EXAMENSARBETEN I MATEMATIK MATEMATISKA INSTITUTIONEN, STOCKHOLMS UNIVERSITET

EXAMENSARBETEN I MATEMATIK

MATEMATISKA INSTITUTIONEN, STOCKHOLMS UNIVERSITET

Linear operators in infinite dimensional vector spaces

av

Kharema Ebshesh

2008 - No 13

Linear operators in infinite dimensional vector spaces

Kharema Ebshesh

Examensarbete i matematik 15 h¨ogskolepo¨ang, f¨ordjupningskurs Handledare: Andrzej Szulkin och Yishao Zhou

Abstract

The theory of linear operators is an extensive area. This thesis is about the linear operators in infinite dimensional vector spaces.

We study elementary properties of Banach spaces, bounded operators, compact operators and spectrum of compact operators. We give an ap- plication to a two-point boundary value problem for a linear ordinary differential equation in the end.

Acknowledgements

I have the pleasure to thank Prof. Andrzej Szulkin and Prof. Yishao Zhou for their help. I enjoyed my work with them, and I learned a lot during the preparation of this thesis.

I would like to thank all the teachers that I had in the Maths Department.

I would like to thank Prof. Mikael Passare for inviting me to Sweden.

Finally I will not forget my teachers in Libya who supported me during my studies.

/Kharema

Contents

Introduction 4

1 Infinite dimensional vector spaces 5

1.1 Normed spaces. Banach spaces . . . 5

1.2 Inner product spaces. Hilbert spaces . . . 16

1.3 Strong and weak convergence . . . 19

2 Linear operators 21 2.1 The domain and range . . . 21

2.2 Bounded and continuous operators . . . 23

2.3 Compact operators . . . 27

2.4 Spectrum of compact operators . . . 30

References 40

Introduction

This report contains two sections. In Section 1, we introduce basic concepts in infinite dimensional vector spaces, such as normed spaces, Banach spaces, inner product spaces and Hilbert spaces. We also study the strong and weak convergence. In Section 2, we study bounded linear operators in infinite dimensional vector spaces. In particular, we study compact operators and their spectrum. Finally, as an application we consider a two-point boundary value problem for a linear ordinary differential equation.

The results in this report are primarily taken from [1].

1 Infinite dimensional vector spaces

In this thesis we shall always assume that X is a vector space such that dim X = ∞ unless otherwise stated.

1.1 Normed spaces. Banach spaces

Definition 1. A normed space is a vector space X in which a function k k : X −→ R is defined and satisfies the following conditions:

-kxk ≥ 0 and = 0 iff x = 0 -kαxk = |α|kxk

-kx + yk ≤ kxk + kyk

for all x, y in X and all scalars α. A normed space X in which every Cauchy sequence has a limit in X is said to be complete. A complete normed space is called a Banach space.

Example 1. (Space l^p). Let 1 ≤ p < ∞ be a fixed real number. By definition, each element in the space l^p is a sequence x = (ξ_j) = (ξ₁, ξ₂, · · · ) of numbers such that |ξ1|^p+ |ξ2|^p+ · · · converges; thus

∞

X

j=1

|ξ_j|^p < ∞,

and the norm is defined by

kxk = (

∞

X

j=1

|ξ_j|^p)^1/p. (1)

Since -kxk = (

∞

X

j=1

|ξ_j|^p)^1/p≥ 0, and = 0 iff x = 0,

-kλxk = (

∞

X

j=1

|λξ_j|^p)^1/p= (

∞

X

j=1

|λ|^p|ξ_j|^p)^1/p= |λ|(

∞

X

j=1

|ξ_j|^p)^1/p= |λ|kxk,

- kx + yk = (

∞

X

j=1

|ξj+ ηj|^p)^1/p ≤ (by the Minkowski inequality, see (12) on

p. 14 in [1]) ≤ (

∞

X

j=1

|ξ_k|^p)^1/p+ (

∞

X

j=1

|η_k|^p)^1/p= kxk + kyk, this is indeed a norm.

(Completeness of l^p). Let (x_n) be any Cauchy sequence in the space l^p, where x_m = (ξ₁^(m), ξ₂^(m), · · · ). Then for every ǫ > 0 there is an N such that for all m, n > N,

kxm− xnk = (

∞

X

j=1

|ξ^(m)_j − ξ_j⁽ⁿ⁾|^p)^1/p< ǫ. (2)

It follows that

|ξ^(m)_j − ξ_j⁽ⁿ⁾| < ǫ

for every j = 1, 2. · · · . For fixed j, (ξ_j⁽ⁿ⁾) is a Cauchy sequence of numbers.

Since the real and complex numbers are complete (see [1]), ξ_j⁽ⁿ⁾ −→ ξ_j as n −→ ∞. We define x = (ξ₁, ξ₂, · · · ) and show that x ∈ l^p and x_m −→ x.

From (2) we have for all m, n > N and a fixed k

k

X

j=1

|ξ_j^(m)− ξ_j⁽ⁿ⁾|^p< ǫ^p.

Letting n −→ ∞, we obtain for m > N

k

X

j=1

|ξ^(m)_j − ξ_j|^p≤ ǫ^p.

Now let k −→ ∞; then for m > N

∞

X

j=1

|ξ^(m)_j − ξj|^p≤ ǫ^p. (3)

This shows that xm− x = (ξ^(m)_j − ξj) ∈ l^p, and x = x_m− (x_m− x) ∈ l^p.

The series in (3) represents kxm− xk^p, hence xm −→ x. Since (xn) was an arbitrary Cauchy sequence in l^p, this proves the completeness of l^p.

Example 2. (Space l^∞). Every element in this space is a complex sequence x = (ξ_j) = (ξ₁, ξ₂, · · · ) such that sup

j

|ξ_j| < ∞, and the norm is defined by kxk = sup

j

|ξ_j|. (4)

The first two conditions for the norm are verified similarly as in the last example. Let us show the triangle inequality.

-kx + yk = sup

j

|ξj+ ηj| ≤ sup

j

(|ξj| + |ηj|) ≤ sup

j

|ξj| + sup

j

|ηj| = kxk + kyk.

