functional analysis problems with solutions...functional analysis problems with solutions anh quang...

Functional Analysis Problems with Solutions

ANH QUANG LE, Ph.D.

September 14, 2013

Contents

Contents 1

1 Normed and Inner Product Spaces 3

2 Banach Spaces 15

3 Hilbert Spaces 27

3.1 Hilbert spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3.2 Weak convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

4 Linear Operators - Linear Functionals 45

4.1 Linear bounded operators . . . . . . . . . . . . . . . . . . . . . . . . 45

4.2 Linear Functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

5 Fundamental Theorems 73

6 Linear Operators on Hilbert Spaces 87

7 Compact Operators 99

8 Bounded Operators on Banach Spaces and Their Spectra 113

9 Compact Operators and Their Spectra 131

10 Bounded Self Adjoint Operators and Their Spectra 143

1

www.MATHVN.com - Anh Quang Le, Ph.D

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2 CONTENTS

Notations:• B(X, Y ): the space of all bounded (continuous) linear operators from X to Y .• Image (T ) ≡ Ran(T ): the image of a mapping T : X → Y .• xn

w−→ x: xn converges weakly to x.• X∗: the space of all bounded (continuous) linear functionals on X.• F or K: the scalar field, which is R or C.• Re, Im: the real and imaginary parts of a complex number.


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Chapter 1

Normed and Inner Product Spaces

Problem 1.Prove that any ball in a normed space X is convex.

Solution.Let B(x0; r) be any ball of radius r > 0 centered at x0 ∈ X, and x, y ∈ B(x0; r).Then

‖x− x0‖ < r and ‖y − x0‖ < r.

For every a ∈ [0, 1] we have

‖ax + (1− a)y − x0‖ = ‖(x− x0)a + (1− a)(y − x0)‖≤ a‖x− x0‖+ (1− a)‖y − x0‖< ar + (1− a)r = r.

So ax + (1− a)y ∈ B(x0; r). ¥.

Problem 2.Consider the linear space C[0, 1] equipped with the norm

‖f‖1 =

∫ 1

0

|f(x)|dx.

Prove that there is no inner product on C[0, 1] agreed with this norm.

3


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4 CHAPTER 1. NORMED AND INNER PRODUCT SPACES

Solution.We show that the norm ‖.‖1 does not satisfy the parallelogram law. Let

f(x) = 1 and g(x) = 2x.

Then

‖f‖1 =

∫ 1

0

1.dx = 1, ‖g‖1 =

∫ 1

0

|2x|dx = 1,

while

‖f − g‖1 =

∫ 1

0

|1− 2x|dx =1

2, ‖f + g‖1 =

∫ 1

0

|1 + 2x|dx = 2.

Thus,

‖f − g‖21 + ‖f + g‖2

1 =17

46= 2(‖f‖2

1 + ‖g‖21) = 4. ¥

Problem 3.Consider the linear space C[0, 1] equipped with the norm

‖f‖ = maxt∈[0,1]

|f(t)|.

Prove that there is no inner product on C[0, 1] agreed with this norm.

Solution.We show that the parallelogram law with respect to the given norm does not holdfor two elements in C[0, 1].Let f(t) = t, g(t) = 1− t, t ∈ [0, 1]. Then f, g ∈ C[0, 1] and

‖f‖ = maxt∈[0,1]

t = 1, ‖g‖ = maxt∈[0,1]

(1− t) = 1,

and‖f + g‖ = max

t∈[0,1]1 = 1, and ‖f − g‖ = max

t∈[0,1]| − 1 + 2t| = 1.

Thus,‖f − g‖2

1 + ‖f + g‖21 = 2 6= 2(‖f‖2

1 + ‖g‖21) = 4. ¥

Problem 4.Prove that:If the unit sphere of a normed space X contains a line segment [x, y] where x, y ∈X and x 6= y , then x and y are linearly independent and ‖x + y‖ = ‖x‖+ ‖y‖ .


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5

Solution.Suppose that the unit sphere contains a line segment [x, y] where x, y ∈ X andx 6= y. Then

‖ax + (1− a)y‖ = 1 for any a ∈ [0, 1].

Choose a = 1/2 then we get ‖12(x + y)‖ = 1, that is ‖x + y‖ = 2. Since x and y

belong to the unit sphere, we have ‖x‖ = ‖y‖ = 1. Hence

‖x + y‖ = ‖x‖+ ‖y‖.Let us show that x, y are linearly independent. Assume y = βx for some β ∈ C. Wehave

1 = ‖ax + (1− a)βx‖ = |a + (1− a)β|.For a = 0 we get |β| = 1 and for a = 1/2 we get |1 + β| = 2. These imply thatβ = 1, and so x = y, which is a contradiction. ¥

Problem 5.Prove that two any norms in a finite dimensional space X are equivalent.

Solution.Since equivalence of norms is an equivalence relation, it suffices to show that anarbitrary norm ‖.‖ on X is equivalent to the Euclidian norm ‖.‖2. Let e1, ..., en bea basis for X. Every x ∈ X can be written uniquely as x =

∑nk=1 ckek. Therefore,

‖x‖ ≤n∑

k=1

|ek|‖ek‖ ≤(

n∑

k=1

|ck|2)1/2 (

n∑

k=1

‖ek‖2

)1/2

≤ A‖x‖2,

where A = (∑n

k=1 |ek|2)1/2is a non-zero constant. This shows that the map x 7→ ‖x‖

is continuous w.r.t. the Euclidian norm. Now consider S = x : ‖x‖2 = 1. This isjust the unit sphere in (X, ‖.‖2), which is compact. The map

S → R defined by x 7→ ‖x‖is continuous, so it attains a minimum m and a maximum M on S. Note that m > 0because S 6= ∅. Thus, for all x ∈ S, we have

m ≤ ‖x‖ ≤ M.

Now, for x ∈ X, x 6= 0, x‖x‖2 ∈ S, so

m ≤ ‖x‖‖x‖2

≤ M.


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That ism‖x‖2 ≤ ‖x‖ ≤ M‖x‖2.

Hence, the two norms are equivalent. ¥

Problem 6.Let X be a normed space.(a) Find all subspaces of X which are contained in some ball B(a; r) of X.(b) Find all subspaces of X which contain some ball B(x0; ρ) of X.

Solution.(a) Let Y be a subspace of X which is contained in some ball B(a; r) of X. Notefirst that the ball B(a; r) must contain the vector zero of X (and so of Y ); otherwise,the question is impossible. For any number A > 0 and any x ∈ Y , we have Ax ∈ Ysince Y is a linear space. By hypothesis Y ⊂ B(a; r), so we have Ax ∈ B(a; r). Thisimplies that ‖Ax‖ < r + ‖a‖. Finally

‖x‖ <r + ‖a‖

A.

A > 0 being arbitrary, it follows that ‖x‖ = 0, so x = 0. Thus, there is only onesubspace of X, namely, Y = 0, which is contained in some ball B(a; r) of X.(b) Let Z be a subspace of X which contain some ball B(x0; ρ) of X. Take anyx ∈ B(0; ρ). Then x + x0 ∈ B(x0; ρ) and so x + x0 ∈ Z since Z ⊃ B(x0; ρ). Now,since x0 ∈ Z, x + x0 ∈ Z and Z is a linear space, we must have x ∈ Z. HenceB(0; ρ) ⊂ Z.Now for any nonzero x ∈ X, we have ρx

2‖x‖ ∈ B(0; ρ) ⊂ Z. Hence x ∈ Z. We canconclude that Z = X. In other words, the only subspace of X which contains someball B(x0; ρ) of X is X itself. ¥

Problem 7.Prove that any finite dimensional normed space :(a) is complete (a Banach space),(b) is reflexive.

Solution.Let X be a finite dimensional normed space. Suppose dim X = d.


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7

(a) By Problem 5, it suffices to consider the Euclidian norm in X. Let e1, ..., edbe a basis for X. For x ∈ X there exist numbers c1, ..., cd such that

x =d∑

k=1

ckek and ‖x‖ =

(d∑

k=1

|ck|2)1/2

.

Let (x(n)) be a Cauchy sequence in X. If for each n, x(n) =∑d

k=1 a(n)k ek then

‖x(n) − x(m)‖ =

(d∑

k=1

|a(n)k − a

(m)k |2

)1/2

→ 0 as n,m →∞.

Hence, for every k = 1, ..., d,

|a(n)k − a

(m)k | → 0 as n,m →∞.

Therefore, each sequence of numbers (a(n)k ) is a Cauchy sequence, so

a(n)k → a

(0)k as n →∞ for every k = 1, 2, ..., d.

Let a =∑d

k=1 a(0)k ek then x(n) → a ∈ X.

(b) Let f ∈ X] where X] is the space of all linear functionals on X. We have

f(x) = f

(d∑

k=1

ckek

)=

d∑

k=1

ckf(ek) =d∑

k=1

ckαk,

where αk = f(ek). Let us define fk ∈ X] by the relation fk(x) = ck, k = 1, ..., d.For any x ∈ X and f ∈ X], we get

f(x) =d∑

k=1

fk(x)αk, i.e., f =d∑

k=1

αkfk.

Hence, dim X] ≤ d.Let

∑dk=1 αkfk = 0. Then, for any x ∈ X,

∑dk=1 αkfk(x) = 0, and by taking

x =∑d

k=1 αkek, we obtain fk(x) = αk, and

d∑

k=1

αkfk(x) =d∑

k=1

|αk|2 = 0.

Hence, αk = 0 for all k = 1, ..., d and thus, dim X] = d. For the space X∗ we haveX∗ ⊂ X], so dim X∗ = n ≤ d and dim(X∗)] = n. From the relation X ⊂ (X∗)∗ ⊂(X∗)] we conclude that d ≤ n. Thus, n = d, and so X = (X∗)∗. ¥


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Problem 8. ( Reed-Simon II.4)(a) Prove that the inner product in a normed space X can be recovered from thepolarization identity:

〈x, y〉 =1

4

[(‖x + y‖2 − ‖x− y‖2)− i(‖x + iy‖2 − ‖x− iy‖2)

].

(b) Prove that a normed space is an inner product space if and only if the normsatisfies the parallelogram law:

‖x + y‖2 + ‖x− y‖2 = 2(‖x‖2 + ‖y‖2).

Solution.(a) For the real field case, the polarization identity is

〈x, y〉 =1

4(‖x + y‖2 − ‖x− y‖2). (∗)

We use the symmetry of the inner product and compute the right hand side of (∗):1

4(‖x + y‖2 − ‖x− y‖2) =

1

4

[〈x + y, x + y〉 − 〈x− y, x− y〉]

=1

2

[〈x, y〉+ 〈y, x〉]

= 〈x, y〉.For the complex field case, we again expand the right hand side, using the relationwe just established:

1

4

[(‖x + y‖2 − ‖x− y‖2)− i(‖x + iy‖2 − ‖x− iy‖2)

]

=1

2

[〈x, y〉+ 〈y, x〉]− i

2

[〈x, iy〉+ 〈iy, x〉]

=1

2〈x, y〉+

1

2〈y, x〉 − i2

2〈x, y〉+

i2

2〈y, x〉

= 〈x, y〉.(b) If the norm comes from an inner product, then we have

‖x + y‖2 + ‖x− y‖2 = 〈x + y, x + y〉+ 〈x− y, x− y〉= 2〈x, x〉+ 2〈y, y〉+ 〈x, y〉+ 〈y, x〉 − 〈x, y〉 − 〈y, x〉= 2(‖x‖2 + ‖y‖2).


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9

Now suppose that the norm satisfies the parallelogram law. Assume the field isC, and define the inner product via the polarization identity from part (a). Ifx, y.z ∈ X, we write

x + y = x +y + z

2+

y − z

2, x + z = x +

y + z

2− y − z

2,

and we have

〈x, y〉+ 〈x, z〉 =1

4

(‖x + y‖2 + ‖x− y‖2 − ‖x− y‖2 − ‖x− z‖2)

+i

4

(‖x + iy‖2 + ‖x + iz‖2 − ‖x− iy‖2 − ‖x− iz‖2)

=1

2

(∥∥∥∥x +y + z

2

∥∥∥∥2

+

∥∥∥∥y + z

2

∥∥∥∥2

−∥∥∥∥x− y + z

2

∥∥∥∥2

−∥∥∥∥y − z

2

∥∥∥∥2)

− i

2

(∥∥∥∥x + iy + z

2

∥∥∥∥2

+

∥∥∥∥iy + z

2

∥∥∥∥2

−∥∥∥∥x− i

y + z

2

∥∥∥∥2

−∥∥∥∥i

y − z

2

∥∥∥∥2)

=1

2

(∥∥∥∥x +y + z

2

∥∥∥∥2

+

∥∥∥∥y + z

2

∥∥∥∥2

−∥∥∥∥y − z

2

∥∥∥∥2

−∥∥∥∥x− y + z

2

∥∥∥∥2)

− i

2

(∥∥∥∥x + iy + z

2

∥∥∥∥2

+

∥∥∥∥iy + z

2

∥∥∥∥2

−∥∥∥∥i

y − z

2

∥∥∥∥2

−∥∥∥∥x− i

y + z

2

∥∥∥∥2)

=1

4

(‖x + y + z‖2 + ‖x‖2 − ‖x− (y + z)‖2 − ‖x‖2)

− i

4

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

= 〈x, y + z〉.This holds for all x, y, z ∈ X, so, in particular,

〈x, ny〉 = n〈x, y〉 for n ∈ N.

And it also satisfies〈x, ry〉 = r〈x, y〉 for r ∈ Q.

Moreover, again by the polarization identity, we have

〈x, iy〉 =1

4

(‖x + iy‖2 − ‖x− iy‖2)− i

4

(‖x− y‖2 − ‖x + y‖2)

= i〈x, y〉.Combining these results we have

〈x, αy〉 = α〈x, y〉 for α ∈ Q+ iQ.


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Now, if α ∈ C, by the density of Q+ iQ in C, there exists a sequence (αn) in Q+ iQconverging to α. It follows that

〈x, αy〉 = α〈x, y〉 for α ∈ C.

Thus the 〈., .〉 is linear.Since ‖i(x− iy)‖ = ‖x− iy‖, we have

〈y, x〉 = 〈x, y〉,and

〈x, x〉 =1

4(‖2x‖2)− i

4

(|1 + i|‖x‖2 − |1− i|2‖x‖2)

= ‖x‖2.

So this shows that the norm is induced by 〈., .〉 and that it is also positive definite,and thus it is an inner product. ¥

Problem 9. (Least square approximation. Reed-Simon II.5)Let X be an inner product space and let x1, ..., xN be an orthonormal set.Prove that ∥∥∥∥∥x−

N∑n=1

cnxn

∥∥∥∥∥is minimized by choosing cn = 〈xn, x〉.

Solution.For every x ∈ X, we write

x =N∑

n=1

〈xn, x〉xn + z, for some z ∈ X. (∗)

We observe that for all n = 1, ..., N ,

〈xn, z〉 = 〈xn, x〉 −N∑

k=1

〈xn, x〉〈xn, xk〉

= 〈xn, x〉 − 〈xn, x〉 = 0.

Therefore z⊥xn. Then due to (∗) we can write

x−N∑

n=1

cnxn =N∑

n=1

(〈xn, x〉 − cn

)xn

︸︷︷︸zN

+z.


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Since z⊥zN , we have

∥∥∥∥∥x−N∑

n=1

cnxn

∥∥∥∥∥

2

= ‖zN‖2 + ‖z‖2

=N∑

n=1

∣∣〈xn, x〉 − cn

∣∣2 + ‖z‖2,

which attains its minimum if

cn = 〈xn, x〉 for all n = 1, ..., N. ¥

Review: Quotient normed space.

• Let X be a vector space, and let M be a subspace of X. We define an equivalence relation onX by

x ∼ y if and only if x− y ∈ M.

For x ∈ X, let [x] = x + M denote the equivalence class of x and X/M the set of all equivalenceclasses. On X/M we define operations:

[x] + [y] = [x + y]α[x] = [αx], α ∈ C.

Then X/M is a vector space.If the subspace M is closed, then we can define a norm on X/M by

‖[x]‖ = infy∈[x]

‖y‖ = infm∈M

‖x + m‖ = infm∈M

‖x− z‖ = d(x,M).

What a ball in X/M looks like?

B([x0]; r) := [x] : ‖[x]− [x0]‖ < r = x + M : ‖x− x0 + M‖ < r.

• Suppose that M is closed in X. The canonical map (the natural projection) is defined by

π : X → X/M, π(x) = [x] = x + M.

It can be shown that ‖π(x)‖ ≤ ‖x‖, ∀x ∈ X, so π is continuous.

Problem 10.Let X be a normed space and M a closed subspace of X. Let π : X → X/Mbe the canonical map. Show that the topology induced by the standard norm onX/M is the usual quotient topology, i.e. that O ⊂ X/M is open in X/M if andonly if π−1(O) is open in X.


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Solution.• If O is open in X/M , then π−1(O) is open in X since π is continuous.

• Now suppose that O ⊂ X/M and that π−1(O) is open in X. We show that O isopen in X/M . Consider an open ball B(0; r), r > 0 in X. Let x ∈ B(0; r). Then‖x‖ < r, and so

‖[x]‖ ≤ ‖x‖ < r.

On the other hand, if ‖[x]‖ < r, then there is an y ∈ M such that ‖x + y‖ < r.Hence x + y ∈ B(0; r), and so

[x] = π(x + y) ∈ π(B(0; r)

).

If [x0] ∈ O, then x0 ∈ π−1(O). Since π−1(O) is open in X, there is an r > 0 suchthat

B(x0; r) ⊂ π−1(O).

This implies that

O = ππ−1(O) ⊃ π(B(x0; r)

)= π

(x0 + B(0; r)

)= x + M : ‖x− x0 + M‖ < r.

The last set is the open ball of radius r > 0 centered at [x0] ∈ O. Thus O is openin X/M. ¥

Problem 11.Let X = C[0, 1], M = f ∈ C[0, 1] : f(0) = 0. Show that X/M = C.

Solution.Given [f ] ∈ X/M , let ϕ([f ]) = f(0). Then the map ϕ : X/M → C is well-defined.Indeed, if [f ] = [g], then (f − g)(0) = 0 so f(0) = g(0). It is clearly linear.If f ∈ X, then g = f − f(0) ∈ M , and so f − g = f(0) is constant, which tells usthat

‖[f ]‖ = |f(0)| = |ϕ([f ])|,so ϕ is an isometry (and thus injective and continuous). Finally, constants are inX = C[0, 1], so ϕ is surjective and thus an isometric isomorphism. ¥

Problem 12.If 0 < p < 1 then `p is a vector space but ‖x‖p = (

∑n |xn|p)1/p is not a norm for

`p.


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13

Solution.Recall that if x = (x1, x2, ...), y = (y1, y2, ...) ∈ `p and α ∈ C then

x + y = (x1 + y1, x2 + y2, ...) and αx = (αx1, αx2, ...).

It is clear that αx ∈ `p. We show that x + y ∈ `p. For t ≥ 0 it not hard to see that(1 + t)p ≤ 1 + tp, 0 < p < 1. This implies that

(a + b)p ≤ ap + bp, 0 < p < 1 and a, b ≥ 0.

Therefore,‖x + y‖p

p ≤ ‖x‖pp + ‖y‖p

p.

Since both ‖x‖pp and ‖y‖p

p are bounded, ‖x + y‖pp is bounded. Hence x + y ∈ `p.

To show ‖.‖p is not a norm for `p, let us take an example: If

x = (1, 0, ...) and y = (0, 1, 0, ...)

then ‖x‖p = ‖y‖p = 1 but ‖x + y‖p = 21/p > 2 since 1/p > 1. Therefore,

‖x + y‖p > ‖x‖p + ‖y‖p. ¥

Problem 13.Suppose that X is a linear space with inner product 〈., .〉. If xn → x and yn → yas n →∞, prove that

〈xn, yn〉 → 〈x, y〉 as n →∞.

Solution.Using the Cauchy-Schwarz and triangle inequalities, we have

|〈xn, yn〉 − 〈x, y〉| ≤ |〈xn − x, yn〉|+ |〈x, yn − y〉|≤ ‖xn − x‖‖yn‖+ ‖x‖‖yn − y‖≤ ‖xn − x‖(‖yn − y‖+ ‖y‖) + ‖x‖‖yn − y‖≤ ‖xn − x‖‖yn − y‖+ ‖xn − x‖‖y‖+ ‖x‖‖yn − y‖.

Since ‖xn − x‖ → 0 and ‖yn − y‖ → 0 as n →∞, we see that

〈xn, yn〉 → 〈x, y〉 as n →∞. ¥


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Problem 14.Prove that if M is a closed subspace and N is a finite dimensional subspace of anormed space X, then M + N := m + n : m ∈ M, n ∈ N is closed.

Solution.Assume dim N = 1, say N = Spanx. The case x ∈ M is trivial. Suppose x /∈ M .Consider the sequence zk := αkx + mk, where mk ∈ M, αk ∈ C, and supposezk ∈ M + N → y as k → ∞. We want to show y ∈ M + N . The sequence (αk)is bounded; otherwise, there exists a subsequence (αk′) such that 0 < |αk′ | → ∞ ask′ →∞. Then

zk′

αk′and

mk′

αk′→ 0 as k′ →∞,

so x must be 0, which is in M . This is a contradiction. Consequently, (αk) is boundedand therefore it has a subsequence (αk′) which is converging to some α ∈ C. Thus

mk′ = zk′ − αk′x → y − αx as k′ →∞.

Hence, y − αx is in M since M is closed. Thus y ∈ M + N .

The solution now follows by induction. ¥


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Chapter 2

Banach Spaces

Problem 15.Let X be a normed space. Prove that X is a Banach space if and only if the series∑∞

n=1 an converges, where (an) is any sequence in X satisfying∑∞

n=1 ‖an‖ < ∞.

Solution.Suppose that X is complete. Let (an) be a sequence in X such that

∑∞n=1 ‖an‖ < ∞.

Let Sn =∑n

i=1 ai be the partial sum. Then for m > n,

‖Sm − Sn‖ =

∥∥∥∥∥m∑

i=n+1

ai

∥∥∥∥∥ ≤m∑

i=n+1

‖ai‖.

By hypothesis, the series∑∞

n=1 ‖an‖ converges, so∑m

i=n+1 ‖ai‖ → 0 as n → ∞.Therefore, (Sn) is a Cauchy sequence in the Banach space X. Thus, (Sn) converges,that is, the series

∑∞n=1 an converges.

Conversely, suppose∑∞

n=1 an converges in X whenever∑∞

n=1 ‖an‖ < ∞. We showthat X is complete. Let (yn) be a Cauchy sequence in X. Then

∃n1 ∈ N : ‖yn1 − ym‖ <1

2whenever m > n1,

∃n2 ∈ N : ‖yn2 − ym‖ <1

22whenever m > n2 > n1.

Continuing in this way, we see that there is a sequence (nk) strictly increasing suchthat

‖ynk− ym‖ <

1

2kwhenever m > nk.

15


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16 CHAPTER 2. BANACH SPACES

In particular, we have

‖ynk+1− ynk

‖ <1

2kfor all k ∈ N.

Set xk = ynk+1− ynk

. Then

n∑

k=1

‖xk‖ =n∑

k=1

‖ynk+1− ynk

‖ <

n∑

k=1

1

2k.

It follows that∑∞

k=1 ‖xk‖ < ∞. By hypothesis, there is an x ∈ X such that∑mk=1 xk → x as m →∞. But we have

m∑

k=1

xk =m∑

k=1

(ynk+1− ynk

)

= ynm+1 − yn1 .

Hence ynm → x + yn1 in X as m → ∞. Thus, the sequence (yn) has a convergentsubsequence and so must itself converges. ¥

Problem 16.Let X be a Banach space. Prove that the closed unit ball B(0; 1) ⊂ X is compactif and only if X is finite dimensional.

Solution.• Suppose dim X = n. Then X is isomorphic to Rn (with the standard topology).The result then follows from the Heine-Borel theorem.

• Suppose that X is not finite dimensional. We want to show that B(0; 1) is notcompact. To do this, we construct a sequence in B(0; 1) which have no convergentsubsequence.

We will use the following fact usually known as Riesz’s Lemma: (See the proof below)Let M be a closed subspace of a Banach space X. Given any r ∈ (0, 1), there exists an x ∈ X suchthat

‖x‖ = 1 and d(x, M) ≥ r.

Pick x1 ∈ X such that ‖x1‖ = 1. Let S1 := Span x1. Then S1 is closed. Accordingto Riesz’s Lemma, there exists x2 ∈ X such that

‖x2‖ = 1 and d(x2, S1) ≥ 1

2.


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17

Now consider the subspace S2 generated by x1, x2. Since X is infinite dimensional,S2 is a proper closed subspace of X, and we can apply the Riesz’s Lemma to findan x3 ∈ X such that

‖x3‖ = 1 and d(x3, S2) ≥ 1

2.

If we continue to proceed this way, we will have a sequence (xn) and a sequence ofclosed subspaces (Sn) such that for all n ∈ N

‖xn‖ = 1 and d(xn+1, Sn) ≥ 1

2.

It is clear that the sequence (xn) is in B(0; 1), and for m > n we have

‖xn − xm‖ ≥ d(xm, Sn) ≥ 1

2.

Therefore, no subsequence of (xn) can form a Cauchy sequence. Thus, B(0; 1) is notcompact. ¥

Proof of Riesz’s Lemma:Take x1 /∈ M . Put d = d(x1,M) = infm∈M ‖m − x1‖. Then d > 0 since M is closed. For anyε > 0, by definition of the infimum, there exists m1 ∈ M such that

0 < ‖m1 − x1‖ < d + ε.

Set x = x1−m1‖x1−m1‖ . Then ‖x‖ = 1 and

‖x−m‖ =1

‖x1 −m1‖∥∥x1 − (m1 + ‖x1 −m1‖m)︸︷︷︸

∈M

∥∥

This implies that

d(x,M) = infm∈M

‖x−m‖ =infm∈M ‖x1 −m‖

‖x1 −m1‖ ≥ d

d + ε.

By choosing ε > 0 small, dd+ε can be arbitrary close to 1. ¥

Problem 17.Let X be a Banach space and M a closed subspace of X. Prove that the quotientspace X/M is also a Banach space under the quotient norm.

Solution.We use criterion established above (in problem 15). Suppose that ([xn]) is any


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sequence in X/M such that∑∞

n=1 ‖[xn]‖ < ∞. We show that

∃[x] ∈ X/M :k∑

n=1

[xn] → [x] as k →∞.

For each n, ‖[xn]‖ = infz∈M ‖xn + z‖, and therefore there is zn ∈ M such that

‖xn + zn‖ ≤ ‖[xn]‖+1

2n

by definition of the infimum. Hence

∞∑n=1

‖xn + zn‖ ≤∞∑

n=1

‖[xn]‖+1

2n< ∞.

But (xn + zn) is a sequence in the Banach space X, and so

∞∑n=1

(xn + zn) = x for some x ∈ X.

Then we have∥∥∥∥∥

k∑n=1

[xn]− [x]

∥∥∥∥∥ =

∥∥∥∥∥k∑

n=1

[xn − x]

∥∥∥∥∥

= infz∈E

∥∥∥∥∥k∑

n=1

(xn − x + z)

∥∥∥∥∥

≤∥∥∥∥∥

k∑n=1

((xn − x) + zn

)∥∥∥∥∥

=

∥∥∥∥∥k∑

n=1

(xn + zn)− x

∥∥∥∥∥ → 0 as k →∞.

Hence∑k

n=1[xn] → [x] as k →∞. ¥


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19

SPACE `p

(Only properties concerning to norms and completeness will be considered. Otherproperties such as duality will be discussed later.)

Problem 18.Show that `p, 1 ≤ p < ∞ equipped with the norm ‖.‖p is a Banach space.

Solution.Let x(i) = (x

(i)1 , ..., x

(i)k , ...) for i = 1, 2, ... be a Cauchy sequence in `p. Then

‖x(i) − x(j)‖p → 0 as i, j →∞.

Since ‖x(i) − x(j)‖p ≥ |x(i)k − x

(j)k | for every k, it follows that

|x(i)k − x

(j)k | → 0 for every k as i, j →∞.

This tells us that the sequence(x

(i)k

)is a Cauchy sequence in F, which is complete,

so that(x

(i)k

)converges to xk ∈ F as i →∞ for each k. Set x = (x1, ..., xk, ...). We

will show that

(∗) ‖x(i) − x‖p → 0 as i, j →∞ and x ∈ `p.

Given ε > 0, for any M ∈ N, there exists N ∈ N such that(

M∑

k=1

|x(i)k − x

(j)k |p

) 1p

≤( ∞∑

k=1

|x(i)k − x

(j)k |p

) 1p

< ε if i, j > N.

Letting j →∞, for i > N we get

(∗∗)(

M∑

k=1

|x(i)k − xk|p

) 1p

< ε.

