Mollifiers. From R. Showalter’s book on PDEs.

(i) Mollifiers are C∞0 approximations of the delta-function.

(ii) Mollification is an operator of convolution with a mollifier:

fε(x) =∫Rnf(x− y)ϕε(y) dy , x ∈ Rn .

(iii) Mollification is a smoothening operator: fε(x) ∈ C∞(Rn).

(iv) Mollification is a bounded linear operator with norm ≤ 1.

(v) Any function in Lp(G), p 6= ∞ or C0(G) can be approximated by itsmollification. Hence C∞0 (G) is dense in Lp(G) and C0(G).

We shall begin with some elementary results concerning the approxima-tion of functions by very smooth functions. The “strong inclusion” K ⊂⊂ Gbetween subsets of Euclidean space Rn means K is compact, G is open,and K ⊂ G. If A and B are sets, their Cartesian product is given byA × B = [a, b] : a ∈ A, b ∈ B. If A and B are subsets of Kn (or anyother vector space) their set sum is A + B = a + b : a ∈ A, b ∈ B. If Gis not open, this definition can be extended if there exists and open set Osuch that O = G, by saying that K ⊂⊂ G if K ⊂⊂ O. For each ε > 0, letϕε ∈ C∞0 (Rn) be given with the properties

ϕε ≥ 0 , supp(ϕε) ⊂ x ∈ Rn : |x| ≤ ε ,

∫ϕε = 1 .

Such functions are called mollifiers and can be constructed, for example, bytaking an appropriate multiple of

ψε(x) =

exp(|x|2 − ε2)−1 , |x| < ε ,0 , |x| ≥ ε .

Let f ∈ L1(G), where G is open in Rn, and suppose that the support of fsatisfies supp(f) ⊂⊂ G. Then the distance from supp(f) to ∂G is a positivenumber δ. We extend f as zero on the complement of G and denote theextension in L1(Rn) also by f . Define for each ε > 0 the mollified function

fε(x) =∫Rnf(x− y)ϕε(y) dy , x ∈ Rn . (0.1)

1

Lemma 0.1 For each ε > 0, supp(fε) ⊂ supp(f) + y : |y| ≤ ε andfε ∈ C∞(Rn).

Proof : The second result follows from Leibniz’ rule and the representation

fε(x) =∫f(s)ϕε(x− s) ds .

The first follows from the observation that fε(x) 6= 0 only if x ∈ supp(f)+y : |y| ≤ ε. Since supp(f) is closed and y : |y| ≤ ε is compact, it followsthat the indicated set sum is closed and, hence, contains supp(fε).

Lemma 0.2 If f ∈ C0(G), then fε → f uniformly on G. If f ∈ Lp(G),1 ≤ p <∞, then ‖fε‖Lp(G) ≤ ‖f‖Lp(G) and fε → f in Lp(G).

Proof : The first result follows from the estimate

|fε(x)− f(x)| ≤∫|f(x− y)− f(x)|ϕε(y) dy

≤ sup|f(x− y)− f(x)| : x ∈ supp(f) , |y| ≤ ε

and the uniform continuity of f on its support. For the case p = 1 we obtain

‖fε‖L1(G) ≤∫ ∫|f(x− y)|ϕε(y) dy dx =

∫ϕε ·

∫|f |

by Fubini’s theorem, since∫|f(x − y)| dx =

∫|f | for each y ∈ Rn and this

gives the desired estimate. If p = 2 we have for each ψ ∈ C0(G)∣∣∣ ∫ fε(x)ψ(x) dx∣∣∣ ≤ ∫ ∫ |f(x− y)ψ(x)| dx ϕε(y) dy

≤∫‖f‖L2(G)‖ψ‖L2(G)ϕε(y) dy = ‖f‖L2(G)‖ψ‖L2(G)

by computations similar to the above, and the result follows since C0(G) isdense in L2(G). (the result for p 6= 1 or 2 is proved as above but using theHolder inequality in place of Cauchy-Schwartz.)

Finally we verify the claim of convergence in Lp(G). If η > 0 we have ag ∈ C0(G) with ‖f − g‖Lp ≤ η/3. The above shows ‖fε − gε‖Lp ≤ η/3 andwe obtain

‖fε − f‖Lp ≤ ‖fε − gε‖Lp + ‖gε − g‖Lp + ‖g − f‖Lp≤ 2η/3 + ‖gε − g‖Lp .

2

For ε sufficiently small, the support of gε−g is bounded (uniformly) andgε → g uniformly, so the last term converges to zero as ε→ 0.

The preceding results imply the following.

Theorem 0.3 C∞0 (G) is dense in Lp(G).

Theorem 0.4 For every K ⊂⊂ G there is a ϕ ∈ C∞0 (G) such that 0 ≤ϕ(x) ≤ 1, x ∈ G, and ϕ(x) = 1 for all x in some neighborhood of K.

Proof : Let δ be the distance from K to ∂G and 0 < ε < ε + ε′ < δ. Letf(x) = 1 if dist(x,K) ≤ ε′ and f(x) = 0 otherwise. Then fε has its supportwithin x : dist(x,K) ≤ ε+ ε′ and it equals 1 on x : dist(x,K) ≤ ε′ − ε,so the result follows if ε < ε′.

3

Distributions. From R. Showalter’s book on PDEs.

(i) Distributions are linear functionals on C∞0 (G).

(ii) For a ‘nice’ function distributions are defined as integrals Tf (φ) =∫G fφ

(iii) A fundamental example of a distribution is the delta-function.

(iv) Distributions are not bounded functionals, but they are closed func-tionals.

(v) A derivative of a distribution is a distribution, well-defined by integra-tion by parts.

(vi) A derivative of a distribution differs from from a regular derivative.

Recall

Cm(G) = f ∈ C(G) : Dαf ∈ C(G) for all |α| ≤ m,

C∞(G) = f ∈ C(G) : Dαf ∈ C(G) for all α,

C∞0 (G) = f ∈ C∞(G) : supp(f) is compact.

All of them are linear spaces. Cm(G) is also a Banach space (check) withthe norm

||f ||Cm(G) = maxα||Dαf ||C(G), ||f ||C(G) = sup

x∈G|f(x)|.

C∞(G) is not a Banach space, but it is a complete metric space with themetric

ρ(f, g) =∞∑n=1

12n

||f − g||Cn(G)

1 + ||f − g||Cn(G).

C∞(G) is not even metrizable, but there is a (nonconstructive) ‘locally com-pact topological vector space’ topology induced by the metric like C∞(G),that makes it complete.

4

0.1

A distribution on G is defined to be a conjugate-linear functional on C∞0 (G).That is, C∞0 (G)∗ is the linear space of distributions on G, and we also denoteit by D∗(G).

Example. The space L1loc(G) = ∩L1(K) : K ⊂⊂ G of locally integrable

functions on G can be identified with a subspace of distributions on G asin the Example of I.1.5. That is, f ∈ L1

loc(G) is assigned the distributionTf ∈ C∞0 (G)∗ defined by

Tf (ϕ) =∫Gfϕ , ϕ ∈ C∞0 (G) , (0.2)

where the Lebesgue integral (over the support of ϕ) is used. Theorem 0.3shows that T : L1

loc(G)→ C∞0 (G)∗ is an injection. In particular, the (equiv-alence classes of) functions in either of L1(G) or L2(G) will be identifiedwith a subspace of D∗(G).

0.2

We shall define the derivative of a distribution in such a way that it agreeswith the usual notion of derivative on those distributions which arise fromcontinuously differentiable functions. That is, we want to define ∂α : D∗(G)→D∗(G) so that

∂α(Tf ) = TDαf , |α| ≤ m , f ∈ Cm(G) .

But a computation with integration-by-parts gives

TDαf (ϕ) = (−1)|α|Tf (Dαϕ) , ϕ ∈ C∞0 (G) ,

and this identity suggests the following.

Definition. The αth partial derivative of the distribution T is the distri-bution ∂αT defined by

∂αT (ϕ) = (−1)|α|T (Dαϕ) , ϕ ∈ C∞0 (G) . (0.3)

Since Dα ∈ L(C∞0 (G), C∞0 (G)), it follows that ∂αT is linear. Every distri-bution has derivatives of all orders and so also then does every function,e.g., in L1

loc(G), when it is identified as a distribution. Furthermore, by thevery definition of the derivative ∂α it is clear that ∂α and Dα are compatiblewith the identification of C∞(G) in D∗(G).

5

0.3

We give some examples of distributions on R. Since we do not distinguishthe function f ∈ L1

loc(R) from the functional Tf , we have the identity

f(ϕ) =∫ ∞−∞

f(x)ϕ(x) dx , ϕ ∈ C∞0 (R) .