(Completeness of l^∞). Let (x_m) be any Cauchy sequence in the space l^∞, where xm = (ξ^(m)₁ , ξ^(m)₂ , · · · ). Then for every ǫ > 0 there is an N such that for all m, n > N

kx_m− x_nk = sup

j

|ξ^(m)_j − ξ_j⁽ⁿ⁾| < ǫ. (5) We shall show that there exists x ∈ l^∞ which is the limit of (xm). By (5) we have for every fixed m, n > N

|ξ_j^(m)− ξ_j⁽ⁿ⁾| < ǫ. (6) Hence for every j the sequence {ξ_j⁽¹⁾, ξ_j⁽²⁾, · · · } is a Cauchy sequence of num- bers. So it converges : ξ_j^(m) −→ ξ_j as m −→ ∞. Let x = (ξ₁, ξ₂, · · · ). We want to show that x ∈ l^∞ and x_m −→ x. From (6) with n −→ ∞ we have for m > N

|ξ^(m)_j − ξ_j| ≤ ǫ. (7)

Since x_m = (ξ_j^(m)) ∈ l^∞, there is a real number k_m such that |ξ_j^(m)| ≤ k_m for all j. Hence by the triangle inequality

|ξj| ≤ |ξj− ξ_j^(m)| + |ξ_j^(m)| ≤ ǫ + km.

This inequality holds for every j. Hence (ξ_j) is a bounded sequence of num- bers. This implies that x = (ξ_j) ∈ l^∞. From (6) we obtain

kxm− xk = sup

j

|ξ_j^(m)− ξj| ≤ ǫ for m > N. This shows that x_n−→ x.

Example 3. (Space C[a, b]). This space is the set of all continuous real- or complex-valued functions x = x(t) defined on a given closed interval J = [a, b], and the norm is defined by

kxk = max

t∈J |x(t)|. (8)

The proof that this is indeed a norm is similar as in Example 2.

(Completeness of C[a, b]). Let (x_m) be any Cauchy sequence in C[a, b]. Then for every ǫ > 0 there is an N such that for all m, n > N,

kxm− xnk = max

t∈J |xm(t) − xn(t)| < ǫ. (9) Hence for any fixed t = t₀ ∈ J,

|x_m(t₀) − x_n(t₀)| < ǫ, m, n > N.

This shows that (x₁(t₀), x₂(t₀), · · · ) is a Cauchy sequence of real or complex numbers, so the sequence converges, say x_m(t₀) −→ x(t₀) as m −→ ∞. In this way we can associate with each t ∈ J a unique number x(t). This defines a function x on J, and we must show that x ∈ C[a, b] and xm −→ x.

From (9) (with n → ∞) we have

max |x_m(t) − x(t)| ≤ ǫ, m > N.

Hence for every t ∈ J,

|x_m(t) − x(t)| ≤ ǫ, m > N. (10) This shows that (x_m(t)) converges to x(t) uniformly on J. Since the x_m’s are continuous on J and the convergence is uniform, the limit function x is con- tinuous on J. Hence x ∈ C[a, b] and x_m−→ x. This proves the completeness of C[a, b].

The following is an example of an incomplete normed space.

Example 4. Let X be the set of all continuous real-valued functions on J = [0, 1], and let

kxk = Z 1

0

|x(t)|dt, (11)

then kxk is a norm, since kx+yk =

Z 1 0

|x(t)+y(t)|dt ≤ Z 1

0

(|x(t)+|y(t)|)dt = Z 1

0

|x(t)|dt+

Z 1 0

|y(t)|dt = kxk + kyk, and the other two conditions are trivially satisfied.

This normed space X is not complete. Let (x_m) be defined by x_m(t) =

(0 if t ∈ [0,¹₂] 1 if t ∈ [a_m, 1] ,

where a_m = ¹₂ + _m¹ and for ¹₂ ≤ t ≤ a_m, (t, x_m(t)) is the linear segment joining (¹₂, 0) and (a_m, 1). For every given ǫ > 0,

kx_m− x_nk < ǫ when m, n > 1 ǫ. Hence (xm) is a Cauchy sequence. For every x ∈ X,

kx_m− xk = Z 1

0

|x_m(t) − x(t)|dt

= Z ¹

2

0

|x(t)|dt + Z a^m

1 2

|x_m(t) − x(t)|dt + Z 1

a^m

|1 − x(t)|dt.

Since each integrand is nonnegative, kxm − xk −→ 0 implies that each integral approaches zero. Since x is continuous, we would have

x(t) =

(0 if t ∈ [0,¹₂) 1 if t ∈ (¹₂, 1].

But this contradicts the continuity of x, hence x /∈ X. This proves that X is not complete.

Definition 2. Let M be a subset in a normed space X, we say M is : - boundedif there is a positive number c such that kxk ≤ c ∀x ∈ M.

- closedif for any sequence (x_n) in M, x_n−→ x implies that x ∈ M.

- compact if any sequence (xn) in M has a convergent subsequence whose limit belongs to M.

We know that in the finite dimensional space R^N, any subset M is com- pact if and only if it is closed and bounded (see [1]). But in infinite dimen- sional normed spaces this is no longer true.

Theorem 1. A compact subset M of a normed space is closed and bounded.

Proof. The proof is the same as for R^N, see [1].

Remark 1. The converse of this theorem is not true in infinite dimensional spaces.

We verify our claim in the space l². Consider the sequence (e_n) in l², where e_n = (δ_nj) has the nth term 1 and all other terms 0. This sequence is bounded since ke_nk = 1. Its terms constitute a point set which is closed because it has no point of accumulation. For the same reason, that point set is not compact. See also Theorem 2 below.

Later we shall need the following lemma:

Riesz’s Lemma. Let Y and Z be subspaces of a normed space X (of any dimension), and suppose that Y is closed and is a proper subset of Z. Then for every real number θ in the interval (0, 1) there is a z ∈ Z such that

kzk = 1, kz − yk ≥ θ for all y ∈ Y.

Proof. Let v ∈ Z − Y and

a = inf

y∈Ykv − yk.

Since Y is closed, a > 0. We now take any θ ∈ (0, 1). By the definition of an infimum there is a y0∈ Y such that

a ≤ kv − y0k ≤ a θ. Let

z = c(v − y₀) where c = 1 kv − y₀k.

