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CHAPTER 2. OPERATORS ON HILBERT SPACES

CHRISTOPHER HEIL

1. Elementary Properties and Examples First recall the basic definitions regarding operators.

Definition 1.1 (Continuous and Bounded Operators). Let X, Y be normed linear spaces, and let L : X → Y be a linear operator.

(a) L is continuous at a point f ∈ X if fn → f in X implies Lfn → Lf in Y .

(b) L is continuous if it is continuous at every point, i.e., if fn → f in X implies Lfn → Lf in Y for every f .

(c) L is bounded if there exists a finite K ≥ 0 such that

∀ f ∈ X, kLfk ≤ K kfk.

Note that kLfk is the norm of Lf in Y , while kfk is the norm of f in X.

(d) The operator norm of L is

kLk = sup

kf k=1kLfk.

(e) We let B(X, Y ) denote the set of all bounded linear operators mapping X into Y , i.e.,

B(X, Y ) = {L: X → Y : L is bounded and linear}.

If X = Y = X then we write B(X) = B(X, X).

(f) If Y = F then we say that L is a functional. The set of all bounded linear functionals on X is the dual space of X, and is denoted

X0 = B(X, F) = {L: X → F : L is bounded and linear}.

We saw in Chapter 1 that, for a linear operator, boundedness and continuity are equivalent.

Further, the operator norm is a norm on the space B(X, Y ) of all bounded linear operators from X to Y , and we have the composition property that if L∈ B(X, Y ) and K ∈ B(Y, Z), then KL∈ B(X, Z), with kKLk ≤ kKk kLk.

Date: February 20, 2006.

These notes closely follow and expand on the text by John B. Conway, “A Course in Functional Analysis,”

Second Edition, Springer, 1990.

1

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Exercise 1.2. Suppose that L : X → Y is a bounded map of a Banach space X into a Banach space Y . Prove that if there exists a c > 0 such that kLfk ≥ c kfk for every f ∈ X, then range(L) is a closed subspace of Y .

Exercise 1.3. Let Cb(Rn) be the set of all bounded, continuous functions f : Rn → F. Let C0(Rn) be the set of all continuous functions f : Rn → F such that lim|x|→∞f (x) = 0 (i.e., for every ε > 0 there exists a compact set K such that |f(x)| < ε for all x /∈ K). Prove that these are closed subspaces of L(Rn) (under the L-norm; note that for a continuous function we have kfk= sup|f(x)|).

Define δ : Cb(Rn)→ F by

δ(f ) = f (0).

Prove that δ is a bounded linear functional on Cb(Rn), i.e., δ ∈ (Cb)0, and find kδk. This linear functional is the delta distribution (see also Exercise 1.26 below).

Example 1.4. In finite dimensions, all linear operators are given by matrices, this is just standard finite-dimensional linear algebra.

Suppose that X is an n-dimensional complex normed vector space and Y is an m- dimensional complex normed vector space. By definition of dimension, this means that there exists a basisBX ={x1, . . . , xn} for X and a basis BY ={y1, . . . , ym} for Y . If x ∈ X, then x = c1x1+· · · + cnxn for a unique choice of scalars ci. Define the coordinates of x with respect to the basis BX to be

[x]BX =

 c1

...

cn

 ∈ Cn.

The vector x is completely determined by its coordinates, and conversely each vector in Cn is the coordinates of a unique x∈ X. The mapping x 7→ [x]BX is a linear mapping of X onto Cn. We similarly define [y]BY ∈ Cm for vectors y ∈ Y .

Let A : X → Y be a linear map (it is automatically bounded since X is finite-dimensional).

Then A transforms vectors x ∈ X into vectors Ax ∈ Y . The vector x is determined by its coordinates [x]BX and likewise Ax is determined by its coordinates [Ax]BY. The vectors x and Ax are related through the linear map A; we will show that the coordinate vectors [x]BX

and [Ax]BY are related by multiplication by an m× n matrix determined by A. We call this matrix the standard matrix of A with respect toBX andBY, and denote it by [A]BX,BY. That is, the standard matrix should satisfy

[Ax]BY = [A]BX,BY [x]BX, x∈ X.

We claim that the standard matrix is the matrix whose columns are the coordinates of the vectors Axk, i.e.,

[A]BX,BY =

 [Ax1]BY · · · [Axn]BY

.

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To see this, choose any x ∈ X and let x = c1x1 +· · · + cnxn be its unique representation with respect to the basis BX. Then

[A]BX,BY [x]BX =

 [Ax1]BY · · · [Axn]BY

 c1

...

cn

= c1[Ax1]BY +· · · + cn[Axn]BY

= [c1Ax1+· · · + cnAxn]BY

= [A(c1x1+· · · + cnxn)]BY

= [Ax]BY.

Exercise 1.5. Extend the idea of the preceding example to show that that any linear mapping L : `2(N) → `2(N) (and more generally, L : H → K with H, K separable) can be realized in terms of multiplication by an (infinite but countable) matrix.

Exercise 1.6. Let A be an m× n complex matrix, which we view as a linear transformation A : Cn → Cm. The operator norm of A depends on the choice of norm for Cn and Cm. Compute an explicit formula forkAk, in terms of the entries of A, when the norm on Cn and Cm is taken to be the `1 norm. Then do the same for the ` norm. Compare your results to the version of Schur’s Lemma given in Theorem 1.23.

