Functional Analysis I

Solutions to Exercises

James C. Robinson

Contents

1 Examples I page 1

2 Examples II 5

3 Examples III 9

4 Examples IV 15

iii

1

Examples I

1. Suppose that

v =n∑j=1

αjej and v =m∑k=1

βkfk.

with αj , βk ∈ K and ej , fk ∈ E. Relabel βk and fk so that fj = ej forj = 1, . . . , n′, and fj /∈ {e1, . . . , en} for j > n′, i.e.

v =n′∑k=1

βkek +m∑

k=n′+1

βkfk,

with the understanding that the first sum is zero if n′ = 0, and the secondzero if n′ = m.

It follows that

n∑j=1

αjej −n′∑k=1

βkek −m∑

k=n′+1

βkfk = 0,

orn′∑j=1

(αj − βj)ej −n∑

k=n′+1

αjej +m∑

k=n′+1

βkfk = 0.

Since E is linearly independent, it follows that αj = βj = 0 for j > n′,from which n′ = n, and that αj = βj for j = 1, . . . , n. So the expansionis unique.

2. If x, y ∈ `lim(C) then x+ y ∈ `lim(C) and αx ∈ `lim(C), since

limn→∞

(xn + yn) = limn→∞

xn + limn→∞

yn and limn→∞

αxn = α limn→∞

xn.

1

2 1 Examples I

3. Write

‖x‖p`p =∑j

|xj |p =∑j

|xj |p−α|xj |α.

Holder’s inequality with exponents a and b yields

‖x‖p`p ≤

∑j

|xj |(p−α)a

1/a∑j

|xj |αb1/b

= ‖x‖p−α`(p−α)a‖x‖α`αb .

We want to choose α, and conjugate exponents a, b (a−1 + b−1 = 1) sothat

(p− α)a = q and αb = r,

so that the factors on the right are powers of ‖x‖`q and ‖x‖`r . Thusb = r/α; since a, b are conjugate it follows that a = r/(r − α), and thenthe condition that (p− α)a = q gives

α = (p− q)r/(r − q), a = (r − q)/(r − p), b = (r − q)/(p− q).

The result follows as stated on substituting for α.4.(i) If there exist a, b with 0 < a ≤ b such that

a‖x‖1 ≤ ‖x‖2 ≤ b‖x‖1

then1b‖x‖2 ≤ ‖x‖1 ≤

1a‖x‖2,

while(ii) if

a‖x‖1 ≤ ‖x‖2 ≤ b‖x‖1

and

α‖x‖2 ≤ ‖x‖3 ≤ β‖x‖2

then

aα‖x‖1 ≤ ‖x‖3 ≤ bβ‖x‖1.

5. Since (by the previous question) being equivalent is an equivalence relation(!), we only need to show that ‖ · ‖2 and ‖ · ‖∞ are equivalent to ‖ · ‖1.

Examples I 3

We have

n∑j=1

|xj |2 ≤

n∑j=1

|xj |

2

≤ nn∑j=1

|xj |2,

which gives ‖x‖2 ≤ ‖x‖1 ≤√n ‖x‖2, and

maxj=1,...,n

|xj | ≤n∑j=1

|xj | ≤ n maxj=1,...,n

|xj |,

i.e. ‖x‖∞ ≤ ‖x‖1 ≤ n‖x‖∞.6. First we show that if f is continuous and K is closed then

f−1(K) = {x ∈ X : f(x) ∈ K}

is closed. (In fact this is in the proof of Corollary 3.12 in the notes.)Take xn ∈ f−1(K), and suppose that xn → x; since f is continuous,f(xn) → f(x). But f(xn) ∈ K, and K is closed, so f(x) ∈ K. It followsthat x ∈ f−1(K), and so f−1(K) is closed.

Now suppose that whenever K is closed, f−1(K) is closed. Take xn ∈ Xwith xn → x, and suppose that f(xn) 6→ f(x). Then for some subsequencexnj , we must have

|f(xnj )− f(x)| > ε.

So f(xnj ) is contained in the closed set Z = Y \B(f(x), ε), where

B(f(x), ε) = {y ∈ Y : ‖y − f(x)‖Y < ε}

is open. It follows that f−1(Z) is closed. Since f(xnj ) ∈ Z, xnj ∈ f−1(Z);since xnj → x and f−1(Z) is closed, it follows that x ∈ f−1(Z), so that

|f(x)− f(x)| > ε,

clearly a contradiction.7. Let Y be closed subset of the compact set K. If yn ∈ Y then yn ∈ K,

so there is a subsequence ynj such that ynj → y ∈ K. Since Y is closed,y ∈ Y , and so Y is compact.

8. If {un} is a Cauchy sequence in U then, since U ⊂ V and the norm on Uis simply the restriction of the map u 7→ ‖u‖V to U , it follows that {un}is also a Cauchy sequence in V . Since V is complete, un → u for someu ∈ V . But since U is closed, we know that if un → u then u ∈ U . So Uis complete.

4 1 Examples I

9. As it stands the question is ambiguous, since one needs to specify a normon c0(K). To be interesting the question requires the `∞ norm; any ele-ment of `p with p <∞ must be an element of c0(K).

To show that c0(K) (with the `∞ norm) is complete, we will first showthat `∞(K) is complete (this case was omitted from the proof of Propo-sition 4.5 in the notes, put the proof is simpler), and then that c0(K) isclosed, which will show that c0(K) is complete by the previous question.