By Minkowski’s inequality,(

M∑

k=1

|xk|p) 1

p

≤(

M∑

k=1

|x(N)k − xk|p

) 1p

+

(M∑

k=1

|x(N)k |p

) 1p

≤(

M∑

k=1

|x(N)k − xk|p

) 1p

+

( ∞∑

k=1

|x(N)k |p

) 1p

< ε +

( ∞∑

k=1

|x(N)k |p

) 1p

.


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Letting M →∞, since the last sum is finite, we see that

‖x‖p =

( ∞∑

k=1

|xk|p) 1

p

< ∞.

This shows that x ∈ `p. Finally letting M →∞ in (∗∗), for i > N we get

‖x(i) − x‖p =

( ∞∑

k=1

|x(i)k − xk|p

) 1p

< ε.

This shows that x(i) → x in `p as required. ¥

Problem 19.(a) Show that `∞ equipped with the norm ‖.‖∞ is a Banach space.(b) Let c0 be the space of sequences converging to 0. Show that c0 is a closedsubspace of `∞.

Solution.(a) We need to show that `∞ is complete. Assume that the sequence (x(n)) is Cauchyin `∞. That is, for all ε > 0, there exists N ∈ N such that

(1) n, m ≥ N ⇒ ‖x(n) − x(m)‖∞ < ε.

For a fixed n, we write x(n) = (x(n)1 , x

(n)2 , ...). Then for N = N(ε) as above,

(2) |x(n)j − x

(m)j | ≤ ‖x(n) − x(m)‖∞ < ε for all j.

So, for a fixed j, the sequence (x(n)j ) is Cauchy, and therefore convergent in C.

Denotexj := lim

n→∞x

(n)j , and x := (x1, x2, ...).

We need to show x ∈ `∞ and x(n) → x as n → ∞. In (2), for a fixed j, lettingn →∞ yields

|xj − x(m)j | < ε for all m ≥ N.

Thereforesup

j|xj − x

(m)j | := ‖x− x(m)‖∞ ≤ ε for all m ≥ N.

That is x(m) → x as m →∞ in `∞. Now for all j ∈ N, n ≥ N ,

|xj| ≤ |xj − x(n)j |+ |x(n)

j | ≤ ‖x− x(n)‖∞ + ‖x(n)‖∞ ≤ ε + ‖x(n)‖∞ < ∞.


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21

This shows that ‖x‖∞ < ∞ and so x ∈ `∞.(b) Of course c0 ⊂ `∞. Assume (x(n)) ∈ c0 that converges in `∞ to x. We have toshow that x ∈ c0. Let ε > 0 be arbitrary. Since x(n) → x in `∞, we can chooseN ∈ N such that

‖x(N) − x‖∞ <ε

2.

Since x(N) = (x(N)1 , x

(N)2 , ...) ∈ c0 we have x

(N)j → 0 as j → ∞. Therefore, choose

J ∈ N such thatj ≥ J ⇒ |x(N)

j | < ε

2.

Then, for j ≥ J ,

|xj| ≤ |xj − x(N)j |+ |x(N)

j | ≤ ‖x− x(N)‖∞ + |x(N)j | < ε

2+

ε

2= ε.

Therefore, x = (x1, x2, ...) ∈ c0 ¥.

Problem 20.(a) Let c00 be the space of sequences such that if x = (xn)n∈N ∈ c00 then xn = 0for all n ≥ n0, where n0 is some integer number. Show that c00 with the norm‖.‖∞ is NOT a Banach space.(b) What is the closure of c00 in `∞?

Solution.We observe that c00 ⊂ c0 ⊂ `∞.(a) Consider the sequence x(n) defined by

x(n) =(1,

1

2,1

3, ...,

1

n, 0, 0, ...

)∈ c00.

Then, for n,m ≥ N ,

‖x(n) − x(m)‖∞ =

1

m+1if n ≥ m,

1n+1

if n ≤ m.

In both cases we have for any N > 0

‖x(n) − x(m)‖∞ ≤ 1

N + 1for n,m ≥ N.

So the sequence x(n) is a Cauchy sequence in c00. Evidently, it is also a Cauchysequence in `∞. Now consider the sequence

x =(1,

1

2,1

3, ...,

1

n,

1

n + 1,

1

n + 2, ...

)∈ `∞.


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We see that x(n) /∈ c00, and

limn→∞

‖x(n) − x‖∞ = limn→∞

1

n + 1= 0.

This tells us that there is a Cauchy sequence in c00 which does not converge tosomething in c00. Therefore, c00 equipped with the ‖.‖∞ norm is not a Banachspace.

(b) We claim that c00 = c0 (closure taken in `∞).According to Problem 19, c0 is closed, so we have c00 ⊂ c0. We show the inverseinclusion. Take an arbitrary sequence x = (x1, x2, ...) ∈ c0. We build a sequence a(n)

from x as follows:

a(n) = (x1, x2, .., xn, 0, 0....).

It is clear that a(n) ∈ c00. Now since the sequence x converges to 0, given any ε > 0,there exists N ∈ N such that

|xi| < ε for i ≥ N.

Then

‖a(n) − x‖∞ = supi≥N

|xi| ≤ ε.

hence a(n) converges to x (in `∞). Hence x ∈ c00. Thus, c0 ⊂ c00. ¥

Problem 21.Prove that:(a) If 1 ≤ p < q < ∞, then `p ⊂ `q and ‖x‖q ≤ ‖x‖p.(b) If x ∈ ⋃

1≤p<∞ `p then ‖x‖p → ‖x‖∞ as p →∞.

Solution.(a) Let x = (x1, x2, ...) ∈ `p. Then, for n large enough, we have |xn| < 1 and hence|xn|q ≤ |xn|p since 1 ≤ p < q < ∞. That implies x ∈ `q. Now we want to show

(∑|xn|q

)1/q

≤(∑

|xn|p)1/p

.

Let an = |xn|p and α = qp

> 1. The above inequality is equivalent to

∑aα

n ≤(∑

an

)α

,


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23

which follows by

∑aα

n ≤ (max an)α−1∑

an ≤(∑

an

)α−1 ∑an =

(∑an

)α

.

(b) Let x = (x1, x2, ...) ∈ `p0 for some p0. Clearly, ‖x‖p ≥ maxn |xn| = ‖x‖∞ for anyfinite p. On the other hand,

‖x‖p =

(∑n

|xn|p)1/p

≤(‖x‖p−p0

∞∑

n

|xn|p0

)1/p

= ‖x‖p−p0

p∞ ‖x‖p0p

p0

p→∞−−−→ ‖x‖∞.

¥

Problem 22.Prove that:(a) If 1 ≤ p < ∞ then `p is separable.(b) `∞ is not separable.

Solution.(a) First we show that E := x ∈ `p : xn = 0, n ≥ N for some N is densein `p. Indeed, if x ∈ `p, x =

∑∞k=1 xkek, where ek is the sequence such that the

k-component is 1 and the others are zero, then∥∥∥∥∥x−

n∑

k=1

xkek

∥∥∥∥∥p

=

( ∞∑

k=n+1

|xk|p)1/p

→ 0 as n →∞.

But∑n

k=1 xkek ∈ E, so E is dense in `p. Now let A ⊂ E consisting of elementsx = (x1, x2, ..., xn, 0, 0, ...) ∈ E such that xk = ak + ibk, ak, bk ∈ Q. Since Q is densein R, A is dense in E. Hence A is dense in `p. Since A is countable, `p is separable.(b) We now show that it is not the case for `∞.Let F := x ∈ `∞ : ∀k ≥ 1, xk = 0 or xk = 1. Then F is uncountable. Notethat for x ∈ F, ‖x‖∞ = 1. Moreover, x, y ∈ F, x 6= y ⇒ ‖x − y‖∞ = 1. Assumethat `∞ is separable. Then there is a set A = a1, a2, ... dense in `∞. So, for allx ∈ F , there exists k ∈ N such that ‖x − ak‖∞ ≤ 1

3. Let F be the family of closed

balls B(x; 13), x ∈ F . If B 6= B′ then B ∩ B′ = ∅. This allows us to construct an

injection f : F → A which maps each B ∈ F with an element a ∈ B ∩ A. This isimpossible since F is uncountable and A is countable. ¥

∗ ∗ ∗∗


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Problem 23. (The space C[0, 1])Let C[0, 1] be the space of all continuous functions on [0, 1].(a) Prove that if C[0, 1] is equipped with the uniform norm

‖f‖ = maxx∈[0,1]

|f(x)|, f ∈ C[0, 1]

then C[0, 1] is a Banach space.(b) Give an example to show that C[0, 1] equipped with the L1-norm

‖f‖1 =

∫ 1

0

|f(x)|dx, f ∈ C[0, 1]

is not a Banach space.

Solution.(a) Let (fn) be a Cauchy sequence in C[0, 1] with respect to the uniform norm. Thenfor any ε > 0, there exists N ∈ N such that

‖fm − fn‖ < ε for m,n ≥ N.

Therefore

(∗) |fm(x)− fn(x)| < ε for m,n ≥ N and x ∈ [0, 1].

This shows that for every x ∈ [0, 1], the sequence(fn(x)

)is a Cauchy sequence of

numbers and therefore converges to a number which depends on x, say, f(x). In(∗), fix n and let m →∞, we have

(∗∗) |f(x)− fn(x)| < ε for n ≥ N and x ∈ [0, 1].

Thus the sequence (fn) converges uniformly to f on [0, 1] so that f is continuous on[0, 1], that is, f ∈ C[0, 1]. From (∗∗) we obtain

maxx∈[0,1]

|f(x)− fn(x)| = ‖f − fn‖ ≤ ε for n ≥ N.

This shows thatlim

n→∞‖f − fn‖ = 0.

(b) For each n ∈ N, consider the function

fn(x) =

nx if 0 ≤ x < 1

n

1 if 1n≤ x ≤ 1.


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25

One can check that the sequence (fn) is a Cauchy sequence with respect to theL1-norm, but it converges to the function

f(x) =

0 if x = 0

1 if 0 < x ≤ 1,

which is not continuous, that is, f /∈ C[0, 1]. Thus the space C[0, 1] is not complete. ¥


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Chapter 3

Hilbert Spaces

3.1 Hilbert spaces

Problem 24.Let M be a closed subspace of a Hilbert space H. Prove that:(a) (M⊥)⊥ = M.(b) If codim M := dim H/M = 1 then dim M⊥ = 1.

Solution.(a) In general, if M is a subset of H then M ⊂ (M⊥)⊥. Indeed,

M⊥ := x ∈ X : x⊥M.

So we have

x ∈ M ⇒ x⊥M⊥ ⇒ x ∈ (M⊥)⊥.

Now suppose M is a closed subspace of H and x ∈ (M⊥)⊥. Since x ∈ H = M⊕M⊥,we have

x = u + v, u ∈ M, v ∈ M⊥.

Since M ⊂ (M⊥)⊥ we have

x = u + v, u ∈ (M⊥)⊥, v ∈ M⊥.

Since x− u ∈ (M⊥)⊥ and v ∈ M⊥ and v = x− u we obtain

v ∈ M⊥ ∩ (M⊥)⊥,

27


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28 CHAPTER 3. HILBERT SPACES

which implies v = 0. Hence, x = u ∈ M .(b) Assume that there are two linearly independent vectors x, y ∈ M⊥. Recall thatM is the zero vector of the linear space X/M . Consider the cosets [x], [y]. Assumeα[x]+β[y] = M , for some scalars α, β. Then αx+βy ∈ M and, since αx+βy ∈ M⊥

as well, we conclude that αx + βy = 0. Hence, α = β = 0 and therefore [x], [y] arelinearly independent. This contradicts the hypothesis codim M = 1. ¥

Problem 25.Let T : H1 → H2 be an isometry of two Hilbert spaces H1 and H2, i.e., ‖Tx‖ =‖x‖ for every x ∈ H1. Prove that

〈Tx, Ty〉 = 〈x, y〉 for every x, y ∈ H1.

Solution.By hypothesis, we have

‖T (x + y)‖2 = ‖x + y‖2 for every x, y ∈ H1.

By opening up the norm using the inner product, we obtain

Re〈Tx, Ty〉 = Re〈x, y〉.

Similarly, ‖T (x + iy)‖2 = ‖x + iy‖2 gives that

Im〈Tx, Ty〉 = Im〈x, y〉.

Hence,〈Tx, Ty〉 = 〈x, y〉. ¥

Problem 26.Let C be a closed convex set in a Hilbert space H. Show that C contains a uniqueelement of minimal norm.

Solution.Let η = infz∈C ‖z‖.


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3.1. HILBERT SPACES 29

• If η = 0, then, by definition of infimum, there is a sequence (zj) in C such that‖zj‖ → 0 as j →∞. Therefore, zj → 0 as j →∞. Since C is closed, 0 ∈ C, and 0is the unique element of minimal norm.• Suppose η > 0. First we show that C contains an element of minimal norm.Take (zj) in C such that ‖zj‖ → η as j → ∞. The convexity of C implies that12(zj + zk) ∈ C, so that

‖zj + zk‖2 = 4.1

4‖zj + zk‖2 ≥ 4η2.

Recall now the parallelogram law:

‖x− y‖2 + ‖x + y‖2 = 2(‖x‖2 + ‖y‖2

).

Applying this we have:

‖zj − zk‖2 = 2(‖zj‖2 + ‖zk‖2

)− ‖zj + zk‖2

≤ 2(‖zj‖2 + ‖zk‖2

)− 4η2 → 4η2 − 4η2 = 0 as j, k →∞.

Thus the sequence (zj) is Cauchy, so converges to some z ∈ H. Since C is closed,z ∈ C. The norm function is continuous, so ‖z‖ = η. This shows that C containsan element of minimal norm.Assume that there are two elements a1, a2 ∈ C such that ‖a1‖ = ‖a2‖ = η. By theabove we have

‖a1 + a2‖2 ≥ 4η2.

By the parallelogram law we have

‖a1 − a2‖2 = 2(‖a1‖2 + ‖a2‖2

)− ‖a1 + a2‖2 ≤ 4η2 − 4η2 = 0.

Hence, a1 = a2. ¥

Problem 27.Let 1 ≤ p < ∞. Prove that `p is a Hilbert space if and only if p = 2.

Solution.• Suppose p = 2. In the space `2 the inner product is defined by

〈x, y〉 =∞∑i=1

xiyi for x = (xi), y = (yi) ∈ `2.


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This inner product gives rise to the norm

‖x‖2 =√〈x, x〉 =

( ∞∑i=1

|xi|2) 1

2

.

According to Problem 18, the normed space `2 is complete. So `2 is a Hilbert space.• Consider the general case where 1 ≤ p < ∞. Assume that `p with the corre-sponding inner product is a Hilbert space. Consider two elements en, em ∈ `p withm 6= n defined as follows:

en = (0, .., 0︸︷︷︸n−1

, 1, 0, ...),

em = (0, .., 0︸︷︷︸m−1

, 1, 0, ...).

Since `p is a Hilbert space, by the parallelogram law we have

‖en + em‖2p + ‖en − em‖2

p = 2(‖en‖2

p + ‖em‖2p

),

That is22/p + 22/p = 22.

The unique solution of this equation is p = 2. We conclude that `p is a Hilbert spaceif and only if p = 2. ¥

Problem 28.Consider the Hilbert space H = L2[−1, 1] equipped with the usual scalar product:

〈x, y〉 =

∫ 1

−1

x(t)y(t)dt, x, y ∈ H.

Let M = x ∈ H :∫ 1

−1x(t)dt = 0.

(a) Show that M is closed in H. Find M⊥.(b) Calculate the distance from y to M for y(t) = t2.

Solution.(a) Let 1 ∈ H be the function 1(t) = 1, ∀t ∈ [−1, 1]. Define the map T : H → Cby

x 7→ 〈1, x〉.


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Then T is linear. We show that T is bounded, so continuous.

|Tx| =∣∣∣∣∫ 1

−1

1(t)x(t)dt

∣∣∣∣ ≤∫ 1

−1

|1(t)| |x(t)|dt

≤(∫ 1

−1

1dt

) 12(∫ 1

−1

|x2(t)|2dt

) 12

=√

2‖x‖2.

By definition, M = Ker T = T−1(0). Since T is continuous, M is closed. Further-more,

x ∈ M ⇔ 〈1, x〉 = 0 ⇔ M = (Span1)⊥.

Since M is closed, (see Problem 23 a).

M⊥ = Span1.(b) The distance from y ∈ H to M is the length of the projection vector of y onM⊥. We have

d(y, M) =|〈1, y〉|‖1‖2

=

(∫ 1

−1

t2dt

)(∫ 1

−1

1dt

)−1/2

=

√2

3. ¥

Problem 29.Consider the Hilbert space H = L2[−1, 1] equipped with the usual scalar product:

〈f, g〉 =

∫ 1

−1

f(t)g(t)dt, f, g ∈ H.

Let E = x ∈ H : f(−t) = f(t), t ∈ [−1, 1].(a) Show that E is closed in H. Find E⊥.(b) Calculate the distance from h to E for h(t) = et.

Solution.(a) Define f(t) := f(−t). Define the map

S : H → H defined by Sf = f .

Clearly, S is linear. S is bounded, so continuous. Indeed,

‖Sf‖2 = ‖f‖2 =

(∫ 1

−1

|f(t)|2dt

) 12

=

(∫ 1

−1

|f(−t)|2dt

) 12

= ‖f‖2.


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In fact, S is an isometry. It follows that I − S is continuous. By definition, E =Ker(I − S), so E is closed.By definition, E consists of all even functions, so E⊥ is the linear subspace of allodd functions. In fact, we have

f(t) =1

2[f(t) + f(−t)]

︸︷︷︸ϕ

+1

2[f(t)− f(−t)]

︸︷︷︸ψ

≡ ϕ(t) + ψ(t),

with ϕ is even, ψ is odd, and

〈ϕ, ψ〉 =

∫ 1

−1

ϕ(t)ψ(t)dt = 0, so that ϕ⊥ψ.

(b) The distance from h ∈ H to E is the length of the projection vector of h on E⊥.By the above expression,

ProjE⊥(h) =1

2[h(t)− h(−t)] =

1

2(et − e−t).

Therefore,

(dist(h,E)

)2=

∥∥∥∥1

2(et − e−t)

∥∥∥∥2

=1

4

∫ 1

−1

|et − e−t|2dt

=1

4(e2 − e−2 − 4).

Thus,

dist(h,E) =1

2

√e2 − e−2 − 4. ¥

Problem 30.Let H be a Hilbert space and M be a closed subspace of H. Denoting by P :H → M the orthogonal projection of H onto M, prove that, for any x, y ∈ H,

〈Px, y〉 = 〈x, Py〉.

(This is telling us that P is self-adjoint).

Solution.We know that if M is a closed subspace of H, then


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• For all u ∈ H, there exist unique uM ∈ M and uM⊥ ∈ M⊥ such that

u = uM + uM⊥ .

• If P : H → M the orthogonal projection of H onto M, then

Pu = uM .

Now for arbitrary x, y ∈ H, we have

x = xM + xM⊥ y = yM + yM⊥

Px = xM Py = yM .

With these, can have

〈Px, y〉 = 〈xM , yM + yM⊥〉 = 〈xM , yM〉,since 〈xM , yM⊥〉 = 0, and

〈x, Py〉 = 〈xM + xM⊥ , yM〉 = 〈xM , yM〉,since 〈xM⊥ , yM〉 = 0. Thus,

〈Px, y〉 = 〈x, Py〉. ¥

Problem 31.Let H be a Hilbert space and A ⊂ H a closed convex non-empty set. Prove thatPA : H → H is non-expansive, i.e.,

‖PA(x)− PA(y)‖ ≤ ‖x− y‖, ∀x, y ∈ H.

(PA is the orthogonal projection on A).

Solution.We claim:

(∗) Re⟨x− PA(x), PA(x)− a

⟩≥ 0, ∀x ∈ H, a ∈ A.

Let xA = PA(x). Then x can be decomposed uniquely as

x = xA + x′A, ∀xA ∈ A, x′A ∈ A⊥.


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We have

2Re⟨x− PA(x), PA(x)− a

⟩= 〈x′A, xA − a〉+ 〈xA − a, x′A〉= 〈x′A, x′A〉+ 〈xA − a, xA − a〉= ‖x′A‖2 + ‖xA − a‖2 ≥ 0.

Hence (∗) is proved.Replacing a with PA(y) in (∗), we obtain

(3.1) Re⟨x− PA(x), PA(x)− PA(y)

⟩≥ 0.

Analogously,

Re⟨y − PA(y), PA(y)− PA(x)

⟩≥ 0.

And therefore,

(3.2) Re⟨PA(y)− y, PA(x)− PA(y)

⟩≥ 0.

Adding 3.1 and 3.2 we get

Re⟨x− y − [PA(x)− PA(y)], PA(x)− PA(y)

⟩≥ 0, i.e.,

Re⟨x− y, PA(x)− PA(y)

⟩≥ ‖PA(x)− PA(y)‖2.(3.3)

Form the Cauchy-Schwarz inequality, we have

Re⟨x− y, PA(x)− PA(y)

⟩≤ |〈x− y, PA(x)− PA(y)〉|≤ ‖x− y‖ ‖PA(x)− PA(y)‖

and from here, replacing in 3.3, we obtain that

‖PA(x)− PA(y)‖2 ≤ ‖x− y‖ ‖PA(x)− PA(y)‖.Thus,

‖PA(x)− PA(y)‖ ≤ ‖x− y‖. ¥

Problem 32.Let X be a Hibert space, and G1 ⊂ G2 ⊂ ... ⊂ Gn ⊂ ... be a sequence of closedlinear subspaces of X. Let

G = Span

(⋃

n∈NGn

).


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(a) Prove that d(x,G) = limn→∞ d(x,Gn), ∀x ∈ X.(b) Prove that PG(x) = limn→∞ PGn(x), ∀x ∈ X. Note: PG is the orthogonalprojection of X on G.

Solution.(a) Let

A = Span

(⋃

n∈NGn

).

Then we have

d(x,G) = d(x, A) = d(x,A).

For any ε > 0 we have

d(x,G) + ε = d(x, A) + ε > d(x,A) = infa∈A

‖x− a‖.

From this, we deduce that there is some aε ∈ A such that

d(x, G) + ε > ‖x− aε‖.

Since aε ∈ Span(⋃

n∈NGn

), we can find λ1, ..., λk ∈ K and x1, ..., xk ∈

⋃n∈NGn such

that

aε = λ1x1 + ... + λkxk.

Then there are n1, ..., nk ∈ N such that x1 ∈ Gn1 , ..., xk ∈ Gnk. Let n = maxn1, ..., nk.

By hypothesis, the sequence (Gn) is increasing , so we have Gn1 , ..., Gnk⊂ Gn, so

x1, ..., xk ∈ Gn. And since Gn is a linear space,

aε = λ1x1 + ... + λkxk ∈ Gn.

That is,

(∗) ∀ε > 0, ∃n ∈ N : d(x,G) + ε > ‖x− aε‖ ≥ d(x,Gn).

From Gn ⊂ G it follows that

(∗∗) d(x,G) ≤ d(x,Gn), ∀n ∈ N.

(∗) and (∗∗) imply that

d(x,G) ≤ d(x,Gn) < d(x,G) + ε.


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Hence,d(x,G) = inf

n∈Nd(x,Gn).

From Gn ⊂ Gn+1, it follows that d(x,Gn+1) ≤ d(x,Gn). The sequence of real

numbers(d(x,Gn)

), which is decreasing and bounded below, must converge to its

infimum. Thuslim

n→∞d(x,Gn) = d(x, G).

(b) Let an = PGn(x), a = PG(x). Then by part (a) we have

limn→∞

‖x− an‖ = limn→∞

d(x,Gn) = d(x,G).

Since Gn ⊂ G, ∀n ∈ N, then (an) ⊂ G and so an → PG(x) in norm, that is,

limn→∞

‖an − a‖ = 0,

which gives thatlim

n→∞PGn(x) = PG(x). ¥

Problem 33.Let H be a Hilbert space.(a) Prove that for any two subspaces M, N of H we have

(M + N)⊥ = M⊥ ∩N⊥.

(b) Prove that for any two closed subspaces E, F of H we have

(E ∩ F )⊥ = E⊥ + F⊥.

Solution.(a) If x ∈ (M + N)⊥, then for every m ∈ M and n ∈ N we have

〈m + n, x〉 = 0

since m + n ∈ M + N . For n = 0 we have 〈m,x〉 = 0. This holds for all m ∈ M , sox ∈ M⊥. Similarly x ∈ N⊥. Thus x ∈ M⊥∩N⊥, and hence (M +N)⊥ ⊂ M⊥∩N⊥.If x ∈ M⊥ ∩N⊥, then we have

〈m,x〉 = 0 and 〈n, x〉 = 0, ∀m ∈ M, n ∈ N.


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Hence〈m + n, x〉 = 0.

This means that x ∈ (M + N)⊥. Hence M⊥ ∩N⊥ ⊂ (M + N)⊥.

(b) From part (a) it follows that

M + N = (M⊥ ∩N⊥)⊥.

Setting E⊥ in the place of M and F⊥ in the place of N , we obtain

E⊥ + F⊥ = (E ∩ F )⊥. ¥Why in question (b) E and F must be closed?

Problem 34.A system xii∈N in a normed space X is called a complete system ifSpan ∑n

i=1 αixi : ∀n ∈ N, αi ∈ F is dense on X.If xii∈N is a complete system in a Hilbert space H and x⊥xi for every i, showthat x = 0.

Solution.Given x ∈ X, if x⊥xi for every i, then x⊥ Span ∑n

i=1 αixi. Let D := Span ∑ni=1 αixi.

By definition, D = X. Then there exists a sequence (xn) in D converging to x andx⊥xn for every n. Hence

0 = 〈x, xn〉 → 〈x, x〉 = ‖x‖2 as n →∞.

Thus x = 0. ¥

Problem 35.Let H be a Hilbert space and ϕi∞i=1 an orthonormal system in H. Show that

‖ϕn − ϕm‖ =√

2 for m 6= n.

Solution.Using orthogonality and the fact that ‖ϕk‖ = 1, ∀k ∈ N, we get for n 6= m,

‖ϕn − ϕm‖2 = 〈ϕn − ϕm, ϕn − ϕm〉= 〈ϕn, ϕn〉 − 〈ϕn, ϕm〉 − 〈ϕm, ϕn〉+ 〈ϕm, ϕm〉= ‖ϕn‖2 + ‖ϕm‖2 = 2. ¥


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Problem 36.Let eii∈N be an orthonormal set in a Hilbert space H. Prove that

∞∑i=1

|〈x, ei〉〈y, ei〉| ≤ ‖x‖ ‖y‖, ∀x, y ∈ H.

Solution.Using the Cauchy-Schwarz inequality for `2 and the Bessel inequality for H, we have

∞∑i=1

|〈x, ei〉〈y, ei〉| ≤( ∞∑

i=1

|〈x, ei〉|2) 1

2( ∞∑

i=1

|〈x, ei〉|2) 1

2

≤ (‖x‖2) 1

2(‖y‖2

) 12 = ‖x‖ ‖y‖.

Problem 37.Let H be a Hilbert space and A and B be any two subsets of H. Show that(a) A⊥ is a closed subspace of H.(b) A ⊂ (A⊥)⊥ := A⊥⊥.(c) A ⊂ B ⇒ B⊥ ⊂ A⊥.

(d) A⊥ = A⊥

= A⊥.

Solution.(a) Let x, y ∈ A⊥ and α, β ∈ F. Then for any a ∈ A,

〈αx + βy, a〉 = 〈αx, a〉+ 〈βy, a〉= α〈x, a〉+ β〈y, a〉 = 0.

Hence αx + βy ∈ A⊥, so A⊥ is a subspace of H.We show now that A⊥ is closed. Suppose that the sequence (xn) in A⊥ convergesto some x ∈ H. Since 〈xn, a〉 = 0 for all a ∈ A, by continuity of the inner product,we have

〈x, a〉 = 〈 limn→∞

xn, a〉 = limn→∞

〈xn, a〉 = 0.

So x ∈ A⊥, and A⊥ is closed.


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(b) For any x ∈ A, we have x⊥A⊥. This implies that x ∈ (A⊥)⊥. Thus

A ⊂ (A⊥)⊥ := A⊥⊥.

(c) Take any x ∈ B⊥. Then x⊥y for all y ∈ B. But A ⊂ B, so x⊥y for all y ∈ A.Hence x ∈ A⊥.

(d) From (a) we have

A⊥ = A⊥,

which is the second equality in (d). Now pick any x⊥A, that is, x ∈ A⊥. If a ∈ A,then there exists a sequence (an) is A such that an → a. Since x⊥A, we have x⊥an

for all n. Hence

〈x, a〉 = 〈x, limn→∞

an〉 = limn→∞

〈x, an〉 = 0.