(a) If f ∈ C1(R), then

∂f(ϕ) = −f(Dϕ) = −∫f(Dϕ ) =

∫(Df)ϕ = Df(ϕ) , (0.4)

where the third equality follows by an integration-by-parts and all othersare definitions. Thus, ∂f = Df , which is no surprise since the definition ofderivative of distributions was rigged to make this so.

(b) Let the ramp and Heaviside functions be given respectively by

r(x) =x , x > 00 , x ≤ 0 , H(x) =

1 , x > 00 , x < 0 .

Then we have

∂r(ϕ) = −∫ ∞

0xDϕ(x) dx =

∫ ∞−∞

H(x)ϕ(x) dx = H(ϕ) , ϕ ∈ C∞0 (G) ,

so we have ∂r = H, although Dr(0) does not exist.(c) The derivative of the non-continuous H is given by

∂H(ϕ) = −∫ ∞

0Dϕ = ϕ(0) = δ(ϕ) , ϕ ∈ C∞0 (G) ;

that is, ∂H = δ, the Dirac functional. Also, it follows directly from thedefinition of derivative that

∂mδ(ϕ) = (−1)m(Dmϕ)(0) , m ≥ 1 .

(d) Letting A(x) = |x| and I(x) = x, x ∈ R, we observe that A = 2r− Iand then from above obtain by linearity

∂A = 2H − 1 , ∂2A = 2δ . (0.5)

Of course, these could be computed directly from definitions.

6

(e) For our final example, let f : R → K satisfy f |R− ∈ C∞(−∞, 0],

f |R

+ ∈ C∞[0,∞), and denote the jump in the various derivatives at 0 by

σm(f) = Dmf(0+)−Dmf(0−) , m ≥ 0 .

Then we obtain

∂f(ϕ) = −∫ ∞

0f(Dϕ)−

∫ 0

−∞f(Dϕ) (0.6)

=∫ ∞

0(Df)ϕ+ f(0+)ϕ(0) +

∫ 0

−∞(Df)ϕ− f(0−)ϕ(0)

= Df(ϕ) + σ0(f)δ(ϕ) , ϕ ∈ C∞0 (G) .

That is, ∂f = Df + σ0(f)δ, and the derivatives of higher order can becomputed from this formula, e.g.,

∂2f = D2f + σ1(f)δ + σ0(f)∂δ ,∂3f = D3f + σ2(f)δ + σ1(f)∂δ + σ0(f)∂2δ .

For example, we have

∂(H · sin) = H · cos ,∂(H · cos) = −H · sin +δ ,

so H · sin is a solution (generalized) of the ordinary differential equation

(∂2 + 1)y = δ .

0.4

Before discussing further the interplay between ∂ and D we remark that toclaim a distribution T is “constant” on R, means that there is a numberc ∈ K such that T = Tc, i.e., T arises from the locally integrable functionwhose value everywhere is c:

T (ϕ) = c

∫R

ϕ , ϕ ∈ C∞0 (R) .

Hence a distribution is constant if and only if it depends only on the meanvalue of each ϕ. This observation is the key to the proof of our next result.

7

Theorem 0.5 (a) If S is a distribution on R, then there exists anotherdistribution T such that ∂T = S.

(b) If T1 and T2 are distributions on R with ∂T1 = ∂T2, then T1 − T2 isconstant.

Proof : First note that ∂T = S if and only if

T (ψ′) = −S(ψ) , ψ ∈ C∞0 (R) .

This suggests we consider H = ψ′ : ψ ∈ C∞0 (R). H is a subspace ofC∞0 (R). Furthermore, if ζ ∈ C∞0 (R), it follows that ζ ∈ H if and only if∫ζ = 0. In that case we have ζ = ψ′, where

ψ(x) =∫ x

−∞ζ , x ∈ R .

Thus H = ζ ∈ C∞0 (R) :∫ζ = 0 and this equality shows H is the kernel of

the functional ϕ 7→∫ϕ on C∞0 (R). (This implies H is a hyperplane, but we

shall prove this directly.)Choose ϕ0 ∈ C∞0 (R) with mean value unity:∫

R

ϕ0 = 1 .

We shall show C∞0 (R) = H⊕K ·ϕ0, that is, each ϕ can be written in exactlyone way as the sum of a ζ ∈ H and a constant multiple of ϕ0. To checkthe uniqueness of such a sum, let ζ1 + c1ϕ0 = ζ2 + c2ϕ0 with the ζ1, ζ2 ∈ H.Integrating both sides gives c1 = c2 and, hence, ζ1 = ζ2. To verify theexistence of such a representation, for each ϕ ∈ C∞0 (G) choose c =

∫ϕ and

define ζ = ϕ− cϕ0. Then ζ ∈ H follows easily and we are done.To finish the proof of (a), it suffices by our remark above to define T on

H, for then we can extend it to all of C∞0 (R) by linearity after choosing,e.g., Tϕ0 = 0. But for ζ ∈ H we can define

T (ζ) = −S(ψ) , ψ(x) =∫ x

−∞ζ ,

since ψ ∈ C∞0 (R) when ζ ∈ H.Finally, (b) follows by linearity and the observation that ∂T = 0 if and

only if T vanishes on H. But then we have

T (ϕ) = T (cϕ0 + ζ) = T (ϕ0)c = T (ϕ0)∫ϕ

and this says T is the constant T (ϕ0) ∈ K.

8

Theorem 0.6 If f : R → R is absolutely continuous, then g = Df definesg(x) for almost every x ∈ R, g ∈ L1

loc(R), and ∂f = g in D∗(R). Conversely,if T is a distribution on R with ∂T ∈ L1

loc(R), then T (= Tf ) = f for someabsolutely continuous f , and ∂T = Df .

Proof : With f and g as indicated, we have f(x) =∫ x

0 g + f(0). Then anintegration by parts shows that∫

f(Dϕ) = −∫gϕ , ϕ ∈ C∞0 (R) ,

so ∂f = g. (This is a trivial extension of (0.4).) Conversely, let g = ∂T ∈L1

loc(R) and define h(x) =∫ x

0 g, x ∈ R. Then h is absolutely continuous andfrom the above we have ∂h = g. But ∂(T − h) = 0, so Theorem 0.5 impliesthat T = h+ c for some constant c ∈ K, and we have the desired result withf(x) = h(x) + c, x ∈ R.

0.5

Finally, we give some examples of distributions on Rn and their derivatives.(a) If f ∈ Cm(Rn) and |α| ≤ m, we have

∂αf(ϕ) = (−1)|α|∫RnfDαϕ =

∫RnDαf · ϕ = (Dαf)(ϕ) , ϕ ∈ C∞0 (Rn) .

(The first and last equalities follow from definitions, and the middle one is acomputation.) Thus ∂αf = Dαf essentially because of our definition of ∂α.

(b) Let

r(x) =x1x2 . . . xn , if all xj ≥ 0 ,0 , otherwise.

Then

∂1r(ϕ) = −r(D1ϕ) = −∫ ∞

0. . .

∫ ∞0

(x1 . . . xn)D1ϕdx1 . . . dxn

=∫ ∞

0. . .

∫ ∞0

x2 . . . xn ϕ(x) dx1 . . . dxn .

Similarly,

∂2∂1r(ϕ) =∫ ∞

0. . .

∫ ∞0

x3 . . . xn ϕ(x) dx ,

and∂(1,1,...,1)r(ϕ) =

∫RnH(x)ϕ(x) dx = H(ϕ) ,

9

where H is the Heaviside function (= functional)

H(x) =

1 , if all xj ≥ 0 ,0 , otherwise.

(c) The derivatives of the Heaviside functional will appear as distribu-tions given by integrals over subspaces of Rn. In particular, we have

∂1H(ϕ) = −∫ ∞

0. . .

∫ ∞0

D1ϕ(x) dx =∫ ∞

0. . .

∫ ∞0

ϕ(0, x2, . . . , xn) dx2 . . . dxn ,

a distribution whose value is determined by the restriction of ϕ to 0×Rn−1,

∂2∂1H(ϕ) =∫ ∞

0. . .

∫ ∞0

ϕ(0, 0, x3, . . . , xn) dx3 . . . dxn ,

a distribution whose value is determined by the restriction of ϕ to 0 ×0 × Rn−2, and, similarly,

∂(1,1,...,1)H(ϕ) = ϕ(0) = δ(ϕ) ,

where δ is the Dirac functional which evaluates at the origin.(d) Let S be an (n − 1)-dimensional C1 manifold in R

n and supposef ∈ C∞(Rn ∼ S) with f having at each point of S a limit from each side ofS. For each j, 1 ≤ j ≤ n, we denote by σj(f) the jump in f at the surfaceS in the direction of increasing xj . (Note that σj(f) is then a function onS.) Then we have

∂1f(ϕ) = −f(D1ϕ) = −∫Rnf(x)D1ϕ(x) dx

=∫Rn(D1f)(ϕ)(x) dx+

∫. . .