So first suppose that xk = (xk1, xk2, · · · ) is a Cauchy sequence in `∞(K).

Then for every ε > 0 there exists an Nε such that

‖xn − xm‖p`∞ = supj|xnj − xmj | < ε for all n,m ≥ Nε. (1.1)

In particular {xkj }∞k=1 is a Cauchy sequence in K for every fixed j. SinceK is complete (recall K = R or C) it follows that for each k ∈ N

xkj → ak

for some ak ∈ R.Set a = (a1, a2, · · · ). We want to show that a ∈ `∞ and also that‖xk−a‖`∞ → 0 as k →∞. First, since {xk} is Cauchy we have from (1.1)that ‖xn − xm‖`∞ < ε for all n,m ≥ Nε, and so

supj|xnj − xmj | ≤ ε.

Letting m→∞ we obtain

supj|xnj − aj | ≤ ε,

and so xk − a ∈ `∞. But since `∞ is a vector space and xk ∈ `∞, thisimplies that a ∈ `∞ and ‖xk − a‖`∞ ≤ ε.

Now take xn ∈ c0(K) such that xn → x in `∞. Suppose that x /∈ c0(K).Then there exists a δ > 0 and a sequence nj → ∞ such that |xnj | > δ

for every j. Now choose N large enough that ‖xn − x‖`∞ < δ/2 for alln ≥ N .

In particular it follows that |xnnj | > δ/2 for all n ≥ N (and every j);but then xn /∈ c0(K), a contradiction.

2

Examples II

1. If x = (x1, x2, . . .) ∈ `2(R), then∞∑j=1

|xj |2 <∞.

So

‖x− (x1, x2, . . . , xn, 0, 0, . . .)‖2`2 =∞∑

j=n+1

|xj |2 → 0

as n→∞. So `f (R) is dense in `2(R).2. If x ∈ X ∩ Y then there exist εX and εY such that

{y ∈ B : ‖y − x‖ < εX} ⊂ X and {y ∈ B : ‖y − x‖ < εY } ⊂ Y.

Taking ε = min(εX , εY ) it follows that

{y ∈ B : ‖y − x‖ < ε} ⊂ X ∩ Y,

and so X ∩ Y is open.Now, given z ∈ B and ε > 0, since X is dense there exists an x ∈ X

such that ‖x− z‖ < ε/2. Since X is open, there is a δ < ε/2 such that

{x′ ∈ B : ‖x′ − x‖ < δ} ⊂ X.

Since Y is dense, there is a y ∈ Y such that ‖y−x‖ < δ/2. By the above,it follows that we also have y ∈ X. So we have found y ∈ X ∩Y such that‖y − z‖ ≤ ‖y − x‖+ ‖x− z‖ < δ/2 + ε/2 < ε, and so X ∩ Y is dense.

A refinement of this argument allows one to prove the powerful BaireCategory Theorem: a countable intersection of open and dense sets isdense.

5

6 2 Examples II

3. We have to assume that (Y, ‖ · ‖Y ) is a Banach space. Then if xn ∈ Xand xn → x (with x ∈ V ) we know that {xn} is Cauchy in V . So, since

‖F (xn)− F (xm)‖Y ≤ L‖xn − xm‖

it follows that {F (xn)} is a Cauchy sequence in Y . Since Y is complete,we know that limn→∞ F (xn) exists and is an element of Y .

If xn → v, and yn → v, then

‖F (xn)− F (yn)‖ ≤ L‖xn − yn‖.

Taking limits as n→∞ on both sides implies that

limn→∞

F (xn) = limn→∞

F (yn).

That F so defined is continuous is clear: for any v1, v2 ∈ V , one canfind sequences in X that converge to v1, v2, and then from the definition

‖F (v1)− F (v2)‖ ≤ L‖v1 − v2‖.

4. Let V = R, X = Q, and expand each q ∈ Q to the interval Iq = (q −d(q), q + d(q)), where d(q) = |q −

√2|/2. Then

√2 /∈ Iq for any q, so⋃

q∈Q Iq does not cover R.

5. Fix n ∈ N, and cover Aj with a collection of intervals I(j)k such that∑

k |I(j)k | < 2−(n+j). Then ∪jAj is covered by⋃

j,k

I(j)k ,

and∑

j,k |I(j)k | <

∑∞j=n+1 2−j = 2−n.

If Pj occurs almost everywhere then it fails on a set Aj of measurezero. Since the union of the Aj still has measure zero, every Pj occurssimultaneously almost everywhere.

6. Use φn ↑ f to mean that φn is an increasing sequence that convergesalmost everywhere to f . (i) is essentially Lemma 5.5. (ii) If f, g ∈ Linc(R)then there are sequences {φn}, {ψn} ∈ Lstep(R) such that φn ↑ f andψn ↑ g. Then {φn + ψn} is an increasing sequence of step functions withφn + ψn ↑ f + g. So f + g ∈ Linc(R), and since the integral is additive onLstep(R),∫

f + g = limn→∞

∫φn + ψn = lim

n→∞(∫φn +

∫ψn) =

∫f +

∫g.

(iii) If {φn} ∈ Lstep(R) with φn ↑ f then λφn ∈ Lstep(R) and λφn ↑ f .The result follows from the fact that

∫λφn = λ

∫φn.