The following example is one that we will return to many times.

Example 1.7. Let {en}n∈N be an orthonormal basis for a separable Hilbert space H. Then we know that every f ∈ H can be written

f =

X

n=1

hf, eni en. Fix any sequence of scalars λ = (λn)n∈N, and formally define

Lf =

X

n=1

λnhf, eni en. (1.1)

This is a “formal” definition because we do not know a priori that the series above will converge—in other words, equation (1.1) may not make sense for every f .

Note that if H = `2(N) and{en}n∈N is the standard basis, then L is given by the formula Lx = (λ1x1, λ2x2, . . . ), x = (x1, x2, . . . ) ∈ `2(N).

We will show the following (the `-norm of the sequence λ is kλk= supnn|).

(a) The series defining Lf in (1.1) converges for each f ∈ H if and only if λ ∈ `. In this case L is a bounded linear mapping of H into itself, andkLk = kλk.

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(b) If λ /∈ `, then L defines an unbounded linear mapping from the domain domain(L) = n

f ∈ H :

X

n=1

nhf, eni|2 <∞o

(1.2) (which is dense in H) into H.

Proof. (a) Suppose that λ∈ `, i.e., λ is a bounded sequence. Then for any f we have

X

n=1

nhf, eni|2

X

n=1

kλk2hf, eni|2 = kλk2kfk2 < ∞,

so the series defining Lf converges (because {en} is an orthonormal sequence). Moreover, the preceding calculation also shows that kLfk2 = P

n=1nhf, eni|2 ≤ kλk2kfk2, so we see thatkLk ≤ kλk. On the other hand, by orthonormality we have Len= λnen (i.e., each en is an eigenvector for L with eigenvalue λn). Since kenk = 1 and kLenk = |λn| kenk = |λn| we conclude that

kLk = sup

kf k=1kLfk ≥ sup

n∈NkLenk = sup

n∈Nn| = kλk. The converse direction will be covered by the proof of part (b).

(b) Suppose that λ /∈ `, i.e., λ is not a bounded sequence. Then we can find a subsequence (λnk)k∈N such that |λnk| ≥ k for each k. Let cnk = k1 and define all other cnto be zero. Then P

n|cn|2 =P

k 1

k2 <∞, so f =P

ncnen converges (and cn=hf, eni). But the formal series Lf =P

nλncnen does not converge, because

X

n=1

|cnλn|2 =

X

k=1

|cnkλnk|2

X

k=1

k2

k2 = ∞.

In fact, the series defining Lf in (1.1) only converges for those f which lie in the domain defined in (1.2). That domain is dense because it contains the finite span of {en}n∈N, which we know is dense in H. Further, that domain is a subspace of H (exercise), so it is an inner- product space. The map L : domain(L)→ H is a well-defined, linear map, so it remains only to show that it is unbounded. This follows from the facts that en ∈ domain(L), kenk = 1,

and kLenk = |λn| kenk = |λn|. 

Exercise 1.8. Continuing Example 1.7, suppose that λ ∈ ` and set δ = infnn|. Prove the following.

(a) L is injective if and only if λn6= 0 for every n.

(b) L is surjective if and only if δ > 0 (if δ = 0, use an argument similar to the one used in part (b) of Example 1.7 to show that range(L) is a proper subset of H).

(c) If δ = 0 but λn 6= 0 for every n then range(L) is a dense but proper subspace of H.

(d) Prove that L is unitary if and only if|λn| = 1 for every n.

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In Example 1.7, we saw an unbounded operator whose domain was a dense but proper subspace of H. This situation is typical for unbounded operators, and we often write L : X → Y even when L is only defined on a subset of X, as in the following example.

Example 1.9 (Differentiation). Consider the Hilbert space H = L2(0, 1), and define an operator D : L2(0, 1) → L2(0, 1) by Df = f0. Implicitly, we mean by this that D is defined on the largest domain that makes sense, namely,

domain(D) = f ∈ L2(0, 1) : f is differentiable and f0 ∈ L2(0, 1) .

Note that if f ∈ domain(D), then Df is well-defined, Df ∈ L2(0, 1), and kDfk2 < ∞.

Thus every vector in domain(D) maps to a vector in L2(0, 1) which necessarily has finite norm. Yet D is unbounded. For example, if we set en(x) = einx then kenk2 = 1, but Den(x) = e0n(x) = ineinx sokDenk2 = n. While each vector Den has finite norm, there is no upper bound to these norms. Since the en are unit vectors, we conclude that kDk = ∞.

The following definitions recall the basic notions of measures and measure spaces. For full details, consult a book on real analysis.1

Definition 1.10 (σ-Algebras, Measurable Sets and Functions). Let X be a set, and let Ω be a collection of subsets of X. Then Ω is a σ-algebra if

(a) X ∈ Ω,

(b) If E ∈ Ω then X \ E ∈ Ω (i.e., Ω is closed under complements),

(c) If E1, E2,· · · ∈ Ω then S Ek ∈ Ω (i.e., Ω is closed under countable unions) The elements of Ω are called the measurable subsets of X.

If we choose F = R then we usually allow functions on X to take extended-real values, i.e., f (x) is allowed to take the values ±∞. An extended-real-valued function f : X → [−∞, ∞]

is called a measurable function if {x ∈ X : f(x) > a} is measurable for each a ∈ R.