Thus we have x⊥A, so x⊥A. Therefore,

A⊥ ⊂ A⊥. (i)

On the other hand, by (b) we have

A ⊂ A =⇒ A⊥ ⊂ A⊥. (ii)

(i) and (ii) complete the proof. ¥

Problem 38.(a) Show that M :=

x = (xn) ∈ `2 : x2n = 0, ∀n ∈ N

is a closed subspace ofthe Hilbert space `2.(b) Find M⊥.

Solution.(a) Take any x = (xn), y = (yn) ∈ M . It is clear that for any scalars α, β,

(αx + βy)2n = αx2n + βy2n = 0.

That gives that αx + βy ∈ M . Hence M is a linear subspace of `2.Let us prove that it is closed. Take x ∈ M . There exists a sequence x(k) = (x

(k)n ) ∈ M

converging to x as k →∞. Since x(k)2n = 0, we obtain

x2n = limk→∞

x(k)2n = 0,


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that is, x ∈ M . Hence M is closed.(b) Now

z ∈ M⊥ ⇐⇒ 〈z, x〉 = 0, ∀x ∈ M

⇐⇒∞∑

n=0

z2n+1x2n+1 = 0 for all scalars x2n+1 such that∞∑

n=0

|x2n+1|2 < ∞

⇐⇒ z2n+1 = 0, ∀n = 0, 1, 2, ...

Therefore

M⊥ = z = (zn) ∈ `2 : z2n+1 = 0, ∀n = 0, 1, 2, .... ¥

Problem 39.Let V be a subspace of a Hilbert space.

(a) Show that V⊥

= V ⊥.(b) Show that V is dense in H if and only if V ⊥ = 0.

Solution.(a) From Problem 37d we get (a).(b) If V is dense, then V = H. Hence

V ⊥ = V⊥

= H⊥ = 0.Conversely, suppose V ⊥ = 0. If V is not dense in H, that is, V ( H, pickx ∈ H \ V . Let x′ = PV x. Then

x− x′ ∈ V⊥

= V ⊥ = 0.Thus x = x′ ∈ V . This is a contradiction. Thus V is dense in H. ¥

3.2 Weak convergence

Problem 40.Prove that in any finite dimensional vector space, strong convergence and weakconvergence are equivalent.


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3.2. WEAK CONVERGENCE 41

Solution.Consider first the case that X = Fd under the Euclidian norm ‖.‖2. Suppose thatthe sequence (xn) converges weakly to x in Fd. Then for each standard basis vectorek, k = 1, 1, .., d, we have

〈xn, ek〉 → 〈x, ek〉 as n →∞.

That is, weak convergence implies componentwise convergence. But since there areonly finitely many components, this implies norm convergence, since

‖xn − x‖22 =

d∑

k=1

|〈xn, ek〉 − 〈x, ek〉|2 → 0 as n →∞.

For the general case, choose any basis e1, e2, ..., ed for X, and then use the fact thatall norms on X are equivalent to define an isomorphism between X and Fd. ¥

Problem 41.Show that if the sequence (xn) in a normed space X is weakly convergent tox0 ∈ X, then

lim infn→∞

‖xn‖ ≥ ‖x0‖.

Solution.If x0 = 0 then ‖x0‖ = 0 and the statement is obviously true. Now assume ‖x0‖ 6= 0.By a well known theorem 1, there is some f ∈ X∗ such that

‖f‖ = 1, f(x0) = ‖x0‖.

Since (xn) converges weakly to x0 and f is continuous, we have

limn→∞

f(xn) = f(x0) = ‖x0‖.

Butf(xn) ≤ |f(xn)| ≤ ‖f‖ ‖xn‖ = ‖xn‖.

1(Kreyszig, p 223) Let X be a normed space and x0 6= 0 be an element in X. Then there existsa bounded linear functional f ∈ X∗ such that

‖f‖ = 1, f(x0) = ‖x0‖.


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Hence,lim infn→∞

‖xn‖ ≥ limn→∞

f(xn) = ‖x0‖. ¥

Note.If X = H is a Hilbert space, using the definition of weak convergence we can havedifferent solution.

‖x0‖2 = 〈x0, x0〉 = limn→∞

〈x0, xn〉.Since 〈x, xn〉 ≤ ‖x0‖ ‖xn‖, so we have

‖x0‖2 = limn→∞

〈x, xn〉 ≤ ‖x0| lim infn→∞

‖xn‖.

Thus‖x0‖ ≤ lim inf

n→∞‖xn‖.

Problem 42.Let X and Y be normed spaces, T ∈ B(X, Y ) and (xn) a sequence in X. Showthat if xn

w−→ x, then Txnw−→ Tx

Solution.

Recall: Definition of weak convergence in a normed space:

xnw−→ x ⇐⇒ f(xn) → f(x), ∀f ∈ X∗.

We must show thatϕ(Txn) → ϕ(Tx), ∀ϕ ∈ Y ∗.

That is,(ϕ T )xn → (ϕ T )x, ∀ϕ ∈ Y ∗.

But ϕ T ∈ X∗, so our hypothesis xnw−→ x guarantees our desired conclusion. ¥

Problem 43.Let H be a Hilbert space and (xn) be a sequence in H. Suppose xn

w−→ x. Showthat

limn→∞

‖xn − x‖ = 0 ⇐⇒ ‖x‖ ≥ lim supn→∞

‖xn‖ (1).


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3.2. WEAK CONVERGENCE 43

Solution.By Problem 41 we have

‖x‖ ≤ lim infn→∞

‖xn‖.So the right hand side of (1) is equivalent to ‖x‖ = limn→∞ ‖xn‖. Now note that

(i) ‖xn − x‖2 = ‖xn‖2 − 2Re〈xn, x〉+ ‖x‖2

(ii)∣∣‖xn‖ − ‖x‖

∣∣ ≤ ‖xn − x‖.

If limn→∞ ‖xn − x‖ = 0 then by (ii) we get

limn→∞

‖xn‖ = ‖x‖.

If limn→∞ ‖xn‖ = ‖x‖ then (i) give that

limn→∞

‖xn − x‖ = 0. ¥.


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Chapter 4

Linear Operators - LinearFunctionals

4.1 Linear bounded operators

Problem 44.Let (X, ‖.‖1) and (Y, ‖.‖2) be norm spaces, and T ∈ B(X,Y ). We define ‖T‖ as

‖T‖ := infx∈X

M : ‖Tx‖2 ≤ M‖x‖1.

(a) Show that

‖T‖ = sup‖x‖1=1

‖Tx‖2 = sup‖x‖1≤1

‖Tx‖2 = supx∈X, x 6=0

‖Tx‖2

‖x‖1

.

(b) Show that‖T‖ = sup

‖x‖1<1

‖Tx‖2.

Solution.(a) We have

‖Tx‖2

‖x‖1

≤ M, for all x 6= 0, x ∈ X.

45


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46 CHAPTER 4. LINEAR OPERATORS - LINEAR FUNCTIONALS

By definition,

‖T‖ := infx∈X

M : ‖Tx‖2 ≤ M‖x‖1 = supx∈X, x6=0

‖Tx‖2

‖x‖1

(i).

Now, let y = x‖x‖1 for x ∈ X, x 6= 0. Then y ∈ X and ‖y‖ = 1. By (i) we have

‖T‖ = supx∈X, x6=0

‖Tx‖2

‖x‖1

= sup‖y‖=1

∥∥∥∥T

(‖x‖1y

‖x‖1

)∥∥∥∥2

= sup‖y‖=1

‖Ty‖2 = sup‖x‖=1

‖Tx‖2 (ii).

From (ii) it follows that

‖T‖ = sup‖x‖1=1

‖Tx‖2 ≤ sup‖x‖1≤1

‖Tx‖2 ≤ supx∈X, ‖x‖1≤1

‖Tx‖2

‖x‖1

≤ supx∈X, x6=0

‖Tx‖2

‖x‖1

= ‖T‖.

Thus,

‖T‖ = sup‖x‖1≤1

‖Tx‖2.

(b) Let

B := x ∈ X : ‖x‖1 ≤ 1 and Bo := x ∈ X : ‖x‖1 < 1.Since ‖T‖ = supx∈B ‖Tx‖2, there exists a sequence (xn) in B such that

‖T‖ = limn→∞

‖Txn‖2.

Consider the sequence (yn) defined by yn = (1− 12n )xn. It is clear that yn ∈ Bo for

all n ∈ N. Moreover,

limn→∞

‖Tyn‖2 = limn→∞

∥∥∥∥T

(1− 1

2n

)xn

∥∥∥∥ = limn→∞

(1− 1

2n

)‖Txn‖2

= limn→∞

(1− 1

2n) lim

n→∞‖Txn‖2 = ‖T‖.

Thus

supx∈Bo

‖Tx‖2 ≥ ‖T‖.

On the other hand,

supx∈Bo

‖Tx‖2 ≤ supx∈B

‖Tx‖2 = ‖T‖.

The proof is complete. ¥


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4.1. LINEAR BOUNDED OPERATORS 47

Problem 46.Let H = `2 be the well known Hilbert space. Consider the left shift defined by

L : `2 → `2, x = (x1, x2, x3, ...) 7→ Lx = (x2, x3, ...).

(a) Show that L is a linear bounded operator. Find ‖L‖.(b) Define the right shift and answer the same question as in part (a).

Solution.We answer only the first part.• It is easy to check the linearity of L.• For any x = (x1, x2, x3, ...) ∈ `2, we have

‖Lx‖2 = ‖(x2, x3, ...)‖2 =∞∑

n=2

|xn|2

≤∞∑

n=1

|xn|2 = ‖x‖2.

Hence‖Lx‖ ≤ ‖x‖, ∀x ∈ `2.

This implies that T is bounded, and

‖L‖ ≤ 1. (∗)On the other hand, consider the sequence e = (0, 1, 0, 0, ...) ∈ `2. We have

‖e‖ = 1 and ‖Le‖ = ‖(1, 0, 0, ...)‖ = 1.

So (see Problem 39)‖L‖ = sup

‖x‖=1

‖Lx‖ ≥ 1. (∗∗)

Combining (∗) and (∗∗) we obtain ‖L‖ = 1. ¥

Problem 47.Let X = C[0, 1] with the max-norm (the uniform norm). We define the integraloperator

K : X → X by Kf(x) =

∫ x

0

f(y)dy.

Show that K is bounded. Find ‖K‖.


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Solution.The operator K is bounded. Indeed,

‖Kf‖ ≤ supx∈[0,1]

∫ x

0

|f(y)|dy ≤∫ 1

0

|f(y)|dy ≤ ‖f‖.

Hence ‖K‖ ≤ 1.In fact, ‖K‖ = 1, since 1 ∈ X, K1 = x and so ‖K1‖ = ‖x‖ = 1. ¥

**Problem 48.Let X = L2[0, 1] with the norm ‖.‖2. We define the integral operator

A : X → X by Af(x) =

∫ x

0

f(y)dy.

Show that A is bounded. Find ‖A‖.

Solution.Warning: It’s completely different from the previous problem!!!

First solution: Using Cauchy-Schwarz inequality, we get

‖Af‖22 =

∫ 1

0

∣∣∣∣∫ t

0

f(s)ds

∣∣∣∣2

dt =

∫ 1

0

∣∣∣∣∣∫ t

0

√cos

πs

2.

f(s)√cos πs

2

ds

∣∣∣∣∣

2

dt

≤∫ 1

0

(∫ t

0

cosπs

2ds

∫ t

0

|f(s)|2cosπs

2

ds

)dt

=2

π

∫ 1

0

(∫ t

0

sinπt

2.|f(s)|2cosπs

2

ds

)dt

=2

π

∫ t

0

(∫ 1

0

sinπt

2

|f(s)|2cosπs

2

dt

)ds

=2

π

∫ t

0

(∫ 1

0

sinπt

2dt

) |f(s)|2cosπs

2

ds

=

(2

π

)2 ∫ t

0

|f(s)|2cosπs

2

ds.

Equality holds for f(s) = cos πs2

. Thus

‖A‖ =2

π. ¥


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Second solution: We find the norm of T = A∗A, then ‖A‖ =√‖T‖.

Since A is compact (we will see this somewhere later) and A∗A is self adjoint, wehave that T is a compact, normal operator. Hence, its spectrum is countable with0 as the only possible cluster point and ‖T‖ is equal to the spectral radius of T .These two facts together imply that

‖T‖ = max|λ| : Tf = λf for some f ∈ X,

We have that A∗ is given by

A∗f(x) =

∫ 1

x

f(y)dy.

Hence

(∗) Tf(x) =

∫ 1

x

∫ y

0

f(z)dz dy.

By the Fundamental Theorem of Calculus, Tf is twice differentiable for all f ∈ X,so any eigenvector of T must satisfy the differential equation

λd2

dx2f(x) =

d2

dx2

∫ 1

x

∫ y

0

f(z)dz dy =d

dx

(−

∫ x

0

f(z)dz

)= −f(x), λ ∈ C \ 0.

By a theorem from differential equations, we know that there are only two linearlyindependent solutions to the above differential equation over C, namely, eiωx ande−iωx, where ω2 = λ, λ, ω 6= 0. Hence the general solution to the above equation is

f(x) = αeiωx + βe−iωx, α, β ∈ C.


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When we apply T , we get

T (αeiωx + βe−iωx) =

∫ 1

x

∫ y

0

(αeiωz + βe−iωz)dz dy

=

∫ 1

x

(αeiωz

iω− βe−iωz

iω

) ∣∣∣y

0dy

=

∫ 1

x

[(αeiωy

iω− βe−iωy

iω

)−

(α

iω− β

iω

)]dy

=

(αeiωy

−ω2− βe−iωy

−ω2

)−

(α

iω− β

iω

)y∣∣∣1

x

=

(αeiω

−ω2− βe−iω

−ω2

)−

(α

iω− β

iω

)

−[(

αeiωx

−ω2− βe−iωx

−ω2

)−

(α

iω− β

iω

)x

]

=1

ω2(αeiωx + βe−iωx) +

(α

iω− β

iω

)x

−(

αeiω

ω2− βe−iω

ω2

)−

(α

iω− β

iω

).

Now since αeiωx + βe−iωx is an eigenvector, we must have that

(α

iω− β

iω

)x−

(αeiω

ω2− βe−iω

ω2

)−

(α

iω− β

iω

)= 0, ∀x ∈ [0, 1],

which implies α = β. Hence

αeiωx + βe−iωx = 2α cos ωx.

Moreover, we must have that Tf(1) = 0 for any eigenvector f by (∗), so 2α cos ω = 0.Since α 6= 0, we must have that

ω =(2n + 1)π

2, n ∈ Z.

Hence the eigenvalues are of the form

λn =1

ω2n

=22

(2n + 1)2π2, n ∈ Z.

Thus

maxλn : n ∈ Z =4

π2(take n = 0,−1).


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We conclude that

‖T‖ =4

π2and ‖A‖ =

2

π. ¥

Problem 49.Let a, b be real numbers such that a < b. Consider the Hilbert space L2[a, b] overR and the operator T : L2[a, b] → R defined by

Tf =

∫ b

a

f(x)dx, f ∈ L2[a, b].

(a) Show that T is bounded. Compute ‖T‖.(b) According to the Riesz’s Theorem, there exists a function g ∈ L2[a, b] suchthat

Tf = 〈f, g〉 for all f ∈ L2[a, b].

Find such a function g and verify that ‖g‖L2 = ‖T‖.

Solution.(a) By Holder’s inequality, we have

|Tf | =

∣∣∣∣∫ b

a

f(x)dx

∣∣∣∣

≤∫ b

a

|1.f(x)|dx

≤(∫ b

a

12dx

)1/2 (∫ b

a

|f(x)|2)1/2

=√

b− a ‖f‖L2 .

Hence, T is bounded, i.e., T ∈ B(L2,R) = (L2)∗, the dual space of L2.From the above we get

‖T‖(L2)∗ ≤√

b− a.

Now consider the function h(x) = 1√b−a

, x ∈ (a, b). It is obviously that h ∈ L2(a, b)and

‖h‖L2 =

(∫ b

a

(b− a)−1

)1/2

= 1 and |Th| =∣∣∣∣∫ b

a

1√b− a

dx

∣∣∣∣ =√

b− a.

Hence, √b− a = |Th| ≤ ‖T‖(L2)∗‖h‖L2 = ‖T‖(L2)∗ .


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Therefore,‖T‖(L2)∗ =

√b− a.

(b) By the Riesz’s Theorem, there exists a function g ∈ L2[a, b] such that Tf =〈f, g〉 for all f ∈ L2[a, b]. Here, functions are real-valued, so we can write

Tf = 〈f, g〉 ⇔∫ b

a

f(x)dx =

∫ b

a

f(x)g(x)dx.

It is evident that the above equation is satisfied for all f ∈ L2[a, b] if we chooseg(x) = 1 on [a, b]. By the uniqueness of this representation guaranteed by Riesz’sTheorem, we can definitely conclude that

g(x) = 1, x ∈ [a, b].

Also, we can verify that

‖g‖L2 =

(∫ b

a

12dx

)1/2

=√

b− a = ‖T‖(L2)∗ . ¥

Problem 50.Let X, Y be normed spaces and T ∈ B(X, Y ). Consider the following statement:

T is an isometry ⇔ ‖T‖ = 1.

Do you agree with it? Why?

Solution.• The direct way (⇒) is correct. Indeed, suppose that T is an isometry; then

‖T‖B(X,Y ) = supx∈X

‖x‖X=1

‖Tx‖Y

= supx∈X

‖x‖X=1

‖x‖X (since ‖Tx‖Y = ‖x‖X)

= 1.

• The other way (⇐) is false. The left shift on `2 is a counter-example (see Problem40). ¥


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Problem 51.Define T : C[0, 1] → C[0, 1] by

(Tx)(t) = t

∫ t

0

x(s)ds.

(a) Prove that T is a bounded linear operator. Compute ‖T‖.(b) Prove that the inverse T−1 : Image(T ) → C[0, 1] exists but not bounded.

Solution.(a) For all x1, x2 ∈ C[0, 1] and all α1, α2 ∈ R and all t ∈ [0, 1] we have

T (α1x1 + α2x2)(t) = t

∫ t

0

(α1x1 + α2x2)(s)ds

= α1t

∫ t

0

x1(s)ds + α2t

∫ t

0

x2(s)ds

= α1T (x1)(t) + α2T (x2)(t)

= [α1T (x1) + α2T (x2)] (t).

Hence,T (α1x1 + α2x2) = α1T (x1) + α2T (x2).

This shows that T is linear.For each x ∈ C[0, 1] we have

‖Tx‖ = maxt∈[0,1]

∣∣∣∣t∫ t

0

x(s)ds

∣∣∣∣ ≤ maxt∈[0,1]

|t|∫ t

0

|x(s)|ds

≤ maxt∈[0,1]

t

∫ t

0

‖x‖ds = maxt∈[0,1]

t2‖x‖ = ‖x‖.

Hence T is a bounded with ‖T‖ ≤ 1. Moreover, if x(t) = 1, t ∈ [0, 1] then ‖x‖ = 1and (Tx)(t) = t

∫ t

0ds = t2, therefore, ‖Tx‖ = maxt∈[0,1] t

2 = 1. Thus,

1 = ‖Tx‖ ≤ ‖T‖‖x‖ and so ‖T‖ ≥ 1.

Hence we have proved that ‖T‖ ≤ 1 and ‖T‖ ≥ 1, so ‖T‖ = 1.(b) Suppose that x ∈ C[0, 1] satisfies

Tx = 0, i.e., t

∫ t

0

x(s)ds = 0, ∀t ∈ [0, 1].


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It follows that∫ t

0x(s)ds = 0, ∀t ∈ (0, 1]. By differentiation w.r.t. t we get x(t) =

0,∀t ∈ (0, 1]. Since x(t) is continuous, we then also have x(0) = 0. Hence, x = 0.We have thus proved

∀x ∈ C[0, 1], T (x) = 0 ⇒ x = 0.

Hence, T−1 exists.Now we show that T−1 is not bounded. Given any n ∈ N, we let xn(t) = tn. Then

xn ∈ C[0, 1] and ‖xn‖ = maxt∈[0,1]

|tn| = 1.

We let

yn(t) = Txn(t) = t

∫ t

0

snds =1

n + 1tn+2.

Then

‖yn‖ = maxt∈[0,1]

∣∣∣∣1

n + 1tn+2

∣∣∣∣ =1

n + 1.

Also by construction, yn Image(T ) and T−1yn = xn; thus

‖T−1yn‖ = ‖xn‖ = 1.

This shows that T−1 cannot be bounded. (For if T−1 were bounded, then we wouldhave

‖T−1yn‖ ≤ ‖T−1‖‖yn‖, i.e., 1 ≤ ‖T−1‖ 1

n + 1, ∀n ∈ N.

This is impossible. ¥

Problem 53.Let 1 < p < ∞ and q be its exponent conjugate. For f ∈ Lp(0,∞), let

(Tf)(x) =1

x

∫ x

0

f(t)dt =

∫ 1

0

f(tx)dt.

Show that(a) Tf is well-defined on (0,∞), and that Tf is continuous with respect to x.(b) Tf ∈ Lp(0,∞).(c) T is a bounded linear operator from Lp(0,∞) to itself. Calculate its norm.(Hint: Use fn = x−1/pχ1≤x≤n.)


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Solution.(a) For f ∈ Lp(0,∞), we need to show ψ : x 7→ ∫ x

0f(t)dt is well-defined and

continuous w.r.t. x. By Holder’s inequality we have∫ x

0

|f(t)|dt ≤ |x|1/q‖f‖p < ∞,

which shows that ψ is well-defined. Also, for x, y ∈ (0,∞), we have

ψ(x)− ψ(y)| =∣∣∣∣∫ y

x

f(t)dt

∣∣∣∣ ≤∫ y

x

|f(t)|dt ≤ |x− y|1/q‖f‖p.

This shows that ψ is continuous.Since (Tf)(x) = ψ(x)

x, T f is also continuous on (0,∞). Moreover we have

|(Tf)(x)| ≤ x−1/p‖f‖p → 0 as x →∞.

So (Tf)(x) → 0 as x →∞.(b) To prove this part (b) we will use the following theorem:

Given a σ-finite measure space (X, A, µ). Let 1 < p < ∞ and q be its exponentconjugate. If |ϕ| < ∞ a.e on X and if

∫X

ϕψdµ exists in C for every ψ ∈ Lq(X),then f ∈ Lp(X).

Now, for any g ∈ Lq(0,∞), by Fubini’s theorem we have

∫

(0,∞)

|(Tf)(x)g(x)|dx ≤∫

(0,∞)

∫ 1

0

|(Tf)(x)||g(x)|dtdx

=

∫

(0,∞)

∫ 1

0

1

t|f(x)||g(x/t)|dtdx

≤ ‖f‖p

∫ 1

0

‖g(./t)‖qdt

t.

But

‖g(./t)‖q =

(∫

(0,∞)

|g(x/t)|qdx

)1/q

= t1/q‖g‖q.

Thus,

∫

(0,∞)

|(Tf)(x)g(x)|dx ≤ ‖f‖p‖g‖q

∫ 1

0

t1/q−1dt = q‖f‖p‖g‖q < ∞.

Hence, Tf ∈ Lp(0,∞).(c) By the last inequality we have

‖Tf‖p ≤ q‖f‖p.


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This shows that T is an bounded operator on Lp(0,∞) with ‖T‖p,p ≤ q. We wouldlike to establish the equality. Consider the function fn, n ∈ N defined by fn =x−1/pχ1≤x≤n. We have

‖fn‖pp =

∫ n

1

(x−1/p

)pdx = ln n.

For x > 1 we have

(Tfn)(x) =1

x

∫ minx,n

1

t−1/pdt = qminx, n1/q − 1

x.

For 0 < ε < 1, since (1− ε)p ≥ 1− pε we have

∫

(0,∞)

[(Tfn)(x)]pdx = qp

(∫ n

1

(1− x−1/q)p 1

xdx + (n1/q − 1)p

∫ ∞

n

x−pdx

)

≥ qp

(ln n− p

q+

1

p− 1

)

≥ qp

(‖fn‖p

p −p

q+

1

p− 1

).

This implies that

limn→∞

‖Tfn‖p

‖fn‖p

≥ q.

Hence, ‖T‖p,p ≥ q. Finally, we get

‖T‖p,p = q. ¥

Problem 54.Let (cj)

∞j=1 be a sequence of complex numbers. Define an operator D on `2 by

Dx = (c1x1, c2x2, ...) for x = (x1, x2, ...) ∈ `2.

Prove that D is bounded if and only if (cj)∞j=1 is bounded, and in this case ‖D‖ =

supj |cj|.

Solution.


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• Suppose (cj)∞j=1 is bounded. Let M = supj |cj| < ∞. Then

‖Dx‖ =

( ∞∑j=1

|cjxj|2)1/2

=

( ∞∑j=1

|cj|2|xj|2)1/2

≤( ∞∑

j=1

M2|xj|2)1/2

= M

( ∞∑j=1

|xj|2)1/2

= M‖x‖.Hence, D is bounded and ‖D‖ ≤ M .• Suppose D is bounded. We want to show that (cj)

∞j=1 is bounded. Consider the

vector ej = (0, 0, .., 0, 1, 0, ...) where the number 1 appears at the j-th coordinate.Clearly ‖ej‖ = 1 and ‖Dej‖ = |cj| for all j = 1, 2, .. Since D is bounded,

|cj| = ‖Dej‖ ≤ ‖D‖ for any j = 1, 2, ...

Hence (cj)∞j=1 is bounded and M = supj |cj| ≤ ‖D‖. Finally

‖D‖ = M. ¥

Problem 55.Prove that B(F, Y ) is not a Banach space if Y is not complete.

Hint: Take a Cauchy sequence (yn) in Y which does not converge and consider the sequenceof operators (Bn) defined by :

Bnλ := λyn; λ ∈ F.

Solution.We follow the suggestion above. It is easy to see that

Bn ∈ B(F, Y ) and ‖Bn‖ = ‖yn‖, ∀n ∈ N.

Since (Bn −Bm)λ = λ(yn − ym), we have

‖Bn −Bm‖ = ‖yn − ym‖, ∀n ∈ N.

Therefore (Bn) is a Cauchy sequence in B(F, Y ). Suppose there exists B ∈ B(F, Y )such that ‖Bn − B‖ → 0 as n →∞. Let y := B1 ∈ Y , where 1 is the unit elementin F. Then

‖yn − y‖ = ‖Bn1−B1‖ ≤ ‖Bn −B‖ → 0 as n →∞,


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i.e., the sequence (yn) converges to y. This contradiction proves that (Bn) cannotbe convergent. Hence, B(F, Y ) is not a Banach space. ¥

Problem 56.Let T1, T2, ... be the following bounded linear operators `1 → `∞:

T1(x1, x2, x3, ...) = (x1, x1, x1, x1, ...)

T2(x1, x2, x3, ...) = (x1, x2, x2, x2, ...)

T3(x1, x2, x3, ...) = (x1, x2, x3, x3, ...)...etc

Prove that the sequence (Tn) is strongly operator convergent. Also prove that(Tn) is not uniformly operator convergent.

Solution.Let T : `1 → `∞ be the bounded linear operator given by

T (x1, x2, x3, ...) = (x1, x2, x3, ...).

T is obviously linear. It is bounded with norm ‖T‖ ≤ 1. Indeed, if x = (x1, x2, x3, ...) ∈`1 then ∞∑

k=1

|xk| = ‖x‖1 < ∞, so |xk| ≤ ‖x‖1, ∀k = 1, 2, ...

Hence,

‖x‖∞ ≤ ‖x‖1.

We claim that (Tn) strongly operator converges to T .For any x = (x1, x2, x3, ...) ∈ `1,

‖Tnx− Tx‖∞ = ‖(0, ..0, xn − xn+1, xn − xn+2, ...)‖∞= sup

j≥1|xn − xn+j|.

Since x = (x1, x2, x3, ...) ∈ `1,∑∞

k=1 |xk| < ∞, so we have limk→∞ |xk| = 0. Hence,given any ε > 0, there is some K ∈ N such that

|xk| ≤ ε, for all k ≥ K.

Then if n ≥ K we have

n + j ≥ K, ∀j ∈ N.


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Hence,

|xn − xn+j| ≤ |xn|+ |xn+j| ≤ 2ε.

Thus, for n ≥ K,

‖Tnx− Tx‖∞ = supj≥1

|xn − xn+j| ≤ 2ε.

In other words,

limn→∞

‖Tnx− Tx‖∞ = 0.

This is true for every x ∈ `1. Hence, the sequence (Tn) is strongly operator conver-gent to T .It follows from this that if (Tn) would be uniformly operator convergent, then thelimit must be equal to T . We show that this is not true. Now we have

(Tn − T )(x) = (0, .., 0, xn − xn+1, xn − xn+2, ...).

