C je kladnakonstanta. Toto lze prepsat na ln ez ln |z| = ln |x | + ln C . Odtud tedy ez = C|xz| = kxy,kde k ∈ R. Nynı proved’me inverznı transformaci k te, co jsme provedli na zacatku(z = y/x). Celkove resenım jsou vsechny funkce y vyhovujıcı podmınce

ey/x = ky.

5. Urcete vsechna resenı diferencialnı rovnice

y ′ = 3y − 7x + 7

3x − 7y − 3.

Resenı: U rovnic tohoto typu zabıra substituce x = ξ + x0, y = η+ y0, kde (x0, y0) jsouresenı nasledujıcı linearnı soustavy

−7x + 3y + 7 = 0,

3x − 7y − 3 = 0.

Rovnici vyresıme Cramerovym pravidlem k tomu potrebujeme nasledujıcı determinanty

D =∣∣∣∣−7 3

3 −7

∣∣∣∣ = 40, Dx =∣∣∣∣−7 3

3 −7

∣∣∣∣, Dy =∣∣∣∣−7 −7−3 −3

∣∣∣∣ .


Odtud x0 = Dx/D = 1 a y0 = Dy/D = 0, nase transformace ma tedy tvar x =ξ + 1, y = η, ktera prevadı nasi rovnici na tvar

dη

dξ= 3η − 7ξ

3ξ − 7η=

3η

ξ− 7

3− 7η

ξ

.

Coz uz je homogennı rovnice a transformace z = η/ξ ji prevadı na

ξdz

dξ+ z = 3z − 7

3− 7z

Po uprave dospejeme k rovnici

7z − 3

7(z2 − 1)dz + dξ

ξ= 0

Po integraci teto rovnice dostavame (z−1)2(z+1)5ξ7 = C 6= 0, po dosazenı z = y/(x−1)a nezbytnych upravach zjistıme, ze

(y − x + 1)2(y + x − 1)5 = K , K ∈ R \ {0}.

6. Urcete resenı Cauchyho ulohy pro linearnı diferencialnı rovnici

y ′ + 2

xy = 0,

vyhovujıcı pocatecnı podmınce y(1) = 2.

Resenı: Separovanım promennych dostavame rovnici

dy

y= −2

xdx .

Integracı teto linearnı rovnice zıskame obecne resenı y = C/x2, kde C ∈ R a x ∈ R\ {0}.Nynı pouzijeme pocatecnı podmınku pro stanovenı konstanty C: y(1) = C/(1)2 = 2.Odtud C = 2. Resenım tedy je funkce y = 2/x2, pro x ∈ R \ {0}.

7. Najdete vsechna resenı nasledujıcı nehomogennı linearnı diferencialnı rovnice

y ′ + 1

xy = 3x .

Resenı: Nejprve vyresıme odpovıdajıcı homogenizovanou rovnici

y ′ + 1

xy = 0.

Metodou separace promennych lze dospet k vysledku

yH =C

x,


kde C ∈ R a x 6= 0. Nynı k resenı homogenizovane rovnice pricteme jedno resenı puvodnırovnice (rıkame mu partikularnı resenı) a zıskame tak obecne resenı nası rovnice. Totoresenı bud’uhodneme (podle tvaru rovnice by to mohl byt polynom a jeho stupen nebudevyssı nez dva) a nebo pouzijeme metodu ,,variace konstant“.

Pouzijme metodu variace konstant. Povazujme nynı C za funkci x (zopakujme, ze tedymame y = C(x)/x). Dosazenım do nası rovnice dostaneme:

− 1

x2 C(x)+ 1

xC ′(x)+ 1

x2 C(x) = 3x

odtudC ′(x) = 3x2.

Dostavame, ze C(x) = x3 (to nenı jedine resenı, dalsı je naprıklad C(x) = x3 + 2. Namstacı jen jedno.). Nase partikularnı resenı tedy je yp = (x3)/x = x2.

Prictenım yp k yH dostaneme celkove resenı nası diferencialnı rovnice

y = yH + yp = x2 + 1

xC,

kde C ∈ R a x 6= 0.

8. Reste linearnı diferencialnı soustavu

x = 3x − 2y

y = 2x − y + 1,

kde y, x jsou funkce promenne t. Vyreste tuto soustavu take s pocatecnı podmınkoux(0) = 0 a y(0) = 1.

Resenı: Nejprve budeme hledat resenı homogenizovaneho systemu, tedy soustavy

x = 3x − 2y

y = 2x − y.

