Some Topics in Advanced Functional Analysis

A Crash Course∗

M.T.Nair

Department of Mathematics, IIT Madras

August 13, 2012

Abstract

In these notes we intend to give a crash course on Some Topics in Advanced Functional

Analysis. The topics covered include the following:

1. Some basic properties of operators,

2. Compact operators,

3. Spectral results,

4. Spectral representation.

∗A series of lectures at IIT Bombay under the NPDE-TCA programme (June 18-19, 2012).

1

1 Some basic properties of operators

1.1 Bounded operators and operators with bounded inverse

In the following, unless stated otherwise, X and Y are normed linear spaces.

Definition 1.1. A linear operator T : X → Y is called a bounded operator if it maps

every bounded subset of X onto a bounded subset of Y . A linear operator which is not

a bounded operator is called an unbounded operator. ♦

Notation:

1. B(X,Y ): The set of all bounded operators from X to Y .

2. B(X) := B(X,X)

3. X ′ := B(X,K), the dual space of X.

Two standard theorems:

THEOREM 1.2. Let T : X → Y be a linear operator. Then the following are equivalent:

1. For every bounded subset S of X, T (S) is bonded in Y .

2. The set Tx : ‖x‖ = 1 is bounded in Y .

3. There exists c > 0 such that ‖Tx‖ ≤ c‖x‖ for all x ∈ X.

4. T is uniformly continuous.

5. T is continuous at 0.

THEOREM 1.3. The set B(X,Y ) is a linear space and

‖T‖ := infc > 0 : ‖Tx‖ ≤ c‖x‖ ∀x ∈ X

defines a norm on B(X,Y ). Further, the quantities

sup‖Tx‖ : ‖x‖ = 1, sup‖Tx‖ : ‖x‖ ≤ 1, sup‖Tx‖‖x‖

: ‖x‖ 6= 0

are equal, and equal to ‖T‖.

THEOREM 1.4. Let T : X → Y be a linear operator. If there exists γ > 0 such that

‖Tx‖ ≥ γ‖x‖ ∀x ∈ X, (1.1)

then

1. T is injective,

2. T−1 : R(T )→ X is a bounded operator, and

3. ‖T−1‖ ≤ 1/γ.

Definition 1.5. A linear operator T : X → Y is said to be bounded below if it satisfies

(1.1) for some γ > 0. ♦

2

Example 1.6. Let X = `p and Y = `r for 1 ≤ p ≤ r <∞. Define

T (x1, x2, . . .) = (0, x1, x2, . . .).

Let x ∈ `p such that ‖x‖p = 1. Since r ≥ p, |xj |r ≤ |xj |p|xj |r−p ≤ |xj |p so that

‖Tx‖rr =

∞∑j=1

|xj |r ≤∞∑j=1

|xj |p = 1.

Thus, ‖Tx‖r ≤ ‖x‖p for every x ∈ `p. Also, ‖Tx‖∞ ≤ ‖x‖p for every x ∈ `p and

‖Te1‖r = ‖e1‖r = 1 = ‖e1‖p. Hence, T is a bounded operator and ‖T‖ = 1. ♦

Example 1.7. Let k(·, ·) be continuous on [a, b]× a, b] and let

(Tx)(s) =

∫ b

a

k(s, t)x(t)dt, x ∈ L1[a, b].

Then

• Tx ∈ C[a, b] for a all x ∈ L1[a, b],

• ‖Tx‖∞ ≤ c‖x‖1 for a all x ∈ L1[a, b], where c := sups,t∈[a,b]

|k(s, t)|.

Thus, T : L1[a, b]→ C[a, b] is a bounded operator.

Since, for each p ∈ [1,∞], there exists cp, dp > 0 such that

‖x‖p ≤ cp‖x‖∞, ‖y‖1 ≤ dp‖y‖p

for all x ∈ C[a, b] and y ∈ Lp[a, b], it follows that T defines a bounded operator from

Lp[a, b] to Lr[a, b] for any p, r ∈ [1,∞]. ♦

Example 1.8. Let X = C1[0, 1] and Y = C[0, 1], both with ‖ · ‖∞, and

Tx = x′, x ∈ C1[0, 1].

Taking xn(t) = tn for t ∈ [a, b], n ∈ N, we have

xn ∈ X, ‖xn‖∞ = 1, ‖Txn‖∞ = n

for every n ∈ N. Hence, T is not a bounded operator.

If X = x ∈ C1[0, 1] : x(0) = 0, then T is bounded below, for in this case, we have

x(t) =

∫ 1

0

x′(s)ds ∀x ∈ X, t ∈ [0, 1],

so that

‖x‖∞ ≤ ‖x′‖∞ = ‖Tx‖∞ ∀x ∈ X.

♦

3

1.2 Closed operators and solvability of operator equations

Suppose T is a linear operator defined on a subspace X0 of X with values in Y . Suppose

T : X0 → Y is a bounded operator. If (xn) in X0 is such that xn → x for some x ∈ X,

then we are not in a position to say that Txn → Tx, as we are not sure that x ∈ X0,

which is the case if X0 is not closed.

Definition 1.9. Let X0 be a subspace of X. A linear operator T : X0 → Y is called a

closed operator if for every (xn) in X0,

xn → x and Txn → y =⇒ x ∈ X0 and Tx = y.

♦

• T : X0 → Y is a closed operator if and only if its graph

G(T ) := (x, Tx) : x ∈ X0

is a closed subspace of X × Y .

Example 1.10. Let X = C[0, 1] = Y with ‖ · ‖∞, X0 = C1[0, 1], and

Tx = x′, x ∈ X0.

We have seen in Example 1.8 that T is a not a bounded operator. It is a closed operator:

To see this, let (xn) is in X0 be such that

‖xn − x‖∞ → 0 and ‖x′n − y‖∞ → 0

for some x, y ∈ X. Then x ∈ X0 and x′ = y (Why?). ♦

Notation:

1. If T is a linear operator defined on a subspace of X taking values in Y , then the

domain of T is denoted by D(T ) and the range of T is denoted by R(T ).

2. C(X,Y ): The set of all closed operators T with D(T ) ⊆ X and R(T ) ⊆ Y .

3. C(X) := C(X,X)

It can be easily seen that

• A bounded operator with a closed domain is a closed operator.

Question: Is a closed operator with a closed domain a bounded operator?

The answer is in affirmative if both X and Y are Banach spaces. That is the essence

of the celebrated closed graph theorem.

THEOREM 1.11. (Closed graph theorem). Let X and Y be Banach spaces. Then

a closed operator T : D(T ) ⊆ X → Y is a bounded operator if and only if D(T ) is a

closed subspace.

4

In the following theorem we list some easily verifiable facts.

THEOREM 1.12. The following hold:

1. Null space of a closed operator is a closed subspace.

2. Inverse of a closed operator is a closed operator.

In applications the following situations occur: Suppose T ∈ C(X,Y ) such that for

every y ∈ Y , there exists a unique x ∈ D(A) such that

Tx = y.

The question is whether x depends continuously on the data y. In other words, one would

like to know whether T−1 is a bounded operator. Thanks to closed graph theorem we

have the following.

THEOREM 1.13. (Bounded inverse theorem) Let X and Y be Banach spaces and

T ∈ C(X,Y ) be injective. Then T−1 : R(T ) → X is a bounded operator if and only if

R(T ) is a closed subspace.