In particular, if x = (0, .., 0, 1, 0, ..) with 1 at the n-th position, then

(Tn − T )(x) = (0, .., 0, 1, 1, ...).

Here ‖x‖1 = 1 and ‖(0, .., 0, 1, 1, ...)‖∞ = 1. Hence,

‖Tn − T‖ ≥ 1, for all n.

Thus, (Tn) is not uniformly operator convergent. ¥

Problem 57.Let H1 and H2 be two Hilbert spaces. Let a1, ..., an be an orthonormal system ofH1 and b1, ...,n be an orthonormal system of H2, and λ1, ..., λn ∈ K. Considerthe operator

U : H1 → H2 defined by U(x) =∑i=1

λibi〈x, ai〉.

Calculate ‖U‖.

Solution.


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From the Pythagoras theorem and the Bessel inequality we have

‖U(x)‖2 =n∑

i=1

|λi|2‖bi‖2|〈x, ai〉|2

=n∑

i=1

|λi|2|〈x, ai〉|2

≤ M2

n∑i=1

|〈x, ai〉|2 ≤ M2‖x‖2,

where M = max1≤i≤n |λi|. Hence, we have that

‖U(x)‖ ≤ M‖x‖, ∀x ∈ H.

Therefore, ‖U‖ ≤ max1≤i≤n |λi|.On the other hand, we have

‖U(ai)‖ ≤ ‖U‖ ‖ai‖ = ‖U‖,U(ai) = λibi, ∀i ∈ 1, ..., n.

It follows that |λi| ≤ ‖U‖, ∀i ∈ 1, ..., n. This implies max1≤i≤n |λi| ≤ ‖U‖. Thus,

‖U‖ = max1≤i≤n

|λi|. ¥

Problem 58.(a) Let X be a Hilbert space. Let a, b ∈ H be two non-zero orthogonal elements.Consider the operator

U : H → H, U(x) = a〈x, b〉+ b〈x, a〉.

Calculate ‖U‖.(b) Consider the operator T : L2[0, π] → L2[0, π] defined by

(Tf)(x) = sin x

∫ π

0

f(t) cos tdt + cos x

∫ π

0

f(t) sin tdt.

Calculate ‖T‖.

Solution.(a) Note that a〈x, b〉 and b〈x, a〉 are two orthogonal vectors. Using the Pythagorastheorem we have

‖U(x)‖2 = ‖a‖2|〈x, b〉|2 + ‖b‖2|〈x, a〉|2.


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From the Bessel inequality we deduce that

‖U(x)‖2 = ‖a‖2‖b‖2

(∣∣∣∣⟨

x,b

‖b‖⟩∣∣∣∣

2

+

∣∣∣∣⟨

x,a

‖a‖⟩∣∣∣∣

2)≤ ‖a‖2‖b‖2‖x‖2.

Hence, ‖U‖ ≤ ‖a‖ ‖b‖.But

U(a) = a〈a, b〉+ b〈a, a〉 = ‖a‖2b.

Therefore,‖U(a)‖ = ‖a‖2‖b‖ ≤ ‖U‖ ‖a‖.

Hence, ‖U‖ ≥ ‖a‖ ‖b‖. Thus,

‖U‖ = ‖a‖ ‖b‖.

(b) Let H = L2[0, π]. Then H, with the usual inner product, is a Hilbert space.The two vectors a = sin x and b = cos x are orthogonal. Indeed,

〈a, b〉 = 〈sin x, cos x〉 =

∫ π

0

sin x cos xdx = 0.

By (a) we get‖T‖ = ‖ sin x‖ ‖ cos x‖.

But

‖ sin x‖2 =

∫ π

0

(sin x)2dx =π

2; ‖ cos x‖2 =

∫ π

0

(cos x)2dx =π

2.

Thus,

‖T‖ =π

2. ¥

Problem 59.(a) Let H be a Hilbert space and e1, e2 ⊂ H an orthonormal system. Let

A =

(a bc d

)

be scalar square matrix. Consider the operators U, V : H → H defined by

U(x) = a〈x, e1〉e1 + b〈x, e2〉e2,

V (x) = c〈x, e1〉e1 + d〈x, e2〉e2.


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Prove that‖U + V ‖2 + ‖U − V ‖2 = 2(‖U‖2 + ‖V ‖2)

if and only if

(max|a+c|,|b+d|)2+(max|a−c|,|b−d|)2=2(max|a|,|b|2+max|c|,|d|2).

(b) Prove that if dim H ≥ 2 then B(H) is not a Hilbert space

Solution.(a) Using Problem 57 we have

‖U‖ = max|a|, |b|; ‖V ‖ = max|c|, |d|.‖U + V ‖ = max|a + c|, |b + d|; ‖U − V ‖ = max|a− c|, |b− d|.

From here we obtain the statement.(b) Since dim H ≥ 2 we can find x, y ∈ H, two linearly independent vectors. Usingthe Gram-Schmidt procedure on x, y we can construct an orthonormal systeme1, e2 ⊂ H. Now construct U, V using the matrix

A =

(2 10 2

).

That is U, V : H → H defined by

U(x) = 2〈x, e1〉e1 + 〈x, e2〉e2, V (x) = 2〈x, e2〉e2.

If B(H) were a Hilbert space, then by the parallelogram law we must have

‖U + V ‖2 + ‖U − V ‖2 = 2(‖U‖2 + ‖V ‖2),

i,e., by (a)

(max2, 3|)2 + (max2, 1)2 = 2(max2, 12 + max0, 22

).

This is not true. ¥


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4.2. LINEAR FUNCTIONALS 63

4.2 Linear Functionals

Review:

Definition 1 Let X be a linear space. A map p : X → R is called sublinear if it is subadditiveand positive homogeneous, i.e.,

p(x + y) ≤ p(x) + p(y), ∀x, y ∈ X.

p(λx) = λp(x), ∀x ∈ X, ∀λ ≥ 0.

Definition 2 Let X be a linear space. A map p : X → K is called seminorm if it is subadditiveand absolutely homogeneous, i.e.,

p(x + y) ≤ p(x) + p(y), ∀x, y ∈ X.

p(λx) = |λ|p(x), ∀x ∈ X, ∀λ ∈ K.

The Hahn-Banach theorems

Theorem 1 (The case K = R). Let X be a real linear space, G ⊂ X a linear subspace, p : X → Ra sublinear functional and f : G → R a linear functional such that f(x) ≤ p(x), ∀x ∈ G. Thenthere is a linear functional f : X → R which extends f , i.e.,

f(x) = f(x), ∀x ∈ G and f(x) ≤ p(x), ∀x ∈ X.

Theorem 2 (The case K = R or C). Let X be a linear space, G ⊂ X a linear subspace,p : X → R a seminorm and f : G → K a linear functional such that |f(x)| ≤ p(x), ∀x ∈ G. Thenthere is a linear functional f : X → K which extends f such that

|f(x)| ≤ p(x), ∀x ∈ X.

Theorem 3 (The normed space case). Let X be a linear space, G ⊂ X a linear subspace andf : G → K a linear and continuous functional. Then there is f : X → K, a linear and continuousfunctional which extends f such that ‖f‖ = ‖f‖. Such an f is called a Hahn-Banach extension forf .

Theorem 4 Let X be a normed space, x0 ∈ X and G ⊂ X a linear subspace such that δ =d(x0, G) > 0. Then there is f : X → K, a linear and continuous functional, such that

f = 0 on G, f(x0) = 1 and ‖f‖ =1δ.


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Three classic problems.

Problem 60.The dual space of `1 is `∞, that is, (`1)∗ = `∞.

Solution.Let (ek) be the standard basis for `1, where ek = δkj. Every x ∈ `1 has uniquerepresentation

x =∞∑

k=1

ξkek.

Norms on `1 and on `∞ are respectively:

‖x‖1 =∞∑

k=1

|ξk|, ‖x‖∞ = supk∈N

|ξk|.

Take f ∈ (`1)∗, that is, f ∈ B(`1,F). For x ∈ `1,

f(x) = f

( ∞∑

k=1

ξkek

)=

∞∑

k=1

ξkγk where γk = f(ek).

Then for all k ∈ N,

|γk| = |f(ek)| ≤ ‖f‖ ‖ek‖ = ‖f‖.So,

(∗) supk∈N

|γk| ≤ ‖f‖.

On the other hand, for every b = (βk) ∈ `∞ we can obtain a corresponding boundedlinear functional g on `1. In fact, we may define g on `1 by

g(x) =∞∑

k=1

ξkβk where x = (ξk) ∈ `1.

Then g : `1 → F is linear, and the boundedness follows from

|g(x)| ≤∞∑

k=1

|ξkβk| ≤ supk∈N

|βk|∞∑

k=1

|ξk| = supk∈N

|βk| ‖x‖1.


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Hence g ∈ (`1)∗.We finally show that the norm of f is the norm on `∞. We have

|f(x)| =∣∣∣∣∣∞∑

k=1

ξkγk

∣∣∣∣∣ ≤ supk∈N

|γk|∞∑

k=1

|ξk| = ‖x‖ supk∈N

|γk|.

Taking the supremum over x such that ‖x‖ = 1, we see that

‖f‖ ≤ supk∈N

|γk| = ‖(γk)‖∞.

From(∗) and this, we obtain‖f‖ = ‖(γk)‖∞.

Thus there is an isometric isomorphism between (`1)∗ and `∞, so that we can write(`1)∗ = `∞. ¥

Problem 61.The dual space of `p is `q, that is, (`p)∗ = `q, here 1 < p < ∞ and 1

p+ 1

q= 1.

Solution.The basis for `p is (ek), where ek = δkj as in the previous problem. Every x ∈ `p

has unique representation

x =∞∑

k=1

ξkek.

Take f ∈ (`p)∗, that is, f ∈ B(`p,F). For x ∈ `p,

(1) f(x) =∞∑

k=1

ξkγk where γk = f(ek).

Let q be the conjugate of p. Consider xn = (ξ(n)k ) with

ξ(n)k =

|γk|qγk

if 1 ≤ k ≤ n and γk 6= 0

0 if k > n or γk = 0.

By substituting this into (1) we obtain

f(xn) =∞∑

k=1

ξ(n)k γk =

n∑

k=1

|γk|q.


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We also have

f(xn) ≤ ‖f‖ ‖xn‖ = ‖f‖(

n∑

k=1

|ξ(n)k |p

) 1p

= ‖f‖(

n∑

k=1

|γk|(q−1)p

) 1p

= ‖f‖(

n∑

k=1

|γk|q) 1

p

.

Together,

f(xn) =n∑

k=1

|γk|q ≤ ‖f‖(

n∑

k=1

|γk|q) 1

p

.

Dividing the last factor and using 1− 1p

= 1q, we get

(n∑

k=1

|γk|q)1− 1

p

=

(n∑

k=1

|γk|q) 1

q

≤ ‖f‖.

Since n is arbitrary, letting n →∞, we obtain

(2)

( ∞∑

k=1

|γk|q) 1

q

≤ ‖f‖.

Conversely, for any b = (βk) ∈ `q we can get a corresponding bounded linear func-tional g on `p. In fact, we may define g on `p by setting

g(x) =∞∑

k=1

ξkβk where x = (ξk) ∈ `p.

Then g is linear. The boundedness follows from the Holder inequality. Hence g ∈(`p)∗. We finally prove that the norm of f is the norm of (γk) in `q. From (1) andthe Holder inequality we have

|f(x)| ≤∣∣∣∣∣∞∑

k=1

ξkγk

∣∣∣∣∣ ≤( ∞∑

k=1

|ξk|p) 1

p( ∞∑

k=1

|γk|q) 1

q

= ‖x‖( ∞∑

k=1

|γk|q) 1

q

.


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Hence

‖f‖ ≤( ∞∑

k=1

|γk|q) 1

q

.

From (2) we obtain

‖f‖ =

( ∞∑

k=1

|γk|q) 1

q

.

This can be written‖f‖ = ‖(γk)‖q where γk = f(ek).

Thus there is an isometric isomorphism between (`p)∗ and `q. ¥

Problem 62.The dual space of c0 is `1, that is, (c0)

∗ = `1.

Solution.Recall that c0 ⊂ `∞, and if x = (λn) ∈ c0 the norm of x in c0 is ‖x‖ = supn∈N |λn|.Let x = (λn) ∈ c0 and (ek), where ek = δkj, be the basis for `∞ as in precedingexamples. Then x has unique representation

x =∞∑

k=1

λkek, and limk→∞

λk = 0.

Consider any f ∈ (c0)∗. Since f is linear,

f(x) =∞∑

k=1

λkγk where γk = f(ek).

For a given N ∈ N, take a special sequence xN0 = (λ

(0)k ) ∈ c0 where

λ(0)k =

γk

|γk| if 1 ≤ k ≤ N and γk 6= 0

0 if k > N or γk = 0.

Note that ‖xN0 ‖ = 1, and we have

|f(xN0 )| =

N∑

k=1

|f(ek)| ≤ ‖f‖ ‖xN0 ‖ = ‖f‖ < ∞.


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Since N is arbitrary,

∞∑

k=1

|f(ek)| =∞∑

k=1

|γk| = ‖f‖ < ∞.

This shows that (γk) ∈ `1 and

(i) ‖(γk)‖ ≤ ‖f‖.

On the other hand,

|f(x)| =∣∣∣∣∣∞∑

k=1

λkγk

∣∣∣∣∣ ≤ supk∈N

|λk|∞∑

k=1

|γk| = ‖x‖ ‖(γk)‖ < ∞.

It follows that(ii) sup

‖x‖=1

|f(x)| = ‖f‖ ≤ ‖(γk)‖.

From (i) and (ii) we obtain ‖f‖ = ‖(γk)‖.Now given (ξk) ∈ `1, we want to construct a linear bounded functional g on c0. Letx = (λk) ∈ c0. Consider the function g : c0 → F defined by

g(x) =∞∑

k=1

λkξk.

It is clear that g is linear. Its boundedness follows from

|g(x)| ≤ supk∈N

|λk|∞∑

k=1

|ξk| = ‖x‖∞∑

k=1

|ξk| < ∞.

Hence g ∈ (c0)∗. Thus there is an isometric isomorphism between (c0)

∗ and `1, sowe can write (c0)

∗ = `1. ¥

∗ ∗ ∗∗

Problem 63.Let X be a normed space and f, g are nonzero linear functionals on X. Showthat

ker(f) = ker(g) ⇐⇒ f = cg for some nonzero scalar c.


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Solution.The reverse way (⇐) is trivial.We show the direct way. Suppose ker(f) = ker(g). Since f 6= 0, there exists somex0 ∈ X such that f(x0) 6= 0, and by rescaling, we can assume that f(x0) = 1. Sincex0 /∈ ker(f) = ker(g), we have g(x0) 6= 0. Given any y ∈ X, we have

f(y − f(y)x0

)= f(y)− f(y)f(x0) = 0.

Therefore, y − f(y)x0 ∈ ker(f) = ker(g). Hence,

g(y)− f(y)g(x0) = g(y − f(y)x0

)= 0.

This implies thatg(y) = g(x0)f(y), ∀y ∈ X.

Hence g = cf with c = g(x0) 6= 0. ¥

Problem 64.Let X be a normed space and f are nonzero linear functional on X. Show thatf is continuous if and only if ker(f) is closed.

Solution.Suppose that f is continuous. Since ker(f) = f−1(0), so ker(f) is closed.Conversely, suppose that ker(f) is closed. Pick an x0 ∈ X such that f(x0) = 1.Assume that f is not continuous, that is, f is not bounded. Then there exists asequence (xn) in X such that

‖xn‖ = 1 and f(xn) ≥ n, ∀n ∈ N.

Define yn = x0 − xn

f(xn). Then

yn ∈ ker(f) for all n and limn→∞

yn = x0.

By hypothesis, ker(f) is closed, so x0 ∈ ker(f), that is, f(x0) = 0. This contradictsour assumption that f(x0) = 1. Thus f is continuous. ¥

Problem 65Let Z be a subspace of a normed space X, and y ∈ X. Let d = d(y, Z). Provethat there exists Λ ∈ X∗ such that ‖Λ‖ ≤ 1, Λ(y) = d and Λ(z) = 0 for allz ∈ Z.


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Solution.If y ∈ Z, then d = 0 and so the zero functional works, so we may assume that y /∈ Z.Consider the subspace Y = Cy + Z ⊂ X. Since y /∈ Z, so for every x ∈ Y , there isa unique α ∈ C and z ∈ Z such that x = αy + z. Define

λ(x) = αd.

We observe that λ(z) = 0 since

x = z ⇒ αy = 0

⇒ α = 0 (otherwise y = 0 so y ∈ Z).

Also we have λ(y) = d since y = 1.y + 0. It is clearly that λ is linear in x and

‖x‖ = |α|‖y +1

αz‖ ≥ αd = |λ(x)|.

Thus, λ is continuous, that is, λ ∈ Y ∗ and ‖λ‖ ≤ 1. By the Hahn-Banach theoremwe may extend λ to an element Λ ∈ X∗ with the same norm. ¥

Problem 66Let X be a normed space and (xn) be a sequence in X. Set V := Span x1, x2, ....Let W be the set of all continuous f ∈ X∗ such that f(xn) = 0, ∀n ∈ N. Provethat

V = x ∈ X : f(x) = 0, ∀f ∈ W.

Solution.Let Y := x ∈ X : f(x) = 0, ∀f ∈ W.• We show that V ⊂ Y . Take any x0 ∈ V . There exists a sequence (uk) in V suchthat uk → x0 as k → ∞. Since V := Span x1, x2, ..., for each k ∈ N, there are

scalars c(k)1 , ..., c

(k)nk such that uk =

∑nk

n=1 c(k)n xn. By linearity of f and by definition

of W , we have

f(uk) =

nk∑n=1

c(k)n f(xn) = 0, ∀f ∈ W.

By continuity of f , we have

f(x0) = 0, ∀f ∈ W.


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Hence x0 ∈ Y , and so V ⊂ Y .• Now given any x0 ∈ Y , we want to show x0 ∈ V . Assume that x0 /∈ V . Thend(x0, V ) = d > 0. Using the result in Problem 65, there is an F ∈ X∗ such that

F (x0) = d and F (x) = 0, ∀x ∈ V .

It follows that

F (x0) = d and F (xn) = 0, ∀n ∈ N.

Hence F ∈ W but F (x0) 6= 0. It means that x0 /∈ Y : a contradiction. Thus x0 ∈ V ,and so Y ⊂ V .The proof is complete. ¥

Problem 67.Let X be a normed space and x1, ..., xn ⊂ X a linearly independent system.Prove that for any α1, ..., αn ∈ K there exists x∗ ∈ X∗ such that

x∗(xi) = αi, ∀i ∈ 1, ..., n.

Solution.Let Y = Spanx1, ..., xn. Since the system x1, ..., xn ⊂ X a linearly independent,it follows that x1, ..., xn is an algebraic basis for Y . Define f : Y → K linear withf(xi) = αi, ∀i ∈ 1, ..., n. Since Y is of finite dimension, f is continuous. By theHahn-Banach theorem there is an extension x∗ ∈ X∗ for f . Then

x∗(xi) = f(xi) = αi, ∀i ∈ 1, ..., n. ¥

Problem 68.Let X be a normed space and Y ⊂ X a linear subspace. For x0 ∈ X \ Y wedefine

f : SpanY, x0 → K, f(y + λx0) = λ, ∀y ∈ Y, ∀λ ∈ K.

(a) Prove that f is well defined and linear.(b) Prove that f is continuous if and only if x0 /∈ Y .(c) Prove that Y =

⋂ker x∗ : x∗ ∈ X∗, Y ⊂ ker x∗.


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Solution.(a) Let us first observe that

SpanY, x0 = y + λx0 : y ∈ Y, λ ∈ K.To show that f is well defined, we show that

(y + λx0 = y′ + λ′x0) ⇒ (y = y′, λ = λ′).

Assume that λ 6= λ′. Then

x0 =y − y′

λ− λ′∈ Y

since y, y′ ∈ Y and Y is a linear subspace. This is a contradiction. Thus, λ = λ′

implies that y = y′. The linearity of f is obvious.(b) Suppose that f is continuous. We will prove that x0 /∈ Y . Assume, for acontradiction, that x0 ∈ Y . Then there is a sequence (yn) ⊂ Y such that yn → x0.Since yn, x0 ∈ Y , yn → x0 and f is continuous, f(yn) → f(x0). By definition off, f(yn) = 0, ∀n ∈ N. It follows that f(x0) = 0, which is false, since by definitionof f , we have f(x0) = 1.Suppose now that that x0 /∈ Y , i.e., x0 ∈ (Y )c, which is open. Then

∃ε > 0 : B(x0; ε) ⊂ (Y )c.

The inclusion is equivalent to Y ∩ B(x0; ε) = ∅. Let x = y + λx0 ∈ SpanY, x0. Ifλ = 0, then

(∗) |f(x)| = 0 ≤ ‖x‖ε

.

If λ 6= 0, then y−λ∈ Y so y

−λ/∈ B(x0; ε), i.e.,

∥∥− yλ− x0

∥∥ ≥ ε. Therefore,

(∗∗) ‖x‖ = |λ| ‖ − y

λ− x0‖ ≥ |λ|ε = ε|f(x)|.

Both cases (*) and (**) show that f is continuous.(c) If x∗ ∈ X∗ and Y ⊂ ker x∗ then Y ⊂ ker x∗ (since ker x∗ is closed) and therefore

Y ⊂⋂ker x∗ : x∗ ∈ X∗, Y ⊂ ker x∗.

Conversely, let x0 ∈⋂ker x∗ : x∗ ∈ X∗, Y ⊂ ker x∗, and assume that x0 /∈ Y .

By (b), there is an f : SpanY, x0 → K linear and continuous such that f |Y =0, f(x0) = 1. Let x∗ ∈ X∗ be a Hahn-Banach extension for f given by the Hahn-Banach theorem. Then x∗ = f on SpanY, x0. In particular, x∗ = 0 on Y . HenceY ⊂ ker x∗ and x∗(x) = f(x0) = 1, that is x /∈ ker x∗ and so

x0 /∈⋂ker x∗ : x∗ ∈ X∗, Y ⊂ ker x∗,

which is a contradiction. ¥


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Chapter 5

Fundamental Theorems

Review some main points:

1. The Baire category theorem:

Definition 3 Let X be a metric space and E be a subset of X.(a) E is called a set of the first category if E is a countable union of nowhere dense (non-dense)sets in X.(b) If E is not a set of the first category, then E is called a set of the second category.

Theorem 5 (Baire category theorem - form I)If X is a complete metric space, then X is a set of the second category.

Let X be a complete metric space. The Baire category theorem tells us that if X =⋃∞

n=1 An, thensome of the set An must have non-empty interior.

Theorem 6 (Baire category theorem - form II)Let X be a complete metric space. If (Un)n∈N is a sequence of open dense subsets of X, then⋂

n∈N Un is dense in X.

It can be shown that these two forms of Baire category theorem are equivalent (try it!).

2. The principle of uniform boundedness (P.U.B.)

Theorem 7 Let X be a Banach space and Y be a normed space. Let F be a family of boundedlinear operators from X to Y . Suppose that for each x ∈ X, ‖Tx‖ : T ∈ F is bounded, then‖T‖ : T ∈ F is bounded.

One of the P.U.B. consequences is:

Theorem 8 (Banach-Steinhaus theorem)Let X be a Banach space and Y be a normed space and (Tn) be a sequence in B(X, Y ). Supposefor every x ∈ X, Tnx → Tx as n → ∞. Then the family ‖Tn‖, n ∈ N is bounded, i.e.,supn∈N ‖Tn‖ < ∞, and T is linear and bounded, i.e., T ∈ B(X,Y ).

73


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74 CHAPTER 5. FUNDAMENTAL THEOREMS

3. The open mapping theorem

Theorem 9 Let X and Y be Banach spaces and T ∈ B(X, Y ). If T is onto (surjective), then Tis an open mapping, (that is, if U is open in X, then T (U) is open in Y ).

Consequence:

Theorem 10 (The inverse mapping theorem)Let X and Y be Banach spaces. If T ∈ B(X,Y ) is bijective, then T−1 exists and T−1 ∈ B(Y, X).

5. The closed graph theorem

Theorem 11 Let X and Y be Banach spaces and T be a linear map from X to Y . If we definethe graph of T by

Γ(T ) := (x, y) ∈ X × Y : y = Tx,then T is bounded (continuous) if and only if Γ(T ) is closed in X × Y .

Problem 69.(a) If X is a normed space, prove that any proper closed linear subspace of X isa nowhere dense set.(b) Prove that c0 is a nowhere dense set of c.

Solution.(a) Let A be a proper closed linear subspace of X. If A is not nowhere dense, then˚A = A 6= ∅ since A is closed. Therefore, A contain an open ball B(a; ε). Takex ∈ B(0; ε) then we have

‖x‖ < ε ⇒ a + x ∈ B(a; ε) ⊂ A.

Since a, a + x ∈ A and A is a linear space, x must be in A. Therefore B(0; ε) ⊂ A.Now take any y ∈ X with y 6= 0, then we have

ε

2‖y‖ y ∈ B(0; ε).

This implies that ε2‖y‖ y ∈ A, and so y ∈ A. Thus X = A. This is a contradiction.

(b) We know that c0 is a proper closed linear subspace of c. So c0 is nowhere densein c. ¥


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75

Problem 70.Show that L2[0, 1] is a subset of the first category in L1[0, 1].

Solution.For short, we write L1 := L1[0, 1] and L2 := L2[0, 1]. We know that L2 ⊂ L1. Foreach n ∈ N, let

An = f : ‖f‖2 ≤ n.Every function in L2 will be in some An, Thus

L2 =⋃

n∈NAn.

We first prove that An is closed. Fix n. Let (fk) be a sequence in An such that‖fk − f‖1 → 0. We show that f ∈ An. Since ‖fk − f‖1 → 0, (fk) convergesin measure to f . Hence there exists a subsequence (fkj

) converging to f almosteverywhere. By definition of An we have

∫ 1

0

f 2kj

(t)dt ≤ n2 for j = 1, 2, ...

Applying Fatou’s lemma, we obtain

∫ 1

0

f(t)dt ≤ n2.

Thus f ∈ An, and An is closed.

Now we show that An is nowhere dense, i.e.(An

)0= (An)0 = ∅. To do this, it

suffices to show that for any open ball Bε(f) in L1, there exists a point g ∈ Bε(f)but g /∈ An. Take

h(t) =ε

4

1√t.

Then h /∈ L2 and ∫ 1

0

h(t)dt =ε

2,

so thath ∈ L1 and ‖h‖1 =

ε

2.

Let g = f + h. Then g /∈ L2 so g /∈ An, and

‖f − g‖1 = ‖h‖1 =ε

2.


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Hence g ∈ Bε(f). Thus An is nowhere dense, and L2 is of the first category inL1. ¥

Problem 71.Let X 6= 0 be a normed space, and A ⊂ X which is not nowhere dense. Wedenote by A′ the derivative set of A, i.e., the set of all accumulation points ofA. Prove that

∃x ∈ X and ε > 0 such that B(x; ε) ⊂ A′.

(Note: A = A ∪ A′).

Solution.Since ˚A 6= ∅,

∃x ∈ X and ε > 0 such that B(x; ε) ⊂ A.

We will prove B(x; ε) ⊂ A′. Let y ∈ B(x; ε). If y /∈ A′, then there is r > 0 such that

B(y; r) ∩ (A \ y) = ∅.

Let δ = minr, 12

(ε − ‖y − x‖) > 0. We claim B(y; δ) ⊂ B(x; ε). Indeed, ifz ∈ B(y; δ) then

‖z − x‖ ≤ ‖z − y‖+ ‖y − x‖< δ + ‖y − x‖≤ 1

2(ε− ‖y − x‖) + ‖y − x‖

=1

2(ε + ‖y − x‖) < ε.

Hence z ∈ B(x; ε).Obviously, B(y; δ) ⊂ B(y; r). So

B(y; δ) ∩ (A \ y) ⊂ B(y; r) ∩ (A \ y) = ∅.

That is B(y; δ) ∩ A ⊂ y. Since y ∈ B(x; ε) ⊂ A, so y ∈ A, from whenceB(y; δ) ∩ A 6= ∅, and therefore

B(y; δ) ∩ A = y.


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77

Since B(y; δ) ∩ A ⊂ B(y; δ) ∩ A and B(y; δ) ⊂ B(x; ε) ⊂ A, i.e., B(y; δ) ∩ A =B(y; δ). Hence, B(y; δ) = y, which is false, since X 6= 0. (Indeed, ∃a ∈ X, a 6=0. Take b = a

‖a‖ ∈ X. Then ‖b‖ = 1 and

y +δ

2b ∈ B(y; δ) = y,

which implies that b = 0 : a contradiction). ¥

Problem 72.Let X is a Banach space and A ⊂ X a dense set. Can we find a functionf : X → R such that, for every x ∈ A, we have limt→x |f(t)| = ∞?