Then kzk = 1, and we show that kz − yk ≥ θ for every y ∈ Y. We have kz − yk = kc(v − y₀) − yk

= ckv − y₀− c⁻¹yk

= ckv − y1k where

y₁ = y₀+ c⁻¹y.

Since y, y0 ∈ Y, y1 ∈ Y. Hence by the definition of a, kv − y1k ≥ a, and kz − yk = ckv − y₁k ≥ ca = a

kv − y0k ≥ a a/θ = θ.

Since y ∈ Y was arbitrary, this completes the proof.

Theorem 2. (Finite dimension). If a normed space X has the property that the closed unit ball M = {x|kxk ≤ 1} is compact, then X is finite dimensional.

Proof. We argue by contradiction. We assume that M is compact but dim X = ∞, and we shall show that this is impossible. Let x₁ be any vector of norm 1, and X₁ be the one dimensional subspace of X generated

by x₁, which is closed because the dimension is finite. Since dim X = ∞, X₁ is a proper subspace of X.

By Riesz’s Lemma there is an x2 ∈ X of norm 1 such that kx₂− x₁k ≥ 1

2.

Let X₂ be two dimensional proper subspace of X generated by x₁, x₂. Again by Riesz’s lemma there is an x₃ ∈ X of norm 1 such that

kx₃− x₁k ≥ 1 2, kx₃− x₂k ≥ 1

2.

Continuing in this way, we obtain a sequence (xn) of elements xn∈ M such that

kx_m− x_nk ≥ 1

2 for all m 6= n.

That is, (x_n) has no convergent subsequence. But M is compact. Hence dim X must be finite.

Definition 3. (Bounded linear functional). A linear functional f is a linear operator with domain D(f ) in a vector space X and range in the scalar field K, where K = R or C. Thus

f : D(f ) −→ K,

and we say the linear functional f is bounded if there exists a real number c such that for all x ∈ D(f ),

|f (x)| ≤ ckxk.

Definition 4. (Dual space). Let X be a normed space. The set of all bounded linear functionals on X constitutes a normed space with the norm defined by

kf k = sup

x∈Xx6=0

|f (x)|

kxk = sup

x∈X kxk=1

|f (x)|. (12)

This set is called the dual space of X and is denoted by X^∗. We also write (f, x) = f (x) for f ∈ X^∗ and x ∈ X.

It is easy to see from (12) that kf k ≥ 0 and kf k = 0 if and only if f = 0, and that kαf k = |α|kf k. Moreover,

kf + gk = sup

x∈Xx6=0

|f (x) + g(x)|

kxk ≤ sup

x∈Xx6=0

|f (x)|

kxk + sup

x∈Xx6=0

|g(x)|

kxk = kf k + kgk.

Hence X^∗ is a normed space.

The following theorem shows the completeness of X^∗.

Theorem 3. The dual space X^∗ of a Banach space X is a Banach space.

Proof. Let (f_n) be a Cauchy sequence in X^∗. Then for each ǫ > 0 there exists N such that

|(f_n, u) − (f_m, u)| = |(f_n− f_m, u)| ≤ kf_n− f_mkkuk < ǫkuk (13) for every n, m > N, u ∈ X. Since (fn, u) is a Cauchy sequence of numbers, it converges to some number c_u and we define f by setting (f, u) = c_u for all u. Hence

n→∞lim(f_n, u) = (f, u).

We must show f is linear, bounded and f_n−→ f. We have

n→∞lim(f_n, α₁u₁+ α₂u₂) = (f, α₁u₁+ α₂u₂), and on the other hand,

n→∞lim(f_n, α₁u₁+ α₂u₂) = lim

n→∞(f_n, α₁u₁) + lim

n→∞(f_n, α₂u₂)

= α1 lim

n→∞(fn, u1) + α2 lim

n→∞(fn, u2)

= α₁(f, u₁) + α₂(f, u₂),

thus f is linear. kf_nk form a Cauchy sequence of positive numbers, hence lim kf_nk = M, and

|(f, u)| = lim

n→∞|(f_n, u)| ≤ lim

n→∞kf_nkkuk = M kuk.

So f is bounded. (13) gives

|(f_n− f, u)| = lim

m→∞|(f_n− f_m, u)| ≤ lim

m→∞kf_n− f_mkkuk ≤ ǫkuk, hence

kfn− f k = sup

u6=0

|(fn− f, u)|

kuk ≤ ǫ for all n > N.

Thus f_n−→ f and X^∗ is complete.

Example 5. Space l¹. The dual space of l¹ is l^∞.

Proof. Let e_k= (δ_kj), then every x ∈ l¹ has a unique representation x =

∞

X

k=1

ξ_ke_k.

Let f ∈ l^1∗, since f is bounded and linear, f (x) =

∞

X

k=1

ξ_kγ_k, γ_k= f (e_k), γ_k’s are uniquely determined by f, and

|γ_k| = |f (e_k)| ≤ kf kke_kk.

Since ke_kk = 1, we have

|γ_k| ≤ kf k for all k.

Hence

sup

k

|γ_k| ≤ kf k. (14)

Thus f ∈ l^∞.

Conversely, let b = (β_k) ∈ l^∞, then we can define the action of g on l¹ by

g(x) =

∞

X

k=1

ξ_kβ_k,

where x = (ξ_k) ∈ l¹. Clearly, g is linear. It remains to show that g is bounded. Note that

|g(x)| ≤

∞

X

k=1

|ξ_kβ_k|

≤ sup

j

|β_j|

∞

X

k=1

|ξ_k|

= kxk sup

j

|β_j|,

hence g is bounded. Therefore g belongs to l^1∗. Finally we show that kf k_l1∗ = kf k_l^∞. We have

|f (x)| = |

∞

X

k=1

ξ_kγ_k| ≤ sup

j

|γ_j|

∞

X

k=1

|ξ_k| = kxk sup

j

|γ_j|.

Since kf k = sup

kxk=1

|f (x)|

kxk , then

kf k ≤ sup

j

|γj|.

From this and (14),

kf k = sup

j

|γ_j|.

This is the norm on l^∞. So we have shown that the l^1∗ norm of f is the norm on l^∞.

Example 6. Space l^p, 1 < p < ∞. The dual space of l^p is l^q, where q is the conjugate of p, that is, 1/p + 1/q = 1.