Examples II 7

(iv) Follows from the same properties for step functions, along similarlines to the above.

7. For x ∈ [n, n+ 1) we have∫ x

0=∫ n

1f(r) dr +

∫ x

nf(r) dr =

n−1∑j=1

(−1)j

j+ (x− n)

(−1)n

n,

and so

limx→∞

∫ x

0f(x) dx =

∞∑j=1

(−1)j

j!<∞.

But the same calculation with |f(x)| gives∫ x

0≥∫ n

1|f(r)|,dr =

n−1∑j=1

1j,

which diverges as n → ∞. [Functions can be ‘integrable’ in some senseeven when they are not ‘Lebesgue integrable’.]

8. Consider fn = n|f |. Then fn is a monotonic sequence, and∫fn =

n∫|f | = 0 for every n. So there exists a g ∈ L1 such that fn → g

almost everywhere. The limit as n → ∞ of fn is zero where f = 0, and+∞ where fn 6= 0. Since g is defined almost everywhere, it follows thatf = 0 almost everywhere.

9. If f ∈ Linc(R) then there exists a sequence {φn} ∈ Lstep(R) such thatφn ↑ f (almost everywhere), and we define∫

f = limn→∞

∫φn.

Since f ≥ φn, ∫|f − φn| =

∫f −

∫φn,

which tends to zero as n→∞ by the definition of∫f .

Now take a step function

φ(x) =n∑j=1

cjχIj (x)

with the endpoints of Ij being aj and bj . Consider the ‘continuous version’

8 2 Examples II

of χI(x), for an interval I with endpoints a, b,

XI(x; δ) =

0 x < a− δ(x− a+ δ)/δ a− δ ≤ x ≤ a1 a < x < b

(b+ δ − x)/δ b ≤ x ≤ b+ δ

0 x > b+ δ.

Now if δ < ε/(nmaxj |cj |) and

φε(x) =n∑j=1

cjXIj (x; δ),

then ∫|φε(x)− φ(x)|dx < ε.

So if g ∈ L1(R), we have g = f1 − f2 ∈ Linc(R). Approximate f1 andf2 to within ε/4 (wrt the L1 norm) by step functions h1 and h2, and thenapproximate h1 and h2 by continuous functions c1 and c2 to within ε/4(again wrt the L1 norm). So c1 − c2 ∈ C0(R) approximates g to within ε,i.e. C0(R) is dense in L1(R).

10. With gn = (∑n

k=1 |fk|)2 we have∫gn =

∫(n∑k=1

|fk|)2 =

∥∥∥∥∥n∑k=1

|fk|

∥∥∥∥∥2

L2

≤

(n∑k=1

‖fk‖L2

)2

≤

( ∞∑k=1

‖fk‖L2

)2

using the definition of the L2 norm and the triangle inequality. It followsthat gn → g almost everywhere for some g ∈ L1. It follows that

∑∞k=1 fk

is absolutely convergent almost everywhere to some f ∈ L2.11. Note that we have∣∣∣∣∣f −

n∑k=1

fk

∣∣∣∣∣2

=

∣∣∣∣∣∞∑

k=n+1

fk

∣∣∣∣∣2

≤

( ∞∑k=n+1

|uk|

)2

≤ |f |2.

So we can use the DCR to deduce that

limn→∞

|f −n∑k=1

fk|2 =∫

limn→∞

|f −n∑k=1

uk|2 = 0.

3

Examples III

1. Expanding the right-hand side gives

(x+ y, x+ y)− (x− y, x− y) + i(x+ iy, x+ iy)− i(x− iy, x− iy)

= ‖x‖2 + (x, y) + (y, x) + ‖y‖2 − ‖x‖2 + (x, y) + (y, x)− ‖y‖2

+ i(‖x‖2 + i(y, x)− i(x, y) + ‖y‖2 − ‖x‖2 + i(y, x)− i(x, y)− ‖y‖2)

= 4 real(x, y) + 4 imag(x, y) = 4(x, y)

2. Write (∫ t

s|f(r)| dt

)2

=(∫ t

sr−1/2r1/2|f(r)|dr

)2

≤(∫ t

sr−1 dr

)(∫ t

sr|f(r)|2 dr

)≤ K(log t− log s)1/2

using the Cauchy-Schwarz inequality. [K2 in examples sheet is a misprint.]3. Suppose that E is orthonormal and that

n∑j=1

αjej = 0

for some αj ∈ K and ej ∈ E. Then taking the inner product of both sideswith ek (with k = 1, . . . , n) gives

(n∑j=1

αjej , ek) = 0 ⇒ αk = 0,

so αk = 0 for each k, i.e. E is linearly independent.

9

10 3 Examples III

4.(i) Suppose that d = 0. Then there exists a sequence yn ∈ Y such that

limn→∞

‖x− yn‖ = 0,

i.e. yn → x. Since Y is a closed linear subspace, it follows that x ∈ Y– but this contradicts the fact that x /∈ Y .

(ii) By definition there exists a sequence yn ∈ Y such that d = limn→∞ ‖x−yn‖. So for n sufficiently large, clearly ‖x − yn‖ < 2d. Let z be onesuch yn.