Solution.Suppose that such a function exists. Since f takes finite values, for every x ∈ X,there is k ∈ N such that |f(x)| ≤ k, i.e.,

X =⋃

k∈NAk where Ak = x ∈ X : |f(x)| ≤ k.

Since X is a Banach space, X is of the second Baire category. So there is k ∈ Nsuch that ˚Ak 6= ∅. Using Problem 74, it follows that there is x ∈ X and ε > 0 suchthat B(x; ε) ⊂ A′

k. Since A is dense, we have

∅ 6= B(x; ε) ∩ A ⊂ A′k ∩ A.

It follows that there is an element a ∈ A which is in A′k. By definition of accumula-

tion points, there is a sequence (xn) ⊂ Ak with xn 6= a, ∀n ∈ N such that xn → a.Since a ∈ A, by hypothesis we have limt→a |f(t)| = ∞. Hence, limn→∞ |f(xn)| = ∞.This is not possible since (xn) ⊂ Ak implies that |f(xn)| ≤ k, ∀k ∈ N. ¥

Problem 73.Show that Q is a subset of the first category of R.

Solution.Since Q is countable, we have

Q =⋃

n∈Nxn and (xn)0 = (xn)0 = ∅. ¥


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Problem 74.Show that the set of piecewise linear functions on R is of the first category.

Solution.Let P denote the set of piecewise linear functions on R. Let Pn be the set of piecewiselinear functions having n intervals of linearity. We have

P =∞⋃

n=1

Pn.

We need to prove that Pn is nowhere dense for every n. Fix an arbitrary n. Letf ∈ Pn. Consider the ball Br(f), r > 0. Let

g(t) =r

2sin(2π4nt) + f(t).

Then

|f(t)− g(t)| = r

2| sin(2π4nt)|.

Hence d(f, g) = r2, and so g ∈ Br(f). We claim that the ball B r

2(g) contains no

element from Pn. Pick h ∈ Br(f) ∩ Pn and suppose d(g, h) < r2. Then

(∗) d(g, h) = sup |g(t)− h(t)| = sup∣∣∣r2

sin(2π4nt) + f(t)− h(t)∣∣∣ <

r

2.

Observing that the term r2sin(2π4nt) oscillates between − r

2and r

24n times on

[0, 1]. Thus the term f(t) − h(t) must also oscillate between negative and positivevalues 4n times for (*) to hold. But this is impossible since the term f(t)− h(t) isa piecewise linear function with at most 2n intervals of linearity. So, the open ballB r

2(g) contains no element from Pn. Since n is arbitrary, we see that Pn is nowhere

dense, and hence P is of the first category. ¥

Problem 75. (Inverse mapping theorem)Let X and Y be Banach spaces and T ∈ B(X,Y ). Suppose T is bijective. Showthat there exist real numbers a, b > 0 such that

a‖x‖ ≤ ‖Tx‖ ≤ b‖x‖, ∀x ∈ X.


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79

Solution.Since T is linear, bijective and bounded, T−1 exists, is linear and bounded by theinverse mapping theorem. Let ‖T−1‖ = 1

aand ‖T‖ = b. Note that T 6= 0, a, b > 0.

Now since T is bounded,

(i) ‖Tx‖ ≤ ‖T‖ ‖x‖ = b‖x‖, ∀x ∈ X.

Also, since T−1 is bounded,

(ii) ‖x‖ = ‖T−1(Tx)‖ ≤ ‖T−1‖ ‖Tx‖ =1

a‖Tx‖, ∀x ∈ X.

(i) and (ii) imply that

a‖x‖ ≤ ‖Tx‖ ≤ b‖x‖, ∀x ∈ X. ¥

Problem 76. Let X = C1[0, 1] be the space of continuously differentiable func-tions on [0, 1] and Y = C[0, 1] . The norm on C[0, 1] and C1[0, 1] is the sup-norm. Consider the map

T : C1[0, 1] → C[0, 1] defined by Tx(t) =dx(t)

dt, t ∈ [0, 1].

Show that the the graph of T is closed but T is not bounded. Does this contradictthe closed graph theorem ?

Solution.• It is clear that T is linear.• We show that Γ(T ) is closed. Suppose xn → x in X = C1[0, 1] and Txn → y inY = C[0, 1]. We must show that y = Tx. For any t ∈ [0, 1], we have

∫ t

0

y(s)ds =

∫ t

0

limn→∞

dxn

dsds, y ∈ C[0, 1]

= limn→∞

∫ t

0

dxn

dsds (uniform convergence)

= limn→∞

(xn(t)− xn(0)

)= x(t)− x(0).

Thus, with y ∈ C[0, 1], we have

x(t) = x(0) +

∫ t

0

y(s)ds, t ∈ [0, 1].


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Hence

x ∈ C1[0, 1] anddx

ds= y on [0, 1].

That is Tx = y, and so (x, y) ∈ Γ(T ), and Γ(T ) is closed.• We show that T is not bounded. Take fn(t) = tn, n ∈ N, t ∈ [0, 1]. Thenfn ∈ C1[0, 1] and Tfn = nfn−1 for n > 1. But we have

‖fn‖ = 1 and ‖Tfn‖ = n,

which shows that T is unbounded. The reason? Is C1[0, 1] with the sup-norm aBanach space? ¥

Problem 77. (P.U.B.)Let H be a separable Hilbert space with an orthonormal basis en∞n=1. Let xnbe a sequence in H. Prove that the following two statements are equivalent:(1) limn→∞〈x, xn〉 = 0, ∀x ∈ H.(2) limn→∞〈em, xn〉 = 0, ∀m ∈ N and ‖xn‖ is bounded.

Solution.• (1) ⇒ (2)Assume that (1) is true . Then the first part of (2) is automatically true. Wehave only to show that ‖xn‖ is bounded. Consider, for each n, the functionalfn(x) = 〈x, xn〉. This is a bounded functional by Schwarz inequality, and, becausefn(x) → 0 for each x, we have that the set ‖fn(x)‖ : n ∈ N is bounded. Theprinciple of uniform boundedness then gives us that the set ‖fn‖ is bounded. But‖fn‖ = ‖xn‖, so ‖xn‖ is bounded.• (2) ⇒ (1)Assume that (2) holds. Let B be the bound of ‖xn‖ and let x ∈ X. We writex =

∑n〈en, x〉en. For every ε > 0, let K be such that

∑m>K

|〈em, x〉|2 <ε

B.

We know that, for m fixed, 〈em, xn〉 → 0 as n →∞. So we may find N such that ifn > N then |〈em, xn〉| < ε

K sup |〈er,x〉| (the denominator is finite because the sequence


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81

〈er, x〉 is an `2-sequence). Then, if n > N , we have

|〈x, y〉| =

∣∣∣∣∣∞∑

m=1

〈em, x〉〈em, xn〉∣∣∣∣∣

≤K∑

m=1

ε

K∑ |〈er, x〉| |〈em, x〉|+

∑m>K

〈em, x〉〈em, xn〉

≤ ε +

( ∑m>K

|〈em, x〉|2)1/2

+

( ∑m>K

|〈em, xn〉|2)1/2

≤ ε +ε

B‖xn‖ ≤ 2ε. ¥

Problem 78. (Closed graph theorem)Let H be a Hilbert space and T : H → H a linear operator which is symmetric,i.e.,

〈Tx, y〉 = 〈x, Ty〉, ∀x, y ∈ H.

Prove that T is continuous.

Solution.Since H is a complete space and T is a linear operator, it is sufficient to prove thatT is a closed graph operator. Let (xn) be a sequence in H such that xn → x ∈ Hand Txn → y ∈ H. We will show that y = Tx .By hypothesis, we have

〈T (xn − x), y〉 = 〈xn − x, Ty〉, ∀y ∈ H.

Hence〈T (xn − x), y〉 → 0, ∀y ∈ H.

That isT (xn − x)

w−→ 0, or, equivalently Txnw−→ Tx.

But Txn → y, so Txnw−→ y. Since the limit is unique, we have Tx = y. Thus, T is

a closed graph operator. ¥

Problem 79. (Banach-Steinhaus theorem)Let a = (a1, a2, ...) = (an)n∈N be a sequence of scalars such that the sequence(anxn)n∈N ∈ c0 for all sequences x = (xn)n∈N ∈ c0. Prove that a ∈ `∞.


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Solution.For every n ∈ N, consider the operator

Un : c0 → c0 defined by Un(x) = Un(x1, x2, ...) = (a1x1, a2x2, ..., anxn, 0, ...).

Then Un is linear, and

‖Un(x1, x2, ...)‖ = ‖(a1x1, a2x2, ..., anxn, 0, ...)‖= max(|a1x1|, ..., |anxn|)≤ ‖x‖max(|a1|, ..., |an|), ∀x ∈ c0.

So Un is continuous, and ‖Un‖ ≤ max(|a1|, ..., |an|).For 1 ≤ k ≤ n, we have

‖Un‖ ≥ ‖Un(ek)‖ = ‖(0, ..., 0, ak, 0, ...)‖ = |ak|,

where ek = (0, .., 0︸︷︷︸k−1

, 1, 0, ...) ∈ c0. Therefore max(|a1|, ..., |an|) ≤ ‖Un‖. Hence,

‖Un‖ = max(|a1|, ..., |an|), ∀n ∈ N.

Now consider the operator

U : c0 → c0 defined by U(x) = (a1x1, a1x2, ...) = (anxn)n∈N.

We have that Un(x) → U(x), ∀x ∈ c0 because

‖Un(x)− U(x)‖ = supk≥n+1

|akxk| → 0.

Now from the Banach-Steinhaus theorem we get that supn∈N ‖Un‖ < ∞, i.e., supn∈N |an| <∞. In other words, a ∈ `∞. ¥

Problem 80. (Similar problem)Let a = (a1, a2, ...) = (an)n∈N be a sequence of scalars such that the sequence (anxn)n∈N ∈ c0 forall sequences x = (xn)n∈N ∈ `∞. Prove that a ∈ c0.

Problem 81. (P.U.B.)Let x = (x1, x2, ...) = (xi)

∞i=1 be a sequence of scalars such that the series∑∞

i=1 xiyi is convergent for all y = (y1, y2, ...) ∈ c0. Prove that x ∈ `1.


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83

Solution.For every n ∈ N, we define the linear operator

Tn : c0 → C, Tn(y) =n∑

i=1

xiyi.

Then we have

|Tn(y)| ≤n∑

i=1

|xiyi| ≤(

n∑i=1

|xi|)‖y‖∞.

This shows that Tn is bounded with

‖Tn‖ ≤n∑

i=1

|xi|.

By hypothesis,∑∞

i=1 xiyi < ∞, the sequence(Tn(y)

)n∈N

converges for every y ∈ co,

and c0 is a Banach space, the principle of uniform boundedness implies that

∃M > 0 : ‖Tn‖ ≤ M, ∀n ∈ N.

Now let x(n) =(x

(n)1 , x

(n)2 , ..., x

(n)n , 0, 0, ...

)be a truncated version of x (so that x(n) ∈

`1). Let

Tn(y) =n∑

i=1

x(n)i yi, y = (y1, y2, ...) ∈ c0.

Define y(n) =(y

(n)1 , y

(n)2 , ...,

)by

y(n)k =

x(n)k

|x(n)k | if x

(n)k 6= 0,

0 if x(n)k = 0.

Then

|Tn(y(n))| =n∑

k=1

|x(n)k | = ‖x(n)‖1 = ‖x(n)‖1‖y(n)‖∞.

Hence

‖Tn‖ ≥ ‖x(n)‖1,

which in turn implies that

‖x(n)‖1 ≤ M, ∀n ∈ N.


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But it is clear from the definition of x(n) that(‖x(n)‖1

)is an increasing sequence of

real numbers. Being bounded above by M , it must converge. Hence

∞∑i=1

|xi| < ∞,

and so x ∈ `1. ¥

Problem 82. (Very similar problem)Let c = (c1, c2, ...) = (ci)∞i=1 be a sequence of scalars such that the series

∑∞i=1 ciai is convergent

for all a = (a1, a2, ...) ∈ `1. Prove that x ∈ `∞.

Problem 83. (Closed graph theorem)Let X, Y and Z be Banach spaces. Suppose that T : X → Y is linear, thatJ : Y → Z is linear, bounded and injective, and that JT ≡ J T : X → Z isbounded.Show that T is also bounded.

Solution.We will show that the graph Γ(T ) is closed. Then by the closed graph theorem, thisimplies that T is bounded (continuous).Let

((xn, yn)

)n∈N be a convergent sequence in X × Y , that is,

xn → x in X, yn → y in Y and Txn = yn.

Since J and JT are continuous,

[yn → y ⇒ Jyn → Jy

]and

[xn → x ⇒ JTxn → JTx

].

Since Txn = yn, we have Jyn → JTx. Since the limit is unique, this gives thatJy = JTx. But by hypothesis J is injective, so we have

Jy = JTx ⇒ y = Tx.

This shows that (x, y) ∈ Γ(T ), and Γ(T ) is closed. ¥


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85

Problem 84. (Closed graph theorem)Let X be a Banach space and E,F two closed subspaces of X such that X =E ⊕ F . Consider the projections on E and on F defined by

PE : X → E, PE(u) = x,

PF : X → F, PF (u) = y, where u = x + y, x ∈ E, y ∈ F.

Use the closed graph theorem to show that PE ∈ B(X,E) and PF ∈ B(X, F ).

Solution.The linearity of these two maps are easy to check. Let us prove that they arebounded by using the closed graph theorem. Denote the graph of PE by ΓE. Wecan write

ΓE = (x, y) ∈ X × E : x− y ∈ F.Let (xn, yn) ∈ ΓE for every n ∈ N. Suppose (xn, yn) → (x, y) as n → ∞. Sincexn − yn ∈ F for every n ∈ N, and F is a closed subspace of X, limn→∞(xn − yn) =x− y ∈ F . It follows that (x, y) ∈ ΓE. Thus, ΓE is closed, and so PE is bounded.The proof for PF is the same. ¥

Problem 85. (Inverse mapping theorem)Let (X, ‖.‖1) and (X, ‖.‖2) be Banach spaces. Suppose that

∃C ≥ 0 : ‖x‖2 ≤ C‖x‖1, ∀x ∈ X.

Show that the two norms ‖.‖1 and ‖.‖2 are equivalent.

Solution.Consider the identity map

id : (X, ‖.‖1) → (X, ‖.‖2), id(x) = x.

It is clear that the identity map is linear and bijective. It is continuous since byhypothesis we have

(∗) ‖id(x)‖2 = ‖x‖2 ≤ C‖x‖1, ∀x ∈ X.

By the inverse mapping theorem, the inverse map id−1 exists and continuous. Thatis

(∗∗) ∃C ′ ≥ 0 : ‖x‖1 ≤ C ′‖x‖2, ∀x ∈ X.


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(∗) and (∗∗) together imply that ‖.‖1 and ‖.‖2 are equivalent. ¥


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Chapter 6

Linear Operators on HilbertSpaces

Review.

1. Definition and elementary properties

Let T : H → K be a linear operator between Hilbert spaces H and K.• The following statements are equivalent:1. T is continuous at 0,2. T is continuous,3. T is bounded on H.• An isomorphism between H and K is a linear surjection U : H → K such that

〈Ux,Uy〉 = 〈x, y〉, ∀x, y ∈ H.

An isomorphism is an isometry and so preserves completeness, but an isometry need not to be anisomorphism.

Proposition 1 Two Hilbert spaces are isomorphic if and only if they have the same dimension.

2. Adjoint of an Operator• Let A ∈ B(H). Then A∗ is called the adjoint operator of A if

〈Ax, x〉 = 〈x,A∗x〉, ∀x ∈ H.

Proposition 2 If A,B ∈ B(H) and α ∈ F, then

(a) (αA + B)∗ = αA∗ + B∗.

(b) (AB)∗ = B∗A∗.

(c) A∗∗ := (A∗)∗ = A.

(d) If A is invertible in B(H) and A−1 is its inverse, then A∗ is invertible and (A∗)−1 = (A−1)∗.

87


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88 CHAPTER 6. LINEAR OPERATORS ON HILBERT SPACES

Proposition 3 It A ∈ B(H) then

‖A‖ = ‖A∗‖ = ‖A∗A‖1/2.

3. Self-adjoint, normal, unitary operators

Definition 4 If A ∈ B(H), then

(a) A is Hermitian or self-adjoint if A∗ = A.

(b) A is normal if AA∗ = A∗A.

(c) A is unitary if it is a surjective isometry.

Proposition 4 Let A ∈ B(H). The following statements are equivalent.

(a) A∗A = AA∗ = I.

(b) A is unitary.

(c) A is a normal isometry.

4. Positive operators

Definition 5 Let H be a Hilbert space. An operator A ∈ B(X) is called positive if

〈Ax, x〉 ≥ 0, ∀x ∈ X.

We write A ≥ 0. If A,B ∈ B(X) and A−B ≥ 0 then we write B ≤ A.

Proposition 5 If A ∈ B(X) then A∗A ≥ 0. In addition, if A ≥ 0, then1. A is self adjoint,2. there exists a unique B ∈ B(X) such that B ≥ 0 and B2 = A. Furthermore, B is also selfadjoint and commutes with every bounded operator which commutes with A. We write B =

√A.

We define |A| = √A∗A.

5. Projection, Orthogonal projection

Definition 6

If P ∈ B(H) and P 2 = P, then P is called a projection.

If P ∈ B(H), P = P 2 and P ∗ = P, then P is called an orthogonal projection.

Proposition 6

If P : H → H is a projection then H = Image P ⊕ ker P .

If H = M ⊕N , where M, N are subspaces of H, then there is a projection P : H → H withImage P = M and kerP = N.

∗ ∗ ∗ ∗ ∗


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89

Problem 86. Let P be an orthogonal projection defined on a Hilbert space H.Show that ‖P‖ = 1.

Solution.If x ∈ H and Px 6= 0, then the use of the Cauchy-Schwarz inequality implies that

‖Px‖ =〈Px, Px〉‖Px‖

=〈x, P 2x〉‖Px‖ (since P ∗ = P )

=〈x, Px〉‖Px‖ ≤ ‖x‖ ‖Px‖

‖Px‖≤ ‖x‖.

Therefore ‖P‖ ≤ 1.Now, if P 6= 0, then there is an x0 ∈ H such that

Px0 6= 0 and ‖P (Px0)‖ = ‖Px0‖.

This implies that ‖P‖ ≥ 1. Thus ‖P‖ = 1. ¥

Problem 87. Given a function φ : [0, 1] → C, consider the operator

P : L2[0, 1] → L2[0, 1] defined by Pf(x) = φ(x)f(x).

Find necessary and sufficient conditions on the function φ for P to be an or-thogonal projection.

Solution.First, in order for P to be a well-defined operator acting on L2[0, 1], the functionφf needs to be in L2[0, 1] for all f ∈ L2[0, 1]. In particular φf is measurable, andtaking f ≡ 1, it follows that φ is a measurable function on [0, 1].Secondly, P is an orthogonal projection if and only if P ∗ = P and P 2 = P . Thelast equality is equivalent to φ2(x)f(x) = φ(x)f(x), ∀f ∈ L2[0, 1]. Again by takingf ≡ 1, we have a2(x) = a(x) for almost every x ∈ [0, 1]. Thus

a(x) = 0 or a(x) = 1 for almost all x ∈ [0, 1].


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In particular φ takes real values. Then

〈Pf, g〉 =

∫ 1

0

Pf(x)g(x)dx

=

∫ 1

0

φ(x)f(x)g(x)dx

=

∫ 1

0

f(x)φ(x)g(x)dx

=

∫ 1

0

f(x)Pg(x)dx

= 〈f, Pg〉,which proves that P is self-adjoint. Since 0 ≤ φ(x) ≤ 1 for a.e. on [0, 1], we havethat

‖Pf‖L2 =

(∫ 1

0

φ(x)2|f(x)|2dx

) 12

≤(∫ 1

0

|f(x)|2dx

) 12

= ‖f‖L2 .

Thus, P is bounded.In conclusion, the necessary and sufficient conditions for P to be an orthogonalprojection is φ a measurable satisfying φ(x) = 0 or φ(x) = 1 for almost all x ∈[0, 1]. ¥

Problem 88. Consider the right-shift on the Hilbert space `2:

S : `2 → `2, S(α1, α2, ...) = (0, α1, α2, ...).

Define its adjoint operator.

Solution.For (αn) = (α1, α2, ...) and (βn) = (β1, β2, ...) in `2,

⟨S∗(αn), (βn)

⟩=

⟨(αn), S(βn)

⟩

= 〈(α1, α2, ...), (0, β1, β2, ...)〉= α2β1 + α3β2 + ...

= 〈(α2, α3, ...), (β1, β2, ...)〉.Thus

S∗(α1, α2, ...) = (α2, α3, ...).

Hence, the adjoint of the right-shift is the left-shift. ¥


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91

Problem 89. Let A : `2 → `2 be defined by

Ax = A(x1, x2, x3, ...) = (0, 0, x3, x4, ...).

Prove A is linear, continuous, self-adjoint and positive. Find√

A.

Solution.With similar argument as the previous problem, we can show that A is linear. Wehave

‖Ax‖2 =∞∑

k=3

|xk|2 ≤∞∑

k=1

|xk|2 = ‖x‖2, ∀x = (x1, x2, x3, ...) ∈ `2.

Therefore,‖Ax‖ ≤ ‖x‖, ∀x ∈ `2.

This shows that A is continuous. Also, for all x, y ∈ `2,

〈Ax, y〉 =∞∑

k=3

xkyk = 〈x,Ay〉.

Hence, A is self-adjoint. And,

〈Ax, x〉 =∞∑

k=3

|xk|2 ≥ 0, ∀x ∈ `2.

So A is positive. Then there exists the square root√

A : `2 → `2. We have

A2x = A(Ax) = A(0, 0, x3, x4, ...) = (0, 0, x3, x4, ...), ∀x ∈ `2.

It follows that A2 = A. Hence,√

A = A. ¥

Problem 90.(Multiplication operator)Let (X, Ω, µ) be a σ-finite measure space. Consider the Hilbert space H =L2(X, Ω, µ) =: L2(µ). If φ ∈ L∞(µ), define

Mφ : L2(µ) → L2(µ) by Mφf = φf.

(a) Show thatMφ ∈ B(H) and ‖Mφ‖ = ‖φ‖∞.

Here ‖φ‖∞ is the µ-essential supremum norm.(b) Show that M∗

φ = Mφ.(c) Show that Mφ is normal. When Mφ is self adjoint? unitary?


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Solution.(a) The linearity of the operator Mφ is evident. We show that Mφ is bounded andcalculate its norm.By definition, ‖φ‖∞ is the infimum of all c > 0 such that |φ(x)| ≤ c a.e. [µ], and so|φ(x)| ≤ ‖φ‖∞ a.e. [µ]. Thus we can assume that φ is a bounded measurable and|φ(x)| ≤ ‖φ‖∞ for all x. If f ∈ L2(µ), then

∫|φf |2dµ ≤ ‖φ‖2

∞

∫|f |2dµ.

That is,Mφ ∈ B

(L2(µ)

)and ‖Mφ‖ ≤ ‖φ‖∞ (∗).

If ε > 0, the σ-finiteness of the measure space implies that

∃∆ ∈ Ω such that 0 < µ(∆) < ∞ and |φ(x)| ≥ ‖φ‖∞ − ε, ∀x ∈ ∆.

If we take f = 1√µ(∆)

χ∆, then we have

f ∈ L2(µ) and ‖f‖2 = 1.

So

‖Mφ‖2 ≥ ‖φf‖2 =1

µ(∆)

∫

∆

|φ|2dµ ≥ (‖φ‖∞ − ε)2.

Letting ε → 0, we get that

‖Mφ‖ ≥ ‖φ‖∞ (∗∗).(*) and (**) give that ‖Mφ‖ = ‖φ‖∞.

(b) For f, g ∈ L2(µ), we have

〈f,Mφg〉 =

∫f

(Mφg

)dµ

=

∫f(φg)dµ

=

∫(φf)gdµ

= 〈Mφf, g〉.This shows that

M∗φ = Mφ.

(c) Every multiplication operator Mφ is normal. Indeed,

MφM∗φ = MφMφ = MφMφ = M∗

φMφ.


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93

Mφ is self-adjoint if and only if φ = φ, that is, φ is real-valued.Mφ is unitary if and only if |φ| = 1 a.e. [µ]. ¥

Problem 91. Let H be a Hilbert space and A ∈ B(H). Show that(a) Image A = (ker A∗)⊥.(b) ker A = (Image A∗)⊥.

Solution.(a) Take any x ∈ Image A. Then there is a y ∈ H such that x = Ay. For anyz ∈ ker A∗, we have

〈x, z〉 = 〈Ay, z〉 = 〈y, A∗z〉 = 〈y, 0〉 = 0.

hence x ∈ (ker A∗)⊥. This proves that Image A ⊂ (ker A∗)⊥. Since (ker A∗)⊥ isclosed, it follows that

(i) Image A ⊂ (ker A∗)⊥.

On the other hand, if x ∈ (Image A)⊥, then for all y ∈ H, we have

0 = 〈Ay, x〉 = 〈y,A∗x〉.

Therefore A∗x = 0, that is, x ∈ ker A∗. This prove that (Image A)⊥ ⊂ ker A∗.Taking orthogonal complements both sides, we obtain

(ii) (ker A∗)⊥ ⊂ Image A ⊂ Image A.

From (i) and (ii) it follows that

Image A = (ker A∗)⊥.

(b) Replacing A by A∗ in (a), we get

(ker A)⊥ = Image A∗.

Taking orthogonal complements both sides and using a result in Problem 39, weobtain

ker A = (Image A∗)⊥ = (Image A∗)⊥. ¥


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Problem 92.(Integral operator)Let (X, Ω, µ) be a measure space. Let k : X ×X → F be an Ω × Ω-measurablefunction for which there are constants c1 and c2 such that

∫

X

|k(x, y)|dµ(y) ≤ c1 a.e.[µ],∫

X

|k(x, y)|dµ(x) ≤ c2 a.e.[µ].

Consider the operator K : L2(µ) → L2(µ) defined by

(Kf)(x) =

∫k(x, y)f(y)dµ(y).

The function k is called the kernel of the operator K.(a) Show that K is a bounded linear operator and ‖K‖ ≤ √

c1c2.

(b) Show that K∗ is the integral operator with kernel k∗(x, y) = k(x, y).

Solution.(a) Linearity of K comes from linearity of the integral

∫. It suffices to show that K

is bounded. Actually it must be shown first that Kf ∈ L2(µ), but this will followfrom the argument that demonstrates the boundedness of K. If f ∈ L2(µ),

|Kf(x)| ≤∫|k(x, y)| |f(y)|dµ(y)

=

∫|k(x, y)|1/2|k(x, y)|1/2|f(y)|dµ(y)

≤[∫

|k(x, y)|dµ(y)

]1/2 [∫|k(x, y)| |f(y)|2dµ(y)

]1/2

≤ √c1

[∫|k(x, y)| |f(y)|2dµ(y)

]1/2

.

Hence∫|Kf(x)|2dµ(y) ≤ c1

∫ ∫|k(x, y)| |f(y)|2dµ(y)dµ(x)

= c1

∫|f(y)|2

∫|k(x, y)|dµ(x)dµ(y)

≤ c1c2‖f‖2.


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95

Now this shows that the formula used to define Kf is finite a.e. [µ], and so

Kf ∈ L2(µ) and ‖Kf‖2 ≤ c1c2‖f‖2.

(b) By definition,

〈Kf, g〉 =

∫k(x, y)f(y)g(y)dµ(y)

=

∫f(y)k(x, y)g(y)dµ(y)

= 〈f,K∗g〉, where K∗g(y) =

∫k(x, y)g(y)dµ(y).

Hence, the kernel of K∗ is k∗(x, y) = k(x, y). ¥

Problem 93. Let H = H ⊕H where H be a Hilbert space. Let A ∈ B(H) andB be the operator defined on H by

B =

(0 iA

−iA∗ 0

)

Prove that ‖A‖ = ‖B‖ and that B is self-adjoint.

Solution.

For any x =

(x1

x2

), y =

(y1

y2

)with x1, x2, y1, y2 ∈ H we have

〈Bx, y〉 =

⟨(0 iA

−iA∗ 0

)(x1

x2

),

(y1

y2

)⟩

= 〈iAx2, y1〉+ 〈−iA∗x1, y2〉= 〈x2,−iA∗y1〉+ 〈x1, iAy2〉=

⟨(x1

x2

),

(0 iA

−iA∗ 0

)(y1

y2

)⟩

= 〈x,By〉.Moreover,

‖Bx‖2 = ‖iAx2‖2 + ‖ − iA∗x1‖2

≤ (max‖A‖, ‖A∗‖)2‖x‖2

= ‖A‖2‖x‖2.