Proof. Let x ∈ l^p and e_k = (δ_kj). Then every x ∈ l^p has a unique represen- tation

x =

∞

X

k=1

ξ_ke_k. Let f ∈ l^p∗, since f is linear and bounded,

f (x) =

∞

X

k=1

ξ_kf (e_k). (15)

Let q be the conjugate of p and consider x_n= (ξ_k⁽ⁿ⁾) with ξ⁽ⁿ⁾_k =

(|γ_k|^q/γ_k if k ≤ n and γ_k6= 0

0 if k > n or γ_k= 0, (16)

where γ_k = f (e_k). By substituting this into (15) we obtain f (xn) =

∞

X

k=1

ξ_k⁽ⁿ⁾γ_k=

n

X

k=1

|γ_k|^q. By (16) and (q − 1)p = q, we have

f (x_n) ≤ kf kkx_nk = kf k(

n

X

k=1

|ξ_k⁽ⁿ⁾|^p)^1/p

= kf k(

n

X

k=1

|γ_k|^(q−1)p)^1/p

= kf k(

n

X

k=1

|γ_k|^q)^1/p.

Then

f (xn) =

n

X

k=1

|γ_k|^q ≤ kf k(

n

X

k=1

|γ_k|^q)^1/p. Dividing by the last factor, we get

(

n

X

k=1

|γ_k|^q)^1−1/p= (

n

X

k=1

|γ_k|^q)^1/q ≤ kf k.

Letting n −→ ∞, we obtain (

∞

X

k=1

|γ_k|^q)^1/q ≤ kf k. (17) This implies that (γ_k) ∈ l^q.

On other hand, from (15) and the H¨older inequality (see (10) on p. 14 in [1]), we have

|f (x)| = |

∞

X

k=1

ξ_kγ_k| ≤ (

∞

X

k=1

|ξ_k|^p)^1/p(

∞

X

k=1

|γ_k|^q)^1/q

= kxk(

∞

X

k=1

|γ_k|^q)^1/q and

kf k ≤ (

∞

X

k=1

|γ_k|^q)^1/q (18)

since kf k = sup_x6=0 |f (x)|

kxk . We have, by (17) and (18), kf k = (

∞

X

k=1

|γ_k|^q)^1/q. (19)

So kf k = kγk_q, where γ = (γ_k) ∈ l^q and γ_k = f (e_k). The mapping of (l^p)^∗ to l^q given by f 7−→ γ is linear and injective. And by (19) it is norm preserving, so it is an isomorphism. Thus we have shown that l^p∗ is isometrically embedded in l^q. We complete the proof by showing that f 7−→ γ is bijective, i. e., for each element of l^q there is a corresponding element of l^p∗. Let (β_k) ∈ l^q, then we can define g on l^p by

g(x) =

∞

X

k=1

ξ_kβ_k,

where x = (ξ_k) ∈ l^p. Then g is linear and by the H¨older inequality, we have

|g(x)| = |

∞

X

k=1

ξ_kβ_k| ≤ (

∞

X

k=1

|ξ_k|^p)^1/p(

∞

X

k=1

|β_k|^q)^1/q = kβkkxk.

Hence g is bounded and thus g belongs to l^p∗. 1.2 Inner product spaces. Hilbert spaces

Definition 5. An inner product space is a vector space X with an inner product ( , ) defined on X. A Hilbert space is a complete inner product space.

Remark 2. The inner product has been defined in [3].

An inner product on X defines a norm on X given by

kxk = (x, x)¹². (20)

A norm on an inner product space satisfies : -Parallelogram equality:

kx + yk²+ kx − yk² = 2(kxk²+ kyk²). (21) -Polarization identity:

Re(x, y) = ¹₄(kx + yk²− kx − yk²)

Im(x, y) = ¹₄(kx + iyk²− kx − iyk²). (22) (22) holds only in complex inner product spaces. By adding

kx + yk²= (x + y, x + y) = kxk²+ (x, y) + (y, x) + kyk², and

kx − yk²= (x − y, x − y) = kxk²+ (x, −y) + (−y, x) + kyk²

= kxk²− (x, y) − (y, x) + kyk² we obtain the parallelogram equality.

To show (22), by (20), the definition of the inner product and since (x, y) + (y, x) = (x, y) + (x, y) = 2Re(x, y), we have

1

4(kx + yk²− kx − yk²) = 1

4[(x + y, x + y) − (x − y, x − y)]

= 1

4[(x, x) + 2Re(x, y) + (y, y) − (x, x) + 2Re(x, y) − (y, y)]

= Re(x, y).

Similarly, we can prove the other equality. Since Im(x, y) = Re(x, iy), we get

1

4(kx + iyk²− kx − iyk²) = 1

4[kxk²+ (x, iy) + (iy, x) + kyk²− (kxk²− (x, iy) − (iy, x) + kyk²)]

= 1

4[2(x, iy) + 2(iy, x)]

= 1

2[(x, iy) + (x, iy)]

= Re(x, iy)

= Im(x, y).

Example 7. Hilbert sequence space l². The inner product is defined by (x, y) =

∞

X

j=1

ξ_jη¯_j,

and the norm is defined by

kxk² = (x, x) = (X

|ξ_j|²)¹². We have shown the completeness of l² in Example 1.

The inner product spaces are normed spaces and the Hilbert spaces are Banach spaces, but the converse is not always true. The following examples show that.

Example 8. The space l^p with p 6= 2 is not an inner product space, hence not a Hilbert space.

Proof. We prove this by showing that the norm does not satisfy the paral- lelogram equality. Let x = (1, 1, 0, 0, · · · ) ∈ l^p and y = (1, −1, 0, 0, · · · ) ∈ l^p, so

kxk = kyk = 2^1/p, kx + yk = kx − yk = 2, hence

8 = kx + yk²+ kx − yk² 6= 2(kxk²+ kyk²) = 4 · 2^2/p if p 6= 2.

Example 9. The space C[a, b] is not an inner product space, hence not a Hilbert space.