Proof. Since T ∈ C(X,Y ) be injective, T−1 : R(T )→ X is a closed operator. Hence, by

closed graph theorem, T−1 : R(T )→ X is a bounded operator if and only if D(T−1) is a

closed subspace, i.e., R(T ) is a closed subspace.

Recall from Theorem 1.4 that:

• A sufficient condition for a linear operator to have a continuous inverse is that it is

bounded below.

Therefore, we have the following.

PROPOSITION 1.14. Let X be a Banach space and T : D(A) ⊆ X → Y be a linear

operator which is bounded below. Then T ∈ C(X,Y ) if and only if R(T ) is closed.

Proof. Since T is bounded below, T−1 : R(T )→ X is a bounded operator. Hence, if R(T )

is closed in Y , then T−1 is a closed operator, consequently, T is also a closed operator.

Conversely, suppose T is a closed operator. Let (xn) in D(A) be such that (Txn)

converges to some y ∈ Y . Then, using the fact that T is bounded below, we see that (xn)

is a Cauchy sequence in X. Since X is a Banach space (xn), converges to some x ∈ X.

Thus, xn → x and Txn → y. Therefore, using the fact that T is a closed operator, we

obtain x ∈ D(A) and y = Tx. Hence, R(T ) is closed in Y .

Example 1.15. Let X = L2[0, 1] = Y ,

X0 = x ∈ L2[0, 1] : x absolutely continuous x(0) = 0, x′ ∈ L2[0, 1]

and

Tx = x′, x ∈ X0.

5

By the fundamental theorem of Lebesgue integration1, we have

x(t) =

∫ t

0

x′(s)ds

for every x ∈ X0 and t ∈ [0, 1] so that

|x(t)| ≤∫ t

0

|x′(s)| ≤ ‖x′‖2.

Hence,

‖x‖2 ≤ ‖x′‖2 ∀x ∈ X0,

that is, T is bounded below. Therefore, T has a bounded inverse from its range. Again,

by fundamental theorem of Lebesgue integration, for every y ∈ L2[0, 1], the function x

defined by

x(t) =

∫ t

0

y(s)ds, t ∈ [0, 1],

belongs to X0 and x′ = y so that R(T ) = L2[0, 1]. Therefore, T−1 is a bounded operator

with closed domain, so that T−1 is a closed operator, and hence T is a also a closed

operator. ♦

Remark 1.16. In Example 1.15, suppose we define

‖x‖0 := ‖x′‖2, x ∈ X0.

Then it can be seen that ‖ · ‖0 is a complete norm on X0, and the inclusion operator

X0 into L2[0, 1] is a bounded operator. In fact, the above norm is induced by the inner

product 〈x, y〉 :=∫ 1

0x′(t)y′(t)dt. ♦

As a corollary to Proposition 1.14, we have

THEOREM 1.17. If X is a Banach space and T ∈ C(X,Y ), then the following are

equivalent:

1. T is bijective and T−1 is continuous,

2. T is bounded below and R(T ) is dense.

THEOREM 1.18. Let X be a Hilbert space and A ∈ C(X) satisfy

|〈Ax, x〉| ≥ γ‖x‖2 ∀x ∈ D(A), (1.2)

for some γ > 0. Then

1. A is bijective,

2. A−1 : X → X is continuous with ‖A−1‖ ≤ 1/γ.

1Fundamental theorem of Lebesgue integration: If y ∈ L1[0, 1], then x defined by x(t) =∫ t

0y(s)ds is

absolutely continuous, differentiable a.e., and x′ = y. Conversely, if x : [0, 1] → K is absolutely continuous,

then it is differentiable a.e., x′ ∈ L1[, b] and x(t) =∫ t

0y(s)ds.

6

Proof. The condition (1.2) implies that A is bounded below. Hence, by Theorem 1.4 and

Proposition 1.14,

• A is injective,

• A−1 : R(A)→ X is continuous with ‖A−1‖ ≤ 1/γ, and

• R(A) closed.

By (1.2 ),

x ∈ R(A)⊥ =⇒ 〈x,Ax〉 = 0 =⇒ x = 0.

Hence, R(A)⊥ = 0 so that by projection theorem, X = R(A) +R(A)⊥ = R(A).

Definition 1.19. Let X be a Hilbert space. An operator A ∈ C(X) is said to be coercive

if there exists γ > 0 such that

|〈Ax, x〉| ≥ γ‖x‖2 ∀x ∈ D(A),

and in that case γ is called a coefficient of coercivity of A. ♦

Remark 1.20. By Theorem 1.18, if a closed operator A in a Hilbert space X is coercive,

then for every y ∈ X, there exists a unique x ∈ X such that

Ax = y

and in that case ‖x‖ ≤ 1γ ‖y‖, where γ is a coefficient of coercivity. In other words, the

equation Ax = y can be solved uniquely and stably. ♦

Example 1.21. Let X be a separable Hilbert space, un : n ∈ N be an orthonormal

basis of X, and (µn) be a sequence of positive real numbers such that µn →∞ as n→∞.

Let

Ax :=

∞∑n=1

µn〈x, un〉un, ∀x ∈ X0,

where

X0 :=x ∈ X :

∞∑n=1

µ2n|〈x, un〉|2 <∞

.

Then we have

〈Ax, x〉 :=

∞∑n=1

µn|〈x, un〉|2 ≥ γ‖x‖2, ∀x ∈ X0,

where γ := infn∈N µn.

Also, for every y ∈ X, x :=∑∞n=1

〈y,un〉µn

un belongs to X0 and satisfies Ax = y, so that

A is onto. Hence, by Proposition 1.14, A is a closed operator. Note that image of the

bounded sequence (un) under A is not a bounded sequence. Hence, A is not a bounded

operator. Thus, A is an unbounded closed operator which is coercive. ♦

Now, we prove and theorem which is important in view of its applications to PDE, the

so called Lax-Milgram theorem, which is an application of Theorem 1.18 together with

Riesz representation theorem. Recall:

7

THEOREM 1.22. (Riesz representation theorem) If X is a Hilbert space, then for

every continuous linear functional f on X, there exists a unique zf ∈ X such that

f(x) = 〈x, zf 〉 ∀x ∈ X,

and in that case ‖f‖ = ‖zf‖.

THEOREM 1.23. (Lax-Milgram theorem) Let X be a Hilbert space and ϕ(·, ·) be a

sesquilinear form on X, i.e., ϕ : X ×X → K is such that x 7→ ϕ(x, y) is linear for each

y ∈ X and y 7→ ϕ(x, y) is conjugate linear for each x ∈ X. Suppose there exist β, γ > 0

such that

|ϕ(x, y)| ≤ β‖x‖ ‖y‖ ∀x, y ∈ X, (i)

|ϕ(x, x)| ≥ γ‖x‖2 ∀x ∈ X. (ii)

Then, for every f ∈ X ′, there exists a unique u ∈ X such that

ϕ(x, u) = f(x) ∀x ∈ X.

Further, ‖u‖ ≤ 1

γ‖f‖.

Proof. Let us settle the uniqueness issue first: Suppose there exist u1, u2 such that

ϕ(x, u1) = f(x) = ϕ(x, u2) ∀x ∈ X.