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Hence, ‖B‖ ≤ ‖A‖. Conversely, one can take x =

(0x2

)and obtain

‖Bx‖ = ‖Ax2‖ ≤ ‖B‖‖x‖ = ‖B‖‖x2‖.Therefore, ‖A‖ ≤ ‖B‖. Finally, we obtain ‖A‖ ≤ ‖B‖. ¥————————–Remark:Note that norm on H ⊕H is

‖(a, b)‖ = ‖a‖+ ‖b‖, a, b ∈ H.

Problem 94. Let T be a self-adjoint operator on a Hilbert space H. Show thatits norm is given by

‖T‖ = sup‖x‖=1

|〈Tx, x〉|.

Solution.For ‖x‖ = 1 we have

|〈Tx, x〉| ≤ ‖Tx‖‖x‖ = ‖Tx‖ ≤ ‖T‖.

Therefore(i) sup

‖x‖=1

|〈Tx, x〉| ≤ ‖T‖.

In order to establish the inverse inequality, we consider the case:

z ∈ H, ‖z‖ = 1, T z 6= 0 and u =1

λTz where λ =

√‖Tz‖.

If we denote by α := sup‖x‖=1 |〈Tx, x〉|, then we have

‖Tz‖2 =⟨T (λz), u

⟩

=1

4

[⟨T (λz + u), λz + u

⟩− ⟨T (λz − u), λz − u

⟩]

≤ α

4

[‖λz + u‖2 + ‖λz − u‖2]

=α

2

[‖λz‖2 + ‖u‖2]

=α

2

[‖λ‖2 + ‖Tz‖2]

= α‖Tz‖.


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97

This implies that, for any z ∈ H with ‖z‖ = 1, we have ‖Tz‖ ≤ α, and hence

(ii) ‖T‖ ≤ α = sup‖x‖=1

|〈Tx, x〉|.

(i) and (ii) completes the proof. ¥

Problem 95Let H be a Hilbert space and A a positive self-adjoint operator on H. Prove thatthe following assertions are equivalent:(i) A(H) is dense in H.(ii) Ker A = 0.(iii) A is positive definite, i.e., 〈Ax, x〉 > 0, ∀x ∈ H \ 0.

Solution.• (i) ⇒ (ii)Suppose Ax = 0. Then, for any y ∈ H,

〈Ax, y〉 = 〈x,Ay〉 = 0 ( since A is self-adjoint)

⇒ x⊥A(H)

⇒ x⊥A(H) = H ( since A(H) is dense)

⇒ x = 0.

• (ii) ⇒ (iii)Since A is positive,

√A = B exists. It is also a self-adjoint operator on H. To show

A is positive definite, we show 〈Ax, x〉 = 0 ⇒ x = 0. Now

0 = 〈Ax, x〉 = 〈B2x, x〉 = 〈B(Bx), x〉 = 〈Bx, Bx〉 = ‖Bx‖2.

This implies that Bx = 0. Therefore,

Ax = B(Bx) = 0.

Since Ker A = 0, we have x = 0.• (iii) ⇒ (i)Assume that A(H) is not dense in H. Then there is x ∈ H \0 such that x⊥A(H).In particular, x⊥Ax, i.e., 〈Ax, x〉 = 0. But A is positive definite, so x = 0, acontradiction. ¥


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Problem 96Let H be a Hilbert space. If A,B : H → H are self-adjoint operators with0 ≤ A ≤ B and B is compact, prove that A is compact.

Solution.Let (xn) be a sequence in the closed unit ball BH . Since B is compact, there is asubsequence (xnk

) such that (Bxnk) converges. From the Cauchy-Schwarz inequality

we have〈Bx, x〉 ≤ ‖Bx‖ ‖x‖, ∀x ∈ H.

It follows that

〈Bxnk−Bxmk

, xmk− xnk

〉 ≤ ‖Bxmk−Bxnk

‖ ‖xmk− xnk

‖≤ ‖Bxmk

−Bxnk‖ (‖xmk

‖+ ‖xnk‖)

≤ 2‖Bxmk−Bxnk

‖.From 0 ≤ A ≤ B we get

〈A(xnk− xmk

), xmk− xnk

〉 ≤ 〈B(xnk− xmk

), xmk− xnk

〉≤ 2‖Bxmk

−Bxnk‖.

On the other hand, we get

‖√

A(x)‖2 = 〈√

A(x),√

A(x)〉= 〈

√A

2(x), x〉

= 〈Ax, x〉≤ ‖Ax‖ ‖x‖, ∀x ∈ H.

Then

‖√

Axmk−√

Axnk‖2 ≤ ‖A(xmk

− xnk)‖ ‖xmk

− xnk‖

≤ ‖A(xmk− xnk

)‖(‖xmk‖+ ‖xnk

‖)≤ 2‖A(xmk

− xnk)‖.

Therefore,‖√

Axmk−√

Axnk‖2 ≤ 〈A(xmk

− xnk), xmk

− xnk〉.

Hence‖√

Axmk−√

Axnk‖2 ≤ 2‖Bxmk

−Bxnk‖.

From this we see that the sequence (√

Axnk) is Cauchy, hence converges. The

operator√

A is compact. And so is the operator A =√

A2. ¥


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Chapter 7

Compact Operators

In this chapter we study general properties of compact operators on Banach and Hilbert spaces.Spectral properties of these operators will be discussed later.

Definition 7 Let X and Y be Banach spaces. An operator T ∈ B(X, Y ) is called compact op-erator if the image of every bounded set in X has compact closure in Y (relatively compact set).Equivalently, T ∈ B(X,Y ) is compact if and only if for every bounded sequence (xn) in X, (Txn)has a convergent subsequence in Y .The set of all compact operators is denoted by B0(X, Y ).

Proposition 7 Let X and Y be Banach spaces. Then B0(X,Y ) is a closed subspace of B(X, Y ).That is, if (Tn) is a sequence of compact operators and T ∈ B(X, Y ) such that ‖Tn−T‖ → 0, thenT ∈ B0(X, Y ).

Definition 8 (Finite rank operators)An operator T : X → Y has finite rank if Image T := T (X) is finite dimensional.

Proposition 8 Let X and Y be Banach spaces. Every finite rank operator from X to Y iscompact.

Proposition 9 Let X and Y be Banach spaces, and T : X → Y be a compact operator. If xnw→ x

then Txn → Tx.

Proposition 10 Let H and K be Hilbert spaces. Then T is compact if and only if T ∗ is compact.

Proposition 11 Let H and K be Hilbert spaces and T ∈ B(H, K). Then T is compact if and onlyif for any sequence (xn) ⊂ H converging weakly to x, the sequence (Txn) converges (strongly) toTx in K.

99


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100 CHAPTER 7. COMPACT OPERATORS

Problem 97. Let X be a Banach space. Prove that if T ∈ B(X) is arbitraryand A ∈ B0(X), then AT and TA are compact operators. (This is called the twosides ideal property for compact operators).

Solution.Suppose (xn) is a sequence in H such that ‖xn‖ ≤ 1 for every n ∈ N. Since T iscontinuous,

‖Txn‖ ≤ ‖T‖ ‖xn‖ ≤ ‖T‖, ∀n ∈ N.

If we set yn = Txn

‖T‖ , and then we have ‖yn‖ ≤ 1 for every n ∈ N. Since A is compact,

the sequence (Ayn) has a convergent subsequence. Now we have

‖T‖Ayn =‖T‖ATxn

‖T‖ = ATxn, ∀n ∈ N.

It follows that the sequence (ATxn) also has a convergent subsequence. Thus AT iscompact. The similar argument for TA ¥

Problem 98.(a) Let X be a Banach space. Show that the identity I : X → X is compact ifand only if X has finite dimensional.(b) Let X, Y be Banach spaces and A ∈ B(X, Y ). Suppose that A has theproperty:

∃c > 0 : ‖Ax‖ ≥ c‖x‖, ∀x ∈ X,

Find condition(s) for X so that A can be a compact operator.

Solution.(a) See Problem 16.

(b) First we note that A is injective. Indeed,

Ax = 0 ⇒ cx = 0 ⇒ x = 0.

Let Z = A(X), then U : X → Z defined by U(x) = A(x) is bijective. Let usconsider U−1 : Z → X. Clearly U−1 is linear. we claim:

(∗) ‖U−1(y)‖ ≤ 1

c‖y‖, ∀y ∈ Z


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101

Proof: Since y ∈ A(X), there is an x ∈ X such that A(x) = U(x) = y. This impliesthat x = U−1(y). By our hypothesis,

‖Ax‖ = ‖U(x)‖ = ‖y‖ ≥ c‖x‖.Thus,

‖y‖ ≥ c‖U−1(y)‖.Hence, (*) is proved. It follows that U−1 is linear and continuous. If U is a compactoperator, then by the ideal property for the compact operators (Problem 97), itfollows that I = U−1U : X → X is compact, which means that X is finite dimen-sional. Conversely, if X is finite dimensional, then every A ∈ B(X, Y ) is compact(in a finite dimensional normed space, a set is compact if and only if it is closed andbounded). Hence A is compact if and only if X is finite dimensional. ¥

Problem 99.Let H and K be Hilbert spaces and A ∈ B(H, K). Show that A is compact ifand only if A∗A is compact.

Solution.• Suppose A ∈ B(H, K) is compact. Let (xn) be a sequence in X converging weaklyto 0. We have

‖A∗Axn‖ ≤ ‖A∗‖ ‖Axn‖.Since A is compact, Axn → 0 (strongly) in Y . Thus A∗Axn → 0, and so A∗A iscompact.

• Reciprocally, suppose A∗A is compact. For any sequence (xn) such that xnw→ 0,

we have‖Axn‖2 = 〈Axn, Axn〉 = 〈xn, A∗Axn〉 ≤ ‖A∗Axn‖ ‖xn‖.

Since ‖x‖ is uniformly bounded and A∗A is compact, A∗Axn → 0. Therefore,Axn → 0, and hence A is compact. ¥

Problem 100.Let X be c0 or `p, 1 ≤ p ≤ ∞. Consider the operator

U : X → X, U(x) = U(x1, x2, ...) = (0, x1, 0, x3, 0, x5, ...).

Prove that U is not compact but U2 is compact.


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Solution.We first note that c0 and `p, 1 ≤ p ≤ ∞ (with appropriate norms) are Banachspaces (see Problems 18,19). We have

U2(x) = U(U(x)) = U(0, x1, 0, x3, 0, x5, ...) = (0, 0, ...).

Thus, U2 = 0, therefore U2 is compact.On the other hand, if

en = (0, ..., 0︸︷︷︸n

, 1, 0, ...) ∈ X

thenU(e2n−1) = e2n, ∀n ∈ N.

Now, we have explicitly

e2n = (0, ..., 0︸︷︷︸2n

, 1, 0, ...), e2(n+k) = (0, ..., 0︸︷︷︸2n

, 0, ..., 0︸︷︷︸2k

, 1, 0, ...),

so thate2(n+k) − e2n = (0, ..., 0︸︷︷︸

2n

,−1, 0, ..., 0︸︷︷︸2k

, 1, 0, ...).

For X = c0 or X = `∞ we have

‖e2(n+k) − e2n‖∞ = 1.

For X = `p, 1 ≤ p < ∞ we have

‖e2(n+k) − e2n‖p = 21/p.

It follows that, in both cases, the sequence(U(e2n−1)

)cannot have any convergent

subsequence. Thus, U is not a compact operator. ¥

Problem 101Let 1 ≤ p < ∞, and λ = (λn)n∈N ⊂ K with supn∈N |λn| < ∞. We define themultiplication operator

Mλ : `p → `p, Mλ(x) = (λ1x1, λ2x2, ...), x = (x1, x2, ...) ∈ `p.

Prove that:(a) Mλ is continuous and ‖Mλ‖ = supn∈N |λn|.(b) Mλ is a compact operator if and only if λ ∈ c0.


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103


|λnxn|p ≤ ‖λ‖p∞|xn|p, ∀n ∈ N.

Since the series∑∞

n=1 |xn|p converges and ‖λ‖∞ < ∞ by hypothesis, the series∑∞n=1 |λnxn|p converges. Moreover,

‖Mλ(x)‖ =

( ∞∑n=1

|λnxn|p)1/p

≤ ‖λ‖∞( ∞∑

n=1

|xn|p)1/p

= ‖λ‖∞‖x‖, ∀x ∈ `p.

This shows that Mλ is continuous and ‖Mλ‖ ≤ ‖λ‖∞. Also, for any n ∈ N,

|λn| = |Mλ(en)| ≤ ‖Mλ‖ ‖en‖ = ‖Mλ‖.Here en = (0, ..., 0︸︷︷︸

n−1

, 1, 0, ...) ∈ `p. Therefore, ‖λ‖∞ = supn∈N |λn| ≤ ‖Mλ‖. Thus,

‖Mλ‖ = supn∈N |λn|.(b) Suppose Mλ is a compact operator. i.e., Mλ(B`p) is relatively compact (B`p theclosed unit ball in `p ). Then

∀ε > 0, ∃nε ∈ N :∞∑

k=nε

|Mλ(x)|p ≤ εp, ∀x ∈ B`p .

Let n ≥ nε. Then for en ∈ B`p we have

∞∑

k=nε

|Mλ(en)|p ≤ εp,

that isn ≥ nε ⇒ |λn| < ε.

So λn → 0, that is, λ ∈ c0.Conversely, if λn → 0, then

∀ε > 0, ∃nε ∈ N : n ≥ nε ⇒ |λn| < ε.

Let x = (x1, x2, ...) ∈ B`p , then

∞∑

k=nε

|Mλ(x)|p =∞∑

k=nε

|λkxk|p ≤ εp

∞∑

k=nε

|xk|p ≤ εp

∞∑

k=1

|xk|p ≤ εp.


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Therefore Mλ(B`p) is relatively compact. ¥

Problem 102Consider the linear operator defined by

T : `2 → `2, x = (ξ1, ξ2, ξ3, ...) 7→ Tx =

(ξ1

1,ξ2

2,ξ3

3, ...

).

Show that T is compact.

Solution.It is clear that T is linear. To show that it is compact, we will show that it is thenorm limit of a sequence of compact operators. Let

Tn : `2 → `2, Tnx =

(ξ1

1,ξ2

2, ...,

ξn

n, 0, 0, ...

).

Then Tn is linear, bounded, and of finite rank so compact. Furthermore,

‖(T − Tn)x‖2 =∞∑

i=n+1

1

i2|ξi|2

≤ 1

(n + 1)2

∞∑i=1

|ξi|2

=‖x‖2

(n + 1)2.

Taking the supremum over all x of norm 1, we see that

‖T − Tn‖ ≤ 1

n + 1.

Hence, Tn → T in norm. Thus, T is compact. ¥

Problem 103Let (cj)

∞j=1 be a sequence of complex numbers. Define an operator D on `2 by

Dx = (c1x1, c2x2, ...), x = (x1, x2, ...) ∈ `2.

Prove that D is compact if and only if limj→∞ cj = 0.


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105

Solution.• We note that D is linear. To show that it is compact, we will show that it is thenorm limit of a sequence of compact operators. Suppose limj→∞ cj = 0. Define Dn

by

Dn = (c1x1, ..., cnxn, 0, 0, ...).

We obtain that

(D −Dn) = (0, ..., 0.cn+1xn+1, cn+2xn+2, ...)

and moreover,

‖D −Dn‖ = supj≥n+1

|cj| → 0 as n →∞.

Since each Dn has finite rank and hence is compact, the operator D is compact.

• Assume that (cj) does not converge to zero as j → ∞. Then, for a given ε > 0,there exists a subsequence (cjk

) such that |cjk| ≥ ε. Consider the sequence of vectors

(ej) of the standard basis. We have ‖ejk‖ = 1 and for any indices m, k we have

‖Dejm −Dejk‖2 = ‖cjmejm − cjk

ejk‖2 = |cjm|2 + |cjk

|2 ≥ 2ε2 > 0.

We conclude that the sequence (Dejk) does not contain a convergent subsequence

and thus the operator D is not compact. ¥

Trick used in problems 102 and 103 is called ”cut off ” method: From the sequence x = (x1, x2, ...)we get the sequence (x1, ..., xn, 0, 0, ...) by cutting off the tail of x.

Problem 104Let g ∈ C[0, 1] be a fixed function. Consider the operator A ∈ B(C[0, 1]) definedby

(Au)(s) := g(s)u(s),

i.e., the operator of multiplication by g. Is this operator compact?

Solution.Note first that C[0, 1], equipped with the sup-norm, is a Banach space.It is clear that if g ≡ 0 then A is compact. Let us prove that if g is not identicallyzero then A is not compact. Indeed, since g is not identically zero, there exists asubinterval [a, b] ⊂ [0, 1] such that

m := mins∈[a,b]

|g(s)| > 0.


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Consider the sequence (un):

un ∈ C[0, 1]; un(s) := sin

(2n s− a

b− aπ

); s ∈ [0, 1], n ∈ N.

It is clear that (un) is a bounded sequence. On the other hand, (Aun) does not haveCauchy subsequences. Indeed, take arbitrary k, n ∈ N with k > n. Let

sn := a +1

2n+1(b− a)

Then sn ∈ [a, b] and

‖Auk − Aun‖ = maxs∈[a,b]

|g(s)(uk(s)− un(s))|≥ m max

s∈[a,b]|uk(s)− un(s)|

≥ m|uk(sn)− un(sn)|= m| sin(2k−n−1π)− sin(π/2)|= m|0− 1| = m > 0.

Hence (Aun) cannot have any convergent subsequence, A is not compact. ¥

Problem 105Given k ∈ L2([0, 1]× [0, 1]), define the operator A : L2([0, 1]) → L2([0, 1]) by

(Af)(x) =

∫ 1

0

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

(a) Show that A is bounded.(b) Under what condition on k, the operator A is self-adjoint.(c) Show that A is compact.

Solution.(Look at Problem 92! They are different!)(a) We estimate ‖A‖ to see A is bounded.

‖Af‖2 =

∫ 1

0

∣∣∣∣∫ 1

0

k(x, y)f(y)dy

∣∣∣∣2

dx

≤∫ 1

0

(∫ 1

0

|k(x, y)|2dy

)dx.

∫ 1

0

|f(y)|2dy (Cauchy-Schwarz)

≤ ‖f‖2.

∫ 1

0

∫ 1

0

|k(x, y)|2dydx.


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107

Since k ∈ L2([0, 1]× [0, 1]),∫ 1

0

∫ 1

0|k(x, y)|2dydx < ∞; hence,

‖A‖ ≤(∫ 1

0

∫ 1

0

|k(x, y)|2dydx

)1/2

< ∞.

Thus A is bounded.

(b) We have

(Af)(x) =

∫ 1

0

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

(A∗g)(x) =

∫ 1

0

k(x, y)g(y)dy.

Therefore,

〈Af, g〉 =

∫ 1

0

∫ 1

0

k(x, y)f(y)dy g(x) dx

=

∫ 1

0

f(y)

∫ 1

0

k(x, y) g(x) dxdy

= 〈f, A∗g〉.

Hence, A is self-adjoint ifk(x, y) = k(y, x).

(c) Let (uj)∞j=1 be an orthonormal basis in L2[0, 1]. Then

k(x, y) =∞∑

j=1

kj(y)uj(x), where kj(y) =

∫ 1

0

k(x, y)uj(x) dx,

for almost all y. Due to the Parseval identity, we have, for almost all y

∫ 1

0

|k(x, y)|2dx =∞∑

j=1

|kj(y)|2,

and

(1)

∫ 1

0

∫ 1

0

|kj(y)|2dxdy =∞∑

j=1

∫ 1

0

|kj(y)|2dy.

We now define the following operator of rank N

kNf(x) =

∫ 1

0

kN(x, y)f(y)dy,


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where kN(x, y) =∑N

j=1 kj(y)uj(x). By Cauchy-Schwarz inequality we obtain

‖(A− kN)f‖2 =

∫ 1

0

∣∣∣∣∫ 1

0

(k(x, y)− kN(x, y))f(y)dy

∣∣∣∣2

dx

≤(∫ 1

0

∫ 1

0

|k(x, y)− kN(x, y)|2dxdy

)(∫ 1

0

|f(y)|2dy

).

≤ ‖f‖2

(∫ 1

0

∫ 1

0


).

Thus by using that the right hand side in (1) is absolutely convergent, we find

‖(A− kN)‖2 ≤∫ 1

0

∫ 1

0


=

∫ 1

0

∫ 1

0

|k(x, y)|2dxdy −∫ 1

0

∫ 1

0

k(x, y)N∑

j=1

kj(y)uj(x) dxdy

−∫ 1

0

∫ 1

0

k(x, y)N∑

j=1

kj(y)uj(x)dxdy +N∑

j=1

∫ 1

0

|kj(y)|2dy

=

∫ 1

0

∫ 1

0

|k(x, y)|2dxdy −N∑

j=1

|kj(y)|2dy → 0 as N →∞. ¥

Problem 106Part IConsider the operator

U : C[0, 1] → C[0, 1] defined by (Uf)(x) =

∫ x

0

etf(t)dt, x ∈ [0, 1],

and the sequence of operators

Un : C[0, 1] → C[0, 1] defined by (Unf)(x) =

∫ x

0

(n∑

k=0

tk

k!

)f(t)dt, x ∈ [0, 1].

Prove that limn→∞ ‖Un − U‖ = 0.

Part II1. Let M be a set of C1-functions f on [0, 1]. Prove that M is relatively compactin C[0, 1] if f satisfies following conditions

|f(0)| ≤ k1 and

∫ 1

0

|f ′(x)|2dx ≤ k2


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109

where k1, k2 are positive constants. (Hint: Use Arzela-Ascoli theorem).2. Show that the operator U in Part I is compact.

Solution.Part IFrom Calculus we know that if φ ∈ C[0, 1] then x 7→ ∫ x

0φ(t)dt is continuous. Hence,

U,Un take their values in C[0, 1]. Using the Taylor expansion et =∑∞

n=0tn

n!we

obtain

(Uf − Unf)(x) =

∫ x

0

( ∞∑

k=n+1

tk

k!

)f(t)dt, ∀x ∈ [0, 1].

Then

|(Uf − Unf)(x)| ≤∫ x

0

( ∞∑

k=n+1

tk

k!

)|f(t)|dt

≤ ‖f‖∞∫ 1

0

( ∞∑

k=n+1

tk

k!

)dt, ∀x ∈ [0, 1].

Thus,

‖Un − U‖ ≤ ‖f‖∞∫ 1

0

( ∞∑

k=n+1

tk

k!

)dt, ∀f ∈ C[0, 1],

≤∫ 1

0

( ∞∑

k=n+1

tk

k!

)dt.

But if un(t) =∑∞

k=n+1tk

k!then un : [0, 1] → R and

|un(t)| ≤∞∑

k=n+1

1

k!→ 0,

that is, un → 0 uniformly on [0, 1]. Hence,∫ 1

0un(t)dt → 0 as n →∞. Thus,

limn→∞

‖Un − U‖ = 0.


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Part II1. For all x, y ∈ [0, 1] with x < y, and all f ∈ M , we have

|f(x)− f(y)| =

∣∣∣∣∫ y

x

|f ′(x)|∣∣∣∣ dt

≤(∫ y

x

|f ′(x)|2dt

)1/2 (∫ y

x

1dt

)1/2

≤ √y − x

(∫ y

x

|f ′(x)|2dt

)1/2

≤√

k2

√y − x.

This shows that M is qui-continuous.Now for all x ∈ M , and all f ∈ M , we have

|f(x)| ≤ |f(x)− f(0)|+ |f(0)|≤ k1 +

√k2.

So M is uniformly bounded. Thus, by Arzela-Ascoli Theorem1, M is relativelycompact in C[0, 1].

2. For ‖f‖ ≤ 1, let g(x) =∫ 1

0etxf(t)dt, ∀x ∈ [0, 1]. Using the differentiation

theorem for the Riemann integral with parameter, we have

g′(x) =

∫ 1

0

∂

∂x(etxf(t))dt =

∫ 1

0

tetxf(t)dt,

|g′(x)| ≤∫ 1

0

tetx|f(t)|dt ≤∫ 1

0

tetxdt ≤ e, ∀x ∈ [0, 1].

It follows that the set of all functions g is uniformly Lipschitz. We also have

|g(0)| =∣∣∣∣∫ 1

0

f(t)dt

∣∣∣∣ ≤∫ 1

0

|f(t)|dt ≤ ‖f‖ ≤ 1, ∀g.

From previous question, it follows that A = Uf : ‖f‖ ≤ 1 is relatively compact.Thus the operator U is compact. ¥

1Arzela-Ascoli Theorem: Let (X, d) be a compact metric space and A ⊂ C(X). Then thefollowing assertions are equivalent:

1. A is relatively compact.

2. A is uniformly bounded and equi-continuous,

3. Any sequence (fn) ⊂ A contains a uniformly convergent subsequence.


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111

Problem 107(a) Let X be an infinite dimensional Banach space, and A be a compact operatoron X. Prove that there is y ∈ X such that the equation A(x) = y has no solution,i.e., A is not surjective.(b) Let 1 ≤ p < ∞, and the operator

U : `p → `p, U(x) =(x1,

x2

2,x3

3, ...

), x = (x1, x2, ...) ∈ `p.

Find an element y ∈ `p for which the equation U(x) = y has no solution.(c) Consider the operator

A : c00 → c00 defined by A(x) = A(x1, x2, ...) =(x1,

x2

2,x3

3, ...

).

Prove that A is compact and bijective. On c00 we have the `∞- norm.

Solution.(a) Let us suppose, for a contradiction, that A is surjective. From the open mappingtheorem it follows that A is an open operator, in particular A(B(0; 1)) ⊂ X is anopen set, i.e.,

∃ε > 0 : B(0; ε) = εB(0; 1) ⊂ A(B(0; 1)),

where B(0; 1) = x ∈ X : ‖x‖ < 1. Since A is compact it follows that A(B(0; 1))is relatively compact. Hence, B(0; ε) is relatively compact, from whence B(0; 1) isrelatively compact, therefore compact. But if B(0; 1) is compact, then, by Problem16, X must be finite dimensional, which is a contradiction. Thus, A is not surjective.

(b) Choose y =(1, 1

2α , 13α , ...

) ∈ `p. If x = (x1, x2, ...) ∈ `p has the property thatU(x) = y, then

1

nxn =

1

nα, ∀n ∈ N,

that is,

xn =1

nα−1, ∀n ∈ N.


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Since x = (x1, x2, ...) ∈ `p, the generalized harmonic series∑∞

n=11

n(α−1)p converges2,therefore, (α− 1)p > 1 ⇔ α > 1 + 1

p. From here, it follows that for

y =

(1,

1

21

p+1

,1

31

p+1

, ...

),

the equation U(x) = y has no solution.

(c) For n ∈ N, consider

An : c00 → c00, An(x) = An(x1, x2, ...) =(x1,

x2

2, ..,

xn

n, 0, ...

).

Then

‖A− An‖ =1

n + 1, ∀n ∈ N.

Therefore An → A in norm. Since every An is a finite rank operator, so An iscompact. The sequence of compact operators (An) converges to A in norm, so Amust be compact. The fact that A is bijective is obvious from the expression whichdefines A. ¥

2 The generalized harmonic series∑∞

n=11

nαp converges if and only if

αp > 1 ⇔ α >1p.


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Chapter 8

Bounded Operators on BanachSpaces and Their Spectra

Review:

1. Definitions

Let X be a Banach space and T ∈ B(X), λ ∈ C.• Resolvent and spectrum of T :Set Tλ = T − λI. The set ρ(T ) of all λ such that Tλ has an inverse Rλ(T ) = (T − λI)−1 is calledthe resolvent of T . The set σ(T ) = C \ ρ(T ) is called the spectrum of T .• Eigenvalues and eigenvectors of T :An x 6= 0 which satisfies Tx = λx for some λ is called an eigenvector of T . The corresponding λis an eigenvalue of T . It is evident that λ ∈ σ(T ).

2. Basic properties

Theorem 12 (Spectrum)If T is a bounded linear operator on a Banach space X, then its spectrum σ(T ) is compact and liesin the disk given by

|λ| ≤ ‖T‖.

Theorem 13 (Resolvent equation)Let T ∈ B(X, X) where X is a Banach space. Then

1. The resolvent Rλ(T ) satisfies the following equation called the resolvent equation

Rµ −Rλ = (µ− λ)RµRλ for µ, λ ∈ ρ(T ).

2. Rλ commutes with any S ∈ B(X, X) which commutes with T .

3. We haveRµRλ = RλRµ for µ, λ ∈ ρ(T ).

113


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114CHAPTER 8. BOUNDED OPERATORS ON BANACH SPACES AND THEIR SPECTRA

3. Classification of spectrum

• λ ∈ C is called a regular point of A ∈ B(X) iff (A− λI)−1 exists and is bounded.