Proof. We show that the norm defined by kxk = max

t∈J |x(t)|, J = [a, b]

cannot be obtained from an inner product since this norm does not satisfy the parallelogram equality. Indeed, if we take x(t) = 1 and y(t) = ^t−a_b−a, we have kxk = 1, kyk = 1 and

x(t) + y(t) = 1 + t − a b − a x(t) − y(t) = 1 − t − a b − a. Hence kx + yk = 2, kx − yk = 1 and

kx + yk²+ kx − yk² = 5 but 2(kxk²+ kyk²) = 4.

This completes the proof.

Definition 6. Let M be a subset of an inner product space X. We say M is an orthogonal set if (x, y) = 0 ∀x, y ∈ M, x 6= y and orthonormal if

(x, y) =

(0 if x 6= y 1 if x = y.

Later we shall need the following theorem:

Riesz’s Representation Theorem (Functionals on Hilbert spaces). Ev- ery bounded linear functional f on a Hilbert space H can be represented in terms of the inner product, namely,

f (x) = (x, z)

where z depends on f, is uniquely determined by f and has norm kzk = kf k.

For a proof, see 3.8-1 in [1].

1.3 Strong and weak convergence

Definition 7. Let (x_n) be a sequence in a normed space X. We say that:

-(x_n) is strongly convergent if there is an x ∈ X such that lim

n→∞kx_n−xk = 0.

This is written lim

n→∞x_n= x or x_n−→ x.

- (x_n) is weakly convergent if there is an x ∈ X such that for every f ∈ X^∗, lim

n→∞f (x_n) = f (x). This is written x_n−→ x or x^w _n⇀ x.

Example 10. Let e_n = (1, 0, 0, · · · ), (0, 1, 0, · · · ), · · · in H = l². Then (e_n) converges weakly to 0, but not strongly. In fact, every f ∈ H^∗ has a Riesz representation z ∈ H. Hence f (e_n) = (e_n, z). Now by the Bessel inequality (see 3.4-6 in [1]),

∞

X

n=1

|(e_n, z)|² ≤ kzk².

Hence the series on the left converges, so that its terms must approach zero as n −→ ∞. This implies

f (e_n) = (e_n, z) −→ 0.

Since f ∈ H^∗ was arbitrary, we see that e_n −→ 0. However, (e^w _n) does not converge strongly because

ke_m− e_nk² = (e_m− e_n, e_m− e_n) = 2, m 6= n.

Lemma 1. Let (x_n) be a weakly convergent sequence in a normed space X, say, x_n−→ x. Then:^w

1. The weak limit x of (x_n) is unique.

2. Every subsequence of (x_n) converges weakly to x.

Proof. 1. If xn w

−→ x, xn w

−→ y, then f (xn) −→ f (x), f (xn) −→ f (y) for f ∈ X^∗. Since (f (x_n)) is a sequence of numbers, its limit is unique.

Hence f (x) = f (y), that is

f (x) − f (y) = f (x − y) = 0

for every f ∈ X^∗. This implies x − y = 0 by Corollary 4.3-4 in [1].

2. (f (xn)) is a convergent sequence of numbers, and every subsequence of (f (x_n)) converges and has the same limit as the sequence.

A subset S of X is said to be fundamental if the closed span of S is X, i. e., if the set of all finite linear combinations of elements of S is dense in X.

Example 11. Let un ∈ X be a bounded sequence. In order that un con- verge weakly to u, it suffices that (f, u_n) converge to (f, u) for all f in a fundamental subset S^∗ of X^∗.

Proof. Let D^∗be the span of S^∗; D^∗ is dense in X^∗. Since f is a finite linear combination of elements of S^∗, (f, un) converges to (f, u) for all f ∈ D^∗. Let ǫ > 0 and g ∈ X^∗ There exists f ∈ D^∗ such that

kg − f k < ǫ.

Since (f, u_n) converges, there is an N such that

|(f, u_n− u_m)| < ǫ for n, m > N.

Thus

|(g, un− um)| = |(g − f, un) + (f, un− um) + (f − g, um)|

≤ |(g − f, un)| + |(f, un− um)| + |(f − g, um)|

≤ M ǫ + ǫ + M ǫ = (2M + 1)ǫ for n, m > N, where M = sup kunk. This shows that (g, un) converges for all g ∈ X^∗.

The relationship between strong and weak convergence is given by

Theorem 4. A sequence (u_n) ⊂ X converges strongly if and only if (f, u_n) converges uniformly for kf k ≤ 1, f ∈ X^∗.

Proof. The ”only if” part follows from

|(f, u_n) − (f, u_m)| ≤ ku_n− u_mkkf k ≤ ku_n− u_mk

because for each ǫ > 0 there exists N such that ku_n− u_mk < ǫ for n, m > N.

To prove the ”if” part, suppose that (f, u_n) converges uniformly for kf k ≤ 1.

This implies that for any ǫ > 0, there exists an N such that

|(f, un− um)| ≤ ǫ if n, m > N and kf k ≤ 1.

Hence

ku_n− u_mk = sup

kf k≤1

|(f, u_n− u_m)| ≤ ǫ for n, m > N by (12).

2 Linear operators

2.1 The domain and range

Let X, Y be normed spaces, and let T be a linear operator defined on a subspace D(T ) of X and taking values in Y. D(T ) is called the domain of T. The range R(T ) of T is defined as the set of all vectors of the form T u with u ∈ D(T ). We write T : D(T ) −→ Y.

Example 12. A finite real matrix (τ_jk), j = 1, · · · , r, k = 1, · · · , n defines a linear operator T : Rⁿ−→ R^r by

y = τ x,

where x = (ξ₁, · · · , ξ_n) and y = (η₁, · · · , η_r). In matrix form we write















 η1

η₂

·

·

· η_r

















=

















t11 t12 · · · t1n

t₂₁ t₂₂ · · · t_2n

· · · ·

· · · ·

· · · · t_r1 t_r2 · · · t_rn

































 ξ₁ ξ2

...

ξ_n

















 .

Example 13. Let X be the vector space of all polynomials on [a, b]. We may define a linear operator T on X by setting

T x(t) = x^′(t)

for every x ∈ X, where the prime denotes differentiation with respect to t.

This operator T maps X onto itself.

Example 14. A linear operator T from C[a, b] into itself can be defined by T x(t) =

Z t a

x(τ )dτ, t ∈ [a, b].

Another linear operator from C[a, b] into itself is defined by T x(t) = tx(t).