If we choose F = C then we require functions on X to take (finite) complex values—there is no complex analogue of±∞. A complex-valued function f : X → C is called a measurable function if its real and imaginary parts are measurable (real-valued) functions.

Definition 1.11 (Measure Space). Let X be a set and Ω a σ-algebra of subsets of X. Then a function µ on Ω is a (positive) measure if

(a) 0≤ µ(E) ≤ +∞ for all E ∈ Ω,

(b) If E1, E2, . . . is a countable family of disjoint sets in Ω, then µ

S

k=1

Ek



=

X

k=1

µ(Ek).

1For example, R. Wheeden and A. Zygmund, “Measure and Integral,” Marcel Dekker, 1977, or G. Folland,

“Real Analysis,” Second Edition, Wiley, 1999.

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In this case, (X, Ω, µ) is called a measure space.

If µ(X) <∞, then we say that µ is a finite measure.

If there exist countably many subsets E1, E2, . . . such that X =S Ek and µ(Ek) <∞ for all k, then we say that µ is σ-finite. For example, Lebesgue measure on Rn is σ-finite.

It is often useful to allow measures to take negative values.

Definition 1.12(Signed Measure). Let X be a set and Ω a σ-algebra of subsets of X. Then a function µ on Ω is a signed measure if

(a) −∞ ≤ µ(E) ≤ +∞ for all E ∈ Ω and µ(∅) = 0,

(b) If E1, E2, . . . is a countable family of disjoint sets in Ω, then µ

S

k=1

Ek

 =

X

k=1

µ(Ek).

Definition 1.13 (Integration). Let (X, Ω, µ) be a measure space.

(a) If f : X → [0, ∞] is a nonnegative, measurable function, then the integral of f over X with respect to µ is

Z

X

f dµ = Z

X

f (x) dµ(x) = sup

 X

j x∈Einfj

f (x) µ(Ej)

 ,

where the supremum is taken over all decompositions E = E1∪ · · · ∪ EN of E as the union of a finite number of disjoint measurable sets Ek (and where we take the convention that

∞ · 0 = 0 · ∞ = 0).

(b) If f : X → [−∞, ∞] and we define

f+(x) = max{f(x), 0}, f(x) = − min{f(x), 0},

then Z

X

f dµ = Z

X

f+dµ − Z

X

fdµ,

as long as this does not have the form∞ − ∞ (in that case the integral would be undefined).

Since|f| = f++ f andR

X |f| dµ always exists (either as a finite number or as ∞), it follows

that Z

X

f dµ exists and is finite ⇐⇒

Z

X|f| dµ < ∞.

(c) If f : X → C, then Z

X

f dµ = Z

X

Re (f ) dµ + i Z

X

Im (f ) dµ, as long as both integrals on the right are defined and finite.

There are many other equivalent definitions of the integral.

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Definition 1.14 (Lp Spaces). Let (X, Ω, µ) be a measure space, and fix 1≤ p < ∞. Then Lp(X) consists of all measurable functions f : X → [−∞, ∞] (if we choose F = R) or f : X → C (if we choose F = C) such that

kfkpp = Z

X|f(x)|pdµ(x) < ∞.

Then Lp(X) is a vector space under the operations of addition of functions and multiplication of a function by a scalar. Additionally, the function k · kp defines a semi-norm on Lp(X).

Usually we identify functions that are equal almost everywhere (we say that f = g a.e. if µ{x ∈ X : f(x) 6= g(x)} = 0), and then k · k becomes a norm on Lp(X).

For p = ∞ we define L(X) to be the set of measurable functions that are essentially bounded, i.e., for which there exists a finite constant M such that|f(x)| ≤ M a.e. Then

kfk = ess sup

x∈X |f(x)| = infM ≥ 0 : |f(x)| ≤ M a.e.

is a semi-norm on L(X), and is a norm if we identify functions that are equal almost everywhere.

For each 1≤ p ≤ ∞, the space Lp(X) is a Banach space under the above norm.

Exercise 1.15 (`p Spaces). Counting measure on a set X is defined by µ(X) = card(E) if E is a finite subset of X, and µ(X) =∞ if E is an infinite subset. Let Ω = P(X) (the set of all subsets of X), and show that (X, Ω, µ) is a measure space. Show that Lp(X, Ω, µ) = `p(X).

Show that µ is σ-finite if and only if X is countable.

Exercise 1.16(The Delta Measure). Let X = Rnand Ω =P(X). Define δ(E) = 1 if 0 ∈ E and δ(E) = 0 if 0 /∈ E. Prove that δ is a measure, and find a formula for

Z

Rn

f (x) dδ(x).

Sometimes this integral is written informally asR

Rnf (x) δ(x) dx, but note that δ is a measure on Rn, not a function on Rn (see also Exercise 1.26 below).

Exercise 1.17. Fix 0 ≤ g ∈ L1(Rn), under Lebesgue measure. Prove that µ(E) =R

Eg(x) dx defines a finite measure on Rn.

With this preparation, we can give some additional examples of operators on Banach or Hilbert spaces.