Then, we have

ϕ(x, u1 − u2) = 0 ∀x ∈ X.

This implies ϕ(u1 − u2, u1 − u2) = 0, which implies, by condition (ii) on ϕ, u1 − u2 = 0.

Now, the rest of the results: By Riesz representation theorem, there exists a unique

v ∈ X be such that

f(x) = 〈x, v〉 ∀x ∈ X,

and in that case we also have ‖f‖ = ‖v‖.Note that, by the condition (i) on ϕ, (x, y) 7→ ϕ(x, y) is continuous on x ×X. Now,

let y ∈ X. Since x 7→ ϕ(x, y) is a continuous linear functional, by Riesz representation

theorem, there exists a unique zy ∈ X such that

ϕ(x, y) = 〈x, zy〉 ∀x ∈ X.

Let Ay := zy, y ∈ X. Then A is a linear operator on X (verify) and

|〈x,Ay〉| = |ϕ(x, y)| ≤ β‖x‖ ‖y‖ ∀x, y ∈ X,

|〈x,Ax〉| = |ϕ(x, x)| ≥ γ‖x‖2 ∀x ∈ X.

In particular, A is a bounded operator and satisfies the assumptions in Theorem 1.18.

Therefore, there exists a unique u ∈ X such that

Au = v and ‖u‖ ≤ 1

γ‖v‖.

8

Thus,

ϕ(x, u) = 〈x,Au〉 = 〈x, v〉 = f(x), ∀x ∈ X,

and

‖u‖ ≤ 1

γ‖v‖ =

1

γ‖f‖.

This completes the proof.

Example 1.24. Consider the Dirichlet problem for the Poisson equation:

−∆u = f on Ω,

u = 0 on ∂Ω,

where Ω is a bounded open subset of Rk, ∂Ω is the boundary of Ω, and f ∈ L2(Ω).

Suppose we are looking for a solution u ∈ C2c (Ω) for the above equation. If such a

solution exists, then we would get,

−∫

Ω

(∆u)v =

∫Ω

fv ∀ v ∈ C∞c (Ω).

Note that ∫Ω

(∆u)v = −∫

Ω

∇u.∇v.

Thus, we have ∫Ω

∇u.∇v =

∫Ω

fv ∀ v ∈ C∞c (Ω),

and therefore, ∫Ω

∇u.∇v =

∫Ω

fv ∀ v ∈ H10 (Ω). (∗)

An element u ∈ H10 (Ω) is called a weak solution of the Dirichlet problem if it satisfies (∗),

i.e., if there exists u ∈ H10 (Ω) such that

ϕ(u, v) = F (v) ∀v ∈ H10 (Ω),

where

ϕ(u, v) :=

∫Ω

∇u.∇v

defines a bilinear form on H10 (Ω)×H1

0 (Ω) and

F (v) :=

∫Ω

fv

is a linear functional on H10 (Ω). In fact, F belongs to the dual of H1

0 (Ω), since

|F (v)| ≤∫

Ω

|fv| ≤ ‖f‖L2‖v‖L2 ≤ ‖f‖L2‖v‖H10.

Here, the Sobolev space H10 (Ω) is the completion of the space C∞c (Ω) with respect to the

norm

‖v‖2H10

= ‖v‖L2 + ‖∇v‖L2 .

9

We also observe that

|ϕ(u, v)| ≤ ‖∇u‖L2‖∇v‖L2 ≤ ‖u‖H10‖v‖H1

0,

|ϕ(v, v)| =∫

Ω

|∇v|2 = ‖|∇v‖2L2 .

By Poicare inequality2, there exists a constant c0 > 0 such that

‖∇v‖L2 ≥ c0‖v‖2L2 ∀ v ∈ H10 (Ω).

Therefore,

‖∇v‖L2 =1

2(‖|∇v‖L2 + ‖|∇v‖L2) ≥ 1

2(‖|∇v‖L2 + c0‖v‖2L2)

so that for every v ∈ H10 (Ω),

‖∇v‖L2 ≥ min1, c02

(‖|∇v‖L2 + ‖v‖2L2) = c1‖v‖H10,

where c1 := min1, c0/2. Thus,

ϕ(v, v) ≥ c1‖v‖2H10.

We have shown that ϕ satisfies the conditions in Lax-Milgram theorem (Theorem 1.23).

Thus, for every f ∈ L2(Ω) (so that F belongs to the in the dual of H10 (Ω)), there exists

a unique u ∈ H10 (Ω) such that (∗) holds. ♦

Proof of Poincare inequality for the case of Ω = (a, b) ⊆ R: For v ∈ C1c (Ω), we have

u(t) =

∫ t

a

u′(s)ds, t ∈ Ω,

so that

|u(t)| ≤∫ t

0

|u′(s)| ≤∫ b

a

|u′(s)|ds ≤√

(b− a) ‖u′‖L2 , , t ∈ Ω,

Hence,

‖u‖2L2 =

∫ b

a

|u(t)|2dt ≤∫ b

a

[√

(b− a)‖u′‖L2 ]2dt = (b− a)2‖u′‖2L2 .

Thus, ‖u′‖2L2 ≥ c0‖u‖2L2 with c0 = 1/(b − a)2. Since this is true for every C∞c (Ω) and

since C∞c (Ω) is dense in H10 (Ω), we obtain

‖u′‖2L2 ≥ c0‖u‖2L2 ∀ v ∈ H10 (Ω).

This is Poincare inequality.

Another consequence of Riesz representation theorem is the definition of adjoint of an

operator.

2Jules Henri Poincare (29 April 1854 17 July 1912) was a French mathematician, theoretical physicist,

engineer, and a philosopher of science.

10

THEOREM 1.25. Let X and Y be a Hilbert spaces and T : D(T ) ⊆ X → Y be linear

operator with D(T ) dense in X. Then there exists a unique linear operator A∗ with

D(T ∗) := y ∈ Y : x 7→ 〈Tx, y〉 continuous

and R(A∗) ⊆ X such that

〈Tx, y〉 = 〈x, T ∗y〉 ∀x ∈ D(T ), y ∈ D(T ∗).

Note that

D(T ∗) = y ∈ Y : ∃ a unqiue zy ∈ X such that 〈Tx, y〉 = 〈x, zy〉 ∀x ∈ D(T ).

Definition 1.26. Let X and Y be a Hilbert spaces and A : D(T ) ⊆ X → Y be a linear

operator with dense domain D(T ). The operator T ∗ obtained in Theorem 1.25 is called

the adjoint of T . ♦

For a linear operator T : D(T ) ⊆ X → Y with dense domain D(T ), we observe that

have the following:

• T ∗ ∈ C(Y,X) with D(T ∗) is dense in Y .

• R(T )⊥ = N(T ∗). In particular, R(T ) is dense ⇐⇒ T ∗ is one-one.

• If A ∈ B(X,Y ) so that D(T ) = X, then D(T ∗) = Y and T ∗ ∈ B(Y,X), and in that

case ‖T ∗‖ = ‖T‖ and ‖T ∗T‖ = ‖A‖2.

Definition 1.27. Let X be a Hilbert space and T ∈ C(X) with D(T ) dense in X. Then

(i) T is called a symmetric operator if

〈Tx, y〉 = 〈x, Ty〉 ∀x ∈ D(T ).