In these examples we can easily verify that the dimension of the range of T is not exceeding the dimension of the domain of T. Let us prove this in the following theorem.

Theorem 5. Let T be a linear operator. If dim D(T ) = n < ∞, then dim R(T ) ≤ n.

Proof. If y₁, . . . , y_n+1 are elements in R(T ), then there are x₁, · · · , x_n+1 in D(T ) such that y₁ = T x₁, · · · , y_n+1 = T x_n+1. Since dim D(T ) = n, this set {x1, · · · , xn+1} must be linearly dependent. Hence

α1x1+ · · · + αn+1xn+1= 0 for some scalars α₁, · · · , α_n+1, not all zero, and

T (α₁x₁+ · · · +_n+1x_n+1) = α₁y₁+ · · · + α_n+1y_n+1= 0.

This shows that {y₁, · · · , y_n+1} is a linearly dependent set. Hence dim R(T ) ≤ n.

2.2 Bounded and continuous operators

Definition 8. (Bounded linear operator). Let X and Y be normed spaces and T : D(T ) → Y a linear operator, where D(T ) ⊂ X. The operator T is said to be bounded if there is a real number c such that for all x ∈ D(T )

kT xk ≤ ckxk.

Example 15. Let X and Y be normed spaces. T : X −→ Y is bounded if and only if T maps bounded sets in X into bounded sets in Y.

Proof. Suppose T is bounded, then there is a number c such that kT xk ≤ ckxk. Let A be a bounded subset, and c₁= max

x∈A kxk, then kT xk ≤ cc₁ ∀x ∈ A, thus the image of any bounded set in X is bounded. Conversely, suppose we have kT xk ≤ M ∀ kxk ≤ 1. Then kT xk ≤ M kxk ∀x, so T is a bounded operator.

Example 16. The operator T : l^∞ −→ l^∞ defined by y = (η_j) = T x, η_j = ξ_j/j, x = (ξ_j), is bounded.

Proof. Let A be any bounded set, then there is c such that kxk ≤ c for all x ∈ A. Then

kT xk = max

j |ξ_j

j | ≤ max

j |ξ_j| = kxk, hence T is bounded.

Example 17. Let T be a bounded linear operator from a normed space X onto a normed space Y. If there is a positive b such that

kT xk ≥ bkxk for all x ∈ X, then T⁻¹: Y −→ X exists and is bounded.

Proof. T is bounded, hence there is c > 0 such that kT xk ≤ ckxk. So bkxk ≤ kT xk ≤ ckxk.

Then T x = 0 iff x = 0, that is, T⁻¹ exists. Thus ∀x ∃y such that the inequality becomes

b ≤ kyk kT⁻¹yk ≤ c.

In particular,

kT⁻¹yk ≤ 1 bkyk.

Hence T⁻¹ is bounded.

Example 18. The inverse T⁻¹ : R(X) −→ X of a bounded linear operator T : X −→ Y need not be bounded.

The operator defined in Example 16 is bounded and T⁻¹ exists. It is given by the formula

T⁻¹(y) = (jηj) = (η1, 2η2, 3η3, · · · ) and is not bounded.

Theorem 6. (Finite dimension). If a normed space X is finite dimensional, then every linear operator on X is bounded.

Proof. Let dim X = n and let {e₁, · · · , e_n} be a basis for X. We take any x =P ξje_j and consider any linear operator T on X. We have

kT xk = k

n

X

j=1

ξ_jT e_jk ≤

n

X

j=1

|ξ_j|kT e_jk ≤ max

k kT e_kk

n

X

j=1

|ξ_j|.

To the last sum we apply Lemma 2.4-1 in [1] which states that X|ξ_j| ≤ 1

ckX

ξ_je_jk = 1 ckxk for some c > 0. So

kT xk ≤ max kT e_kkX

|ξ_j| ≤ 1

cmax kT e_kkkxk.

Let γ = ¹_cmax

k kT e_kk, then

kT xk ≤ γkxk and T is bounded.

Definition 9. (Continuous linear operator). Let X, Y be normed spaces, D(T ) ⊂ X, and let T : D(T ) −→ Y be a linear operator. T is continuous at x₀ ∈ D(T ) if for every ǫ > 0 there is a δ > 0 such that

kT x − T x0k < ǫ for all x ∈ D(T ) satisfying kx − x0k < δ.

In fact there is an immediate relation between bounded and continuous operators. We show this in next theorem.

Theorem 7. T is continuous if and only if T is bounded.

Proof. For T = 0 the statement is trivial. Let T 6= 0 be bounded, we consider any x₀∈ D(T ). Let ǫ > 0, then for every x ∈ D(T ) such that

kx − x₀k < δ where δ = ǫ kT k we obtain

kT x − T x₀k = kT (x − x₀)k ≤ kT kkx − x₀k < kT kδ = ǫ.

Since x₀∈ D(T ) was arbitrary, this shows that T is continuous.

Conversely, if T is continuous at an arbitrary x₀ ∈ D(T ), then for every ǫ > 0 there is a δ > 0 such that

kT x − T x0k ≤ ǫ for all x ∈ D(T ) satisfying kx − x0k ≤ δ.

We now take any y 6= 0 in D(T ) and set x = x₀+ δ

kyky.

Hence kx − x₀k = δ and

kT x − T x0k = kT (x − x0)k = kT ( δ

kyky)k = δ

kykkT yk ≤ ǫ.

Thus

kT yk ≤ ǫ δkyk.

Hence T is bounded.

Example 19. Let T be a bounded linear operator. Then x_n−→ x implies T xn−→ T x where xn, x ∈ D(T ). By the last theorem, as n −→ ∞ we have

kT x_n− T xk = kT (x_n− x)k ≤ kT kkx_n− xk −→ 0.

Let X, Y be Banach spaces. We denote by B(X, Y ) the set of all bounded operators from X to Y.

Lemma 2. B(X, Y ) is a Banach space under the norm kT k = sup

x6=0

kT xk kxk = sup

kxk=1

kT xk.

Proof. That kT k is a norm can be seen in the same way as for kf k in (12).