Example 1.18 (Multiplication Operators). Let (X, Ω, µ) be a measure space, and let φ ∈ L(X) be a fixed measurable function. Then for any f ∈ L2(X) we have that f φ is measurable, and

kfφk22 = Z

X|f(x) φ(x)|2dx ≤ Z

X|f(x)|2kφk2dx = kφk2kfk22 < ∞,

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so f φ ∈ L2(X). Therefore, the multiplication operator Mφ: L2(X) → L2(X) given by Mφf = f φ is well-defined, and the calculation above shows that kMφfk2 ≤ kφkkfk2. Therefore Mφ is bounded, and kMφk ≤ kφk.

If we assume that µ is σ-finite, then we can show that kMφk = kφk, as follows. Choose any ε > 0. Then by definition of L-norm, the set E = {x ∈ X : |φ(x)| > kφk − ε}

has positive measure. Since X is σ-finite, we can write X = ∪Fm where each µ(Fm) < ∞.

Since E =∪(E ∩ Fm) is a countable union, we must have µ(E ∩ Fm) > 0 for some m. Let F = E∩ Fm, and set f = µ(F )11/2 χF. Thenkfk2 = 1, butkMφfk2 ≥ (kφk− ε) kfk2. Hence kMφk2 ≥ kφk− ε.

Exercise: Find an example of a measure µ that is not σ-finite and a function φ such that kMφk < kφk.

Exercise 1.19. Let (X, Ω, µ) be a measure space, and let φ be a fixed measurable function.

Prove that if f φ∈ L2(X) for every f ∈ L2(X), then we must have φ∈ L(X).

Solution. Assume φ /∈ L(X). Set

Ek = {x ∈ X : k ≤ |φ(x)| < k + 1}.

The Ek are measurable and disjoint, and since φ is not in L(X) there must be infinitely many Ek with positive measure. Choose any Enk, k ∈ N, all with positive measure and let E =∪Enk. Define

f (x) =

 1

k µ(Enk)1/2, x∈ Enk,

0, x /∈ E.

Then

Z

X|f|2dµ =

X

k=1

Z

Enk

1

k2µ(Enk) =

X

k=1

1

k2 < ∞, but

Z

X|fφ|2dµ ≥

X

k=1

Z

Enk

k2

k2µ(Enk) =

X

k=1

1 = ∞,

which is a contradiction. 

Exercise 1.20. Continuing Example 1.18, do the following.

(a) Determine a necessary and sufficient condition on φ which implies that Mφ: L2(X)→ L2(X) is injective.

(b) Determine a necessary and sufficient condition on φ which implies that Mφ: L2(X)→ L2(X) is surjective.

(c) Prove that if Mφ is injective but not surjective then Mφ−1: range(Mφ) → L2(X) is unbounded.

(d) Extend from the case p = 2 to any 1≤ p ≤ ∞.

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Example 1.21 (Integral Operators). Let (X, Ω, µ) be a σ-finite measure space. An integral operator is an operator of the form

Lf (x) = Z

X

k(x, y) f (y) dµ(y). (1.3)

This is just a formal definition, we have to provide conditions under which this makes sense, and the following two theorems will provide such conditions. The function k that determines the operator is called the kernel of the operator (not to be confused with the kernel/nullspace of the operator!).

Note that an integral operator is just a generalization of matrix multiplication. For, if A is an m× n matrix with entries aij and u∈ Cn, then Au∈ Cm, and its components are given by

(Au)i =

n

X

j=1

aijuj, i = 1, . . . , m.

Thus, the values k(x, y) are analogous to the entries aij of the matrix A, and the values Lf (x) are analogous to the entries (Au)i.

The following result shows that if the kernel is square-integrable, then the corresponding integral operator is bounded. Later we will define the notion of a Hilbert–Schmidt operator.

For the case of integral operators mapping L2(X) into itself, it can be shown that L is a Hilbert–Schmidt operator if and only if the kernel k belongs to L2(X × X).

Theorem 1.22 (Hilbert–Schmidt Integral Operators). Let (X, Ω, µ) be a σ-finite measure space, and choose a kernel k∈ L2(X× X). That is, assume that

kkk22 = Z

X

Z

X|k(x, y)|2dµ(x) dµ(y) < ∞.

Then the integral operator given by (1.3) defines a bounded mapping of L2(X) into itself, and kLk ≤ kkk2.

Proof. Although a slight abuse of the order of logic (technically we should show Lf exists before trying to compute its norm), the following calculation shows that L is well-defined and is a bounded mapping of L2(X) into itself:

kLfk22 = Z

X|Lf(x)|2dµ(x)

= Z

X

Z

X

k(x, y) f (y) dµ(y)

2

dµ(x)

≤ Z

X

Z

X|k(x, y)|2dµ(y)

 Z

X|f(y)|2dµ(y)

 dµ(x)

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= Z

X

Z

X|k(x, y)|2dµ(y)kfk22dµ(x)

= kkk22kfk22,

where the inequality follows by applying Cauchy–Schwarz to the inner integral. Thus L is

bounded, and kLk ≤ kkk2. 

The following result is one version of Schur’s Lemma. There are many forms of Schur’s Lemma, this is one particular special case.

Exercise: Compare the hypotheses of the following result to the operator norms you calculated in Exercise 1.6.

Theorem 1.23. Let (X, Ω, µ) be a σ-finite measure space, and Assume that k is a measurable function on X× X which satisfies the “mixed-norm” conditions

C1 = ess sup

x∈X

Z

X|k(x, y)| dµ(y) < ∞ and C2 = ess sup

y∈X

Z

X|k(x, y)| dµ(x) < ∞.