(ii) T is called self adjoint if D(T ∗) = D(T ) and T ∗ = T .

♦

• Every densely defined self adjoint operator is a closed operator.

Example 1.28. (i) Let X = `2 and T ∈ B(`2) be the right shift operator on `2, i.e.,

T (x1, x2, . . .) = (0, x1, x2, . . .).

Then T ∗ is the left shift operator,

T ∗(x1, x2, . . .) = (x2, x3, . . .).

(ii) Let k(·, ·) be continuous on [a, b]× a, b] and let T ∈ B(L2[a, b]) be defined by

(Tx)(s) =

∫ b

a

k(s, t)x(t)dt, x ∈ L2[a, b].

11

Then T ∗ is given y

(T ∗x)(s) =

∫ b

a

k(t, s)x(t)dt, x ∈ L2[a, b].

Hence, T is self adjoint if and only if k(t, s) = k(s, t) for all s, t ∈ [a, b].

(iii) Let T : D(T ) ⊆ L2[a, b]→ L2[a, b] defined by

Tx = x′, x ∈ D(T ),

where

D(T ) := x ∈ L2[a, b] : x absolutely continuous, x(a) = 0 and x′ ∈ L2[a, b].

It can shown that D(T ) is dense in L2[a, b]. Taking

Y0 := y ∈ L2[a, b] : y absolutely continuous, y(b) = 0 and y′ ∈ L2[a, b]

we see that for x ∈ D(T ) and y ∈ Y0,

〈Tx, y〉 =

∫ b

a

x′(t)y(t)dt =[y(t)x(t)

]ba−∫ b

a

y′(t)x(t)dt = 〈x, z〉,

where z = −y′. Thus, T ∗y = −y′ with D(T ∗) := Y0.

(iii) Let T : D(T ) ⊆ L2[a, b]→ L2[a, b] defined by

Tx = ix′, x ∈ D(T ),

where

D(T ) := x ∈ L2[a, b] : x absolutely continuous, x(a) = x(b) and x′ ∈ L2[a, b].

It can shown that D(T ) is dense in L2[a, b] and T is a self adjoint operator.

♦

Example 1.29. Let A be as in Example 1.21, that is,

Ax :=

∞∑n=1

µn〈x, un〉un, ∀x ∈ X0,

where

X0 :=x ∈ X :

∞∑n=1

µ2n|〈x, un〉|2 <∞

,

X is a separable Hilbert space, un : n ∈ N is an orthonormal basis of X, and (µn) is a

sequence of positive real numbers such that µn →∞ as n→∞. Note that for x ∈ D(T )

and y ∈ X,

〈Ax, y〉 =

∞∑n=1

µn〈x, un〉〈un, y〉.

Thus, for y ∈ Y , there exists z ∈ X such that 〈Ax, y〉 = 〈x, z〉 if and only if y ∈ D(T )

and in that case z =∑∞n=1 µn〈y, un〉, so that A∗y = z = Ay, showing that A is a self

adjoint operator. ♦

12

2 Compact operators

Unbounded closed operators occur naturally in PDE and its applications. In many cases,

such operators can be represented as in the case of Example 1.21, that is,

Ax :=

∞∑n=1

µn〈x, un〉un, ∀x ∈ X0,

where

X0 :=x ∈ X :

∞∑n=1

µ2n|〈x, un〉|2 <∞

,

X is a separable Hilbert space, un : n ∈ N is an orthonormal basis of X, and (µn) is

a sequence of positive real numbers such that µn →∞ as n→∞. Such representations

are possible as they are inverses of certain compact operators.

Definition 2.1. Let X and Y be normed linear spaces and T : X → Y be a linear

operator. Then T is said to be a compact operator if

clTx : ‖x‖ ≤ 1

is a compact subset of Y . ♦

We observe the following facts:

• Every compact operator is a bounded operator.

• Every finite rank bounded operator is a compact operator.

• Identity operator on a normed linear space is a compact operator if and only if the

space is finite dimensional.

• If Y is a Banach space, the norm limit of a sequence of compact operators from X

to Y is a compact operator.

NOTATION: K(X,Y ): the set of all compact operators from X to Y .

THEOREM 2.2. Let X and Y be normed linear spaces. If T : X → Y is an injective

compact operator of infinite rank, then T−1 : R(T )→ X is not continuous.

Proof. Let T : X → X be an injective compact operator. If T−1 : R(T ) → X is

continuous, then I = T−1T is a compact operator, and hence X is finite dimensional;

consequently, A is of finite rank.

THEOREM 2.3. Let X and Y be Banach spaces and T ∈ K(X,Y ). If T is of infinite

rank, then R(T ) is not closed.

Proof. Suppose T is of infinite rank. Let T : X/N(T )→ Y be defined by

T (x+N(T )) = Tx, x ∈ X.

It can be seen that T is an injective compact operator. Then by Theorem 2.2, inverse of

T is not continuous. Hence, by closed graph theorem, R(T ) is not closed.

13

• If T is a compact operator of infinite rank, then the operator equation

Tx = y

cannot be solved in a stable manner.

Example 2.4. Let k(·, ·) be a continuous function defined on [a, b]×[a, b]. For x ∈ L1[a, b],

let

(Tx)(s) =

∫ b

a

k(s, t)x(t) dµ(t), s ∈ [a, b].

Then Tx ∈ C[a, b], and T : L1[a, b] → C[a, b] is a compact operator with respect to the

norms ‖ · ‖1 and ‖ · ‖∞ on L1[a, b] and C[a, b] respectively. To see this, first observe that

for x ∈ L1[a, b] and s, τ ∈ [a, b],

(Tx)(s)− (Tx)(τ) =

∫ b

a

[k(s, t)− k(τ, t)]x(t) dµ(t)

so that

|(Tx)(s)− (Tx)(τ)| ≤(

supt∈[a,b]

|k(s, t)− k(τ, t)|)‖x‖1.

From this, it follows that Tx ∈ C[a, b] for every x ∈ C[a, b] and

Tx : x ∈ L1[a, b], ‖x‖1 ≤ 1

is bounded and equi-continuous in C[a, b]. Therefore, by Arzela-Ascoli theorem3, the

operator T : L1[a, b]→ C[a, b] is compact with respect to the norms ‖ · ‖1 and ‖ · ‖∞ on

L1[a, b] and C[a, b] respectively.

Since the inclusion operators Lp[a, b] → L1[a, b] and C[a, b] → Lr[a, b] are bounded

operators for any p, r ∈ [1,∞], the operator T defined above is a compact operator from

X to Y , where X are Y any of the spaces Lp[a, b], Lr[a, b], C[a, b]. ♦

Example 2.5. Let X and Y Hilbert spaces, un : n ∈ N and vn : n ∈ N be orthonor-

mal sets in X and Y , respectively, and (λn) be a sequence of nonzero scalars such that

λn → 0 as n→∞. Let

Tx =

∞∑n=1

λn〈x, un〉vn ∀x ∈ X.

For k ∈ N, let Tk : X → X be defined by

Tkx =

k∑n=1

λn〈x, un〉vn ∀x ∈ X.

Then each Tk is a finite rank bounded operator and

‖(T − Tk)x‖2 ≤ maxn>k|λn|2‖x‖2 ∀x ∈ X.