∫σ1(f)ϕ(s) dx2 . . . dxn

where s = s(x2, . . . , xn) is the point on S which (locally) projects onto(0, x2, . . . , xn). Recall that a surface integral over S is given by∫

SF ds =

∫AF · sec(θ1) dA

when S projects (injectively) onto a region A in 0 × Rn−1 and θ1 is theangle between the x1-axis and the unit normal ν to S. Thus we can writethe above as

∂1f(ϕ) = D1f(ϕ) +∫Sσ1(f) cos(θ1)ϕ dS .

10

However, in this representation it is clear that the integral is independent ofthe direction in which S is crossed, since both σ1(f) and cos(θ1) change signwhen the direction is reversed. We need only to check that σ1(f) is evalu-ated in the same direction as the normal ν = (cos(θ1), cos(θ2), . . . , cos(θn)).Finally, our assumption on f shows that σ1(f) = σ2(f) = · · · = σn(f), andwe denote this common value by σ(f) in the formulas

∂jf(ϕ) = (Djf)(ϕ) +∫Sσ(f) cos(θj)ϕ dS .

These generalize the formula (1.6).(e) SupposeG is an open, bounded and connected set in Rn whose bound-

ary ∂G is a C1 manifold of dimension n − 1. At each point s ∈ ∂G thereis a unit normal vector ν = (ν1, ν2, . . . , νn) whose components are directioncosines, i.e., νj = cos(θj), where θj is the angle between ν and the xj axis.Suppose f ∈ C∞(G) is given. Extend f to Rn by setting f(x) = 0 for x /∈ G.In C∞0 (Rn)∗ we have by Green’s second identity

( n∑j=1

∂2j f

)(ϕ) =

∫Gf

( n∑j=1

D2j ϕ

)=∫G

n∑j=1

(D2j f)ϕ

+∫∂G

(f∂ϕ

∂ν− ϕ∂f

∂ν

)dS , ϕ ∈ C∞0 (Rn) ,

so the indicated distribution differs from the pointwise derivative by thefunctional

ϕ 7→∫∂G

(f∂ϕ

∂ν− ϕ∂f

∂ν

)dS ,

where ∂f∂ν = ∇f · ν is the indicated (directional) normal derivative and

∇f = (∂1f, ∂2f, . . . , ∂nf) denotes the gradient of f . Hereafter we shall alsolet

∆n =n∑j=1

∂2j

denote the Laplace differential operator in D∗(Rn).

11

Sobolev spaces. From R. Showalter’s book on PDEs.

(i) Sobolev spaces Wm,p(G) are completions of Cm(G) with respect toLp-norms.

(ii) Sobolev spaces Wm,p0 (G) are completions of Cm0 (G) with respect to

Lp-norms.

(ii) Negative Sobolev spacesW−m,q(G) are bounded distributions onWm,p0 (G),

p <∞.

0.6

Hilbert space case: p = 2. Let G be an open set in Rn and m ≥ 0

an integer. Recall that Cm(G) is the linear space of restrictions to G offunctions in Cm0 (Rn). On Cm(G) we define a scalar product by

(f, g)Hm(G) =∑∫

GDαf ·Dαg : |α| ≤ m

and denote the corresponding norm by ‖f‖Hm(G).

Define Hm(G) to be the completion of the linear space Cm(G) with thenorm ‖ · ‖Hm(G). Hm(G) is a Hilbert space which is important for much ofour following work on boundary value problems. We note that the H0(G)norm and L2(G) norm coincide on C(G), and that we have the inclusions

C0(G) ⊂ C(G) ⊂ L2(G) .

Since we have identified L2(G) as the completion of C0(G) it follows thatwe must likewise identify H0(G) with L2(G). Thus f ∈ H0(G) if and only ifthere is a sequence fn in C(G) (or C0(G)) which is Cauchy in the L2(G)norm and fn → f in that norm.

Let m ≥ 1 and f ∈ Hm(G). Then there is a sequence fn in Cm(G)such that fn → f in Hm(G), hence Dαfn is Cauchy in L2(G) for eachmulti-index α of order ≤ m. For each such α, there is a unique gα ∈ L2(G)such that Dαfn → gα in L2(G). As indicated above, f is the limit of fn,so, f = gθ, θ = (0, 0, . . . , 0) ∈ Rn. Furthermore, if |α| ≤ m we have from anintegration-by-parts

(Dαfn, ϕ)L2(G) = (−1)|α|(fn, Dαϕ)L2(G) , ϕ ∈ C∞0 (G) .

12

Taking the limit as n→∞, we obtain

(gα, ϕ)L2(G) = (−1)|α|(f,Dαϕ)L2(G) , ϕ ∈ C∞0 (G) ,

so gα = ∂αf . That is, each gα ∈ L2(G) is uniquely determined as the αth

partial derivative of f in the sense of distribution on G. These remarksprove the following characterization.

Theorem 0.7 Let G be open in Rn and m ≥ 0. Then f ∈ Hm(G) if and

only if there is a sequence fn in Cm(G) such that, for each α with |α| ≤ m,the sequence Dαfn is L2(G)-Cauchy and fn → f in L2(G). In that casewe have Dαfn → ∂αf in L2(G).

Corollary Hm(G) ⊂ Hk(G) ⊂ L2(G) when m ≥ k ≥ 0, and if f ∈ Hm(G)then ∂αf ∈ L2(G) for all α with |α| ≤ m.

We shall later find that f ∈ Hm(G) if ∂αf ∈ L2(G) for all α with |α| ≤ mCase 1 ≤ p ≤ ∞. When p 6= 2 then the norm of the Sobolev spaces is

not induced by a scalar product, the construction is based on completion ofCm(G) functions and its derivatives in Lp(G)-space. If we use

(f, ϕ)L2(G) = Tf (ϕ) =∫Gfϕdx

when ϕ ∈ C∞(G), then all the computations are literally the same as in thecase of Hm(G) Similar to Hm(G) case, f ∈ Wm,p(G) if and only if there isa sequence fn in Cm(G) such that, for each α with |α| ≤ m, the sequenceDαfn is Lp(G)-Cauchy and fn → f in Lp(G). The Wm,p-norm is definedas: [

‖f‖Wm,p(G)

]p=

∑|α|≤m

∫G|Dαf |p : |α| ≤ m

0.7

Case p = 2. The case p 6= 2 is identical unless mentioned explicitly. Wedefine Hm

0 (G) to be the closure in Hm(G) of C∞0 (G). Generally, Hm0 (G) is

a proper subspace of Hm(G). Note that for any f ∈ Hm(G) we have

(∂αf, ϕ)L2(G) = (−1)|α|(f,Dαϕ)L2(G) , |α| ≤ m , ϕ ∈ C∞0 (G) .

We can extend this result by continuity to obtain the generalized integration-by-parts formula

(∂αf, g)L2(G) = (−1)|α|(f, ∂αg)L2(G) , f ∈ Hm(G) , g ∈ Hm0 (G) , |α| ≤ m .

13

This formula suggests that Hm0 (G) consists of functions in Hm(G) which

vanish on ∂G together with their derivatives through order m− 1. We shallmake this precise in the following.

Since C∞0 (G) is dense in Hm0 (G), each element of Hm

0 (G)′ determines(by restriction to C∞0 (G)) a distribution on G and this correspondence is aninjection. Thus we can identify Hm

0 (G)′ with a space of distributions on G,and those distributions are characterized as follows.

Theorem 0.8 Hm0 (G)′ is (identified with) the space of distributions on G

which are the linear span of the set

∂αf : |α| ≤ m , f ∈ L2(G) .

Proof : If f ∈ L2(G) and |α| ≤ m, then

|∂αf(ϕ)| ≤ ‖f‖L2(G)‖ϕ‖Hm0 (G) , ϕ ∈ C∞0 (G) ,

so ∂αf has a (unique) continuous extension to Hm0 (G). Conversely, if T ∈

Hm0 (G)′, there is an h ∈ Hm

0 (G) such that

T (ϕ) = (h, ϕ)Hm(G) , ϕ ∈ C∞0 (G) .

But this implies T =∑|α|≤m(−1)|α|∂α(∂αh) and, hence, the desired result,

since each ∂αh ∈ L2(G).

Remark. The space H−m(G)′ is identified by definition with Hm0 (G).

What happens when p 6= 2? One in general has be very careful with theL∞-L1 duality, because

(i) The dual of L1(G) is L∞(G), but the dual of L∞(G) is not L1(G).

(ii) C∞0 is not dense in L∞(G), moreover L∞(G) is not a separable space.

Hence, we define W−m,q(G) spaces as duals of Wm,p0 (G) only when p 6=∞.

Another question is why we don’t define W−m,q(G) as a dual of Wm,p(G)?The reason is that C∞0 is not dense in Wm,p(G), therefore Wm,p(G)′ maycontain objects which are not distributions!