Then the integral operator given by (1.3) defines a bounded mapping of L2(X) into itself, and kLk ≤ (C1C2)1/2.

Proof. Choose any f ∈ L2(X). Then, by applying the Cauchy–Schwarz inequality, we have kLfk22 =

Z

X |Lf(x)|2dµ(x)

= Z

X

Z

X

k(x, y) f (y) dµ(y)

2

dµ(x)

≤ Z

X

Z

X|k(x, y)|1/2 |k(x, y)|1/2|f(y)| dµ(y)

2

dµ(x)

≤ Z

X

Z

X|k(x, y)| dµ(y)

 Z

X |k(x, y)| |f(y)|2dµ(y)

 dµ(x)

≤ Z

X

C1

Z

X|k(x, y)| |f(y)|2dµ(y) dµ(x)

= C1

Z

X|f(y)|2 Z

X|k(x, y)| dµ(x) dµ(y)

≤ C1

Z

X|f(y)|2C2dµ(y)

= C1C2kfk22,

where we have used Tonelli’s Theorem to interchange the order of integration (here is where we needed the fact that µ is σ-finite). Thus L is bounded and kLk ≤ (C1C2)1/2. 

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Exercise 1.24. Consider what happens in the preceding example if we take 1 ≤ p ≤ ∞ instead of p = 2. In particular, in part b, show that if C1, C2 <∞ then L: Lp(X)→ Lp(X) is a bounded mapping for each 1≤ p ≤ ∞ (try to do p = 1 or p = ∞ first).

Exercise 1.25 (Volterra Operator). Define L : L2[0, 1]→ L2[0, 1] by Lf (x) =

Z x 0

f (y) dy.

Show directly that L is bounded. Then show that L is an integral operator with kernel k : [0, 1]2 → F defined by

k(x, y) =

(1, y≤ x, 0, y > x.

Observe that k∈ L2([0, 1]2), so L is compact. This operator is called the Volterra operator.

Exercise 1.26(Convolution). Convolution is one of the most important examples of integral operators. Consider the case of Lebesgue measure on Rn. Given functions f , g on Rn, their convolution is the function f ∗ g defined by

(f ∗ g)(x) = Z

Rn

f (y) g(x− y) dy,

provided that the integral makes sense. Note that with g fixed, the mapping f 7→ f ∗ g is an integral operator with kernel k(x, y) = g(x− y).

(a) Let g ∈ L1(Rn) be fixed. Use Schur’s Lemma (Theorem 1.23) to show that Lf = f∗ g is a bounded mapping of L2(Rn) into itself. In fact, use Exercise 1.24 to prove Young’s Inequality: If f ∈ Lp(Rn) (1≤ p ≤ ∞) and g ∈ L1(Rn), then f ∗ g ∈ Lp(Rn), and

kf ∗ gkp ≤ kfkpkgk1. In particular, L1(Rn) is closed under convolution.

(b) Note that we cannot use the Hilbert–Schmidt condition (Theorem 1.22) to prove Young’s Inequality, since

Z

Rn

Z

Rn|g(x − y)|2dx dy = ∞, even if we assume that g ∈ L2(Rn).

(c) Prove that convolution is commutative, i.e., that f ∗ g = g ∗ f.

(d) Prove that there is no identity element in L1(Rn), i.e., there is no function g ∈ L1(Rn) such that f ∗ g = f for all f ∈ L1(Rn). This is not trivial—it is easier to do if you make use of the Fourier transform on Rn, and in particular use the Riemann–Lebesgue Lemma to derive a contradiction.

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(e) Some texts do talk informally about a “delta function” that is an identity element for convolution, defined by the conditions

δ(x) =

(∞, x = 0,

0, x6= 0, and Z

Rn

δ(x) dx = 1,

but no such function actually exists. In particular, the function δ defined on the left-hand side of the line above is equal to zero a.e., and hence is the zero function as far as Lebesgue integration is concerned. That is, we have R

Rnδ(x) dx = 0, not 1. The “delta function” is really just an informal use of the delta distribution (see Exercise 1.3) or the delta measure (see Exercise 1.16). Show that if we define the convolution of a function f with the delta measure δ to be

(f ∗ δ)(x) = Z

Rn

f (x− y) dδ(y), (1.4)

then f ∗ δ = f for all f ∈ L1(Rn). Note that in the “informal” notation of Exercise 1.16, (1.4) reads

(f ∗ δ)(x) = Z

Rn

f (x− y) δ(y) dy, which perhaps explains the use of the term “delta function.”

Exercise 1.27. Prove that L1(Rn) is not closed under pointwise multiplication. That is, prove that there exist f , g ∈ L1(Rn) such that the pointwise product h(x) = (f g)(x) = f (x)g(x) does not belong to to L1(Rn).

Exercise 1.28 (Convolution Continued). (a) Consider the space Lp[0, 1], where we think of functions in Lp[0, 1] as being extended 1-periodically to the real line. Define convolution on the circle by

(f ∗ g)(x) = Z 1

0

f (y) g(x− y) dy,

where the periodicity is used to define g(x− y) when x − y lies outside [0, 1] (equivalently, replace x− y by x − y mod 1, the fractional part of x − y). Prove a version of Young’s Inequality for Lp[0, 1].