Najdeme vlastnı cısla a vlastnı vektory matice teto soustavy

A =(

3 −22 −1

).

Bohuzel tato matice ma jednu dvojnasobnou vlastnı hodnotu λ = 1, k nı prıslusny vlastnıvektor je

u =(

11

).

Najdeme tedy vektor v (nekdy mu rıkame ,,zobecneny vlastnı vektor“), ktery se operatoremA− λE zobrazı na vlastnı vektor u. Resıme tedy linearnı soustavu(A− λE)v = u, odtuddostaneme, ze

v =(

321

).

Obecne resenı homogenizovaneho systemu tedy je(

xy

)

H= c1u eλt + c2(v + tu) eλt = c1

(11

)et + c2

( 32 + 1t1+ 1t

)et .


Nynı musıme urcit partikularnı resenı nası puvodnı soustavy. Zamyslıme-li se nad tım,vidıme, ze x a y by mohly byt konstantnı funkce naprıklad x = A a y = B , potom stacıdoresit soustavu

0 = 3A− 2B

0 = 2A− B + 1.

Resenı je A = −2 a B = −3. Pricteme-li toto partikularnı resenı k obecnemu resenıhomogenizovaneho systemu, dostaneme obecne resenı nası soustavy.

(xy

)= c1

(11

)et + c2

(32 + 1t1+ 1t

)et +

(−2−3

).

Pro stanovenı konstant c1, c2 pouzijeme pocatecnı podmınku, resıme soustavu

x(0) = c1 e0 + c2(32 + 1 · 0) e0 − 2 = 0

y(0) = c1 e0 + c2(1+ 1 · 0) e0 − 3 = 1.

Dostavame c1 = 8 a c2 = −4. Resenı rovnice s pocatecnı podmınkou je(

xy

)= 8

(11

)et − 4

(32 + 1t1+ 1t

)et +

(−2−3

).

9. Vyreste nasledujıcı linearnı diferencialnı rovnici

y ′′ + y = 0.

Resenı: Teorie nam radı provest substituci y ′ = z a prıklad dopocıtavat jako soustavurovnic prvnıho radu, to my udelame, ale preskocıme uvodnı kroky a prejdeme rovnouk charakteristickemu polynomu matice tohoto systemu. Tento polynom se napadne podoba(a nenı to nahoda) nası puvodnı rovnici, tedy

λ2 + 1 = 0.

Tento polynom ma bohuzel pouze komplexnı koreny λ1,2 = ±i. Resenım samozrejmejsou linearnı kombinace funkcı f1 = eix a f2 = e−ix , pokud bychom chteli zapsat resenıpomocı realnych funkcı nenı nic jednodussıho, nez v prostoru resenı (linearnı obal funkcıf1, f2) zvolit jina (realna) nezavisla resenı treba 1

2 ( f2−i f1) = sin x a 12 ( f2+i f1) = cos x .

Resenım tedy je

y = c1 sin x + c2 cos x ekvivalentne y ∈ [[sin x, cos x]].

Obdobne resıme soustavy rovnic, pokud nam vychazejı komplexnı vlastnı cısla.

Cvicenı

1. Rozhodnete, ktere z nasledujıcıch funkcı jsou resenım direfencialnı rovnice

x(x − 1)y ′ + 2xy = 1.

a) y = cos x; b) y = ex ;


c) y = (x − ln x)/(x − 1)2; d) y = x2.

2. Rozhodnete, ktere z nasledujıcıch funkcı jsou resenım diferencialnı rovnice

yy ′′′ − y ′y ′′ = 0.

a) y = ex ; b) y = x2;c) y = x; d) y = sin x;e) y = cos x; f) y = 0.

3. Najdete diferencialnı rovnici radu k tak, aby resenıma) byla funkce y = x2 a k = 1;b) byly vsechny funkce tvaru y = x − c/x , kde c ∈ R a k = 1;c) byly vsechny funkce tvaru y = c1x + c2, kde c1, c2 ∈ R a k = 2;d) byly vsechny funkce tvaru y = c1x + c2 ex , kde c1, c2 ∈ R a k = 2;e) byly vsechny funkce y = c ec/x , kde c ∈ R a k = 1.

4. V diferencialnı rovnici xy ′ = y proved’te trasnformacia) z = yx2; b) z = ey;c) z2 = y; d) z = √y;e) z2 = y + 1; f) z = xy.