Examples.a. Operator div p for p ∈ L2(Ω) as a bounded linear functional in

H−1(Ω)

14

Since C∞0 (Ω) is dense in H10 (Ω)

div p(φ) = −∫

Ωp · ∇φdxφ ∈ C∞0 (Ω)

By Cauchy-Schwartz this function is dominated by the sublinear function

||p||L2 ||φ||H1

Hence we can uniquely extend it to the whole H10 (Ω) by Hahn-Banach the-

orem with||div p||H−1 = sup

||φ||H1=1

∫Ωp · ∇φdx ≤ ||p||L2

b. Similarly operator curl v for p ∈ L2(Ω) as a bounded linear functionalin H−1(Ω)

(curl v(φ))ij = −∫

Ω

(vi∂φ

∂xj− vj

∂φ

∂xi

)dx

We shall have occasion to use the two following results, each of whichsuggests further that Hm

0 (G) is distinguished from Hm(G) by boundaryvalues.

Theorem 0.9 Hm0 (Rn) = Hm(Rn). (Note that the boundary of Rn is empty.)

Proof : Let τ ∈ C∞0 (Rn) with τ(x) = 1 when |x| ≤ 1, τ(x) = 0 when|x| ≥ 2, and 0 ≤ τ(x) ≤ 1 for all x ∈ Rn. For each integer k ≥ 1, defineτk(x) = τ(x/k), x ∈ Rn. Then for any u ∈ Hm(Rn) we have τk ·u ∈ Hm

0 (Rn)and (exercise) τk · u → u in Hm(Rn) as k → ∞. Thus we may assumeu has compact support. Letting G denote a sphere in R

n which containsthe support of u, we have from Lemma 0.2 of Section 1.1 that the mollifiedfunctions uε → u in L2(G) and that (Dαu)ε = Dα(uε)→ ∂αu in L2(G) foreach α with |α| ≤ m. That is, uε ∈ C∞0 (Rn) and uε → u in Hm(Rn).

Theorem 0.10 Suppose G is an open set in Rn with sup|x1| : (x1, x2, . . . , xn)∈ G = K <∞. Then

‖ϕ‖L2(G) ≤ 2K‖∂1ϕ‖L2(G) , ϕ ∈ H10 (G) .

15

Proof : We may assume ϕ ∈ C∞0 (G), since this set is dense in H10 (G). Then

integrating the identity

D1(x1 · |ϕ(x)|2) = |ϕ(x)|2 + x1 ·D1(|ϕ(x)|2)

over G by the divergence theorem gives∫G|ϕ(x)|2 = −

∫Gx1(D1ϕ(x) · ϕ(x) + ϕ(x) ·D1ϕ(x)) dx .

The right side is bounded by 2K‖D1ϕ‖L2(G)‖ϕ‖L2(G), and this gives theresult. In the case p 6= 2 use

D1(x1 · |ϕ(x)|p) = |ϕ(x)|p + x1 ·D1(|ϕ(x)|p).

Since|ϕ(x)|p = ϕ(x)ϕ(x)p−1 = ϕ(x)ϕ(x)p−1,

in the divergence theorem we’ll use

D1(|ϕ(x)|p) = pϕ(x)p−1 ·D1ϕ(x)

and by Holder with q = p/(p− 1)∫G|ϕ(x)|p−1|D1ϕ(x)| ≤

[ ∫G|ϕ(x)|p

](p−1)/p[ ∫G|D1ϕ(x)|p]1/p

=[||ϕ(x)||Lp

]p−1||D1ϕ(x)||Lp

we get [||ϕ(x)||Lp

]p≤ 2K

[||ϕ(x)||Lp

]p−1||D1ϕ(x)||Lp

or||ϕ(x)||Lp ≤ 2K||D1ϕ(x)||Lp .

16

Sobolev embedding theorem.

(i) A compact operator maps weakly convergent sequences to stronglyconvergent sequences

(ii) If fn converges weakly in Wm+1,p(G), then it converges strongly inWm,p(G).

Definition: Let V and W be two Banach spaces with V ⊂W . We saythe space V is continuously embedded in W and write

V →W

if||f ||W ≤ C||f ||V for any f ∈ V.

We say the space V is compactly embedded in W and write

V →→W

if it is continuously embedded in W and each bounded sequence in V has aconvergent subsequence in W . Equivalent in reflexive spaces: each weaklyconvergent sequence in V is convergent in W .

Remark: Sometimes it is useful to think about embeddings as linearoperators A : V →W .

Definition: Let V and W be two Banach spaces. Then the operatorA : V →W is continuous if

||Af ||W ≤ C||f ||V for any f ∈ V.

The operator A is compact if every bounded sequence fn has a weaklyconvergent subsequence Afn.

Example of compact operators: Integral operators

K(f) =∫Gk(x, y)f(y)dyx ∈ G

are compact under certain smooth conditions on the kernel k(x, y). Forexample, if |k(x, y)− k(x, y)| ≤ |x− x|+ |y − y|, then they are compact onLp 1 ≤ p ≤ ∞ (check!).

17

Sobolev embedding theoremSmoothness of the boundary, Lipschitz domain: On Lipschitz

domains the Sobolev embedding estimates hold. Basically all domains youencounter are Lipschitz.

Definition: A function is said to be (m,β)-Holder continuous f(x) ∈Cm,β(G), G is closed, 0 < β ≤ 1, if f(x) ∈ Cm(G) and there exists aconstant C such that for any for any x ∈ G

supx,x∈G

|Dαf(x)−Dαf(x)| ≤ C|x− x|β, |α| = m.

The Holder norm is defined by

||f ||Cm,β (G) = ||f ||Cm,β (G) +∑|α|=m

supx,x∈G

|Dαf(x)−Dαf(x)||x− x|β

If m = 0 then the function is said to be β-Holder continuous. If β = 1, thenthe function is said to be m-Lipschitz continuous. If m = 0, β = 1, then thefunction is said to be Lipschitz continuous.

Definition: Let G be open and bounded in Rn, and let V denote afunction space on Rn−1. We say ∂G is of class V if for each point x ∈ ∂G,there exist an r > 0 and a function g ∈ V such that upon a transformationof the coordinate system if necessary, we have

G ∩B(x, r) = x ∈ B(x, r)|xn > g(x1, . . . , xn−1).

Here, B(x, r) denotes the n-dimensional ball centered at x with radius r. Inparticular, when V consists of Lipschitz continuous functions, we say G is aLipschitz domain.

Examples All (curvilinear) polygons are Lipschitz domains; however adisk with a crack:

B(0, 1)− [0, 1)

is not a Lipschitz domain.Sobolev embedding theorem. G is an open bounded Lipschitz

domain in Rn and u ∈W j+m,p(G)

(i) mp < n. Find q, such that 1/q = 1/p−m/n , then

W j+m,p(G) →W j,q(G),

and for any 1 ≤ q∗ < q

W j+m,p(G) →→W j,q∗(G),

18

(ii) mp = n. Then for any 1 ≤ q∗ <∞

W j+m,p(G) →→W j,q∗(G),

(iii) mp > n. ThenW j+m,p(G) →→ Cj(G).

Remark: Holder space embedding. In part (iii) one can show thatif k = m− [n/p]− 1, β = [n/p] + 1− n/p. Then

W j+m,p(G) → Cj+k,β(G) if n/p 6= integer,

and for any 0 ≤ β∗ < β

W j+m,p(G) →→ Cj+k,β∗(G).

Remark In part (ii) by analogy q = ∞ should be a continuous em-bedding, however it is not for the standard reason - C∞0 is not dense inL∞.

We give here a proof for the part (i) under some additional assumptionsto simplify our task:

(i) We assume that j = 0, the result extends to the case j > 0 if we workwith all ∇αu, |α| = j instead of u.

(ii) We assume that our functions have compact support in G, that isu ∈ Wm,p

0 (G). This assumption can be easily removed, by the ex-tension operator construction - if the boundary is Lipschitz thenany function in u ∈ Wm,p(G) can be extended to a function (withslightly larger norm) in Ω with compact support, for some slightlylarger domain Ω, G ⊂ Ω.

(iii) We assume that G is rectangular domain, say a cube [0, 1]×[0, 1]×. . .×[0, 1]. This simplification saves us some estimates at the boundary.

(iv) We assume that k = 1, the result for k > 1 follows from the result fork = 1, the case of W 1,p

0 (G) by iteration:

Lq →W 1,q1 →W 2,q2

where1q

=1q1− 1n,

1q1

=1q2− 1n,

19

and hence1q

=1q2− 2n

The proof relies on four main observations:

(i) Sobolev inequality: If 1 ≤ p < n and q = npn−p , then there is a constant

C = C(n, p) such that

||u||Lq ≤ C||∇u||Lp , u ∈ C10 (Rn) (0.7)

and since C10 (G) is dense in W 1,p

0 (G) inequality (0.7) holds for anyu ∈W 1,p

0 (G), the continuous embedding follows immediately.