(iii) We have

‖x− y‖ =∥∥∥∥ x− z‖x− z‖

− y∥∥∥∥ =

1‖x− z‖

∥∥∥∥x− (z + ‖x− z‖y)∥∥∥∥ > 1

2dd =

12,

since z + ‖x− z‖y ∈ Y and d = inf{‖x− y‖ : y ∈ Y }.5. Take x1 ∈ B with ‖x1‖ = 1. Let Y1 be the linear subspace {αx1 : α ∈ K}.

Using the result of question 7 there exists an x2 ∈ B with ‖x2‖ = 1 suchthat ‖x2−x1‖ > 1

2 . Now let Y2 = Span(x1, x2), and use question 7 to findx3 ∈ B with ‖x3‖ = 1 with ‖x3 − y‖ > 1

2 for every y ∈ Span(x1, x2) – inparticular

‖x3 − x1‖ >12

and ‖x3 − x2‖ >12.

Continuing in this way we obtain a sequence {xn} ∈ B such that ‖xn‖ = 1and

‖xn − xm‖ >12

for all n 6= m. It follows that the unit ball is not compact, since nosubsequence of the {xn} can be a Cauchy sequence.

6. The unit ball is closed and bounded. If B is finite-dimensional then thisimplies that the unit ball is compact. We have just shown, conversely,that if B is infinite-dimensional then the unit ball is not compact.

7. Take x, y ∈ A and suppose that λx + (1 − λ)y ∈ A for all λ ∈ [0, 1] suchthat 2kλ ∈ N. Then given µ with 2k+1µ ∈ N, 2kµ /∈ N we have

µ = l2−(k+1)

with l odd, so that

µ =(l − 1)

22−(k+1) +

(l + 1)2

2−(k+1).

With λ1 = (l− 1)2−(k+1) and λ2 = (l+ 1)2−(k+1) we have 2kλ1, 2kλ2 ∈ N,and so

µx+ (1− µ)y =12

[(λ1x+ (1− λ1)y) + (λ2x+ (1− λ2)y)] ,

Examples III 11

with λjx+ (1− λj)y ∈ A, i.e. µx+ (1− µ)y ∈ A.It follows by induction that λx + (1 − λ)y ∈ A for any λ ∈ [0, 1] such

that 2kλ ∈ N for some k ∈ N. Since any λ ∈ [0, 1] can be approximatedby a sequence λj such that 2jλj ∈ N, it follows since A is closed thatλx+ (1− λy) ∈ A for all λ ∈ [0, 1], i.e. A is convex.

8. We have x ∈ (X⊥)⊥ if for every y ∈ X⊥, (x, y) = 0. So clearly X ⊆(X⊥)⊥. If we do not have equality here then there exists a z /∈ X with(z, y) = 0 for every y ∈ X⊥. But we know that z = x+ ξ with x ∈ X andξ ∈ X⊥; but then (x+ ξ, ξ) = 0, which implies that ‖xi‖2 = 0, i.e. x ∈ X.So (X⊥)⊥ = X as claimed.

9. We have

e′4 = x3 −

(x3,

√58

(3x2 − 1)

)√58

(3x2 − 1)−

(x3,

√32x

)√32x

−(x3,

1√2

)1√2

= x3 − 58

(3x2 − 1)∫ 1

−1(3t5 − t3) dt− 3x

2

∫ 1

−1t4 dt− 1

2

∫ 1

−1t3 dt

= x3 − 3x5.

Then

‖e′4‖2 =∫ 1

−1

(t3 − 3t

5

)2

dt =[t7

7− 6t5

25+

3t3

25

]1

−1

=8

7× 25

and so

e4 =

√78

(5x3 − 3x).

10. Approximation of sinx by a third degree polynomial f3(x): note that(sinx, e1) and (sinx, e3) will be zero since sinx is odd and e1 and e3 areeven, so

f3(x) =3x2

∫ 1

−1t sin tdt+

7(5x3 − 3x)8

∫ 1

−1(5t3 − 3t) sin tdt.

Now, ∫ 1

−1t sin t dt = [−t cos t]1−1 +

∫ 1

−1cos tdt = [sin t]1−1 = 2 sin 1

12 3 Examples III

and ∫ 1

−1t3 sin t dt =

[−t3 cos t

]1−1

+∫ 1

−13t2 cos t dt

=[3t2 sin t

]1−1− 6

∫ 1

−1t sin t dt

= −6 sin 1,

giving ∫ 1

−1(5t3 − 3t) sin tdt = −36 sin 1;

it follows that

f3(x) = sin 1[3x+

7(27x− 45x3)2

]= sin 1

[195x− 315x3

2

].

11. First way: given orthonormal polynomials ej(x) on [−1, 1] put

ej(x) =√

2 ej(2x− 1),

since then∫ 1

0ej(x)ek(x) dx = 2

∫ 1

0ej(2x− 1)ek(2x− 1) dx =

∫ 1

−1ej(y)ek(y) dy.

Doing this gives

e1(x) = 1, e2(x) =√

3 (2x− 1),

e3(x) =√

52

[3(4x2 − 4x+ 1)− 1] =√

5 (6x2 − 6x+ 1),

and

e4(x) =√

72[5(8x3 − 12x2 + 6x− 1)− 3(2x− 1)

]=√

7 [20x3 − 30x2 + 12x− 1].