5. Najdete alespon jedno resenı nasledujıcıch diferencialnıch rovnica) y ′ + y = 0; b) y ′ + y = sin x .c) y ′ + y = 1; d) y ′ + y = 2x + 2;e) y ′ + y = x cos x; f) y ′ + y = e3x ;g) y ′ + y = ex + e−x + 1.

6. Vyreste nasledujıcı diferencialnı rovnice (vhodnou metodou se jevı metoda separacepromennych)

a) y ′ = 3y; b) y ln y − xy ′ = 0, y(1) = 1;c) y tan x + y ′ = 0; d) y ′(1+ x2) = 1+ y2, y(0) = 1;e) y cos x − y ′ sin x = 0; f) y ′ = 2

√y ln x, y(e) = 1.

7. Vyreste nasledujıcı homogennı rovnicea) 2x2y ′ = x2 + y2; b) y ′xy = x2 + y2;c) xy ′ = y +

√y2 + x2; d) y ′x = y + x ey/x ;

e) y ′y(1− x2) = x(1− y2), y(2) = 2; f) y ′(y − x) = x + y;g) xy ′ = y ln(x/y); h) xy ′ = x + 2y;i) x + yy ′ = 2y; j) (x4 + y4)y ′ = x3y.

8. Vyreste diferencialnı rovnicea) 2y + 3x − 1+ (4y + 6x − 5)y ′ = 0; b) y + 2 = (2x + y − 4)y ′;c) xy ′′ + 2y ′ + 4xy = 0;d) 2yy ′ − y2/x = x (premyslejte o substituci y(x) = √z(x)).

9. Najdete vsechna resenı nasledujıcıch linearnıch diferencialnıch rovnicea) y ′ + 2xy = 4x; b) xy ′ + 2y = 0;c) y ′(x cos y + sin(2y)) = 1; d) y ′ + y

√x = 3x2;

e) y ′ − 2xy = 2x ex2; f) y ′ + y/x = 3x;

g) 3y2y ′ = ex ; h) y ′ cos y = 1;i) y ′ + y/ tan x = sin x; j) y ′ + xy = xy3;k) y ′ − y = 1; l) y ′ = (y + 1) sin x;

m) (cos x)y ′ + (sin x)y = 1; n) 4y ′′ + 4y ′ + λy = 0, kde λ ∈ (0, 1).

10. Reste nasledujıcı soustavy linearnıch diferencialnıch rovnica) x = 5x + 2y + et

y = x − 6y + e2t ;b) x = 7x + 6y − cos t

y = 2x + 6y − sin t;c) x = x + y

y = −5x − y;d) x = −x + 5y

y = −x + y + 8t;e) x = 5x − 10y − 20z

y = 5x + 5y + 10zz = 2x + 4y + 9z;

f) x = x − y + zy = x + y − zz = − y + 2z;

g) x = yy = x + 3y − 4zz = x + 2y − z;

h) x = 3x + y − zy = −x + 2y + zz = x + y + z.

11. Reste nasledujıcı linearnı diferencialnı rovnice vyssıch radua) y ′′ − 9y = 0; b) y ′′ + 2y ′ + y = 0;c) y ′′′ − y ′ = 0; d) y ′′′′ − 5y ′′ + 4y = 0;e) y ′′′ − y ′′ = 0; f) y ′′′′ − y ′′′ + y ′′ = 0;g) y ′′ − 7y ′ + 10y = 0; h) y ′′ − 7y ′ + 10y = 40;i) y ′′ − 7y ′ + 10y = 6 e2x ; j) 3y ′′ − 4y ′ = 3 e4;k) y ′′ − 9y = 0; l) y ′′ + 2y + y = 0;

m) y ′′ + y ′ = 0; n) y ′′′ − 13y ′′ + 12y ′ = 0;o) y ′′′ − 2y ′′ = 0; p) y ′′′ − 3y ′′ + 3y ′ − y = 0.

Vysledky

1. a) ne, b) ne, c) ano, d) ne. 2. a) ano, b) ne, c) ano, d) ano, e) ano, f) ano. 3. a) y ′ = 2y/x ,b) 2x − y − xy ′ = 0, c) y ′′ = 0, d) y − x(y ′ − y ′′) − y ′′ = 0, e) y − (x/ ln y ′) ey′ = 0.4. a) x2(z′−2xy) = z/x , b) z′x− z ln z = 0, c) 2xz′ = z, d) 2xz′ = z, e) 2xz′− z2 = −1,f) z′ = 2z/x . 5. a) y = k e−x , b) y = 1