(b) Consider the sequence space `p(Z). Define convolution on Z by (x∗ y)n = X

m∈Z

xmyn−m. Prove a version of Young’s Inequality for `p(Z).

Prove that `1(Z) contains an identity element with respect to convolution, i.e., there exists a sequence in `1(Z) (typically denoted δ) such that δ∗ x = x for every x ∈ `p(Z).

(c) Identify the essential features needed to define convolution on more general domains, and prove a version of Young’s Inequality for that setting.

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Exercise 1.29 (Convolution and the Fourier Transform). Let F be the Fourier transform on the circle, i.e., it is the isomorphism F : L2[0, 1]→ `2(Z) given by Ff = ˆf = { ˆf (n)}n∈Z, where

f (n) =ˆ hf, eni = Z 1

0

f (x) e−2πinxdx, en(x) = e2πinx.

(a) Prove that the Fourier transform converts convolution in to multiplication. That is, prove that if f , g ∈ L2[0, 1], then (f ∗ g) = ˆf ˆg, i.e.,

(f ∗ g)(n) = ˆf (n) ˆg(n), n∈ Z.

(b) Note that if g ∈ L2[0, 1], then g ∈ L1[0, 1], so by Young’s Inequality we have that f ∗ g ∈ L2[0, 1]. Holding g fixed, define an operator L : L2[0, 1] → L2[0, 1] by Lf = f ∗ g.

Since {en}n∈Z is an orthonormal basis for L2[0, 1], we have

f = X

n∈Z

f (n) eˆ n, f ∈ L2[0, 1].

Show that

Lf = f∗ g = X

n∈Z

ˆ

g(n) ˆf (n) en, f ∈ L2[0, 1].

Thus, in the “Fourier domain,” convolution acts by changing or adjusting the amount that each “component” or “frequency” encontributes to the representation of the function in this basis: the weight ˆf (n) for frequency n is replaced by the weight ˆg(n) ˆf (n). Explain why this says that L is analogous to multiplication by a diagonal operator. In engineering parlance, convolution is also referred to as filtering. Explain why this terminology is appropriate.

Compare this operator L to Example 1.7.

2. The Adjoint of an Operator

Example 2.1. Note that the dot product on Rn is given by x · y = xTy, while the dot product on Cn is x· y = xTy.¯

Let A be an m× n real matrix. Then x 7→ Ax defines a linear map of Rn into Rm, and its transpose AT satisfies

∀ x ∈ Rn, ∀ y ∈ Rm, Ax· y = (Ax)Ty = xTATy = x· (ATy).

Similarly, if A is an m× n complex matrix, then its Hermitian or adjoint matrix AH = AT satisfies

∀ x ∈ Cn, ∀ y ∈ Cm, Ax· y = (Ax)Ty = x¯ TATy = x¯ · (AHy).

Theorem 2.2 (Adjoint). Let H and K be Hilbert spaces, and let A : H → K be a bounded, linear map. Then there exists a unique bounded linear map A: K → H such that

∀ x ∈ H, ∀ y ∈ K, hAx, yi = hx, Ayi.

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Proof. Fix y ∈ K. Then Lx = hAx, yi is a bounded linear functional on H. By the Riesz Representation Theorem, there exists a unique vector h∈ H such that

hAx, yi = Lx = hx, hi.

Define Ay = h. Verify that this map A is linear (exercise). To see that it is bounded, observe that

kAyk = khk = sup

kxk=1|hx, hi|

= sup

kxk=1|hAx, yi|

≤ sup

kxk=1kAxk kyk

≤ sup

kxk=1kAk kxk kyk = kAk kyk.

We conclude that A is bounded, and thatkAk ≤ kAk.

Finally, we must show that A is unique. Suppose that B ∈ B(K, H) also satisfied hAx, yi = hx, Byi for all x ∈ H and y ∈ K. Then for each fixed y we would have that hx, By − Ayi = 0 for every x, which implies By − Ay = 0. Hence B = A.  Exercise 2.3 (Properties of the adjoint).

(a) If A∈ B(H, K) then (A) = A.

(b) If A, B∈ B(H, K) and α, β ∈ F, then (αA + βB) = ¯αA+ ¯βB. (c) If A∈ B(H1, H2) and B ∈ B(H2, H3), then (BA) = AB.

(d) If A∈ B(H) is invertible in B(H) (meaning that there exists A−1 ∈ B(H) such that AA−1 = A−1A = I), then A is invertible in B(H) and (A−1) = (A)−1.

Remark 2.4. Later we will prove the Open Mapping Theorem. A remarkable consequence of this theorem is that if X and Y are Banach spaces and A : X → Y is a bounded bijection, then A−1: Y → X is automatically bounded.

Proposition 2.5. If A∈ B(H, K), then kAk = kAk = kAAk1/2 =kAAk1/2.

Proof. In the course of proving Theorem 2.2, we already showed thatkAk ≤ kAk. If f ∈ H, then

kAfk2 = hAf, Afi = hAAf, fi ≤ kAAfk kfk ≤ kAk kAfk kfk. (2.1) Hence kAfk ≤ kAk kfk (even if kAfk = 0, this is still true). Since this is true for all f we conclude that kAk ≤ kAk. Therefore kAk = kAk.