3Arzela-Ascoli theorem: If S is a pointwise bounded and equicontinuous subset of C[a, b], then its closure,

with respect to ‖ · ‖∞, is compact.

14

Hence,

‖T − Tk‖ ≤ maxn>k|λn| → 0 as k →∞.

Thus, T is a compact operator. ♦

Example 2.6. Let A be as in Example 1.21. Then we see that

A−1x =

∞∑n=1

〈x, un〉µn

un ∀x ∈ X.

This operator is of the form T in Example 3.5, and hence A−1 is a compact operator. ♦

In fact, we shall see that every compact operator between Hilbert spaces is an operator

of the form as in Example 2.5.

Definition 2.7. Let X and Y be normed linear spaces. A linear operator T : X → Y is

called completely continuous if for every sequence (xn) in X which converges weakly

to an element x ∈ X, (Txn) converges to Tx. ♦

We may recall that (xn) converges weakly to x if and only if

f(xn)→ f(x) as n→∞ ∀ f ∈ X ′.

• If (xn) converges to x, then (xn) converges weakly; but the converse in not true.

• If (xn) converges weakly to x, then (xn) is bounded; but the converse in not true.

Therefore:

• If T is completely continuous, then T is continuous; but the converse in not true.

THEOREM 2.8. If T : X → Y is a compact operator, then it is completely continuous.

Proof. Suppose T : X → Y is a compact operator. Let (xn) in X converges weakly to

x ∈ X. We have to show that Txn → Tx. Let yn = Txn for every n ∈ N. It is enough to

show that every subsequence of (yn) has a subsequence which converges to Tx (Why?).

So, let (yn) be a subsequence of(yn), and let xn ∈ X be such that T xn = yn. Since (xn)

is bounded4 and T is a compact operator, there exists a subsequence (xn) of (xn) such

that (T xn) converges to some y ∈ Y . It is enough to show that y = Tx.

Note that, for every f ∈ Y ′, f(T xn)→ f(y). Let g = f T . Then g ∈ X ′. Since (xn)

converges weakly to x, g(xn)→ g(x); that is,

f(T xn)→ f(Tx) as n→∞.

Thus, we obtain f(Tx) = f(y) for every f ∈ Y ′. As a consequence of Hahn-Banach

extension theorem, y = Tx.

4Every weakly convergent sequence is bounded - a consequence of UBP

15

• Converse of Theorem 2.8 does not hold.

It is known that in `1, every weakly convergent sequence is convergent5. Hence, identity

operator on `1 is completely continuous, but not compact. However,

• If X is a reflexive Banach space, then every completely continuous operator from

X to Y is a compact operator.

This follows (how?) from the fact that, in a reflexive Banach space, every bounded

sequence has a weakly convergent subsequence6.

3 Spectrum of an operator

Definition 3.1. Let X be a normed linear space, A be a linear operator with D(A) and

R(A) subspaces of X. Then the resolvent set of A is the set

ρ(A) = λ ∈ K : A− λI is bijective and (A− λI)−1 ∈ B(X).

The spectrum is the set

σ(A) := K \ ρ(A).

♦

Recall that λ ∈ K is an eigenvalue of a linear operator A : X → X if there exists a

nonzero x ∈ X such that

Ax = λx.

Clearly, the eigen spectrum σeig(A), the set of all eigenvalues of A, is a subset of σ(A).

• In view of bounded inverse theorem which is a consequence of closed graph theorem,

if X is a Banach space and A ∈ C(X), then

σ(A) = λ ∈ K : A− λI is not bijective.

Definition 3.2. Let X be a normed linear spaceX0 be a subspace of X and A : X0 → X

be a linear operator. The sets

σapp(A) = λ ∈ K : A− λI is not bounded below,

and

σcom(A) = λ ∈ K : R(A− λI) is not dense

are called the approximate eigenspectrum and compression spectrum, respec-

tively. ♦

5This result is known as Schur’s theorem.6This fact is known as Eberlein Smulian theorem

16

• λ ∈ σapp(A) if and only if there exists a sequence (xn) in D(A) such that ‖xn‖ = 1

for every n ∈ N and ‖Axn − λxn‖ → 0.

In view of Theorem 1.17, we have the following.

THEOREM 3.3. Let X be a Banach space and A ∈ C(X) and λ ∈ K. Then

σ(A) = σapp(A) ∪ σcom(A),

Note that

σeig(A) ⊆ σapp(A).

THEOREM 3.4. Let A : D(A) ⊆ X → X be a linear operator. Then σapp(A) is a

closed set.

Proof. Let (λn) be a sequence in σapp(A) such that λn → λ for some λ ∈ K. Assume for

a moment that λ 6∈ σapp(A). Then there exists c > 0 such that ‖Ax − λx‖ ≥ c‖x‖ for

every x ∈ D(A). Then, for every x ∈ D(A),

‖Ax− λnx‖ = ‖(A− λx)− (λn − λ)x‖ ≥ ‖A− λx‖ − |λn − λ| ‖x‖ ≥ c− |λn − λ|)‖x‖.

If N ∈ N is such that |λn − λ| ≤ c/2, then we obtain

‖Ax− λNx‖ ≥c

2‖x‖ ∀x ∈ D(A).

This contradicts the fact that λN ∈ σapp(A).

Example 3.5. Let X be a Hilbert space, un : n ∈ N be an orthonormal set in X and

(λn) be a sequence of nonzero scalars. Let

Ax =

∞∑n=1

λn〈x, un〉un ∀x ∈ X0,

where

X0 = x ∈ X :

∞∑n=1

|λn|2|〈x, un〉|2 <∞.

We show that

σeig(A) = λn : n ∈ N and σ(A) = clλn : n ∈ N = σapp(A).

Note that

Aun = λnun ∀n ∈ N

so that λn : n ∈ N ⊆ σeig(A), and for x ∈ X0,

Ax = 0 ⇐⇒ x ∈ un : n ∈ N⊥

so that N(A) = un : n ∈ N⊥. Note that for x ∈ X0 ∩N(A)⊥,

〈x, un〉 = 0 ∀n ∈ N=⇒x = 0

17

so that un : n ∈ N is a orthonormal basis for X0 ∩ N(A)⊥. Now, let x ∈ X0. By

projection theorem7, there a exists unique pair u, v) with u ∈ N(A)⊥, v ∈ N(A) such

that x = u+ v. Then, u ∈ X0 and 〈x, un〉 = 〈u, un〉 for all n ∈ N so that

u =

∞∑n=1

〈u, un〉un =

∞∑n=1

〈x, un〉un.

Hence, for λ ∈ K and

Ax− λx = (Au− λu) + (Av − λv) = (Au− λu)− λv,

where

Au− λu =

∞∑n=1

(λn − λ)〈x, un〉un.

Thus, for a nonzero x ∈ X0,

Ax = λx ⇐⇒ λ ∈ λn : n ∈ N.