(ii) Interpolation inequality: for any 1 ≤ q∗ < q, there exists 0 < λ ≤ 1(note: λ is strictly positive) such that

||u||Lq∗ ≤ ||u||λL1 ||u||1−λLq . (0.8)

The interpolation inequality implies that it is sufficient to show strongconvergence for L1 and use continuous embedding result (0.7).

(iii) If a domain G is subdivided into a finite number of smaller domains Qδ,then weak convergence of un (f, un−u)→ 0 implies strong convergenceof averages

|〈un〉(x)Qδ − 〈u〉Qδ | → 0 (0.9)

Where we denote 〈u〉Qδ the average of u over a domain Qδ.

〈u〉Qδ =1

Vol(Qδ)

∫Qδ

udx

Let 〈u〉(x) = 〈u〉Qδ if x ∈ Qδ. Note that 〈u〉(x) takes only finitelymany values - one for each Qδ, therefore

||〈un〉(x)− 〈u〉(x)||L1(G) ≤ Vol(G) maxx∈G|〈un〉(x)− 〈u〉(x)| → 0.

(iv) Poincare-type inequality (Rellich-Kondrachov) Assume that Qδ ⊂ Rnis a cube [0, δ]× [0, δ]× . . .× [0, δ]. If u ∈ C1(Q) then for any p ≥ 1

||u− 〈u〉Qδ ||L1(Qδ) ≤ δ||∇u||L1(Qδ). (0.10)

Hence||u− 〈u〉||L1(Q) ≤ δ||∇u||L1(Q) ≤ δC||∇u||Lp(Q).

20

Combining (iii) and (iv) we have

||u− un||L1(Q) ≤ Vol(G) maxx∈G|〈un〉(x)− 〈u〉(x)|

+δC||∇u||Lp + δC||∇un||Lp → 0.

Proof of Sobolev inequality (0.7): Generalized Holder inequality(by induction):

m∑i=1

1pi

= 1, fi ∈ Lpi(G)

then ∫Gf1f2 . . . fmdx ≤ ||f1||Lp1 ||f2||Lp2 . . . ||fm||Lpm

Case p = 1.

|u(x)| ≤ |∫ x

−∞Diu(x)dxi| ≤

∫R|Diu(x)|dxi

Hence

|u(x)|nn−1 ≤Π

n

i=1

[ ∫R|Diu(x)|dxi

] 1n−1 .

Integrate over Rn and use generalized Holder:∫Rn|u(x)|

nn−1dx ≤

∫Rn

Πn

i=1

[ ∫R|Diu(x)|dxi

] 1n−1dx

=∫Rn−1

[ ∫R

[ ∫R|D1u(x)|dx1

] 1n−1Π

n

i=2

[ ∫R|Diu(x)|dxi

] 1n−1dx1

]dx2 . . . dxn

=∫Rn−1

[ ∫R|D1u(x)|dx1

] 1n−1[ ∫

RΠ

n

i=2

[ ∫R|Diu(x)|dxi

] 1n−1dx1

]dx2 . . . dxn

(by generalized Holder with pi = pj = n− 1 )

≤∫Rn−1

[ ∫R|D1u(x)|dx1

] 1n−1Π

n

i=2

[ ∫R2|Diu(x)|dxidx1

] 1n−1dx2 . . . dxn

(similarly for other variables x2, . . . , xn)

. . . ≤Πn

i=1

[ ∫Rn|Diu(x)|dx

] 1n−1 =

[Π

n

i=1

∫Rn|Diu(x)|dx

] 1n−1

21

Since for nonnegative ai:

Πn

i=1ai ≤1n

( n∑i=1

ai)n,

we have ∫Rn|u(x)|n/(n−1)dx ≤

[ 1n

( n∑i=1

∫Rn|Diu(x)|dx

)n]1/(n−1)

= n−1/(n−1)( n∑i=1

∫Rn|Diu(x)|dx

)n/(n−1),

hence||u||Ln/(n−1) ≤ C||∇u||L1

Case p > 1. In general apply scaling argument- take the power α ‘properly’so that the of scaling uα and ∇(uα) is the same in corresponding norms:

||uα||Ln/(n−1) ≤ C||uα−1∇u||L1

≤ C||uα−1||Ls ||∇u||Lp , 1/s+ 1/p = 1.

Equating the norms of u we choose α

αn/(n− 1) = (α− 1)s

we haveq = αn/(n− 1)

||u||αLq ≤ C||u||α−1Lq ||∇u||Lp

Proof of interpolation inequality (0.8): Use Holder inequality:

||u||Lq∗ =[||uq∗||L1

]1/q∗=[||uλq∗u(1−λ)q∗||L1

]1/q∗≤[||uλq∗||Ls ||u(1−λ)q∗||Lt

]1/q∗,1s

+1t

= 1.

We want(λq∗)s = 1, ((1− λ)q∗)t = q,

1s

+1t

= 1, (0.11)

22

because then ||uλq∗||Ls = ||u||1/sL1 and ||u(1−λ)q∗||Lt = ||u||q/sLq . Such λ, t ands can be found from (0.11), and this gives

1q

(1− λ) + λ =1q∗

which implies 0 < λ ≤ 1. Combining all the terms

||u||Lq∗ ≤[||u||1/sL1 ||u||q/sLq

]1/q∗= ||u||1/sq∗L1 ||u||q/sq∗Lq = ||u||λL1 ||u||1−λLq .

Proof of strong convergence of averages (0.9): Subdivide G intocubes Qδ, where each

Qδ = Qδ(m1,m2, . . . ,mn) = [δm1, δ(m1 + 1)]× . . .× [δmn, δ(mn + 1)].

Then if ||uk||W 1,p ≤ C, by (0.7) ||uk||L1 ≤ C, then by the Alaoglu theo-rem, there is a subsequence (denoted hereafter by uk) which is weaklyconvergent in L1, this means that if we take the trial linear functional to beintegration over each Qδ we have |〈u〉Qδ − 〈un〉Qδ | → 0

Proof of Poincare type inequality (0.10): Since u ∈ C1(Qδ)) thereis a point x = (x1, . . . , xn) such that 〈u〉Qδ = u(x) then

u(x)− u(x) =n∑j=1

∫ xj

xj

Dju(x1, x2, . . . , xj−1, s, xj+1, . . . , xn)ds.

Therefore for any x ∈ Qδ

|u(x)− 〈u〉Qδ | ≤n∑j=1

∫ δ

0|Dju|dxj .

Note that each term on the right-hand side depends only on n−1 variables,therefore integrating with respect to all variables

||u− 〈u〉Qδ ||L1(Qδ) ≤ δn∑j=1

∫Qδ

|Dju|dx. ≤ δ||∇u||L1(Qδ).

23

Trace. From R. Showalter’s book on PDEs.

(i) Trace γ0 is a bounded linear functional γ0 : H1(G)→ L2(∂G).

(ii) u ∈ H10 (G) iff γ0(u) = 0.

(iii) Higher order traces a normal derivatives: “γj = ∂j

∂νj|∂G”.

We shall describe the sense in which functions in Hm(G) have “boundaryvalues” on ∂G when m ≥ 1. Note that this is impossible in L2(G) since ∂Gis a set of measure zero in Rn. First, we consider the situation where G isthe half-space Rn+ = (x1, x2, . . . , xn) : xn > 0, for then ∂G = (x′, 0) : x′ ∈Rn−1 is the simplest possible (without being trivial). Also, the general case

can be localized as in Section 2.3 to this case, and we shall use this in ourfinal discussion of this section.

0.8

We shall define the (first) trace operator γ0 when G = Rn+ = x = (x′, xn) :

x′ ∈ Rn−1, xn > 0, where we let x′ denote the (n−1)-tuple (x1, x2, . . . , xn−1).For any ϕ ∈ C1(G) and x′ ∈ Rn−1 we have

|ϕ(x′, 0)|2 = −∫ ∞

0Dn(|ϕ(x′, xn)|2) dxn .

Integrating this identity over Rn−1 gives

‖ϕ(·, 0)‖2L2(Rn−1) ≤∫Rn+

[(Dnϕ · ϕ+ ϕ ·Dnϕn )] dx

≤ 2‖Dnϕ‖L2(Rn+)‖ϕ‖L2(Rn+) .

The inequality 2ab ≤ a2 + b2 then gives us the estimate

‖ϕ(·, 0)‖2L2(Rn−1) ≤ ‖ϕ‖2L2(Rn+) + ‖Dnϕ‖2L2(Rn+) .

Since C1(Rn+) is dense in H1(Rn+), we have proved the essential part of thefollowing result.