2 (sin x−cos x), c) y = 1+k e−x , d) y = 2x−k e−x ,e) y = 1

2 (x − 1) sin x + 12 x cos x , f) y = 1

4 e3x , g) y = ex + x e−x + 1. 6. a) y = 3x + c,b) y = 1, c) y = k cos x , k ∈ R \ {0}, d) y = tan(arctan x + π/4), e) y = c sin x ,f) y = (x(ln x−1)+1)2. 7. a) y = x+ x/( 1

2 ln |x |+ c); c) y = x sin(ln |cx |) c ∈ R\ {0},d) y = −x ln(c− ln |x |), e) y2 = x2, f) x2 + 2xy − y2 = c, h) y = cx2 − x c ∈ R \ {0},i) (y/x−1) ey/x = x , j) y e6y6/x6 = cx2. 8. a) 4

10 (y− x)− 12 − 3

5 ln |20y+30x−22| = c,

b) y = 12

√(x − 3)5 + c

√x − 3− 2. 9. a) y = c e−x2 + 2, b) y = c/x , c) c esin y − 2(1+

sin y) = x , d) y = c e−2√

x3/3 + 3√

x3 − 9/2, e) y = (x2 + c) ex2, f) y = k/x + x2.

10. a) x = c12 e−4t + c2 e−7t + 740 et + 2

27 e2t , y = c1 e−4t + c2 e−7t + 140 et + 7

27 e2t ,c) x = c12 cos 2t+c2 sin 2t , y = c1(− cos 2t−2 sin 2t)+c2(2 cos 2t−sin 2t). 11. n) y =c1+c2 ex+c3 e12x , o) y = c1+c2x+c3 e

√2x+c4 e−

√2x , p) y = c1 ex+c2x ex+c3x2 ex .

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References

[1] K. Rektorys a spol.: Prehled uzite matematiky, SNTL, Praha 1968.

[2] V. Jarnık: Diferencialnı pocet I,II, CSAV, Praha 1963.

[3] V. Jarnık: Integralnı pocet I,II, CSAV, Praha 1963.

[4] W. Rudin: Analyza v realnem a komplexnım oboru, Academia, Praha 1997.

[5] M. Spivak: Matematiceskij analyz na mnogoobrazijach, Mir, Moskva 1968.

[6] J. Kurzweil: Obycejne diferencialnı rovnice, SNTL, Praha 1978.

[7] M. Gregus, M. Svec, V. Seda: Obycajne diferencialne rovnice, Alfa-SNTL, Bratisla-va 1985.

[8] D. Krupka, Uvod do analyzy na varietach, SPN, Praha 1986.

[9] V. Arnold Obyknovennyje differencialnyje uravnenija, Nauka, Moskva 1971.

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226

Notations

[A]. . . 136≻ . . . 12∼ . . . 13≈. . . 39‖a‖. . . 9∂A. . . 124(, θ). . . 145;. . . 152;;. . . 153f . . . 22x . . . 130An(A). . . 147Br (x). . . 10BdA. . . 13BS. . . 38◦Br(x). . . 10

e1, . . . , en . . . 109C. . . 145ch. . . 156comp. . . 54CR(A). . . 147Cube Rn . . . 83△. . . 62Dh f (x). . . 24detB A. . . 136degω,deg p. . . 117,62

DifC. . . 146Difn . . . 52div F . . . 143discontQ f . . . 91dxi . . . 118EV. . . 48Fix f . . . 38grad f (x). . . 31gr f . . . 76I k . . . 123Ia(t). . . 165Imz. . . 145Inj(X). . . 78IntA . . . 84Iso(X,Y ). . . 39J f (x). . . 30JMeas. . . 95L (X,Y ). . . 17Lsym(

n X; Y ). . . 60lin A. . . 28LipA k. . . 37Lk(X). . . 109Lk

sym(X). . . 109LP f . . . 84Locmin f . . . 693k(X). . . 112Mfk(Rn). . . 129

Mfk∂(R

n). . . 135NS. . . 9Or Mfk . . . 137or X . . . 136Rn(X,Y ). . . 62Pn(X,Y ). . . 63Na A. . . 74Null. . . 87Rez. . . 145resa f . . . 157rot F . . . 143Sk . . . 137sgnσ . . . 113sh. . . 156Sol(a,b)(∗). . . 162Sol(∗)t0,x0 . . . 163span A. . . 28supp f . . . 140Sym. . . 59Tx M . . . 131tsa f . . . 155UP f . . . 84vol A. . . 83�k(A). . . 117�A f . . . 86ωx f . . . 89

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