Also, for λ 6∈ clλn : n ∈ N,

‖Ax− λx‖2 = ‖Au− λu‖2 + ‖λv‖2 ≥ ‖Au− λu‖2 =

∞∑n=1

|λn − λ|2|〈x, un〉|2 ≥ d2‖x‖2,

where 0 < d ≤ |λ − λn| for all n ∈ N. Thus, A − λI is bounded below. Also, for every

y ∈ X, if we take

x =

∞∑n=1

〈y, un〉λn − λ

un,

then we have∞∑n=1

|〈y, un〉|2

|λn − λ|2≤ ‖y‖

2

d2

so that x ∈ X0 and Ax−λx = y. Thus, A−λI is onto as well. Hence, A−λI is bijective

with A− λI)−1 ∈ B(X) so that λ ∈ ρ(A). Thus, we obtain

σ(A) ⊆ clλn : n ∈ N ⊆ σapp(A).

Thus, in this case, we have

σeig(A) = λn : n ∈ N and σ(A) = clλn : n ∈ N = σapp(A).

♦

Remark 3.6. In the above example:

• If (λn) is a sequence in R, then A is self adjoint.

• If (λn) is a bounded sequence, then A ∈ B(X) and A is a normal operator, i.e.,

A∗A = AA∗.

7Projection theorem: If Y is a closed subspace of a Hilbert space X, then X = Y + Y ⊥.

18

• If λn → 0, then A ∈ K(X).

• If (λn) is an unbounded sequence, then A is an unbounded operator, not necessarily

a closed operator.

♦

THEOREM 3.7. Let X be a Hilbert space and A ∈ C(X) be densely defined. Then

(i) σ(A) = σapp(A) ∪ λ : λ ∈ σeig(A∗),

(ii) σapp(A) ⊆ cl〈Ax, x〉 : x ∈ D(A), ‖x‖ = 1.

In particular, if A is self adjoint, then

σ(A) = σapp(A) ⊆ R.

Proof. (i) We have already observed that σ(A) = σapp(A) ∪ σcom(A). Hence, we have

σ(A) = σapp(A) ∪ λ : λ ∈ σeig(A∗).(ii) Let λ ∈ σapp(A), and let xn ∈ X be such that ‖xn‖ = 1 for all n ∈ N and

‖Axn − λxn‖ → 0. Then we have

|〈Axn, xn〉 − λ| = |〈Axn − λxn, xn〉| ≤ ‖Axn − λxn‖ → 0.

Thus, λ ∈ cl〈Ax, x〉 : x ∈ D(A), ‖x‖ = 1 ⊆ R.

The particular case is obvious.

THEOREM 3.8. Let X be a Banach space and A ∈ C(X). Then σ(A) is a closed set.

For the proof of the above theorem we shall make use of the following proposition.

PROPOSITION 3.9. Let X be a Banach space and A ∈ B(X). If ‖A‖ < 1, then I−Ais bijective.

Proof. Suppose ‖A‖ < 1. Then, for every x ∈ X,

‖x−Ax‖ ≥ (1− ‖A‖)‖x‖

so that I −A is bounded below. In particular, I −A is injective and R(I −A) is closed.

Assume for a moment that R((I − A) 6= X. Then, by a consequence of Hahn-Banach

extension theorem, there exists a continuous linear functional f on X such that ‖f‖ = 1

and

f(x−Ax) = 0 ∀x ∈ X.

Hence,

|f(x)| = ‖f(Ax)‖ ≤ ‖f‖ ‖A‖ ‖x‖ = ‖A‖ ‖x‖ ∀x ∈ X.

Thus, ‖f‖ ≤ ‖A‖ < 1, which is a contradiction to the fact that ‖f‖ = 1.

19

Proof of Theorem 3.8. It is enough to prove that ρ(A) is an open set. For this, let

λ ∈ ρ(A). Note that , for any µ ∈ K,

A− µI − (A− λI)− (µ− λ) = [I − (µ− λ)(A− λI)−1](A− λI). (∗)

Now, suppose that |µ − λ| < 1/‖(A − λI)−1‖. Then by Proposition 3.9, the operator

I − (µ− λ)(A− λI)−1 is bijective. Hence, from the relation (∗), it follows that A− µI is

bijective. Thus,

µ ∈ K : |µ− λ| < 1/‖(A− λI)−1‖ ⊆ ρ(A).

This is true for every λ ∈ ρ(A). hence, ρ(A) is an open set.

Remark 3.10. For a linear operator,

• the spectrum can be empty,

• the spectrum can be the whole of the scalar field.

♦

THEOREM 3.11. Let A : X → X be a self adjoint operator on a Hilbert space. Then

(i) A is a bounded operator,

(ii) ∃λ ∈ σ(A) such that |λ| = ‖A‖.

3.1 Spectral results for compact operators

Let X be a normed linear space. As a consequence of Theorem 2.2, we have the following.

THEOREM 3.12. If X is infinite dimensional and A ∈ K(X), then 0 ∈ σapp(A).

THEOREM 3.13. Let A ∈ K(X). Then the following hold.

(i) σapp(A) \ 0 = σeig(A) \ 0,

(ii) N(A− λI) is finite dimensional for every nonzero λ ∈ σeig(A).

(iii) 0 is the only possible limit point of σeig(A), and σeig(A) is a countable set,

Proof. (i) Its enough to prove that σapp(A)\0 ⊆ σeig(A)\0. So, let λ ∈ σapp(A)\0and let (xn) in X such that ‖xn‖ = 1 for every n ∈ N and ‖Axn − λxn‖ → 0 as n→∞.

Since A is a compact operator, there exists a subsequence (xkn) such that Axkn → y for

some y ∈ X. Hence,

xkn =1

λ[Axkn − (Axkn − λxkn)]→ y

λ

so that we have

Axkn →Ay

λ.

Since Axkn → y, we obtain Ay = λy. Also, y 6= 0, since ‖xkn‖ = 1 and λ 6= 0. Thus, we

have proved that λ ∈ σeig(A).

20

(ii) Let λ be a nonzero eigenvalue of A. Suppose N(A − λI) is infinite dimensional.

Let xn : n ∈ N be a linearly independent subset of N(A − λI). Let X0 = 0 and for

n ∈ N, let

Xn := spanx1, . . . , xn.

Since xn : n ∈ N is linearly independent, by Riesz lemma8, for each n ∈ N, there exists

un ∈ Xn such that

‖un‖ = 1 and dist (un, Xn−1) ≥ 1

2.

Note that for n,m ∈ N with n > m, um ∈ Xn−1 and

‖Aun −Aum‖ = ‖λ(un − um)‖ ≥ |λ|2.

Thus, (Aun) does not have a Cauchy subsequence so that (Aun) does not have a conver-

gent subsequence.

(iii) For r > 0, let

∆r := λ ∈ σeig(A) : |λ| ≥ r.

It is enough to prove that prove that ∆r is a finite set. Assume for a moment that ∆r

is an infinite set for some r > 0. Let (λn) be a sequence of distinct elements from ∆r.

For each n ∈ N, let xn be an eigen vector of A corresponding to the eigenvalue λn. Let

X0 = 0 and for n ∈ N, let

Xn := spanx1, . . . , xn.

Since xn : n ∈ N is linearly independent, by Riesz lemma, for each n ∈ N, there exists

un ∈ Xn such that

‖un‖ = 1 and dist (un, Xn−1) ≥ 1

2.

Note that for n,m ∈ N with n > m,

Aun −Aum = λnun + (Aun − λnun)−Aum,

where

Aun − λnun ∈ Xn−1 and Aum ∈ Xn−1.