Theorem 0.11 The trace function γ0 : C1(G)→ C0(∂G) defined by

γ0(ϕ)(x′) = ϕ(x′, 0) , ϕ ∈ C1(G) , x′ ∈ ∂G ,

(where G = Rn+) has a unique extension to a continuous linear operator

γ0 ∈ L(H1(G), L2(∂G)) whose range is dense in L2(∂G), and it satisfies

γ0(β · u) = γ0(β) · γ0(u) , β ∈ C1(G) , u ∈ H1(G) .

24

Proof : The first part follows from the preceding inequality and the Rietzrepresentation theorem for Hilbert spaces. If ψ ∈ C∞0 (Rn−1) and τ is thetruncation function defined in the proof of Theorem 0.9, then

ϕ(x) = ψ(x′)τ(xn) , x = (x′, xn) ∈ Rn+defines ϕ ∈ C1(G) and γ0(ϕ) = ψ. Thus the range of γ0 contains C∞0 (Rn−1).The last identity follows by the continuity of γ0 and the observation that itholds for u ∈ C1(G).

Theorem 0.12 Let u ∈ H1(Rn+). Then u ∈ H10 (Rn+) if and only if γ0(u) =

0.

Proof : If un is a sequence in C∞0 (Rn+) converging to u in H1(Rn+), thenγ0(u) = lim γ0(un) = 0 by Theorem 0.11.

Let u ∈ H1(Rn+) with γ0u = 0. If τj : j ≥ 1 denotes the sequence oftruncating functions defined in the proof of Theorem 0.9, then τju → u inH1(Rn+) and we have γ0(τju) = γ0(τj)γ0(u) = 0, so we may assume that uhas compact support in Rn.

Let θj ∈ C1(R+) be chosen such that θj(s) = 0 if 0 < s ≤ 1/j, θj(s) = 1if s ≥ 2/j, and 0 ≤ θ′j(s) ≤ 2j if (1/j) ≤ s ≤ (2/j). Then the extension ofx 7→ θj(xn)u(x′, xn) to all of Rn as 0 on Rn− is a function in H1(Rn) withsupport in x : xn ≥ 1/j, and (the proof of) Theorem 0.9 shows we mayapproximate such a function from C∞0 (Rn+). Hence, we need only to showthat θju→ u in H1(Rn+).

It is an easy consequence of the Lebesgue dominated convergence the-orem that θju → u in L2(Rn+) and for each k, 1 ≤ k ≤ n − 1, that∂k(θju) = θj(∂ku) → ∂ku in L2(Rn+) as j → ∞. Similarly, θj(∂nu) → ∂nuand we have ∂n(θju) = θj(∂nu) + θ′ju, so we need only to show that θ′ju→ 0in L2(Rn+) as j →∞.

Since γ0(u) = 0 we have u(x′, s) =∫ s

0 ∂nu(x′, t) dt for x′ ∈ Rn−1 ands ≥ 0. From this follows the estimate

|u(x′, s)|2 ≤ s∫ s

0|∂nu(x′, t)|2 dt .

Thus, we obtain for each x′ ∈ Rn−1∫ ∞0|θ′j(s)u(x′, s)|2 ds ≤

∫ 2/j

0(2j)2s

∫ s

0|∂nu(x′, t)|2 dt ds

≤ 8j∫ 2/j

0

∫ s

0|∂nu(x′, t)|2 dt ds .

25

Interchanging the order of integration gives∫ ∞0|θ′j(s)u(x′, s)|2 ds ≤ 8j

∫ 2/j

0

∫ 2/j

t|∂nu(x′, t)|2 ds dt

≤ 16∫ 2/j

0|∂nu(x′, t)|2 dt .

Integration of this inequality over Rn−1 gives us

‖θ′ju‖2L2(Rn+) ≤ 16∫Rn−1×[0,2/j]

|∂nu|2 dx

and this last term converges to zero as j →∞ since ∂nu is square-summable.

0.9

We can extend the preceding results to the case where G is a sufficientlysmooth region in Rn. Suppose G is given as in Section 2.3 and denote byGj : 0 ≤ j ≤ N, ϕj : 1 ≤ j ≤ N, and βj : 0 ≤ j ≤ N the opencover, corresponding local maps, and the partition-of-unity, respectively.Recalling the linear injections Λ and λ constructed in Section 2.3, we areled to consider function γ0 : H1(G)→ L2(∂G) defined by

γ0(u) =N∑j=1

(γ0((βju) ϕj)

) ϕ−1

j

=N∑j=1

γ0(βj) · (γ0(u ϕj)ϕ−1j )

where the equality follows from Theorem 0.11. This formula is preciselywhat is necessary in order that the following diagram commutes.

H1(G) Λ−→ H10 (G)×H1(Q+) × · · ·× H1(Q+)yγ0

y yγ0

L2(∂G) λ−→ L2(Q0) × · · ·× L2(Q0)

Also, if u ∈ C1(G) we see that γ0(u) is the restriction of u to ∂G. Theseremarks and Theorems 0.11 and 0.12 provide a proof of the following result.

26

Theorem 0.13 Let G be a bounded open set in Rn which lies on one sideof its boundary, ∂G, which we assume is a C1-manifold. Then there existsa unique continuous and linear function γ0 : H1(G) → L2(∂G) such thatfor each u ∈ C1(G), γ0(u) is the restriction of u to ∂G. The kernel of γ0 isH1

0 (G) and its range is dense in L2(∂G).

This result is a special case of the trace theorem which we briefly discuss.For a function u ∈ Cm(G) we define the various traces of normal derivativesgiven by

γj(u) =∂ju

∂νj

∣∣∣∣∣∂G

, 0 ≤ j ≤ m− 1 .

Here ν denotes the unit outward normal on the boundary of G. WhenG = R

n+ (or G is localized as above), these are given by ∂u/∂ν = −∂nu|xn=0.

Each γj can be extended by continuity to all of Hm(G) and we obtain thefollowing.

Theorem 0.14 Let G be an open bounded set in Rn which lies on one sideof its boundary, ∂G, which we assume is a Cm-manifold. Then there is aunique continuous linear function γ from Hm(G) into

∏m−1j=0 Hm−1−j(∂G)

such that

γ(u) = (γ0u, γ1u, . . . , γm−1(u)) , u ∈ Cm(G) .

The kernel of γ is Hm0 (G) and its range is dense in the indicated product.

The Sobolev spaces over ∂G which appear in Theorem 0.14 can bedefined locally. The range of the trace operator can be characterized bySobolev spaces of fractional order and then one obtains a space of bound-ary values which is isomorphic to the quotient space Hm(G)/Hm

0 (G). Suchcharacterizations are particularly useful when considering non-homogeneousboundary value problems and certain applications, but the preceding resultswill be sufficient for our needs.

27

Compact operators in Banach spaces.

(i) Spectrum is a generalization of the set of eigenvalues.

(ii) Spectrum is compact. Resolvent set is open.

(iii) For a compact operator and any ε > 0, there are finitely many eigen-values |λ| ≥ ε and their eigenspaces are finite-dimensional.

When X is a finite-dimensional linear space and

A : X → X

is linear, then the equation Af = g has a well-developed solvability theory interms of eigenvalues. We wish to extend these results to compact operators.We study here equations of the second kind:

λf +Af = g.

Let us generalize the concept of an eigenvalue for a bounded (not necessarilycompact) linear operator A on a Banach space X. The range (or image)

R(T ) = y ∈ X : ∃x ∈ X, y = Ax ⊂ X.

The nullspace (or kernel)

N(A) = x ∈ X : Ax = 0 ⊂ X.

If A is bounded then N(A) is closed (check!), however R(A) may not beclosed.Example. Suppose

f(x) ∈ X = f ∈ C[0, 1], f(0) = 0.

A : f(x)→∫ x

0f(t)dt

Then A is one-to-one, R(A) is dence in X but R(A) is not closed

R(A) = f ∈ C1[0, 1], f(0) = 0, f ′(0) = 0.

If Aλ = A − λI, λ ∈ C be invertible, then such λ is said to be in theresolvent set of T ρ(A) ⊂ A, and the inverse A−1

λ is called the resolventoperator. The spectrum of T σ(T ) is the complement of the resolventset σ(T ) = C− ρ(T ). By open mapping theorem λ ∈ σ(T ) if Aλ is not bothone-to-one and onto. The spectrum is subdivided into three parts:

28

(i) point spectrum

σp(A) = λ ∈ C : Aλ is not one-to-one,

If λ ∈ σp(A), then λ is called an eigenvalue of A.