• If λ is not a regular point, it is called a spectrum point. All such points form the spectrumσ(A).

• Every λ ∈ C with |λ| > ‖A‖ is a regular point.

• Classification of spectrum:

1. The point spectrum: σp(A) is the set of eigenvalues of A.

2. The continuous spectrum: λ ∈ σc(A) iff λ ∈ σ(A)\σp(A) and Image(A−λI) is densein X.

3. The residual spectrum: σr(A) = σ(A) \ (σp(A) ∪ σc(A)). If λ ∈ σr(A) then

Image(A− λI) 6= X and ker(A− λI) = 0

.

4. Spectral radius

Definition 9 Let T ∈ B(X,X) where X is a Banach space. The spectral radius rσ(T ) of T is theradius of the smallest closed disk centered at the orgin and containing σ(T ).

rσ(T ) := supλ∈σ(T )

|λ|.

Formula for spectral radius:It can be shown that

rσ(T ) = limn→∞

n√‖Tn‖.

5. Spectral mapping theorem

Theorem 14 (Spectral theorem for polynomials)Let T ∈ B(X, X) where X is a Banach space, and

p(λ) = αnλn + αn−1λn−1 + ... + α0, αn 6= 0.

Thenσ(p(T )) = p(σ(T )),

that is, the spectrum of the operator

p(T ) = αnTn + αn−1Tn−1 + ... + α0I

consists precisely of all those values which the polynomial p assumes on the spectrum σ(T ) of T .

∗ ∗ ∗ ∗ ∗


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115

Problem 108Let X be a Banach space. Suppose that A ∈ B(X) is an invertible operator.Show that

σ(A−1) = λ−1 : λ 6= 0, λ ∈ σ(A).

Solution.For λ 6= 0, we can write

A−1 − λ−1I = (λI − A)λ−1A−1.

From this equality we conclude that A−1 − λ−1I is invertible if and only if A − λIis invertible. Hence, we have

σ(A−1) = λ−1 : λ ∈ σ(A). ¥

Problem 109Let X be a Banach space, let A ∈ B(X) and λ ∈ C. Assume that there exists asequence (xn) in X such that

‖xn‖ = 1, ∀n ∈ N and Axn − λxn → 0 as n →∞.

Prove that λ ∈ σ(A).

Solution.Assume that A− λI is invertible. Then, there exists a number c > 0 such that (seeproblem 75)

‖(A− λI)x‖ ≥ c‖x‖, ∀x ∈ X.

Replace x by xn with ‖xn‖ = 1 for all n, we have

‖Axn − λxn‖ = ‖(A− λI)xn‖ ≥ c‖xn‖ = c.

This contradicts the condition in the statement of the problem. ¥


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Problem 110Let (an) and (bn) be complex sequences such that

|an−1| > |an| (n→∞)−−−−→ 0 and |bn−1| > |bn| (n→∞)−−−−→ 0.

Consider the operator T : `2 → `2 defined by

Tx = (a1x1, a2x2 + b1x1, a3x3 + b2x2, ...), x = (xj) ∈ `2.

(a) Show that T is compact.(b) Find all eigenvalues and eigenvectors of T .

Solution.(a) Let the sequence of operators Tn : `2 → `2, n = 1, 2, ... be defined by for anyx ∈ `2:

(Tnx)j =

(Tx)j if 1 ≤ j ≤ n

0 if j > n.

All the Tn’s are operators of finite rank and hence compact. Moreover, we have

‖Tnx− Tx‖ ≤ (|an+1|+ |bn|)‖x‖,which implies that

‖Tn − T‖ → 0 as n →∞.

But a uniform limit of a sequence of compact operators is compact. Hence, T iscompact.(b) Suppose λ ∈ C is an eigenvalue of T and that x 6= 0 is the correspondingeigenvector. Then

0 = Tx− λx =((a1 − λ)x1, (a2 − λ)x2 + b1x1, ..., (an − λ)xn + bn−1xn−1, ...

).

If λ coincides with none of the an’s, then x = 0: impossible. So it is necessary thatλ = an for some n. In this case, xn can be chosen arbitrary, and x1 = ... = xn−1 = 0,and for k = 1, 2, ... we have 0 = (an+k−λ)xn+k +bn+k−1xn+k−1. If we choose xn = 1then we get

xn+k =bn+k−1

λ− an+k

bn+k−2

λ− an+k−1

...bn

λ− an+1

, k = 1, 2, ...

Thus, for any n, λ = an is a simple eigenvalue. The corresponding eigenvector isx = (0, ..0, xn, xn+1, ...) defined as above. ¥


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117

Problem 111Let X be a Banach space and let A ∈ B(X) such that An = 0 for some n ∈ N(A is nilpotent). Find σ(A).

Solution.The spectral mapping theorem implies

λn : λ ∈ σ(A) = σ(An) = σ(0) = 0.Therefore,

λ ∈ σ(A) ⇔ λn = 0 ⇔ λ = 0.

Thus, σ(A) = 0. ¥

Problem 112Let P ∈ B(X) be a projection, i.e., a linear operator on X such that P 2 = P .Construct the resolvent R(P ; λ) of P

Solution.If P = 0, then obviously

P − λI = −λI; σ(P ) = 0; R(P ; λ) := (P − λI)−1 = −λ−1I.

If P = I, then

P − λI = (1− λ)I; σ(P ) = 1; R(P ; λ) := (P − λI)−1 = (1− λ)−1I.

Suppose P is non- trivial, i.e., P 6= 0, I. Take any λ 6= 0, 1. Then using the equalitiesP 2 = P ; Q2 = Q and QP = PQ where Q = I − P , we obtain

((1− λ)−1P − λ−1Q

)(P − λI) =

((1− λ)−1P − λ−1Q

)((1− λ)P − λQ)

= P + Q = I.

Similarly, we have

(P − λI)((1− λ)−1P − λ−1Q

)= I.

Thus,

R(P ; λ) = (1− λ)−1P − λ−1Q = λ−1((1− λ)−1P − I

). ¥


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Problem 113Let CR be the space of all continuous and bounded functions x(t) on R with norm‖x‖ = supR |x(t)|. On the space CR we define the operator A by

(Ax)(t) = x(t + c),

where c ∈ R is a constant. Prove that

σ(A) = λ ∈ C : |λ| = 1.

Solution.Notice that

‖Ax‖ = supt∈R

|x(t + c)| = supτ∈R

|x(τ)| = ‖x‖.

It follows that ‖A‖ = 1 and therefore all of the point of λ ∈ C : |λ| > 1 areregular points of A. The operator A is invertible since the operator defined by

(A−1x)(t) = x(t− c)

is bounded and is the inverse of A. Next ‖A−1‖ = 1 and hence all of the point ofλ ∈ C : |λ| > 1 are regular points of A−1. From Problem 65 we deduce that allof the points λ ∈ C : |λ| < 1 are regular points of A.Consider |λ| = 1. This means that

λ = eiϕ, 0 ≤ ϕ ≤ 2π.

Set a = iϕc

and xa(t) = eat. We obtain xa ∈ CR and

(Axa)(t) = ea(t+c) = eateac = λxa.

This means that λ is an eigenvalue of A. Thus,

σ(A) = σp(A) = λ ∈ C : |λ| = 1. ¥

Problem 114(a) Let H be a Hilbert space and a, b ∈ H. Consider the rank-one operator

U : H → H, U(x) = 〈x, a〉b, x ∈ H.


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119

Calculate ‖U‖ and the spectral radius r(U). Show that

r(U) = ‖U‖ ⇔ a, b are linearly independent.

(b) Let X be a Banach space, x∗ ∈ X∗ and y ∈ X. Consider the rank-oneoperator

V : X → X, V (x) = x∗(x)y, x ∈ X.

Calculate ‖V ‖ and the spectral radius r(V ). Show that

r(V ) = ‖V ‖ ⇔ |x∗(y)| = ‖x∗‖ ‖y‖.


‖U(x)‖ = |〈x, a〉| ‖b‖ ≤ ‖a‖ ‖b‖ ‖x‖, ∀x ∈ H.

Hence, ‖U‖ ≤ ‖a‖ ‖b‖. We also have

‖U‖ ‖a‖ ≥ ‖U(a)‖ = |〈a, a〉| ‖b‖ = ‖a‖2 ‖b‖.

Therefore, ‖U‖ ≥ ‖a‖ ‖b‖. Thus,

‖U‖ = ‖a‖ ‖b‖.

For x ∈ H, let y = U(x). Then

U2(x) = U(y) = 〈y, a〉b, where 〈y, a〉 = 〈Ux, a〉 = 〈x, a〉〈a, b〉.

Denoting λ = 〈a, b〉, we have

U2(x) = λ〈x, a〉b = λU(x), ∀x ∈ H.

ThereforeU2 = λU.

From here, by induction, we get

Un = λn−1U, ∀n ∈ N.

Nowr(U) = lim

n→∞(‖U‖n)

1n = lim

n→∞

(|λ|n−1

n ‖U‖ 1n

)= |λ| = |〈a, b〉|.


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The last assertion is a consequence of the fact proved above and the fact that inthe Cauchy-Schwarz inequality we have equality if and only if a, b are linearlyindependent.

r(U) = ‖U‖ ⇔ ‖a‖ ‖b‖ = |〈a, b〉| ⇔ a, b are linearly independent.

(b) We have‖V (x)‖ = |x∗(x)| ‖y‖ ≤ ‖x∗‖ ‖y‖ ‖x‖, ∀x ∈ X.

Therefore, ‖V ‖ ≤ ‖x∗‖ ‖y‖.If y = 0 then ‖V ‖ = 0. Suppose y 6= 0. For x ∈ X, we have

|x∗(x)| ‖y‖ = ‖V (x)‖ ≤ ‖V ‖ ‖x‖.Then

|x∗(x)| ≤ ‖V ‖‖y‖ ‖x‖ ⇒ ‖x∗‖ ≤ ‖V ‖

‖y‖⇒ ‖V ‖ ≥ ‖x∗‖ ‖y‖.

Thus,‖V ‖ = ‖x∗‖ ‖y‖.

To calculate r(V ) we use the same procedure as in (a). For x ∈ H, let z = V (x).Then

V 2(x) = V (z) = x∗(z)y

z = V (x) = x∗(x)y,

x∗(z) = x∗(x)x∗(y).

Therefore,V 2(x) = x∗(y)x∗(x)y = λx∗(x)y, λ = x∗(y).

Thus, V 2 = λV . And from here, by induction, we get

V n = λn−1V, ∀n ∈ N.

And as above we obtainr(V ) = |λ| = |x∗(y)|. ¥

Problem 115Let k ∈ C([0, 1]× [0, 1]) be a given function. Consider the operator

B ∈ B(C[0, 1]) defined by (Bu)(s) =

∫ s

0

k(s, t)u(t)dt.

Find σ(B) and r(B).


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121

Solution.Let us prove by induction that

(∗) |(Bnu)(s)| ≤ Mn

n!sn‖u‖∞, ∀s ∈ [0, 1], ∀n ∈ 0, 1, 2, ...

where

M := max(s,t)∈[0,1]2

|k(s, t)|.

For n = 0, then B0 = I, and (*) is trivial. Suppose (*) holds for n = k. Then forn = k + 1 we have

∣∣(Bk+1u)(s)∣∣ =

∣∣∣∣∫ s

0

k(s, t)(Bku)(t)dt

∣∣∣∣

≤∫ s

0

|k(s, t)|∣∣(Bku)(t)

∣∣ dt

≤ M

∫ s

0

∣∣(Bku)(t)∣∣ dt

≤ M

∫ s

0

Mk

k!tk‖u‖∞dt

=Mk+1

k!‖u‖∞

∫ s

0

tkdt

=Mk+1

(k + 1)!sk+1 ‖u‖∞, ∀s ∈ [0, 1].

Hence, (*) is proved by induction.It follows from (*) that

‖Bnu‖∞ ≤ Mn

n!‖u‖∞, ∀u ∈ C[0, 1],

i.e.,

‖Bn‖ ≤ Mn

n!, ∀n ∈ 0, 1, 2, ....

Therefore

r(B) = limn→∞

‖Bn‖1/n ≤ limn→∞

M

(n!)1/n= 0.

Since r(B) = 0, σ(B) cannot contain nonzero elements. Taking into account thatσ(B) is nonempty, we conclude that σ(B) = 0. ¥


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Problem 116Let X be a Banach space and A,B ∈ B(X). Suppose AB = BA. Prove that

r(A + B) ≤ r(A) + r(B),

where r(T ) is the spectral radius of an operator T ∈ B(X).

Solution.Recall

r(T ) := sup|λ| : λ ∈ σ(T ) and r(T ) = limn→∞

‖T n‖1/n.

Take an arbitrary ε > 0. The spectral radius formula implies that

‖An‖ ≤ (r(A) + ε

)n, ‖Bn‖ ≤ (

r(B) + ε)n

for sufficiently large n ∈ N. Therefore there exists a constant M ≥ 1 such that

‖An‖ ≤ M(r(A) + ε

)n, ‖Bn‖ ≤ M

(r(B) + ε

)n, ∀n ∈ N.

Since AB = BA, we have

(A + B)n =n∑

k=0

n!

k!(n− k)!An−kBk.

Hence

‖(A + B)n‖ ≤n∑

k=0

n!

k!(n− k)!‖An−k‖ ‖Bk‖

≤ M2

n∑

k=0

n!

k!(n− k)!

(r(A) + ε)n−k(r(B) + ε

)k

= M2(r(A) + r(B) + 2ε

)n, ∀n ∈ N.

Consequently,

r(A + B) = limn→∞

‖(A + B)n‖1/n ≤ r(A) + r(B) + 2ε, ∀ε > 0.

This implies thatr(A + B) ≤ r(A) + r(B). ¥


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123

Problem 117Let K ⊂ C be an arbitrary non-empty compact set. Construct an operatorB ∈ B(`p), 1 ≤ p ≤ ∞, such that σ(B) = K.

Solution.Let λkk∈N be a dense subset of K. (Recall that every metric compact space isseparable, that is, it contains a countable dense subset.) Consider the operatorB : `p → `p defined by

Bx = (λ1x1, λ2x2, ...), ∀x = (x1, x2, ...) ∈ `p.

Then B is a bounded operator and λk’s are its eigenvalues. Consequently,

λkk∈N ⊂ σ(B).

Since σ(B) is closed and λkk∈N is dense in K,

K ⊂ σ(B).

On the other hand, let λ ∈ C \K. Then d := infk∈N |λk − λ| > 0 and B − λI has abounded inverse (B − λI)−1 : `p → `p defined by

(B − λI)−1x =

(1

λ1 − λx1, ...,

1

λk − λxk, ...

), ∀x = (x1, x2, ...) ∈ `p.

Hence, λ /∈ σ(B). Therefore, σ(B) ⊂ K. Finally,

σ(B) = K. ¥

Problem 118Let g ∈ C[0, 1] be a fixed function and A ∈ B(C[0, 1]) be defined by

(Af)(t) = g(t)f(t), t ∈ [0, 1].

Find σ(A) and construct effectively the resolvent R(A; λ). Find the eigenvaluesand eigenvectors of A.


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Solution.Let λ ∈ C, λ /∈ g([0, 1]) := g(t); t ∈ [0, 1]. Then, since g ∈ C[0, 1],

1

g − λ∈ C[0, 1]

and A− λI has an inverse

R(A; λ) = (A− λI)−1 ∈ B(C[0, 1])

defined by

R(A; λ)f(t) = (g(t)− λ)−1f(t), t ∈ [0, 1].

Hence, σ(A) ⊂ g[0, 1].Suppose now λ ∈ g([0, 1]), i.e., λ = g(t0) for some t0 ∈ [0, 1]. Then

(A− λI)f(t0) = (g(t0)− λ)f(t0) = 0,

i.e., Image(A− λI) consists of functions vanishing at t0.Consequently, Image(A − λI) 6= C[0, 1] and A − λI is not invertible. Therefore,g([0, 1]) ⊂ σ(A). Finally,

σ(A) = g([0, 1]).

Take an arbitrary λ ∈ g([0, 1]). Let g−1(λ) := τ ∈ [0, 1] : g(τ) = λ. The equationAf = λf , i.e., (g(t) − λ)f(t) = 0 is equivalent to f(t) = 0, ∀t ∈ [0, 1] \ g−1(λ) . Ifg−1(λ) contains an interval of positive length, then it is easy to see that the set

f ∈ C[0, 1] \ 0 : f(t) = 0, ∀t ∈ [0, 1] \ g−1(λ)

is non-empty and coincides with the set of all eigenvectors corresponding to theeigenvalues λ. If g−1(λ) does not contain an interval of positive length, then [0, 1] \g−1(λ) is dense in [0, 1] and f(t) = 0, ∀t ∈ [0, 1] \ g−1(λ) implies by continuity thatf ≡ 0. In this case λ is not an eigenvalue. ¥

Problem 119Let X be a Banach space and A,B ∈ B(X). Show that for any λ ∈ ρ(A)∩ ρ(B),

R(B; λ)−R(A; λ) = R(B; λ)(A−B)R(A; λ).


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125

Solution.

R(B; λ)(A−B)R(A; λ) = R(B; λ)[(A− λI)− (B − λI)

]R(A; λ)

=[R(B; λ)(A− λI)−R(B; λ)(B − λI)

]R(A; λ)

= R(B; λ)(A− λI)R(A; λ)−R(B; λ)(B − λI)R(A; λ)

= R(B; λ)−R(A; λ). ¥

Problem 120Let k ∈ C([0, 1]× [0, 1]) be given. Consider the operator B ∈ B(C[0, 1]) definedby

(Bu)(s) =

∫ s

0

k(s, t)u(t)dt.

Find the spectral radius of B. What is the spectrum of B. (Hint: Prove byinduction that

|(Bnu)(s)| ≤ Mn

n!sn‖u‖∞, ∀n ∈ N,

for some constant M > 0).

Solution.Let us first prove by induction that

|(Bnu)(s)| ≤ Mn

n!sn‖u‖∞, ∀s ∈ [0, 1], ∀n ∈ N, (1)

where

M := max(s,t)∈[0,1]2

|k(s, t)|.

(1) is true for n = 1. Indeed,

|(Bu)(s)| ≤ M

∫ s

0

|u(t)|dt ≤ M‖u‖∞∫ s

0

dt =M1

1!s1‖u‖∞.


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Suppose (1) holds for n = k. Then for n = k + 1 we have

|(Bk+1u)(s)| =

∣∣∣∣∫ s

0

k(s, t)(Bku)(t)dt

∣∣∣∣

≤∫ s

0

|k(s, t)||(Bku)(t)|dt

≤ M

∫ s

0

|(Bku)(t)|dt

≤ M

∫ s

0

Mk

k!tk‖u‖∞dt

=Mk+1

k!‖u‖∞

∫ s

0

tkdt

=Mk+1

(k + 1)!sk+1‖u‖∞, ∀s ∈ [0, 1].

Hence, (1) is true for n = k + 1. Thus (1) is proved by induction.It follows from (1) that

‖Bnu‖∞ ≤ Mn

n!‖u‖∞, ∀u ∈ C[0, 1].

It follows that

‖Bn‖ ≤ Mn

n!, ∀n ∈ N.

Therefore,

r(B) = limn→∞

‖Bn‖1/n ≤ limn→∞

M

(n!)1/n= 0.

Since r(B) = 0, σ(B) cannot contain non-zero elements. Taking into account thatσ(B) is not empty, we conclude that σ(B) = 0. ¥

Problem 121Determine the spectra of the left and the right shift operators on `2:

R(x1, x2, x3, ....) = (0, x1, x2, ...),

L(x1, x2, x3, ....) = (x2, x3, x4, ....).

Classify them into point, continuous, and residual spectrum.

Solution.We have shown that ‖R‖ = ‖L‖ = 1 (problem 46). It follows that

λ ∈ C : |λ| > 1 ⊂ ρ(R) and λ ∈ C : |λ| > 1 ⊂ ρ(L).


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127

Now I prove the following four claims, and then I will state the conclusion.

Claim(1): R− λI is injective (one-to-one) for all λ ∈ C such that |λ| ≤ 1.

Proof.If λ = 0, then

Rx = 0 ⇒ xi = 0, ∀i ∈ N.

Hence x = 0, and so R− λI is injective.Suppose 0 < |λ| ≤ 1. Then

(R− λI)x = 0 ⇒ (0, x1, x2, ...) = (λx1, λx2, λx3, ....).

Since λ 6= 0, this implies that x1 = x2 = ... = 0. Hence x = 0, and R − λI isinjective. ¤Claim(2): R− λI is not surjective (onto) for all λ ∈ C such that |λ| ≤ 1.

Proof.Note that if λ = 0, then e1 = (1, 0, ...) /∈ Image(R− λI) = Image R.Suppose 0 < |λ| ≤ 1. Then

(R− λI)x = e1 ⇒ xn = − 1

λn, ∀n ∈ N

⇒ ‖x‖2 =∞∑

n=1

|xn|2 =∞∑

n=1

(1

|λ|2)n

.

The above series cannot be convergent because 0 < |λ| ≤ 1 implies that 1|λ|2 ≥ 1.

And hence e1 /∈ Image(R− λI). Therefore, R− λI is not surjective. ¤Claim(3): L− λI injective (one-to-one) for all λ ∈ C such that |λ| = 1.

Proof.Suppose it was not injective. There would be some nonzero x ∈ `2 such that(L− λI)x = 0. Then

(x2, x3, x4, ....) = (λx1, λx2, λx3, ....).

Hencexn = λn−1x1, ∀n ∈ N.

Since x 6= 0, x1 6= 0. Since |λ| = 1, we have

‖x‖ =∞∑

n=0

|x1|2|λ|2n =∞∑

n=0

|x1|2.

This sum cannot be finite since x1 6= 0, but this is impossible. So x must be zero,and hence L− λI injective. ¤


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Claim(4): L− λI is not injective (one-to-one) for all λ ∈ C such that |λ| < 1.

Proof.By a similar argument as above, we see that any nonzero x that satisfies the equation(L− λI)x = 0 is of the form

x = (x1, λx1, λ2x1, λ3x1, ....).

Choose x1 = 1, then

‖x‖ =∞∑

n=0

(|λ|2)n.

The series is convergent since |λ| < 1, so x ∈ `2 is nonzero and satisfies the equation(L− λI)x = 0. Thus L− λI is not injective. ¤Conclusion:

• Claims (1) and (2) show that

σ(R) = λ ∈ C : |λ| ≤ 1.Recall that Image(R−λI) is dense if and only if ker(R−λI)∗ = ker(L− λI) = 0.Since |λ| = |λ|, claims (3) and (4) show that Image(R − λI) is dense if and only if|λ| = 1. Therefore

σc(R) = λ ∈ C : |λ| = 1 and σr(R) = λ ∈ C : |λ| < 1.Also from the above results we get

σp(R) = ∅.

• Note that claim (4) shows us that

σp(L) = λ ∈ C : |λ| < 1.Since σ(L) is a closed set, we get

σ(L) = λ ∈ C : |λ| ≤ 1.(As from the above we know that |λ| > 1 implies that λ ∈ ρ(L).)For |λ| = 1 we know that ker(L − λI)∗ = ker(R − λI) = 0 by claim (1). HenceImage(L− λI) is dense. Therefore

σc(L) = λ ∈ C : |λ| = 1 and σr(L) = ∅. ¥

∗ ∗ ∗∗


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129

Alternate solution.

Consider R, L : `2 → `2 defined by

Rx = (0, x1, x2, ...); Lx = (x2, x3, ...); x = (x1, x2, x3, ...) ∈ `2.

It is clear that ‖R‖ = ‖L‖ = 1. So, every λ ∈ C with |λ| > 1 is a regular point forboth of the operators R and L. Concerning the eigenvalues of these operators, weobtain the following:

Lx = λx (x 6= 0) ⇒ x2 = λx1; x3 = λx2; ...

⇒ x = (1, λ, λ2, λ3, ...)x1.

Such a vector belongs to `2 iff |λ| < 1. Hence,

σp(L) = λ ∈ C : |λ| < 1.

From the above result we also have dim ker(L− λI) = 1.For R we have

Rx = λx (x 6= 0) ⇒ 0 = λx1; x1 = λx2; x2 = λx3; ...

⇒ x1 = x2 = x3 = ... = 0...

⇒ x = 0 : a contradiction.

Hence, σp(R) = ∅.Next, since L∗ = R and R∗ = L, we obtain

(1) Image(R− λI)⊥ = ker(L− λI),

(2) Image(L− λI)⊥ = ker(R− λI).

For |λ| < 1 the relation (1) yields

codim Image(R− λI) = dim ker(L− λI) = 1.

Hence, λ ∈ C : |λ| < 1 ⊂ σr(R). Since the spectrum of an operator is closed, weconclude that

λ ∈ C : |λ| = 1 ⊂ σ(L) and λ ∈ C : |λ| = 1 ⊂ σ(R).

Moreover, for |λ| = 1, form (1) and (2) we have

Image(R− λI) = Image(L− λI) = `2.


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Hence, σc(R) = σc(L) = λ ∈ C : |λ| = 1.Conclusion:

σ(R) = σ(L) = λ ∈ C : |λ| ≤ 1σp(R) = σp(L) = λ ∈ C : |λ| < 1

σr(L) = ∅, σr(R) = λ ∈ C : |λ| < 1σc(L) = σc(R) = λ ∈ C : |λ| = 1. ¥


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Chapter 9

Compact Operators and TheirSpectra

As bounded linear operators, compact operators share spectral properties of bounded linear oper-ators. Besides, compact operators have some more particular spectral properties.Let T ∈ B(X) be a compact operator on a Banach space X. Suppose dimX = ∞.

1. 0 ∈ σ(T ). Every spectral value λ 6= 0 is an eigenvalue.

2. For λ 6= 0, dim ker(Tλ) ≡ dimker(T − λI) < ∞.

3. For λ 6= 0, the range of Tλ ≡ T − λI is closed.

4. The set of eigenvalues of T , namely σp(T ), is at most countable. The value λ = 0 is the onlypossible point of accumulation of that set.

Problem 122Let T ∈ B(X) be a compact operator on a Banach space X. Suppose dim X = ∞.Show that

(a) dim ker(T nλ ) < ∞ ∀n ∈ N,

(b) 0 = ker(T 0λ ) ⊂ ker(T 1

λ ) ⊂ ker(T 2λ ) ⊂ ...

Solution.Since Tλ is linear, Tλ0 = 0. By induction we get

T nλ x = 0 ⇒ T n+1

λ x = 0, ∀n ∈ N,

131


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132 CHAPTER 9. COMPACT OPERATORS AND THEIR SPECTRA

and so (b) follows.We now prove (a). By the binomial formula,

T nλ = (T − λI)n =

n∑

k=0

(n

k

)T k(−λ)n−k

= (−λ)nI + T

n∑

k=1

(n

k

)T k−1(−λ)n−k

︸︷︷︸S

.

This can be written

T nλ = W − µI, where µ = −(−λ)n.

Note that T is compact and S is bounded, so W = TS = ST is compact. Theproperty 2 above gives that dim ker(T n

λ ) < ∞. ¥

Problem 123Let T ∈ B(X) be a compact operator on a Banach space X. Suppose dim X = ∞.Show that

0 ∈ σ(T ).

Solution.If 0 /∈ σ(T ) then T is invertible, and we have TT−1 = I. But T and T−1 are compact,so I is compact. This requires that the dimension of X is finite (problems 16, 98):a contradiction. Thus 0 ∈ σ(T ). ¥

Problem 124Let T : X → X be a compact operator on a normed space X and let λ 6= 0. Thenthere exists a smallest integer r (depending on λ) such that from n = r on, thekernels ker(T n

λ ) are equal, and if r > 0, the inclusions

ker(T 0λ ) ⊂ ker(T 1

λ ) ⊂ ... ⊂ ker(T rλ)

are all proper (strict).


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133

Solution.For simplicity, we let Nn := ker(T n

λ ).• We know that Nm ⊂ Nm+1 (Problem 122). Suppose that Nm = Nm+1 for no m.Then Nn is a proper subspace of Nn+1 for every n. Since these kernels are closed,by Riesz lemma, there is a sequence (yn) in Nn such that

‖yn‖ = 1 and ‖yn − x‖ ≥ 1

2∀x ∈ Nn−1.

We show that

(∗) ‖Tyn − Tym‖ ≥ 1

2|λ| for m < n,

so that the sequence (Tyn) has no convergence subsequences. This contradicts thecompactness of T .From Tλ = T − λI we have T = Tλ + λI and

Tyn − Tym = λyn − x where x = Tλym + λym − Tλyn.