Hence,

‖Aun −Aum‖ ≥ dist (λnun, Xn−1) = |λn|dist (un, Xn−1) ≥ r

2.

Thus, (Aun) does not have a Cauchy subsequence so that (Aun) does not have a conver-

gent subsequence.

8Riesz lemma: If Y is a closed proper subspace of a normed linear spaceX and 0 < r < 1, then there exists

x ∈ X \ Y such that ‖x‖ = 1 and dist (x, Y ) ≥ r.

21

4 Spectral representations

4.1 For compact self adjoint operators

Recall from Theorem 3.13 that if A is a compact operator, then its

1. eigen spectrum is countable,

2. 0 is the only possible limit point of he eigen spectrum, and

3. eigenspace associated with every nonzero eigenvalue is finite dimensional.

We shall make use of the following lemma.

LEMMA 4.1. Let X be a Hilbert space and A ∈ B(X) be a self adjoint operator. If X0

is a closed subspace such that A(X0) ⊆ X0, then A(X⊥0 ) ⊆ X⊥0 .

THEOREM 4.2. (Spectral representation) Let X be a Hilbert space and A : X → X

be a nonzero compact self adjoint operator. Let λj : j ∈ Λ be the set of all (distinct)

nonzero eigenvalues of A, where Λ = 1, . . . , n for some n ∈ N if σeig(A) is a finite set

and Λ = N if σeig(A) is an infinite set. Then

Ax =∑i∈Λ

λi

ni∑j=1

〈x, vij〉vij ,

where vij : j = 1, . . . , ni is an orthonormal basis of N(A− λiI) for each i ∈ Λ.

Proof. Let X0 = cl span⋃i∈Λ

vij : j = 1, . . . , ni. Clearly,⋃i∈Λ

vij : j = 1, . . . , ni is an

orthonormal basis for the closed subspace X0. Let x ∈ X. Then by projection theorem,

x can be written uniquely as

x = u+ v with u ∈ X0, v ∈ X⊥0 .

Note that

〈x, vij〉 = 〈u, vij〉 ∀ i ∈ Λ, j = 1, . . . , ni,

so that

u =∑i∈Λ

ni∑j=1

〈u, vij〉vij =∑i∈Λ

ni∑j=1

〈x, vij〉vij .

Thus,

Ax = Au+Av =∑i∈Λ

λi

ni∑j=1

〈x, vij〉vij +Av.

Hence, it is enough to prove that Av = 0. By Lemma 4.1, X⊥0 is invariant under A.

Therefore, B := A|X⊥

0

is a compact self adjoint operator on X⊥0 . We, show that B = 0,

which would imply that Av = 0.

Suppose B 6= 0. Then B will have a nonzero eigenvalue, say λ, which would be an

eigenvalue of A as well. Let y ∈ X⊥0 be a corresponding eigenvector. Then we have

Ay = λy so that the λ = λi for some i ∈ Λ, and hence y ∈ N(A − λiI) ⊆ X0. This is a

contradiction to the fact that y 6= 0.

22

THEOREM 4.3. Under the assumptions of Theorem 4.2

A =∑i∈Λ

λiPi,

where Pi is the orthogonal projection defined by

Pix =

ni∑j=1

〈x, vij〉vij , i ∈ Λ.

Proof. It is enough to prove the theorem for the case Λ = N. In this case, by Theorem

4.2,

Ax =

∞∑i=1

λi

ni∑j=1

〈x, vij〉vij ,

for all x ∈ X. Hence,

‖Ax−n∑i=1

λiPix‖2 =

∞∑i=n+1

|λi|2ni∑j=1

|〈x, vij〉|2,

Thus,

‖Ax−n∑i=1

λiPix‖2 ≤ (maxi>n|λi|2)‖x‖2 ∀x ∈ X.

Thus,

‖A−n∑i=1

λiPi‖ ≤ maxi>n|λi|.

Since |λn| → 0 as n→∞, it follows that

A =

∞∑i=1

λiPi.

COROLLARY 4.4. If A is a compact self adjoint operator on a Hilbert space, then

every nonzero spectral value of A is an eigenvalue of A.

THEOREM 4.5. (Singular value representation) Let X and Y Hilbert spaces and

T : X → Y be a compact operator of infinite rank. Then there exists an orthonormal basis

un : n ∈ N for N(T )⊥, an orthonormal basis vn : n ∈ N for R(T ) and a sequence

(sn) of positive real numbers with sn → 0 as n→∞ such that

Tx =

∞∑n=1

sn〈x, un〉vn ∀x ∈ X.

Proof. Note that T ∗T is a compact self adjoint operator so that by Th-spect-cpt, there

exists an orthonormal set un : n ∈ N in X and a sequence (µn) of nonzero real numbers

such that µn → 0 and

T ∗Tx =

∞∑n=1

µn〈x, un〉un, x ∈ X.

23

Note that

T ∗Tun = µnun ∀n ∈ N.

Hence,

µn = 〈µnun, un〉 = 〈T ∗Tun, un〉 = 〈Tun, Tun〉 = ‖Tun‖2 ≥ 0.

Taking

sn =õn and vn =

Tu

sn∀n ∈ N,

we have

Tun = snvn and T ∗vn = snun ∀n ∈ N.

We also know that

〈x, un〉 = 0 ∀n ∈ N=⇒T ∗Tx = 0 =⇒x ∈ N(T ∗T ) = N(T )

so that un : n ∈ N is an orthonormal basis of N(T )⊥. Also, for every x ∈ X,

〈Tx, vn〉 = 〈x, T ∗vn〉 = 〈x, snun〉 = sn〈x, un〉

so that

〈Tx, vn〉 = 0 ∀n ∈ N=⇒〈x, un〉 = 0 ∀n ∈ N=⇒x ∈ N(T ) =⇒Tx = 0.

Hence, vn : n ∈ N is an orthonormal basis of R(T ). Therefore, for every x ∈ X,

Tx =

∞∑n=1

〈Tx, vn〉vn =

∞∑n=1

〈x, T ∗vn〉vn =

∞∑n=1

sn〈x, un〉vn.

This completes the proof.

If sn, un, vn are as in the above theorem, hen we have

Tun = snvn, T ∗vn = snun.

COROLLARY 4.6. Let X and Y Hilbert spaces and T : X → Y be a compact operator.

Then there exists sequence of finite rank bounded operators Tn such that ‖T − Tn‖ → 0

as n→∞. In fact,

Tnx :=

n∑j=1

sj〈x, uj〉vj ∀x ∈ X,

where (sn, un, vn) : n ∈ N is as in Theorem 4.5, and

‖T − Tn‖ ≤ supj>n

sj .

24

4.2 For self adjoint operators

THEOREM 4.7. Let A ∈ B(X) be a self adjoint operator on a Hilbert space X and

a = inf W (A), b = supW (A). Then there exists a family Eλ : a ≤ λ ≤ b of orthogonal

projections such that

(i) Eλ = 0 ∀λ < a and Eλ = I ∀λ > b,

(ii) λ ≤ µ =⇒ 〈Eλx, x〉 ≤ 〈Eµx, x〉 ∀x ∈ X,

(iii) limλ→λ+

Eλx = Eλx ∀x ∈ X,

and

〈Ax, x〉 =

∫ b

a

λd〈Eλx, x〉 ∀x ∈ X,

where the integral is in the sense of Riemann-Stieltjes.