(ii) continuous spectrum

σc(A) = λ ∈ C : Aλ is one-to-one and R(Aλ) is dense in X,

but A−1λ is not bounded,

(iii) residual spectrum

σr(A) = λ ∈ C : Aλ is one-to-one and R(Aλ) is not dense in X,

Proposition. The point, continuous, and residual spectra are disjoint andtheir union is σ(A).Proof: Suppose λ 6∈ σp(A) ∪ σc(A) ∪ σr(A). Then Aλ is one-to-one, hasdense range, and A−1

λ is bounded. We need to show that Aλ is onto, thenλ ∈ ρ(A). By the extension theorem (A−1

λ is bounded and R(Aλ) is densein X)

A−1λ : R(Aλ)→ X

can be extended toA−1λ : X → X.

Now for any y such that the sequence yn, yn ∈ R(Aλ) converges to y.Since A−1

λ is bounded, then xn, xn = A−1λ yn converges to x, and since Aλ

is continuousyn = Aλxn → Aλx.

Thus Aλx = y, and Aλ is onto.Example: Consider A : X → X where X = L2([0, 1]) or X = C([0, 1]).Define A(f(t)) = tf(t). We have Aλ(f(t)) = (t−λ)f(t) whence A−1

λ (f(t)) =1t−λf(t). The spectrum of A consists of all λ for which t − λ is zero, henceσ(A) = [0, 1]. For any λ Aλ is one-to-one, therefore σp = 0.Suppose X = L2([0, 1]). Then R(Aλ) is dense for all λ ∈ [0, 1]. Henceσ(A) = σc(A) = [0, 1].Suppose C([0, 1]). Then R(Aλ) is not dense for any λ ∈ [0, 1]. Henceσ(A) = σr(A) = [0, 1].Note: If λ

∫σc(A), this means that (A − λ)f = g is ill-posed. If λ

∫σr(A),

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this often means that (A−λ)f = g is ill-posed; however if R(Aλ) is closed inX, then the problem (A− λ)f = g is well-posed, but it requires ‘solvabilitycondition on the right-hand side’: g ∈ R(Aλ).Example: Let X = l2, define the averaging operator

A : (x1, x2, x3, . . .)→ (x1, (x1 + x2)/2, (x2 + x3)/2, . . .).

Let λ = 1

Aλ : (x1, x2, x3, . . .)→ (0, (x1 − x2)/2, (x2 − x3)/2, . . .),

1 6∈ σp(A) because then x1 = x2 = x3 = . . ., but (1, 1, 1, . . .) 6∈ l2.Proposition. For a bounded linear operator A, ρ(A) ∈ C is open, andσ(A) ∈ C is compact. Moreover σ(A) ⊂ BR, where R = ||A||.Proof: Let R = ||A||, For any λ, |λ| > R by the geometric series theorem

A−1λ (A− λI)−1 = − 1

λ(I − A

λ)−1 = − 1

λ

∞∑n=0

(A

λ)n

is a well-defined, bounded operator. Hence σ(A) ⊂ BR. If there exists A−1λ ,

let 0 < ε ≤ 1/||A−1λ ||, then (again by the geometric series theorem) for any

µ, |µ| < ε

A−1λ+µ = (Aλ − µI)−1 = A−1

λ (I − µA−1λ )−1 = A−1

λ

∞∑n=0

(µA−1λ )n

is a well-defined, bounded operator. Hence ρ(A) is open. Therefore σ(A) isclosed and bounded, therefore it is compact.Definition. An operator

A : X → Y

is compact if every bounded sequence xn is mapped to a sequence yn =Axn that has a convergent subsequence.Proposition. If A is compact, xn converges weakly to x then yn = Axnconverges strongly to y = Ax.Proof: Let us first show that yn converges weakly to y. Let g ∈ Y ∗ thendefine f(z) = g(Az). Then f ∈ X∗, therefore f(xn) → f(x). Henceg(yn) → g(y). Suppose yn does not converge strongly to y. Then ynhas a subsequence such that ||y − ynk || ≥ ε, ∀k. But ||xnk || ≤ C thereforethere must be a subsequence of ynkl such that ||y − ynkl || → 0.

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An extensive class of compact operators g(t) = Af(t) where f(t) is a func-tion defined on an interval [0, 1] can be written in the form

g(t) =∫ 1

0k(t, s)f(s)ds. (0.12)

Proposition. Formula (0.12) defines a compact operator on C([0, 1]) ifk(s, t) is bounded and all points of discontinuity of k(s, t) lie on a finitenumber of curves t = φk(s).Proposition. If A is compact, and B is bounded then AB and BA arecompact.Proof: If M is a bounded set then BM is also bounded. Therefore AB iscompact by definition. Similarly, if xn → x then Byn → By if B is bounded.That proves the second part of the statement.Corollary. A compact operator cannot have a bounded inverse.Proof: If there is an inverse A−1, which is bounded then the identity oper-ator I = AA−1 is compact, which is a contradiction.Proposition. If A is a compact operator on a Banach space X then σp(A) iscountable and its only possible accumulation point is 0. Moreover, if λ 6= 0,then N(Aλ) is finite-dimensional.Note that here A : X → X otherwise the operator Aλ is not defined.Proof: The proof is by contradiction, i.e. assume that there exists aninfinite number of eigenvectors

Axn = λnxn, |λn| ≥ ε > 0, n = 1, 2, . . .

Consider E the subspace of X spanned by xn. Recall that it means that wetake finite linear combinations of xn. By construction E is invariant underthe map A, i.e.

A : E → E.

In this space the operator A has a bounded inverse such that

||A−1x|| ≤ 1ε||x||.

Therefore A−1 can be extended to a bounded linear operator to the closureE. Therefore we can view A as a bounded linear operator on the Banachspace E. Since we have shown that there it is invertible, therefore A, cannotbe compact.

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Fredholm alternative.

Spectrum of compact operator is point spectrum except 0. Thegoal is to show that is λ 6= 0 and λ ∈ σ(A) then λ is an eigenvalue, thatisλ ∈ σ(A).Lemma. If λ 6= 0 then R(Aλ) is closed.Proof: By contradiction. Suppose xn ∈ X such that Aλxn → y, buty 6∈ R(Aλ). y 6= 0, because 0 ∈ R(Aλ), therefore we can assume that xn 6∈N(Aλ). Let ρn = ρ(xn, N(Aλ)). Let zn ∈ N(Aλ) such that ||xn−zn|| ≤ 2ρn.For the sequence an = ||xn − zn|| there are only two options: either there isa subsequence of it which is bounded or an →∞.Consider the first option, then ||xn − zn|| contains a bounded subsequencesuch that A(xn − zn) is convergent. But

xn − zn =1λ

(A(xn − zn)−Aλ(xn − zn))

Since both terms on the righthand side are convergent (the latter convergesto y 6∈ R(Aλ) because Aλzn = 0) it implies that xn−zn itself has a convergentsubsequence, say

xn − zn → x

Then

Aλ(xn − zn) = Aλ(xn)→ Aλx,Aλ(xn − zn) = Aλ(xn)→ y,

and hence y = Ax.Suppose now an →∞. Normalize

wn =xn − znan

,

then ||wn|| = 1 is bounded and

Aλwn =1anAλxn → 0,

since Aλxn converges to y. But again

wn =1λ

(Awn −Aλwn).

Hence there is a convergent subsequence

wn → w.

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We need to have w ∈ N(Aλ), since Aλwn → 0. Finally, let

yn = zn + anw ∈ N(Aλ)

then ρn ≤ ||xn − zn||, but xn − zn = xn − yn − anw = an(wn − w), and wehave

ρn ≤ ||xn − yn|| ≤ 2ρn||wn − w||

therefore||wn − w|| >

12.

Contradiction.For simplicity of notation, assume that λ = 1 and hence we study theoperator A1 = Ax− x. The general case reduces to this one by rescaling.Proposition. Let A be a compact operator

A : X → X

X is a Banach space. If the equation

Ax− x = y (0.13)

is solvable for all y, then the equation

Ax− x = 0

has no solutions distinct from the zero solution.Proof: By contradiction, assume there is x1 such that Ax1−x1 ≡ A1x1 = 0.Let

En = x ∈ X : An1x = 0

We haveE1 ⊂ E2 ⊂ E3 . . . (0.14)

Let us show that the sequence En is strictly increasing. Since (0.13) has asolution for any y, there exists x2, x3, . . .

A1x2 = x1, A1x3 = x2, . . .

Then xn ∈ En but xn 6∈ En−1. Therefore En is strictly increasing. Allthe subspaces En are linear and closed (they are the (finite-dimensional)nullspaces of corresponding bounded operators), therefore there is an ele-ment yn+1 ∈ En+1 such that

||yn+1|| = 1 and ρ(yn+1, En) ≥ 1/2,

33

where ρ(yn+1En) = inf ||yn+1 − x||, x ∈ En. Indeed, suppose ρ(xn+1, En) =α, where xn+1 is constructed above. Then there exists x such that ||xn+1 −x|| < 2α; but

ρ(xn+1 − x, En) = ρ(xn+1, En) = α.

Letyn+1 =

xn+1 − x||xn+1 − x||

.