Let m < n. We show that x ∈ Nn−1. Since m ≤ n− 1, we clearly have λym ∈ Nm ⊂Nn−1. Also ym ∈ Nm implies

0 = Tmλ ym = Tm−1

λ (Tλym),

that is, Tλym ∈ Nm−1 ⊂ Nn−1. Similarly, yn ∈ Nn implies Tλyn ∈ Nn−1. Together,x ∈ Nn−1. Also x = 1

λx ∈ Nn−1. Hence

‖λyn − x‖ = |λ| ‖yn − x‖ ≥ 1

2|λ|.

Thus we have (∗). Therefore, we must have Nm = Nm+1 for some m.• We now prove that

(∗∗) Nm = Nm+1 =⇒ Nn = Nn+1 for all n > m.

Assume that this does not hold. Then Nn is a proper subspace of Nn+1 for somen > m. We consider an x ∈ Nn+1 \Nn. By definition,

T n+1λ x = 0 but T n

λ x 6= 0.

Set z = T n−mλ x. Then

Tm+1λ z = T n+1

λ x = 0 but Tmλ z = T n

λ x 6= 0.

Hencez ∈ Nm+1 but z /∈ Nm.

So Nm is a proper subspace of Nm+1. This contradicts (∗∗). The first statement isproved, where r is the smallest n such that Nn = Nn+1. Consequently, if r > 0, theinclusions in the theorem are strict. ¥


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Problem 125Let T : X → X be a compact operator on a Banach space X and let λ 6= 0.Then there exists a smallest integer q (depending on λ) such that from n = qon, the ranges T n

λ (X) are equal, and if q > 0, the inclusions

T 0λ (X) ⊃ T 1

λ (X) ⊃ ... ⊃ T qλ(X)

are all proper (strict).

Solution.For simplicity, we let Rn := T n

λ (X). Suppose that Rs = Rs+1 for no s. Then Rn+1 isa proper subspace of Rn for every n. Since these ranges are closed, by Riesz lemma,there is a sequence (xn) in Rn such that

‖xn‖ = 1 and ‖xn − x‖ ≥ 1

2∀x ∈ Rn+1.

Let m < n. Since T = Tλ + λI, we can write

(i) Txm − Txn = λxm − (−Tλxm + Tλxn + λxn).

On the right hand side, λxm ∈ Rm, xm ∈ Rm, so that Tλxm ∈ Rm+1. Since n > m,also Tλxn + λxn ∈ Rn ⊂ Rn+1. Hence (i) is of the form

Txm − Txn = λ(xm − x) with x ∈ Rm+1.

Consequently,

‖Txm − Txn‖ = |λ| ‖xm − x‖ ≥ 1

2|λ|.

This contradicts the fact that (Txn) has a convergent subsequence since (xn) isbounded and T is compact. Thus, Rs = Rs+1 for some s. Let q be the smallest ssuch that Rs = Rs+1. Then, if q > 0, the inclusions in the theorem are proper. Fur-thermore, Rq+1 = Rq means that Tλ maps Rq onto itself. Hence repeated applicationof Tλ gives Rn+1 = Rn for every n > q. ¥

Problem 126Let A be an invertible operator, and let K be a compact operator in a Banachspace. Prove that(a) dim(ker(A + K)) < ∞.(b) codim(Image(A + K)) < ∞.


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135

Solution.(a) Since A is invertible, we can write

A + K = A(I + A−1K).

The operator A−1K is compact (see note above), so dim(ker(I +A−1K)) < ∞. Thisimplies that

A invertible ⇒ ker(AB) = ker(B), ∀B ∈ B(X).

Indeed,

x ∈ ker(B) ⇒ ABx = A0 = 0

⇒ x ∈ ker(AB).

And

x ∈ ker(AB) ⇒ ABx = 0

⇒ Bx = 0

⇒ x ∈ ker(B).

It follows that

dim(ker(A + K)) = dim(A(I + A−1K))

= dim(I + A−1K) < ∞.

(b) One can writeA + K = (I + KA−1)A.

The operator KA−1 is compact, so codim(I + KA−1) < ∞. This implies that if Ais invertible, and then Image(BA) = Image(B). Indeed,

x ∈ Image(B) ⇒ x = By

⇒ x = (BA)A−1y ∈ Image(BA).

And

x ∈ Image(BA) ⇒ x = BAy

⇒ x = B(Ay) ∈ Image(B).

Thus, we obtain

codim(Image(A + K)) = codim(Image((I + KA−1)A))

= codim(Image(I + KA−1)) < ∞. ¥


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Problem 127Consider the operator K : L2([0, 1]) → L2([0, 1]) defined by

(Kf)(t) =

∫ 1

0

k(t, s)f(s)ds

where k(s, t) = mint, s : t, s ∈ [0, 1].(a) Prove that K is a compact self-adjoint operator.(b) Find the spectrum σ(K) and the norm ‖K‖.

Solution.(a) Since k(t, s) = k(s, t) and k(t, s) is a continuous function, the operator is self-adjoint and compact (see Problem 105).(b) Since K is compact and self-adjoint, the spectrum of K consists of zero and realeigenvalues1. Assume that λy = Ky. This means that

(1) λy(t) =

∫ t

0

mint, sy(s)ds +

∫ 1

t

mint, sy(s)ds

=

∫ t

0

sy(s)ds +

∫ 1

t

ty(s)ds

=

∫ t

0

sy(s)ds + t

∫ 1

t

y(s)ds.

Taking the derivative twice, we obtain

(2) λy′(t) = ty(t) +

∫ 1

t

y(s)ds− ty(t) =

∫ 1

t

y(s)ds

λy′′(t) = −y(t).

Clearly, λ 6= 0; otherwise, y = 0 so ker(K) = 0. We have the differential equation

(3) λy′′ + y = 0 with b.v.c. y′(0) = y(0) = 0

because of (1). Let us prove that λ > 0, which means that the operator K is positive.Multiplying (3) by y and integrating we obtain

λ

∫ 1

0

y′′(t)y(t)dt + ‖y‖2 = 0.

1If T ∈ B(H) is self adjoint, all its eigenvalues are real. (We will see this in the next chapter.)


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137

Integrating by parts we obtain

λ

(y′y|10 −

∫ 1

0

|y′|2dt

)+ ‖y‖2 = 0.

The b.v.c. yield

−λ

∫ 1

0

|y′|2dt + ‖y‖2 = 0.

Hence λ > 0.The solution of the differential equation is

y = C1 cost√λ

+ C2 sint√λ

.

From the b.v.c. it follows that C1 = 0 and C2√λ

cos 1√λ

= 0. Therefore, the eigenvaluesof K are

λk =4

π2(2k − 1)2, k = 1, 2, ...

Since K is seft-adjoint, we obtain

‖K‖ = maxk∈N

|λk| = |λ1| = 4

π2. ¥

Problem 128(Similar problem)Consider the operator K : L2([0, 1]) → L2([0, 1]) defined by

(Kf)(t) =∫ 1

0

k(t, s)f(s)ds

where k(s, t) = maxt, s : t, s ∈ [0, 1].(a) Prove that K is a compact self-adjoint operator.(b) Find the spectrum σ(K) and the norm ‖K‖.(c) Is K a positive operator?

Problem 129Let S be the operator defined on C[0, 1] by

(Sf)(x) =

∫ x

0

f(y)dy.

(a) Compute the spectrum of S.(b) Show that S is compact.


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Solution.(a) First we show that S is continuous. For f, g ∈ C[0, 1] we have

‖Sf − Sg‖ = ‖S(f − g)‖ =

∣∣∣∣∫ x

0

(f(y)− g(y))dy

∣∣∣∣

≤∫ x

0

|f(y)− g(y)|dy

≤ supx∈[0,1]

∫ x

0

|f(y)− g(y)|dy

≤ |[0, 1]|.‖f − g‖ = ‖f − g‖.

Hence, S is Lipschitz continuous with constant 1. Thus, ‖S‖ ≤ 1. Next, we showthat ‖Sn‖1/n → 0 as n →∞. Observe that

|(Sf)(x)| ≤∫ x

0

|f(t)|dt ≤ ‖f‖x = ‖f‖ x1

1!, x ∈ [0, 1].

By induction

|(Sn+1f)(x)| ≤∫ x

0

|Snf(t)|dt

≤ ‖f‖∫ x

0

tn

n!dt

= ‖f‖ xn+1

(n + 1)!, x ∈ [0, 1], n = 1, 2, ...

Thus,

‖Snf‖ ≤ ‖f‖ 1

n!, n = 1, 2, ...

So

‖Sn‖1/n ≤(

1

n!

)1/n

→ 0 as n →∞.

Recall that the spectral radius of S is given by

r(S) = limn→∞

‖Sn‖1/n.

This implies that r(S) = 0. Thus σ(S) = 0.(b) Suppose F ⊂ C[0, 1] is a bounded subset. Put

‖F‖ := supf∈F

‖f‖.


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139

Then SF is equi-continuous by the Fundamental Theorem of the Calculus:

∀f ∈ F, |(Sf)′(x)| ≤ ‖f‖ ≤ ‖F‖.

SF is bounded since:

∀f ∈ F, |(Sf)(x)| ≤ ‖f‖ ≤ ‖F‖.

By Ascoli-Arzela theorem SF is relatively compact. Thus S is compact. ¥

Problem 130Set

(Tf)(x) =

∫ 1−x

0

f(y)dy, f ∈ C[0, 1], x ∈ [0, 1].

(a) Prove that T is a linear bounded and compact operator on C[0, 1].(b) Calculate σ(T ) and the eigenvalues of T .

Solution.(a)

• Linearity of T : trivial.

• Boundedness of T :

|(Tf)(x)| =∣∣∣∣∫ 1−x

0

f(y)dy

∣∣∣∣ ≤ ‖f‖,

where ‖f‖ = maxx∈[0,1] |f(x)|. Hence,

‖T‖ ≤ 1.

• Compactness of T : let (fn)∞n=1 be a bounded sequence in C[0, 1]. Hence,

‖fn‖ ≤ M ∀n ∈ N

for some M ≥ 0. By Arzela-Ascoli theorem, it suffices to show that A :=Tfn; n ∈ N is bounded and equicontinuous. We have

‖Tfn‖ ≤ ‖T‖‖fn‖ ≤ ‖fn‖ ≤ M ∀n ∈ N.


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Given any ε > 0, without lost of generality, we can assume x < y, then

|(Tfn)(x)− (Tfn)(y)| =∣∣∣∣∫ 1−x

1−y

f(y)dy

∣∣∣∣ ≤ M |x− y|.

Thus,

|(Tfn)(x)− (Tfn)(y)| < ε ∀n ∈ N provided |x− y| < ε

M.

(b) First we see that λ = 0 is an eigenvalue. Assume that λ 6= 0 is an eigenvalue,i.e.,

λg(x) = (Tg)(x) =

∫ 1−x

0

g(y)dy, x ∈ [0, 1]

for some 0 6= g ∈ C[0, 1]. This implies that

Tg ∈ C1[0, 1] and λg′(x) = −g(1− x), x ∈ [0, 1].

Moreover, we have g(1) = 0. But g ∈ C1[0, 1] implies that g ∈ C2[0, 1]. Bydifferentiating once more we get

λg′′(x) =g(x)

λ, x ∈ [0, 1] and g(1) = g′(0) = 0.

Henceg(x) = A cos(x/λ) with g(1) = 0.

This gives that

λk =1

π2

+ kπ, k ∈ Z.

Check if all these λ’s are eigenvalues. We calculate

(Tgk)(x) =

∫ 1−x

0

cos(t/λk)dt

= λk[sin(t/λk)]1−x0

= λk sin[(π

2+ kπ

)(1− x)

]

= λk(−1)kgk(x).

Hence, λ = λ2l, l ∈ Z are the eigenvalues of T , i.e.,

σp(T ) = λ2l, l ∈ Z.We know that σ(T ) is closed and σ(T ) \ 0 ⊂ σp(T ). This yields

σ(T ) = 0 ∪ σp(T ). ¥


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141

Problem 131Let T be a compact operator on a Hilbert space H and (λn) be a sequence ofcomplex numbers. Suppose there exists a nested sequence of distinct subspaces(Mn) such that for all n ∈ N

(i) Mn (Mn+1

(ii) (T − λnI)Mn+1 ⊂ Mn.

Prove that limn→∞ λn = 0.

Solution.Since Mn is a subspace of Mn+1, we can write Mn+1 = Mn⊕(M⊥

n )M , where (M⊥n )M is

the orthogonal complement of Mn in Mn+1. For short we write (M⊥n )M = Mn+1ªMn.

Let en be a sequence of unit vectors defined by

e1 ∈ M1, en+1 ∈ Mn+1 ªMn, ∀n ∈ N.

Clearly, that is an orthonormal system. Moreover,

⟨(T − λnI)en, en

⟩= 0, for all n ≥ 2,

which implies that

‖Ten‖ ≥ |〈Ten, en〉| =∣∣⟨(T − λnI)en, en

⟩∣∣ + |〈λnen, en〉|= 0 + |〈λnen, en〉| = |λn|.

Since T is compact and enw→ 0, it follows that limn→∞ Ten = 0. Thus

limn→∞

λn = 0. ¥

Problem 132Let T be a compact operator on a Hilbert space H and any C > 0. Provethat there is a finite number of linearly independent eigenvectors x1, ..., xn of Tcorresponding to eigenvalues λ1, ..., λn such that λi > C for all i = 1, ..., n.

Solution.We can rescale to get ‖xi‖ = 1 for all i = 1, ..., n. Suppose to the contrary that


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there is an infinite sequence xn of unit vectors, and a sequence of eigenvalues λnsatisfying

λn > C and Txn = λnxn, ∀n ∈ N.

Let Mn = Spanx1, ..., xn, then Mn is a nested sequence of subspaces of H andthe inclusions Mn ⊂ Mn+1 are strict. Let x ∈ Mn, then there are c1, ..., cn ∈ C suchthat x =

∑ni=1 cixi. So we have

(T − λnI)x = (T − λnI)n∑

i=1

cixi

=n∑

i=1

ci(T − λnI)xi

=n∑

i=1

ci(Txi − λnxi)

=n∑

i=1

ci(λi − λn)xi ∈ Mn−1.

This implies that(T − λnI)Mn ⊂ Mn−1, n ≥ 2.

From Problem 130, we obtainlim

n→∞λn = 0.

This contradicts the assumption λn > C for all n ∈ N. ¥

Note: Argument in problems 130, 131 is the proof of Proposition 4 in the review at the beginningof this chapter.


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Chapter 10

Bounded Self Adjoint Operatorsand Their Spectra

Review some main points.

Bounded self adjoint operators on Hilbert spaces were defined and considered before. This chapteris devoted to their spectral properties.

Definition:

Let T ∈ B(H) where H is a complex Hilbert space. The adjoint operator of T is the operatorT ∗ : H → H defined by

〈Tx, y〉 = 〈x, T ∗y〉, ∀x, y ∈ H.

T is said to be self adjoint if T = T ∗. We can say that T is self adjoint if and only if

〈Tx, y〉 = 〈x, Ty〉, ∀x, y ∈ H.

Another equivalent condition is:T is self adjoint if and only if 〈Tx, x〉 is real for all x ∈ H.

Let T ∈ B(H) be a bounded self adjoint operator on the complex Hilbert space H.

Proposition 12 (Eigenvalues and eigenvectors)1. All eigenvalues of T (if they exist) are real.2. Eigenvectors corresponding to different eigenvalues of T are orthogonal.

Proposition 13 (Spectrum)1. σ(T ) ⊂ R.2. σ(T ) ⊂ [m,M ] where m = inf‖x‖=1〈Tx, x〉 and M = sup‖x‖=1〈Tx, x〉.3. m,M ∈ σ(T ).

Proposition 14 (Norm)‖T‖ = max|m|, |M | = sup‖x‖=1 |〈Tx, x〉|.

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144CHAPTER 10. BOUNDED SELF ADJOINT OPERATORS AND THEIR SPECTRA

Problem 133Let T : H → H be a bounded self-adjoint linear operator on a complex Hilbertspace H. Prove that the residual spectrum of T is empty, that is,

σr(T ) = ∅.

Solution.Assume that σr(T ) 6= ∅. Take λ ∈ σr(T ). By the definition of σr(T ), T−1

λ exists butits domain is not dense in H. Hence, by the projection theorem, there is a y 6= 0 inH such that y is perpendicular to the domain D(T−1

λ ) of T−1λ . But D(T−1

λ ) is therange of Tλ, hence

〈Tλx, y〉 = 0, ∀x ∈ H.

Since λ is real and T is self-adjoint, we obtain

〈x, Tλy〉 = 0, ∀x ∈ H.

Taking x = Tλy, we get ‖Tλy‖2 = 0, so that

Tλy = Ty − λy = 0.

Since y 6= 0, this shows that λ is an eigenvalue of T . But this contradicts λ ∈σr(T ). ¥

Second solution:By Problem 91, noting that T is self adjoint, for any λ ∈ C we have

Image(T − λI)⊥ = ker(T ∗ − λI) = ker(T − λI).

And hence, if Image(T − λI) 6= H, then λ is an eigenvalue of T . Since T is selfadjoint, λ is real. Thus λ = λ is an eigenvalue of T . Therefore λ does not belong tothe residual spectrum of T. ¥

Problem 134Let T : H → H be a bounded self-adjoint linear operator on a complex Hilbertspace H. Prove that

(∗) λ ∈ ρ(T ) ⇐⇒ ∃c > 0 : ∀x ∈ H, ‖Tλx‖ ≥ c‖x‖.


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Solution.• If λ ∈ ρ(T ) then Rλ := T−1

λ : H → H exists and is bounded, say ‖Rλ‖ = k > 0.Now since I = RλTλ, we have for every x ∈ H

‖x‖ = ‖RλTλx‖ ≤ ‖Rλ‖ ‖Tλx‖ = k‖Tλx‖.

This gives

‖Tλx‖ ≥ c‖x‖, where c =1

k.

• Conversely, suppose (∗) holds. We shall show:

(a) Tλ : H → Tλ(H) is bijective;

(b) Tλ(H) is dense in H;

(c) Tλ(H) is closed in H;

so that Tλ(H) = H and Rλ := T−1λ is bounded by the bounded inverse theorem.1

(a) By (7.1), we have for x1, x2 ∈ H

‖Tλx1 − Tλx2‖ = ‖Tλ(x1 − x2)‖ ≥ c‖x1 − x2‖.

Therefore,Tλx1 = Tλx2 =⇒ x1 = x2.

Thus Tλ : H → Tλ(H) is bijective.(b) We show that x0⊥Tλ(H) implies x0 = 0, so that Tλ(H) = H by the projectiontheorem.2 Let x0⊥Tλ(H) Then for all x ∈ H we have

0 = 〈Tλx, x0〉 = 〈Tx, x0〉 − λ〈x, x0〉.

Since T is self-adjoint,〈Tx, x0〉 = 〈x, Tx0〉.

It follows that〈x, Tx0〉 = λ〈x, x0〉 = 〈x, λx0〉.

Thus Tx0 = λx0. So x0 = 0 since otherwise, λ = λ would be an eigenvalue of T ,and Tλx0 = 0, which would imply

0 = ‖Tλx0‖ ≥ c‖x0‖ > 0 : a contradiction.

1A bounded linear operator T from a Banach space X onto a Banach space Y is an openmapping. Hence if T is bijective, then T is continuous and thus bounded.

2If Y is a closed subspace of H, then H = Y ⊕ Y ⊥.


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(c) To show Tλ(H) is closed we show

y ∈ Tλ(H) =⇒ y ∈ Tλ(H).

Let y ∈ Tλ(H). There is a sequence (yn) in Tλ(H) which converges to y. For everyn we have yn = Tλxn for some xn ∈ H. By (7.1),

‖xn − xm‖ ≤ 1

c‖Tλ(xn − xm)‖ =

1

c‖yn − ym‖.

Hence (xn) is Cauchy. Since H is complete, (xn) converges, say, xn → x. SinceTλ is continuous, yn = Tλxn → Tλx. Since the limit is unique, Tλx = y. Hencey ∈ Tλ(H). Thus Tλ(H) is closed in H. ¥

Problem 135(a) Let A ∈ B(X) where X is a Banach space. Suppose there exists m > 0 suchthat

‖Ax‖ ≥ m‖x‖, ∀x ∈ X.

Show that Image A is closed in X.(b) Let A ∈ B(H) be self adjoint, where H is a Hilbert space. Let λ ∈ C suchthat Imλ 6= 0. Prove that

‖Ax− λx‖ ≥ |Imλ| ‖x‖, ∀x ∈ H.

Prove that λ is a regular point of A.

Solution.(a) Let (xn) be a sequence in X. Suppose Axn → y as n →∞. From the hypothesiswe get

‖Axn − Axm‖ ≥ m‖xn − xm‖ for n 6= m.

Since (Axn) is a Cauchy sequence in X, it follows that (xn) is also a Cauchy sequencein X. Hence x → x as n →∞. But A is continuous, so Axn → Ax as n →∞. Byuniqueness of the limit, we obtain y = Ax. This shows that Image A is closed.

(b) Let λ = a + ib with a, b ∈ R and b = Imλ 6= 0. We have for all x ∈ H

‖Ax− λx‖2 = 〈Ax− (a + ib)x,Ax− (a + ib)x〉= 〈(A− aI)x− ibx, (A− aI)x− ibx〉= ‖(A− aI)x‖2 + b2‖x‖2,


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147

which implies that‖Ax− λx‖ ≥ |b|‖x‖, ∀x ∈ H.

By part (a), Image(A − λI) is closed and therefore λ /∈ σc(A). By Problem 91σr(A) = ∅, thus λ is a regular point of A. ¥

Problem 136Let A ∈ B(H) be self adjoint, where H is a Hilbert space.(a) Prove that

‖A‖ = supx6=0

|〈Ax, x〉|‖x‖2

.

(b) Prove that at least one of ‖A‖ or −‖A‖ is an element of σ(A).

Solution.(a) Using the Cauchy-Schwarz inequality, we have

|〈Ax, x〉| ≤ ‖Ax‖ ‖x‖ ≤ ‖A‖ ‖x‖2, ∀x ∈ H.

Hence

supx 6=0

|〈Ax, x〉|‖x‖2

≤ ‖A‖. (i)

Now we establish the reverse. Notice first that for all x, y ∈ H we have

⟨A(x + y), x + y

⟩− ⟨A(x− y), x− y

⟩= 2

[〈Ax, y〉+ 〈Ay, x〉].

Using the triangle inequality, we get

2∣∣〈Ax, y〉+ 〈Ay, x〉

∣∣ ≤∣∣〈A(x + y), x + y〉

∣∣ +∣∣〈A(x− y), x− y〉

∣∣.

If we let C = supx 6=0|〈Ax,x〉|‖x‖2 , then the Parallelogram Law gives that

∣∣〈Ax, y〉+ 〈Ay, x〉∣∣ ≤ 1

2C

(‖x + y‖2 + ‖x− y‖2)

= C(‖x‖2 + ‖y‖2

). (∗)

Now let x be any vector with ‖x‖ = 1 and let y = Ax‖Ax‖ ( the case Ax = 0 does not

give the supremum, hence we may assume that Ax 6= 0). Then ‖y‖ = 1. From (∗)we get ∣∣∣∣

〈Ax,Ax〉‖Ax‖ +

〈Ax,Ax〉‖Ax‖

∣∣∣∣ ≤ 2C.


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Hence ‖Ax‖ ≤ C. This holds for all x ∈ H with ‖x‖ = 1. Thus ‖A‖ ≤ C. (ii)Combine (i) and (ii) we obtain

‖A‖ = supx6=0

|〈Ax, x〉|‖x‖2

.

(b) If we take x arbitrary with ‖x‖ = 1, then we get

‖A‖ = sup‖x‖=1

|〈Ax, x〉|. (∗∗)

Let (xn) be a sequence in H with ‖xn‖ = 1 such that |〈Axn, xn〉| converges to‖A‖ (this is possible by (∗∗)). Let 〈Axn, xn〉 → λ (it may be necessary to pass tosubsequence). Clearly, λ = ±‖A‖. Now,

0 ≤ ‖Axn − λxn‖2 = ‖Axn‖2 − 2λ〈Axn, xn〉+ λ2‖xn‖2

≤ 2λ2 − 2λ〈Axn, xn〉,which converges to 0 as n →∞. Thus λ ∈ σ(A) (see Problem 109). ¥

∗ ∗ ∗∗To close this chapter, we introduce a well known theorem relative to compact self adjoint operatorson Hilbert spaces:

The spectral theorem for compact self adjoint operators on Hilbert spaces.

Problem 137Let T ∈ B(H) be a compact self adjoint operator on a Hilbert space H.(a) There exists a system (finite or infinite) of orthonormal eigenvectorse1, e2, ... of T and corresponding eigenvalues λ1, λ2, ... of T such that |λ1| ≥|λ2| ≥ .... If the system is infinite then λn → 0 as n →∞.(b) Eigenvectors and eigenvalues mentioned above satisfy the following equation:

Tx =∞∑

k=1

λk〈x, ek〉ek, ∀x ∈ H.

Solution.(a) We use Proposition 13-3 (that we proved in Problem 136b) repeatedly for con-structing eigenvalues and eigenvectors.


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149

Let H1 = H and T1 = T . Then by Proposition 13-3, there exists an eigenvalue λ1

of T1 and a corresponding eigenvector e1 such that

‖e1‖ = 1 and |λ1| = ‖T1‖.

Now Spane1 is a closed subspace of H1 hence, by the projection theorem,

H1 = Spane1 ⊕ Spane1⊥.

Let H2 = Spane1⊥. Clearly H2 is a closed subspace of H1 and T (H2) ⊂ H2.Indeed, if x ∈ H2 then x⊥e1, hence Tx = λx ⇒ Tx⊥e1. Let T2 be the restrictionof T1 on H2, that is, T2 = T1|H2 = T |H2 . Then T2 is a compact and self adjointoperator in B(H2). If T2 = 0, then there is nothing to prove. Assume that T2 6= 0.Then by Proposition 13-3, there exists an eigenvalue λ2 of T2 and a correspondingeigenvector e2 such that

‖e2‖ = 1 and |λ2| = ‖T2‖.

Since T2 is a restriction of T1,

|λ2| = ‖T2‖ ≤ ‖T1‖ = λ1.

By construction e1 and e2 are orthonormal.Now let H3 = Spane2, e2⊥. Clearly H3 ⊂ H2 and T (H3) ⊂ H3. The operatorT3 = T |H3 is compact and self adjoint....If we continue to proceed in this way, eitherafter some stage, say n, we get Tn = 0 or there exists an infinite sequence (λn) ofeigenvalues of T and corresponding eigenvectors (en) satisfying

‖en‖ = 1, ∀n ∈ N and |λ1| ≥ |λ2| ≥ ...

If the sequence (λn) is infinite, we show that λn → 0 as n → ∞. Suppose thatλn 9 0 as n → ∞. Then there exists ε > 0 such that |λn| > ε for infinitely manyn. For n 6= m, we have

‖Ten − Tem‖2 = ‖λnen − λmem‖2 = λ2n + λ2

m > ε2.

This shows that the sequence (Ten) has no convergent subsequence, a contradictionto the compactness of T . Hence λn → 0 as n →∞.

(b) There are two cases to consider:Case 1. Tn = 0 for some n.Let xn = x − ∑n

k=1〈x, ek〉ek for all x ∈ H. Then xn⊥ek for 1 ≤ k ≤ n, sincee1, e2, ... is an orthonormal system and

〈xn, ek〉 = 〈x, ek〉 − 〈x, ek〉 = 0.


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Hence

0 = Tnxn = Tx−n∑

k=1

〈x, ek〉Tkek =n∑

k=1

λk〈x, ek〉ek.

That is,

Tx =n∑

k=1

λk〈x, ek〉ek =∞∑

k=1

λk〈x, ek〉ek, ∀x ∈ H.

Case 2. Tn 6= 0 for infinitely many n.For x ∈ H, by Case 1, we have

∥∥∥∥∥Tx−n∑

k=1

λk〈x, ek〉ek

∥∥∥∥∥ = ‖Tnxx‖ ≤ ‖Tn‖ ‖xn‖

= |λn| ‖xn‖ → 0 as n →∞.

Hence

Tx =∞∑

k=1

λk〈x, ek〉ek, ∀x ∈ H. ¥

THANK YOU and GOOD LUCK!

Anh [email protected]


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Bibliography

[1] Conway, J.B. A course in Functional Analysis. Second edition. Springer. NewYork, 1990.

[2] Kreyszig. E. Introductory Functional Analysis with Applications. John Wileyand Son. New York, 1989.

[3] Reed, M.; Simon, B. Methods of modern Mathematics Physics. I. FunctionalAnalysis. Second edition. Academic Press, Inc. New York, 1980.

[4] Royden. H.L. Real Analysis. Third edition. Prentice Hall. NJ, 1988.

[5] Rudin. W Functional Analysis. McGraw-Hill, Inc. New York, 1991.

[6] Yeh, J. Real Analysis. Theory of measure and integration. Second edition. WorldScientific. NJ, 2006.

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