Or the painful way: e1(x) = 1, then

e′2(x) = x−∫ 1

0tdt = x− 1

2,

and then

‖e′2‖2 =∫ 1

0(x− 1

2)2 dx =

[(x− 1

2)3

3

]1

0

=112

Examples III 13

which gives e2(x) =√

3 (2x− 1). Now set

e′3(x) = x2 − 3(2x− 1)(∫ 1

0t2(2t− 1) dt

)−∫ 1

0t2 dt

= x2 − 3(2x− 1)[t4

2− t3

3

]1

0

− 13

= x2 − 2x− 12− 1

3= x2 − x+

16,

and then

‖e′3‖2 =∫ 1

0(t2 − t+

16

)2 dt =∫ 1

0t4 − 2t3 +

4t2

3− t

3+

136

dt

=[t5

5− t4

2+

4t3

9− t2

6+

t

36

]1

0

=1

180,

and so e3(x) =√

5(6x2 − 6x+ 1).Finally, we have

e′4(x) = x3 − 5(6x2 − 6x+ 1)(∫ 1

0t3(6t2 − 6t+ 1) dt

)−3(2x− 1)

(∫ 1

0t3(2t− 1) dt

)−∫ 1

0t3 dt

= x3 − 6x2 − 6x+ 14

− 920

(2x− 1)− 14

=20x3 − 30x2 + 12x− 1

20

with

‖e′4‖2 =1

400

∫ 1

0(20t3 − 30t2 + 12t− 1)2 dt

=1

400

∫ 1

0400t6 + 900t4 + 144t2 + 1− 1200t5 + 480t4 − 40t3 − 720t3

+60t2 − 24t dt

=1

400

[4007

+9005

+1443

+ 1− 12006

+4805− 40

4− 720

4+

603− 24

2

]=

1400× 7

,

and so e4(x) =√

7 (20x3 − 30x2 + 12x− 1) as above.

14 3 Examples III

12. Given x, y ∈ `2w, set xj = w1/2j xj and yj = w

1/2j yj . Then

(x, y)`2w = (x, y)`2 ,

so the fact that (·, ·)`2w is an inner product on `2w follows from the factthat (·, ·)`2 is an inner product on `2. Similarly the completeness of `2wfollows from the completeness of `2 – if {xn} is a Cauchy sequence in `2wthen {xn} is a Cauchy sequence in `2, so there exists a y ∈ `2 such that

xn → y. Setting zj = yj/w1/2j gives a z ∈ `2w such that xn → z in `2w.

4

Examples IV

1. If ‖x‖X ≤ 1 then

‖Ax‖Y = ‖x‖X∥∥∥∥A x

‖x‖X

∥∥∥∥ ≤ ∥∥∥∥A x

‖x‖X

∥∥∥∥ ,and so since ‖(x/‖x‖X)‖X = 1 we have

sup‖x‖X≤1

‖Ax‖Y ≤ sup‖z‖X=1

‖Az‖Y ≤ sup‖z‖X≤1

‖Az‖Y .

Rearranging the above equality we have

‖Ax‖Y‖x‖X

=∥∥∥∥A x

‖x‖X

∥∥∥∥Y

,

and so, again since ‖(x/‖x‖X)‖X = 1,

supx 6=0

‖Ax‖Y‖x‖X

= sup‖x‖=1

‖Ax‖Y .

2. Given x ∈ `2, we have

‖Tx‖2`2 =∑j

|xj |2

j2≤∑j

|xj |2,

so ‖T‖ ≤ 1. Now, for any fixed n one can take xn = (1, · · · , 1, 0, · · · ), anelement of `2 whose first n entries are 1 and whose others are zero. Then

Txn = (1, 1/2, 1/3, . . . , 1/n, 0, · · · )

which is in `2. {Txn} forms a sequence in `2 that converges to

(1, 1/2, 1/3, 1/4, . . .),

which is an element of `2 since∑

n(1/n2) is finite. However, the ‘preimage’

15

16 4 Examples IV

of this would be the sequence consisting all of 1s, which is not an elementof `2. So the range of T is not closed.

3. Clearly fφ is linear. Given u ∈ C0([a, b]) we have

|fφ(u)| =∣∣∣∣∫ b

aφ(t)u(t) dt

∣∣∣∣≤

(maxt∈[a,b]

|u(t)|)∫ b

a|φ(t)|dt

= ‖u‖∞∫ b

a|φ(t)| dt, (4.1)

and so ‖fφ‖ ≤∫ ba |φ(t)| dt. Now since φ is continuous we can find a

sequence of continuous functions un that approximate the sign of φ (+1if φ > 0, −1 if φ < 0) increasingly closely in the L2 norm: if φ(s) 6= 0then there is an interval around s on which φ has the same sign. Sincethis interval has non-zero length, and the union of these intervals lieswithin [a, b], there are at most a countable number, {Ij}∞j=1. Let un bethe continuous function defined on Ij = (lj , rj) as

un(t) = sign(φ(t))×

(t− lj)/δj,n lj ≤ t ≤ lj + δj,n

1 lj + δj,n < t < rj − δj,n(rj − t)/δj,n rj − δj,n ≤ t ≤ rj

where δj,n = min(2−(n+j), (rj − lj)/2), and zero when φ(t) = 0. Then∫ b

a|un(t)− sign(φ(t))|2 dt ≤

∞∑j=1

2−(n+j) = 2−(n−1),

i.e. un → sign(φ) in L2(a, b). Note that ‖un‖∞ = 1. It follows that

|fφ(un)| =∣∣∣∣∫ b

aφ(t)un(t) dt

∣∣∣∣= ≥

∣∣∣∣∫ b

a|φ(t)|dt−

∫ b

aφ(t)(un(t)− sign(φ(t))) dt

∣∣∣∣≥

∫ b

a|φ(t)| dt−

∣∣∣∣∫ b

aφ(t)(un(t)− sign(φ(t))) dt

∣∣∣∣≥

∫ b

a|φ(t)| dt− ‖φ‖L2‖un − sign(φ)‖L2 .