Remark 4.8. A family of orthogonal projections as in Theorem 4.7 is called a resolution

of identity on [a, b].

The integral in Theorem 4.7 is usually written as

A =

∫ b

a

λdEλ

which is also meaningful in the sense that the resolution of identity Eλ : a ≤ λ ≤ bgives rise to a measure Ω 7→ E(Ω) on all Borel subsets of [a, b]. ♦

THEOREM 4.9. Let A be a densely defined self adjoint operator in a Hilbert space X.

Then there exists a family Eλ : λ ∈ R of orthogonal projections such that

(i) limλ→−∞

Eλx = 0 and limλ→∞

Eλx = x ∀x ∈ X,

(ii) λ ≤ µ =⇒ 〈Eλx, x〉 ≤ 〈Eµx, x〉 ∀x ∈ X,

(iii) limλ→λ+

Eλx = Eλx ∀x ∈ X,

and

〈Ax, x〉 =

∫ ∞−∞

λd〈Eλx, x〉 ∀x ∈ D(A),

where the integral is in the sense of improper Riemann-Stieltjes, and

D(A) = x ∈ X :

∫ ∞−∞|λ|2d |〈Eλx, x〉|2 <∞,

Remark 4.10. The integral in Theorem 4.9 is same as A =

∫ ∞−∞

λdEλ, where Ω 7→ E(Ω)

is the Borel measure on R induced by the resolution of identity Eλ : λ ∈ R. ♦

References

[1] Nair, M. T. (2002): Functional Analysis: A First Course, Printice-Hall of India

(Third print: PHI learning, 2010), New Delhi: .

25

5 Problems

1. Verify the equivalence of statements in Theorem 1.2.

2. Prove Theorems 1.3 and 1.4.

3. Let T : X → Y be a linear operator. Prove that T is injective and T−1 : R(T )→ X

is continuous if and only if T is bounded below.

4. Let T : C[a, b]→ C[a, b] be the identity operator, i.e., Tx = x for every x ∈ C[a, b].

(a) Let X = C[a, b] with ‖·‖∞ and Y = C[a, b] with ‖·‖1. Show that T ∈ B(X,Y ),

but T 6∈ B(Y,X).

(b) Let X = C[a, b] with ‖ · ‖2 and Y = C[a, b] with ‖ · ‖1. Show that T ∈ B(X,Y ),

but T 6∈ B(Y,X).

(c) For 1 ≤ p < r ≤ ∞, let Xp = C[a, b] with ‖ · ‖p and Xr = C[a, b] with ‖ · ‖r.Show that T ∈ B(Xr, Xp), but T 6∈ B(Xp, Xr).

5. Let X and Y be Hilbert spaces and A : X → Y be a linear operator.

(a) Prove that, if there exists β > 0 such that ‖〈Ax, y〉| ≤ β‖x‖ ‖y‖ for all x ∈ Xand y ∈ Y , then A ∈ B(X,Y ).

(b) Prove that, if Y 6= 0 and there exists γ > 0 such that ‖〈Ax, y〉| ≥ γ‖x‖ ‖y‖ for

all x ∈ X and y ∈ Y , then A is bounded below.

(c) Prove that, if Y = X 6= 0 and there exists γ > 0 such that |〈Ax, x〉| ≥ γ‖x‖2

for all x ∈ X, then A is bounded below.

6. Let k(·, ·) be continuous on [a, b]× a, b], p, r ∈ [1,∞] and let

(Tx)(s) =

∫ b

a

k(s, t)x(t)dt, x ∈ L1[a, b].

Prove that, Tx ∈ Lr[a, b] for every x ∈ Lp[a, b] and T : Lp[a, b] → Lr[a, b] is a

bounded operator.

7. Prove that a bounded operator with a closed domain is a closed operator.

8. Prove Theorem 1.12.

9. Supply proof for Theorem 1.17.

10. Prove Theorem 1.25.

11. Let X and Y be a Hilbert spaces and T : D(T ) ⊆ X → Y be a linear operator with

dense domain D(T ). Justify the following statements:

(a) If T is a liner operator from X to Y with D(T ) dense in X, then

D(T ∗) = y ∈ Y : ∃ a unqiue zy ∈ X such that 〈Tx, y〉 = 〈x, zy〉 ∀x ∈ D(T ).

(b) If T ∈ C(X,Y ) and D(T ) dense in X, then T ∗ ∈ C(Y,X) and D(T ∗) is dense

in Y .

26

(c) If T ∈ B(X,Y ) so that D(T ) = X, then D(T ∗) = Y and T ∗ ∈ B(Y,X), and in

that case ‖T ∗‖ = ‖T‖ and ‖T ∗T‖ = ‖T‖2.

12. Let X be a separable Hilbert space, un : n ∈ N is an orthonormal basis of X, and

(µn) is a sequence of positive real numbers such that µn →∞ as n→∞. Let A be

defined by

Ax :=

∞∑n=1

µn〈x, un〉un, ∀x ∈ X0,

where

X0 :=x ∈ X :

∞∑n=1

µ2n|〈x, un〉|2 <∞

.

Show that A is a self adjoint operator.

13. Prove the following:

(a) Every compact operator is a bounded operator.

(b) Every finite rank bounded operator is a compact operator.

(c) Identity operator on a normed linear space is a compact operator if and only if

the space is finite dimensional.

(d) If Y is a Banach space, the norm limit of a sequence of compact operators from

X to Y is a compact operator.

14. Let T be defined as in Problem 6. Show that T : Lp[a, b] → Lr[a, b] is a compact

operator for every p, r ∈ [1,∞].

15. Let X = x ∈ C1[0, 1] : x(0) = 0 with ‖x‖X := ‖x′‖∞ and Y = C[a, b] with

‖x‖Y := ‖x‖∞. Prove that the inclusion operator from X to Y is a compact

operator.

16. Justify the following statements:

(a) If (xn) converges to x, then (xn) converges weakly; but the converse in not

true.

(b) If (xn) converges weakly to x, then (xn) is bounded; but the converse in not

true.

(c) If T is completely continuous, then T is continuous; but the converse in not

true.

17. let A be as in Example 3.5. Prove the following.

(a) If (λn) is a sequence in R, then A is self adjoint.

(b) If (λn) is a bounded sequence, then A ∈ B(X) and A is a normal operator, i.e.,

A∗A = AA∗.

(c) If λn → 0, then A ∈ K(X).

(d) If (λn) is an unbounded sequence, then A is an unbounded operator, not nec-

essarily a closed operator.

27

18. Let A : X → X be a self adjoint operator on a Hilbert space. Then prove that A is

a bounded operator.

19. Let X be a Hilbert space and A ∈ B(X) be a self adjoint operator. Prove the

following.

(a) If X0 is a closed subspace such that A(X0) ⊆ X0, then A(X⊥0 ) ⊆ X⊥0 .

(b) If λ1, . . . , λn ⊆ σeig(A) and X0 = N(A − λ1I) + · · · + N(A − λnI), then

A(X0) ⊆ X0 and

σeig(A1) = λ1, . . . , λn and σeig(A2) = σeig(A) \ λ1, . . . , λn,

where A1 and A2 are restrictions of A to X0 and X⊥0 , respectively.

28