Let (Tn) be a Cauchy sequence of elements of B(X, Y ). Then (Tnu) is a Cauchy sequence in Y for all u ∈ X, hence for each ǫ > 0 there exists N such that

kT_nu − T_muk ≤ kT_n− T_mkkuk < ǫkuk. (23) Since Y is complete, there is v ∈ Y such that

Tnu −→ v

and we can define v = T u. As in the proof of Theorem 3, we see that T is linear. We show that T is bounded and T_n−→ T. Since kT_nk form a Cauchy sequence of positive numbers, then kT_nk ≤ M for some M and all n, and

kT uk = lim

n→∞kT_nuk ≤ lim

n→∞kT_nkkuk ≤ M kuk for all u, so T is bounded. Now (23) gives

k(T_n− T )uk = lim

m→∞k(T_n− T_m)uk ≤ lim

m→∞kT_n− T_mkkuk ≤ ǫkuk.

Since this holds for all u ∈ X,

n→∞lim kT_n− T k = 0.

Definition 10. (Convergence of sequences of operators). Let X and Y be normed spaces. A sequence (T_n) of operators T_n∈ B(X, Y ) is said to be:

- uniformly operator convergentif (T_n) converges to some T in the norm on B(X, Y ), i.e.,

kT_n− T k −→ 0.

We use the notation T_n−→ T.

- strongly operator convergentto T if (Tnx) converges strongly in Y for every x ∈ X, i. e.,

kT_nx − T xk −→ 0.

This is denoted by T_n−→ T.^s

- weakly operator convergentto T if (T_nx) converges weakly in Y for every x ∈ X, i. e.,

|f (T_nx) − f (T x)| −→ 0 for all f ∈ Y^∗.

This is denoted by T_n −→ T. T is called the uniform, strong and weak^w operator limit of (Tn), respectively.

Example 20. Let u_n∈ X and T_n∈ B(X). If u_n−→ u and T^s _n−→ T, then^s T_nu_n−→ T u. If u^s _n−→ u and T^s _n−→ T, then T^w _nu_n−→ T u.^w

Proof. If u_n→ u and T^s _n→ T then,^s

kT_nu_n− T uk = kT_nu_n− T u_n+ T u_n− T uk

≤ kT_nu_n− T u_nk + kT u_n− T uk

= k(T_n− T )u_nk + kT (u_n− u)k

≤ kT_n− T kku_nk + kT kku_n− uk −→ 0.

If u_n→ u and T^s _n→ T, then for all f ∈ X^{w ∗} we have

|f (T_nu_n) − f (T u)| = |f (T_nu_n) − f (T_nu) + f (T_nu) − f (T u)|

≤ |f (T_nu_n) − f (T_nu)| + |f (T_nu) − f (T u)|

≤ kf kkT_nkku_n− uk + |f (T_nu) − f (T u)| −→ 0.

2.3 Compact operators

Definition 11. Let X and Y be normed spaces. A linear operator T : X −→ Y is called a compact linear operator if for every bounded subset M of X, the set T (M ) is compact.

Lemma 3. Let X and Y be normed spaces. Then:

1. Every compact linear operator T : X −→ Y is bounded, hence contin- uous.

2. If dim X = ∞, the identity operator I : X −→ X is not compact.

Proof. 1. The unit sphere U = {x ∈ X| kxk = 1} is bounded. Since T is compact, T (U ) is compact, and is bounded, so that

sup

kxk=1

kT xk < ∞.

Hence T is bounded and Theorem 7 shows that it is continuous.

2. The closed unit ball M = {x ∈ X| kxk ≤ 1} is bounded. If M is compact, then dim X < ∞ by Theorem 2, which contradicts that X is of infinite dimension. Thus M cannot be compact, showing that I(M ) = M = M is not compact, i.e. I is not compact.

From this lemma it follows that the identity operator in a normed space is compact if and only if X is finite dimensional.

Theorem 8. Let X and Y be normed spaces and T : X −→ Y a linear operator. Then T is compact if and only if it maps every bounded sequence (x_n) in X onto a sequence (T x_n) in Y which has a convergent subsequence.

Proof. If T is compact and (x_n) is bounded, then (T x_n) is compact and Definition 2 shows that (T x_n) contains a convergent subsequence.

Conversely: Let B be any bounded subset in X, and let (y_n) be any se- quence in T (B). Then y_n= T x_nfor some x_n∈ B, and (x_n) is bounded since B is bounded. By assumption, (T xn) contains a convergent subsequence.

Hence T (B) is compact, this shows that T is compact.

Example 21. If T1 and T2 are compact linear operators from a normed space X into a normed space Y, then T₁+ T₂ is a compact linear operator.

Proof. Let (x_n) be any bounded sequence in X. Since T₁ is compact, by Theorem 8 (T₁x_n) has a convergenct subsequence (T₁x_n1) in Y and similarly, (T₂x_n1) has a convergent subsequence (T₂x_n2) in Y. Then ((T₁+ T₂)x_n2) is a convergent subsequence of ((T₁ + T₂)x_n). Hence T₁ + T₂ is a compact operator.

Theorem 9. (Finite dimensional domain or range). Let X and Y be normed spaces and T : X −→ Y a linear operator. Then:

1. If T is bounded and dim T (X) < ∞, the operator T is compact.

2. If dim X < ∞, T is compact.

Proof. 1. Let (x_n) be any bounded sequence in X. Then the inequality kT x_nk ≤ kT kkx_nk shows that (T x_n) is bounded. Hence (T x_n) is compact since dim T (X) < ∞. It follows that (T x_n) has a convergent subsequence. Since (xn) was an arbitrary bounded sequence in X, the operator T is compact by Theorem 8.

2. By Theorem 6 T is bounded, by Theorem 5 dim T (X) ≤ dim X. Hence by 1. T is compact.

Theorem 10. (Sequence of compact linear operators). Let (T_n) be a se- quence of compact linear operators from a normed space X into a Banach space Y. If (T_n) is uniformly operator convergent, say, kT_n− T k −→ 0, then the limit operator T is compact.

Proof. Let (x_m) be any bounded sequence in X, then it has subsequences (x_1,m), (x_2,m), · · · such that (T₁x_1,m), (T₂x_2,m), · · · are Cauchy. Here (x_1,m) is a subsequence of (x_m), (x_2,m) is a subsequence of (x_1,m) etc. Let (y_m) = (x_m,m). This is a subsequence of (x_m) such that for every fixed n the sequence (Tnym) is Cauchy. Since kxmk ≤ c for some c > 0, also kymk ≤ c for all m.