Since φ ∈ C0([a, b]) we know that ‖φ‖L2 < ∞, and we have just shown

Examples IV 17

that ‖un− sign(φ)‖L2 → 0 as n→∞. It follows since ‖un‖∞ = 1 that wecannot improve the bound in (4.1) and therefore

‖fφ‖ =∫ b

a|φ(t)| dt.

4. Take x, y ∈ L2(0, 1), then

(Tx, y) =∫ 1

0

(∫ t

0K(t, s)x(s) ds

)y(t) dt

=∫ 1

0

∫ t

0K(t, s)x(s)y(t) ds dt

=∫ 1

0

∫ 1

tK(t, s)x(s)y(t) dt ds

=∫ 1

0

(∫ 1

tK(t, s)y(t) dt

)x(s) ds

= (x, T ∗y)

(draw a picture to see how the limits of integration change) where

T ∗(y)(s) =∫ 1

tK(t, s)y(t) dt.

5. If x ∈ (range(T ))⊥ then

(x, y) = 0 for all y ∈ range(T ),

i.e.

(x, Tz) = 0 for all z ∈ H.

Since (T ∗)∗ = T , this is the same as (T ∗x, z) = 0 for all z ∈ H. Thisimplies that T ∗x = 0, i.e. that x ∈ Ker(T ∗). This argument can bereversed, which gives the required equality.

Now – with apologies for the mistake in the question: we will show thatλ is an eigenvalue of T ∗ – suppose that range(T − λI) 6= H. Then thereexists some non-zero u ∈ (range(T − λI))⊥. By the equality we have justproved, this gives a non-zero u in Ker((T − λI)∗), i.e. a non-zero u suchthat

(T − λI)∗u = 0.

Since (T − λI)∗ = T ∗ − λI, this gives a non-zero u with T ∗u = λu. So λis an eigenvalue of T ∗.

18 4 Examples IV

6.(i) If T ∈ B(H,H) then T ∗ ∈ B(H,H) and ‖T ∗‖op = ‖T‖op. So if {xn} isa bounded sequence in H, with ‖xn‖ ≤M , say, it follows that

‖T ∗xn‖ ≤ ‖T ∗‖op‖xn‖ ≤M‖T ∗‖op.

So {T ∗xn} is also a bounded sequence in H. Since T is compact, itfollows that {T (T ∗xn)} = {(TT ∗)xn} has a convergent subsequence,which shows that TT ∗ is compact.

(ii) Following the hint, we have

‖T ∗x‖2 = (T ∗x, T ∗x) = (TT ∗x, x) ≤ ‖TT ∗x‖‖x‖.

Now, if {TT ∗xn} is Cauchy then given any ε > 0, there exists a N suchthat for all n,m ≥ N ,

‖TT ∗xn − TT ∗xm‖ = ‖TT ∗(xn − xm)‖ ≤ ε.

It follows that for all n,m ≥ N ,

‖T ∗xn − T ∗xm‖2 = ‖T ∗(xn − xm)‖2 ≤ ‖TT ∗(xn − xm)‖‖xn − xm‖.

Since {xn} is a bounded sequence, with ‖xn‖ ≤M , say, we have

‖T ∗xn − T ∗xm‖2 ≤Mε for all n,m ≥ N.

It follows that {T ∗xn} is Cauchy.(iii) So suppose that {xn} is a bounded sequence in H. Part (i) shows

that TT ∗ is compact, so {TT ∗xn} has a subsequence {TT ∗xnj} that isCauchy. But part (ii) shows that this implies that {T ∗xnj} is Cauchytoo. So T ∗ is compact.

7.(i) If for every z ∈ H we have

(xn, z)→ (x, z) and (xn, z)→ (y, z)

then clearly (x, z) = (y, z) for every z ∈ H. But then (x − y, z) = 0for every z ∈ H; in particular we can take z = x− y, which shows that‖x− y‖2 = 0, i.e. that x = y.

(ii) Take y ∈ H, and suppose that (en, y) does not tend to zero as n →∞. Then for some ε > 0, thee exists a sequence nj → ∞ such that|(enj , y)| > ε. But then

‖y‖2 =∑n

|(y, ej)|2 ≥∑j

|(y, enj )|2 =∞,

contradicting the fact that y ∈ H with {ej} an orthonormal basis.

Examples IV 19

(iii) For some fixed z ∈ H, consider the map f : H → K given by

u 7→ (Tu, z).

This map is clearly linear,

u+ λv 7→ (T (u+ λv), z) = (Tu, z) + λ(Tv, z),

and it is bounded,

|(Tu, z)| ≤ ‖Tu‖‖z‖ ≤(‖T‖op‖z‖

)‖u‖.

So f ∈ H∗. It follows that there exists a y ∈ H such that

(Tu, z) = f(u) = (u, y)

for every u ∈ H. So if un ⇀ u, in particularly for this choice of y wehave

(un, y)→ (u, y) ⇒ (Tun, z)→ (Tu, z).