Consider the sequence Ayk. If m > n

||Aym −Ayn|| = ||ym − (yn +A1ym −A1yn)|| ≥ 1/2,

because yn + A1ym − A1yn ∈ Em−1. Therefore Ayk does not have aconvergent subsequence, but it is bounded. Hence we have a contradictionto compactness.Let us discuss the adjoint operator of A1.Definition. Suppose

A : X → Y

is a bounded linear operator, and X and Y are Banach spaces. Then theadjoint operator is defined as

A∗ : Y ∗ → X∗

by the formula[A∗(f)](x) = f [A(x)].

So A∗ is defined on the space of bounded linear functionals “in the otherdirection”.Proposition. If

A : X → X

is compact then A∗ is compact.Proof: Suppose ||gn|| ≤ C. Since by Alaoglu theorem a ball in X∗ is weak-*compact, there is a subsequence of gn, denoted thereafter as gn, suchthat this subsequence converges to some g ∈ X∗ weak-*, that is for anychosen x ∈ X

(gn − g)(y)→ 0

We want to show that if

fn(x) = gn(A(x)). (0.15)

34

then ||fn − f ||X∗ → 0. By (0.15) if y = Ax

(fn − f)(x)→ 0

Suppose ||fn−f ||X∗ 6→ 0. Then there is a subsequence of fn again denotedas fn and a sequence xn ∈ X such that

|(fn − f)(xn)| ≥ δ||xn||∀n.

By rescaling, choose ||xn|| = 1. But fn(xn) = gn(A(xn)). Since A is com-pact, then since ||xn|| = 1 there is a subsequence of xn again denoted asxn such that ||A(xn − xm)|| → 0 as n,m → ∞. For any δ/2 > ε > 0choose m = m(ε) such that

2C||A(xn − xm)|| ≤ ε

2∀n ≥ m

where max ||gn|| ≤ C. Choose n0 = n0(m) such that

|(f − fn)xm| ≤ε

2∀n ≥ n0

Then for all n ≥ maxn0,m we have Since ||gn|| ≤ C, then

|(fn − f)(xn)| ≤ |(fn − f)(xm)|+ |(fn − f)(xm − xn)| = |(fn − f)(xm)|

+|(gn − g)A(xn − xm)| ≤ |(fn − f)(xm)|+ 2C||A(xn − xm)|| ≤ ε

This is the contradiction. Therefore ||fn − f ||X∗ → 0.Combining two previous propositions we have:Proposition. If the equation

h = A∗f − f

is solvable for all h, then the equation 0 = A∗f − f has only zero solution.Lemma. For for a compact operator A and any x ∈ X define the distance

ρ(x) = ρ(x,N(A1)) = infy∈N(A1)

||x− y||.

Then there is a M > 0 such that

ρ(x) ≤M ||A1x||.

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Proof Suppose not, i.e. there is a sequence xn ∈ X such that xn 6∈ N(A1)with

ρ(xn)||A1xn||

→ ∞.

Since N(A1) is finite-dimensional there is wn ∈ N(A1) such that

||xn − wn|| = ρ(xn)

Rescalevn =

xn − wnρ(xn)

,

then ||vn|| = 1 and

A1vn =A1xnρ(xn)

→ 0.

We can writevn = Avn −Avn + vn = Avn −A1vn.

And since A1vn → 0 and A is compact there is a subsequence which weagain denote vn such that

vn → v.

Then v ∈ N(A1), because A1vn → 0, but

||vn − v|| = ||xn − wn − d(xn)v

d(xn)|| ≥ 1,

because −wn − d(xn)v ∈ N(A1) and d(xn) ≤ ||xn − y|| for any y ∈ N(A1).Hence vn does not converge to v. Contradiction.Lemma If y ∈ R(A1) there is x ∈ X such that y = A1x and ||x|| ≤M ||y||.Proof: For any y ∈ R(A1) there is some x0 such that y = A1x0. By theprevious lemma there is w ∈ N(A1) such that ρ(x0) = ρ(x0, N(A1)) =||x0 − w||. Let x = x0 − w, then

||x|| = ρ(x0) ≤M ||A1x0|| = M ||A1x|| = M ||y||

Proposition. For some fixed g the equation A∗f − f = g is solvable if andonly if g(x) = 0 for all x ∈ X such that Ax− x = 0.Proof: If A∗f − f = g is solvable then

g(x) = A∗f(x)− f(x) = f(Ax− x) = 0

36

if Ax− x = 0.In the other direction assume g(x) = 0 for all x that satisfy Ax − x = 0.We are going to show that A∗f − f = g is solvable. We will construct thefunctional f , that solves it. For any y = Ax−x let f(y) = g(x). The valuesof this functional are defined uniquely, because if Ax1−x1 = Ax2−x2, thenx1 − x2 ∈ N(A1), and therefore g(x1) = g(x2). So far we have defined f(y),if y ∈ R(A1). This is a bounded linear functional, because

|g(y)| = |h(x)| ≤ ||h||M ||y||

where x is chosen by the previous lemma. Hence, by Hahn-Banach it canbe extended to the whole space and therefore we are done.Corollary. If Ax − x = 0 has no nonzero solutions, then A∗f − f = g issolvable for all g.Proposition For some fixed y the equation Ax − x = y is solvable if andonly if f(y) = 0 for all f ∈ X∗ such that A∗f − f = 0.Proof: If Ax− x = y is solvable. Then for any f such that A∗f − f = 0 wehave

f(y) = f(A(x))− f(x) = A∗f(x)− f(x) = [A∗f − f ](x) = 0.

In the other direction, suppose y 6∈ R(A1) - it cannot be represented in theform y = Ax− x. For each f such that A∗f − f = 0 consider the nullspacesN(f). They are closed subsets of X. Hence ∩fN(f) is also closed. Then weneed to prove that

y 6∈ ∩fN(f).

It is sufficient to show that we can construct a bounded linear functional fsuch that

f(y) 6= 0, A∗f − f = 0.

Consider the linear subspace of X

E = µy +R(A1)

and for all x ∈ E define

f(µy +R(A1)) = µ.

Then f is a bounded linear functional, because R(A1) is closed (check whybounded in more detail). Hence by Hahn-Banach theorem there is a boundedextension of this function to X. Hence we are done.

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Hence we have:Proposition. If Ax − x = 0 has no nonzero solutions, then Ax − x = y issolvable for all y.Proof: If Ax−x = 0 has no nonzero solutions, then A∗f −f = g is solvablefor all g, then A∗f − f = 0 has no nonzero solutions, then Ax − x = y issolvable for all y.The last proposition allows us finally to formulate theFredholm alternative:For an equation Ax− λx = y λ 6= 0 there are only two cases possible:1) It is solvable for all y, i.e. the operator A− λI has a bounded inverse.2) It has a nonzero solution for y = 0, i.e. λ ∈ σp(A), i.e. it is an eigenvalue.Proof The fact that A − λI has a bounded inverse follows from the openmapping theorem - by all the propositions we showed that if A− λI is one-to-one, then it is also onto, therefore it has a bounded inverse. Part 2) isimmediate.One can actually say more about the solvability of the problem

Ax− λx = f,

If λ is an eigenvalue.Proposition(without proof) The dimension of N(Aλ) equals to the dimen-sion of N(A∗λ).The space of compact operators is closed under the operator norm:Proposition. If An is a sequence of compact operators from a Banach spaceto a Banach space that converges in operator norm to an operator A, thenA is a compact operator.Proof: Use the diagonal argument: for any sequence xn there is a con-vergent subsequence A1xn1, then the latter sequence, has a convergentsequence A2xn2, and so on. Form the diagonal sequence and show that itis convergent for A.Definition. A bouned operator

An : X → Y

where X and Y are Banach spaces is said to be finite-dimensional (or finite-rank) of dimesion n, if R(An) has dimension n.Examples. Any functional is a finite-dimensional operator of dimension 1.On R

∞ a projection operator Pn on the first n coordinates is a finite-dimensional operator of dimension n.A very rich class of compact operators comes from the following idea: every

38

finite-dimensional operator is compact. Let’s take a closure with respect tothe operator norm, all elements of this closure by the previous theorem arecompact.Question Do we get this way all compact operators, i.e. can any compactoperator be approximated arbitrarily well by a finite-dimensional operator?The answer is ‘yes’ in a Hilbert space, but the answer is ‘no’ in general (lookfor work of P.Enflo if you are interested).Since L2 is a Hilbert space then A is compact if (check!)

A : L2[0, 1]→ L2[0, 1], A(f)(x) =∫ 1

0k(x, y)f(y)dy,

and k(x, y) ∈ L2[0, 1]× [0, 1] can be approximated by integral operators witha degenerate kernel function

kn(x, y) =n∑

m=1

αm(x)βm(y),

by, for example, taking finite trigonometric polynomials.

39