Next, we have kAAk ≤ kAk kAk = kAk2. But also, from the calculation in (2.1), we have kAfk2 ≤ kAAfk kfk. Taking the supremum over all unit vectors, we obtain

kAk2 = sup

kf k=1kAfk2 ≤ sup

kf k=1kAAfk kfk = kAAk.

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Consequently kAk2 = kAAk. The final equality follows by interchanging the roles of A

and A. 

Exercise 2.6. Prove that if U ∈ B(H, K), then U is an isomorphism if and only if U is invertible and U−1 = U.

Exercise 2.7. (a) Let λ = (λn)n∈N ∈ `(N) be given and let L be defined as in Example 1.7.

Find L.

(b)Prove that the adjoint of the multiplication operator Mφ defined in Exercise 1.18 is the multiplication operator Mφ¯.

Exercise 2.8. Let L and R be the left- and right-shift operators on `2(N), i.e., L(x1, x2, . . . ) = (x2, x3, . . . ) and R(x1, x2, . . . ) = (0, x1, x2, . . . ).

Prove that L = R.

Example 2.9. Let L be the integral operator defined in (1.3), determined by the kernel function k. Assume that k is chosen so that L : L2(X)→ L2(X) is bounded. The adjoint is the unique operator L: L2(X)→ L2(X) which satisfies

hLf, gi = hf, Lgi, f, g ∈ L2(X).

To find L, let A : L2(X)→ L2(X) be the integral operator with kernel k(y, x), i.e., Af (x) =

Z

X

k(y, x) f (y) dµ(y).

Then, given any f and g∈ L2(X), we have hf, Lgi = hLf, gi =

Z

X

Lf (x) g(x) dµ(x)

= Z

X

Z

X

k(x, y) f (y) dµ(y) g(x) dµ(x)

= Z

X

f (y) Z

X

k(x, y) g(x) dµ(x) dµ(y)

= Z

X

f (y) Z

X

k(x, y) g(x) dµ(x) dµ(y)

= Z

X

f (y) Ag(y) dµ(y)

= hf, Agi.

By uniqueness of the adjoint, we must have L = A.

Exercise: Justify the interchange in the order of integration in the above calculation, i.e., provide hypotheses under which the calculations above are justified.

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Exercise 2.10. Let{en}n∈Nbe an orthonormal basis for a separable Hilbert space H. Define T : H → `2(N) by T (f ) ={hf, eni}n∈N. Find a formula for T: `2(N)→ H.

Definition 2.11. Let A∈ B(H).

(a) We say that A is self-adjoint or Hermitian if A = A. (b) We say that A is normal if AA = AA.

Example 2.12. A real n× n matrix A is self-adjoint if and only if it is symmetric, i.e., if A = AT. A complex n× n matrix A is self-adjoint if and only if it is Hermitian, i.e., if A = AH.

Exercise 2.13. Show that every self-adjoint operator is normal. Show that every unitary operator is normal, but that a unitary operator need not be self-adjoint. For H = Cn, find examples of matrices that are not normal. Are the left- and right-shift operators on `2(N) normal?

Exercise 2.14. (a) Show that if A, B ∈ B(H) are self-adjoint, then AB is self-adjoint if and only if AB = BA.

(b) Give an example of self-adjoint operators A, B such that AB is not self-adjoint.

(c) Show that if A, B ∈ B(H) are self-adjoint then A + A, AA, AA, A + B, ABA, and BAB are all self-adjoint. What about A− A or A− B? Show that AA − AA is self-adjoint.

Exercise 2.15. (a) Let λ = (λn)n∈N∈ `(N) be given and let L be defined as in Example 1.7.

Show that L is normal, find a formula for L, and prove that L is self-adjoint if and only if each λn is real.

(b) Determine a necessary and sufficient condition on φ so that the multiplication operator Mφ defined in Exercise 1.18 is self-adjoint.

(c) Determine a necessary and sufficient condition on the kernel k so that the integral operator L defined in (1.23) is self-adjoint.

The following result gives a useful condition for telling when an operator on a complex Hilbert space is self-adjoint.

Proposition 2.16. Let H be a complex Hilbert space (i.e., F = C), and let A ∈ B(H) be given. Then:

A is self-adjoint ⇐⇒ hAf, fi ∈ R ∀ f ∈ H.

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Proof. ⇒. Assume A = A. Then for any f ∈ H we have

hAf, fi = hf, Afi = hAf, fi = hAf, fi.

Therefore hAf, fi is real.

⇐. Assume that hAf, fi is real for all f. Choose any f, g ∈ H. Then hA(f + g), f + gi = hAf, fi + hAf, gi + hAg, fi + hAg, gi.

Since hA(f + g), f + gi, hAf, fi, and hAg, gi are all real, we conclude that hAf, gi + hAg, fi is real. Hence it equals its own complex conjugate, i.e.,

hAf, gi + hAg, fi = hAf, gi + hAg, fi = hg, Afi + hf, Agi. (2.2) Similarly, since

hA(f + ig), f + igi = hAf, fi − ihAf, gi + ihAg, fi + hAg, gi we see that

−ihAf, gi + ihAg, fi = −ihAf, gi + ihAg, fi = ihg, Afi − ihf, Agi.

Multiplying through by i yields

hAf, gi − hAg, fi = −hg, Afi + hf, Agi. (2.3) Adding (2.2) and (2.3) together, we obtain

2hAf, gi = 2hf, Agi = 2hAf, gi.