Since this holds whatever our choice of z, it follows that Tun ⇀ Tu.(iv) **This part requires you to know that any weakly convergence sequence

is bounded, which is not at all obvious (but true).Suppose that Txn 6→ Tx. Then there exists a δ > 0 and a subsequenceTxnj such that

‖Txnj − Tx‖ > δ for all j = 1, 2, 3, . . . . (4.2)

Since T is compact and {xnj} is bounded there exists a further subse-quence such that Txnj → w for some w ∈ H. But since convergenceimplies weak convergence, we must have Txnj ⇀ w. But we alreadyknow that Txnj ⇀ Tx, so it follows from the uniqueness of weak limitsthat w = Tx, and so Txnj → Tx, contradicting (4.2).

8. (i) For an x with K(x, ·) ∈ L2(a, b), use the fact that {φj} is a basis forL2 to write

K(x, y) =∞∑i=1

ki(x)φi(y).

Since

ki(x) =∫ b

aK(x, y)φi(y) dy,

20 4 Examples IV

we have∫ b

a|ki(x)|2 dx =

∫ b

a

∣∣∣∣∫ b

aK(x, y)φi(y) dy

∣∣∣∣2 dx

≤∫ b

a

(∫ b

a|K(x, y)|2 dy

)(∫ b

a|φi(y)|2 dy

)dx

=∫ b

a|φi(y)|2 dy ×

∫ b

a

∫ b

a|K(x, y)|2 dy dx,

and so ki ∈ L2(a, b).(ii) Since ki ∈ L2(a, b), we can write ki(x) =

∑j κijφj , where κij =∫ b

a ki(x)φj(x) dx, i.e.

K(x, y) =∑i,j

κijφi(y)φj(x),

and so {φi(y)φj(x)} is a basis for L2((a, b)× (a, b)).9. Clearly

Tnx =n∑j=1

λnj (x, ej)ej .

If |λj | < 1 for all j = 1, . . . , n then ‖T‖ < 1, and so (I − T )−1 = I + T +T 2 + · · · , i.e.

(I − T )−1 = I +n∑j=1

λj(x, ej)ej +n∑j=1

λ2j (x, ej)ej + · · ·

= I +n∑j=1

(λj + λ2j + · · · )(x, ej)ej

= I +n∑j=1

λj1− λj

(x, ej)ej .

10.(i) Since x = f + αTx, the solution is given by x = (I − αT )−1f . So if‖αT‖op < 1 we can write

x = (I + αT + α2T 2 + α3T 3 + · · · )f =∞∑j=0

(αT )jf.

So this expansion is valid for |α| < 1/‖T‖op.(ii) Suppose that

(Tn−1x)(t) =∫ b

aKn−1(t, s)x(s) ds.

Examples IV 21

Then

(Tnx)(t) =∫ b

aK(t, s)(Tn−1x)(s) ds

=∫ b

aK(t, s)

(∫ b

aKn−1(s, r)x(r) dr

)ds

=∫ b

a

∫ b

aK(t, s)Kn−1(s, r)x(r) dr ds

=∫ b

a

(∫ b

aK(t, s)Kn−1(s, r) ds

)x(r) dr,

i.e.

(Tnx)(t) =∫ b

aKn(t, r)x(r) dr.

with

Kn(t, r) =∫ b

aK(t, s)Kn−1(r, s) ds

as claimed.11. We have

(Tu, v) =

∞∑j=1

λj(u, ej)ej , v

=∞∑j=1

λj(u, ej)(ej , v)

=

u, ∞∑j=1

λj(ej , v)ej

=

u, ∞∑j=1

λj(v, ej)ej

= (u, Tv)

and so T is self-adjoint. Now consider Tnu defined by

Tnu =n∑j=1

λj(u, ej)ej .

Since the range of T is contained in the span of {ej}nj=1 it is finite-dimensional, and so Tn is compact. Since

‖Tnu− Tu‖2 =

∥∥∥∥∥∥∞∑

j=n+1

λj(u, ej)ej

∥∥∥∥∥∥2

=∞∑

j=n+1

|λj |2|(u, ej)|2,

22 4 Examples IV

we have from the fact that λn → 0 that given any ε > 0 there exists anN such that for all n ≥ N we have |λj | < ε, and hence

‖Tnu− Tu‖2 ≤ ε2∞∑

j=n+1

|(u, ej)|2 ≤ ε2∞∑j=1

|(u, ej)|2 = ε2‖u‖2,

i.e.

‖Tn − T‖op ≤ ε,

and so Tn → T . It follows from Theorem 13.8 that T is compact.12. We have

Tu =∫ b

aK(x, y)u(y) dy

=∫ b

a

∞∑j=1

λjej(x)ej(y)u(y) dy

=∞∑j=1

λjej(ej , u),

and so Tek = λkek.To show that these are the only eigenvalues and eigenvectors, if u ∈

L2(a, b) with u = w +∑∞

k=1(u, ek)ek and Tu = λu then, since w ⊥ ej forall j,

Tu = Tw +∞∑k=1

(u, ek)Tek

=∞∑

j,k=1

∫ b

aλjej(x)ej(y)(u, ek)ek(y) dy

=∞∑j=1

λj(u, ej)ej(x)

and

λu =∞∑j=1

λ(u, ej)ej(x).