Since kT_n− T k −→ 0, ∃p such that kT − T_pk < _3c^ǫ. Since (T_py_m) is Cauchy,

∃N such that

kT_py_j− T_py_kk < ǫ

3 for all j, k > N.

Hence for j, k > N

kT y_j− T y_kk ≤ kT y_j− T_py_jk + kT_py_j− T_py_kk + kT_py_k− T y_kk

≤ kT − T_pkky_jk +ǫ

3+ kT_p− T kky_kk

< ǫ 3cc + ǫ

3+ ǫ 3cc = ǫ.

Hence (T y_m) is Cauchy and therefore convergent, that is, T is compact.

Example 22. Prove compactness of T : l² −→ l² defined by y = (η_j) = T x, where η_j = ξ_j/j for j = 1, 2, · · · .

Proof. Let T_n: l² −→ l² be defined by

T_nx = (ξ₁, ξ₂/2, ξ₃/3, · · · , ξ_n/n, 0, 0, · · · ).

T_nis bounded, and it is compact by Theorem 9 because dim R(T_n) = n < ∞.

Furthermore,

k(T − Tn)xk²=

∞

X

j=n+1

|ηj|²=

∞

X

j=n+1

1 j²|ξj|²

≤ 1

(n + 1)²

∞

X

j=n+1

|ξ_j|² ≤ kxk² (n + 1)², hence

supk(T − T_n)xk

kxk ≤ 1

n + 1 and

kT − T_nk ≤ 1 n + 1. Hence Tn−→ T, and T is compact by Theorem 10.

Example 23. Let Y be a Banach space and let T_n: X −→ Y, n = 1, 2, · · · , be bounded operators of finite rank. If (T_n) is uniformly operator convergent, then the limit operator is compact.

Proof. Since dim T_n(X) = rank T_n< ∞, it follows from Theorem 9 that T_n is compact for all n. Hence by Theorem 10 T is compact.

2.4 Spectrum of compact operators

Definition 12. Let X be a normed space and T : X −→ X a bounded linear operator. Denote

T_λ = T − λI, (24)

where λ is a complex number and I is the identity operator on X. If T_λ has no bounded inverse on X, then λ is called a spectral value, denoted λ ∈ σ(T ). If T_λx = 0 for some x 6= 0, then λ is said to be an eigenvalue and x an eigenvector.

If λ is an eigenvalue, then λ ∈ σ(T ), but the converse need not be true.

Theorem 11. (Eigenvalues). The set of eigenvalues of a compact linear operator T : X −→ X on a normed space X is countable (perhaps finite or even empty), and the only possible point of accumulation is λ = 0.

Proof. We suppose the opposite, and we get a contradiction. Let k₀ > 0, and suppose there is an infinite sequence (λ_n) of distinct eigenvalues such that |λn| > k0 for all n.

For each n, let T_λnx_n= 0, x_n6= 0. The set of all the x_n’s is linearly indepen- dent (see Thm. 7.4-3 in [1] or Proposition 2 in [3]). Let Mn=span{x1, · · · , xn}.

Then each x ∈ M_n has a unique representation x = α₁x₁+ · · · + α_nx_n. We apply T − λnI and use T xj = λjxj:

(T − λ_nI)x = α₁(λ₁− λ_n)x₁+ · · · + α_n−1(λ_n−1− λ_n)x_n−1, hence

(T − λ_nI)x ∈ M_n−1 for all x ∈ M_n.

The M_n’s are finite dimensional and therefore closed (see Thm. 2.4-3 in [1]). By Riesz’s Lemma there is a sequence (y_n) such that y_n ∈ M_n and kynk = 1, kyn− xk ≥ ¹₂ for all x ∈ Mn−1.We show that

kT y_n− T y_mk ≥ 1

2k₀ for all n > m,

so that (T y_n) has no convergent subsequence because k₀ > 0. Since T is compact and (y_n) is bounded, we get a contradiction.

Now let m < n, then we have λ_my_m∈ M_n−1and (T − λ_nI)y_n∈ M_n−1. Thus x = 1

λ_n(T y_m− T y_n) + y_n∈ M_n−1. Hence

kT y_n− T y_mk = kλ_ny_n− λ_nxk

= |λ_n|ky_n− xk

≥ 1

2|λ_n| ≥ 1 2k₀.

Theorem 12. (Null space). Let T : X −→ X be a compact linear operator on a normed space X. Then for every λ 6= 0 the null space N (T_λ) of T_λ = T − λI is finite dimensional.

Proof. We shall apply Theorem 2 from which it follows that if the closed unit ball M in N (T_λ) is compact, then dim N (T_λ) < ∞. Now we show that M is compact.

Let (x_n) be in M, then

kxnk ≤ 1.

T is compact and (xn) is bounded, hence (T xn) has a convergent subse- quence (T x_nk) by Theorem 8.

Since x_n∈ M ⊂ N (T_λ),

T_λx_n= T x_n− λx_n= 0, so that

xn= λ⁻¹T xn. Thus

x_nk = λ⁻¹T x_nk

is a convergent subsequence of (x_n), and the limit is in M because M is closed. This proves that M is compact.

Example 24. Let T : X −→ X be a compact linear operator on a normed space. If dim X = ∞, then 0 ∈ σ(T ).

Proof. We assume 0 /∈ σ(T ). Similarly as in the last theorem, we have that if the closed unit ball M in X is compact, then dim X < ∞. Let (x_n) be in M. Since T is compact, there is a convergent subsequence (T xnk) of (T x_n), T x_nk −→ y. T⁻¹ exists because 0 /∈ σ(T ), hence x_nk −→ T⁻¹y.

From Theorem 2 it follows that dim X < ∞ which is a contradiction. Hence 0 ∈ σ(T ).

Lemma 4. Let T : X −→ X be a compact linear operator on a normed space X. If λ 6= 0 is not eigenvalue, then R(T_λ) is a closed subset of X.

Proof. We have to prove that for any convergent sequence (T_λx_n) the limit of it is in R(T_λ).

Suppose

T_λx_n= T x_n− λx_n−→ y.