Taking the inner product with each ek yields

λ(u, ek) = λk(u, ek),

so either (u, ek) = 0 or λ = λk.

Examples IV 23

(i) Let (a, b) = (−π, π) and consider K(t, s) = cos(t−s). Then K is clearlysymmetric, so the corresponding T is self-adjoint. We have

cos(t− s) = cos t cos s− sin t sin s;

recall that cos t and sin t are orthogonal in L2(−π, π). We have

T (sin t) =∫ π

−πK(t, s) sin s ds

=∫ π

−πcos t cos s sin s− sin t sin2 s ds

= −2π sin t

and

T (cos t) =∫ π

−πK(t, s) cos sds

=∫ π

−πcos t cos2 s− sin t sin s cos s ds

= 2π cos t.

(ii) Let (a, b) = (−1, 1) and let

K(t, s) = 1− 3(t− s)2 + 9t2s2.

Then in fact

K(t, s) = 4

(√32t

)(√32s

)+

85

(√58

(3t2 − 1)

)(√58

(3s2 − 1)

),

and so, since {√

32 t,√

58(3t2−1)} are orthonormal (they are some of the

Legendre polynomials from Chapter 6) the integral operator associatedwith K(t, s) has

T (t) = 4t and T (3t2 − 1) =85

(3t2 − 1).

13. This equation is the equation in Theorem 14.2 with p = 1 and q = 0. Soit suffices to show that (i) u1(x) = x satisfies the equation −d2u/dx2 = 0with u(0) = 0 (which is clear), (ii) that u2(x) = (1− x) satisfies the sameequation with u(1) = 0 (which is also clear), and (iii) that Wp(u1, u2) = 1:

Wp(u1, u2) = [u′1u2 − u′2u1] = 1(1− x)− (−1)x = 1.

Or one could follow the proof of Theorem 4.12 with this particular choiceof G to show that u(x) as defined does indeed satisfy −d2u/dx2 = f .

24 4 Examples IV

14.(i) If x 6= 0 then ‖x‖ > 0, from which it follows that ‖Tx‖ > 0, and sox /∈ Ker(T ). So Ker(T ) = {0} and T−1 exists for all y ∈ range(T ).Then we have

‖T (T−1x)‖2 ≥ α‖T−1x‖2,

i.e. ‖T−1x‖2 ≤ α−1‖x‖2, so T−1 is bounded.(ii) Let λ = α+ iβ. Then

‖(T − λI)x‖2 = (Tx− αx− iβx, Tx− αx− iβx)

= ‖(Tx− αx)‖2 − (Tx− αx, iβx)− (iβx, Tx− αx) + |β|2‖x‖2

= ‖Tx− αx‖2 + i(Tx− αx, βx)− i(βx, Tx− αx) + |β|2‖x‖2.

Now, since T is self-adjoint and α is real, it follows that

(Tx− αx, βx) = (βx, Tx− αx),

and so

‖(T − λI)x‖2 = ‖Tx− αx‖2 + β2‖x‖2 ≥ β2‖x‖2,

i.e. T − λI is bounded below as claimed.(iii) If λ ∈ C and T − λI has a densely defined inverse then (T − λI)−1 ∈

B(H,H), i.e. λ ∈ R(T ). So σc(T ) ⊆ R.(iv) If the range of T − λI is not dense in H then by Question 3 we know

that λ is an eigenvalue of T ∗. If T is self-adjoint, we must have λ aneigenvalue of T . But all eigenvalues of T are real, so in fact λ is aneigenvalue of T . So λ ∈ σp(T ) rather than σr(T ), which implies thatσr(T ) is empty.

15.(i) If y ∈ `q then

|l(x)| =

∣∣∣∣∣∣∞∑j=1

xjyj

∣∣∣∣∣∣ ≤ ‖x‖p‖y‖qusing Holder’s inequality. So ‖l‖p∗ ≤ ‖y‖q as claimed.

(ii) Any x = (x1, x2, x3, . . .) ∈ `p can be written as

x =∞∑j=1

xjej ,

for the elements ej defined in the question. Since l is bounded it is

Examples IV 25

continuous, so

l(x) = l

limn→∞

n∑j=1

xjej

= lim

n→∞l

n∑j=1

xjej

= lim

n→∞

n∑j=1

xjl(ej)

=∞∑j=1

xjl(ej)

=∞∑j=1

xjyj ,

if we define yj = l(ej). [Should you come across something similar inan exam you should at least mention continuity of l as the justificationfor taking it inside the infinite sum.]

(iii) We have

‖x(n)‖pp =n∑j=1

∣∣∣∣ |yj |qyj

∣∣∣∣p =n∑j=1

|yj |pq − 1 =n∑j=1

|yj |q,

and so, as claimed,

‖x(n)‖p =

n∑j=1

|yj |q1/p

.

Also

l(x(n)) =n∑j=1

|yj |q

yjyj =

n∑j=1

|yj |q.

Since |l(x(n)| ≤ ‖l‖p∗‖x(n)‖p, we have

n∑j=1

|yj |q ≤ ‖l‖p∗

n∑j=1

|yj |q1/p

,

26 4 Examples IV

from which it follows, since 1− 1/p = 1/q, that n∑j=1

|yj |q1/q

≤ ‖l‖p∗ .

Since this holds for every n, we must have y ∈ `q with ‖y‖q ≤ ‖l‖p∗ .