Since this is true for every f and g, we conclude that A = A.  Example 2.17. The preceding result is false for real Hilbert spaces. After all, if F = R then hAf, fi is real for every f no matter what A is. Therefore, any non-self-adjoint operator provides a counterexample. For example, if H = Rn then any non-symmetric matrix A is a counterexample.

The next result provides a useful way of calculating the operator norm of a self-adjoint operator.

Proposition 2.18. If A∈ B(H) is self-adjoint, then kAk = sup

kf k=1|hAf, fi|.

Proof. Set M = supkf k=1|hAf, fi|.

By Cauchy–Schwarz and the definition of operator norm, we have M = sup

kf k=1|hAf, fi| ≤ sup

kf k=1kAfk kfk ≤ sup

kf k=1kAk kfk kfk = kAk.

To get the opposite inequality, note that if f is any nonzero vector in H then f /kfk is a unit vector, so Akf kf , kf kf ≤ M. Rearranging, we see that

∀ f ∈ H, hAf, fi ≤ M kfk2. (2.4)

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Now choose any f , g ∈ H with kfk = kgk = 1. Then, by expanding the inner products, canceling terms, and using the fact that A = A, we see that

= 2hAf, gi + 2 hg, Afi

= 4 RehAf, gi.

Therefore, applying (2.4) and the Parallelogram Law, we have

4 RehAf, gi ≤ |hA(f + g), f + gi| + |hA(f − g), f − gi|

≤ M kf + gk2+ Mkf − gk2

= 2M kfk2+kgk2

= 4M.

That is, RehAf, gi ≤ M for every choice of unit vectors f and g. Write hAf, gi = |hAf, gi| e. Then eg is another unit vector, so

M ≥ Re hAf, e−iθgi = Re ehAf, gi = |hAf, gi|.

Hence

kAfk = sup

kgk=1|hAf, gi| ≤ M.

Since this is true for every unit vector f , we conclude that kAk ≤ M.  The following corollary is a very useful consequence.

Corollary 2.19. Assume that A∈ B(H).

(a) If F = R, A = A, andhAf, fi = 0 for every f, then A = 0.

(b) If F = C and hAf, fi = 0 for every f, then A = 0.

Proof. Assume the hypotheses of either statement (a) or statement (b). In the case of statement (a), we have by hypothesis that A is self-adjoint. In the case of statement (b), we can conclude that A is self-adjoint because hAf, fi = 0 is real for every f. Hence in either case we can apply Proposition 2.18 to conclude that

kAk = sup

kf k=1|hAf, fi| = 0. 

Lemma 2.20. If A∈ B(H), then the following statements are equivalent.

(a) A is normal, i.e., AA = AA.

(b) kAfk = kAfk for every f ∈ H.

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Proof. (b) ⇒ (a). Assume that (b) holds. Then for every f we have (AA− AA)f, f

= hAAf, fi − hAAf, fi

= hAf, Afi − hAf, Afi

= kAfk2− kAfk2 = 0.

Since AA− AA is self-adjoint, it follows from Corollary 2.19 that AA− AA = 0.

(a) ⇒ (b). Exercise. 

Corollary 2.21. If A∈ B(H) is normal, then ker(A) = ker(A).

Exercise 2.22. Suppose that A∈ B(H) is normal. Prove that A is injective if and only if range(A) is dense in H.

Exercise 2.23. If A∈ B(H), then the following statements are equivalent.

(a) A is an isometry, i.e., kAfk = kfk for every f ∈ H.

(b) AA = I.

(c) hAf, Agi = hf, gi for every f, g ∈ H.

Exercise 2.24. If H = Cn and A, B are n× n matrices, then AB = I implies BA = I.

Give a counterexample to this for an infinite-dimensional Hilbert space. Consequently, the hypothesis AA = I in the preceding result does not imply that AA = I.

Exercise 2.25. If A∈ B(H), then the following statements are equivalent.

(a) AA = AA = I.

(b) A is unitary, i.e., it is a surjective isometry.

(c) A is a normal isometry.

The following result provides a very useful relationship between the range of A and the kernel of A.

Theorem 2.26. Let A∈ B(H, K).

(a) ker(A) = range(A). (b) ker(A)= range(A).

(c) A is injective if and only if range(A) is dense in H.

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Proof. (a) Assume that f ∈ ker(A) and let h ∈ range(A), i.e., h = Ag for some g ∈ K.

Then since Af = 0, we have hf, hi = hf, Agi = hAf, gi = 0. Thus f ∈ range(A), so ker(A)⊆ range(A).

Now assume that f ∈ range(A). Then for any h ∈ H we have hAf, hi = hf, Ahi = 0.

But this implies Af = 0, so f ∈ ker(A). Thus range(A) ⊆ ker(A).

(b), (c) Exercises. 

3. Projections and Idempotents: Invariant and Reducing Subspaces Definition 3.1. a. If E ∈ B(H) satisfies E2 = E then E is said to be idempotent.

b. If E ∈ B(H) satisfies E2 = E and ker(E) = range(E) then E is called a projection.

Exercise 3.2. If E ∈ B(H) is an idempotent operator, then ker(E) and range(E) are closed subspaces of H. Further, ker(E) = range(I − E) and range(E) = ker(I − E).

Lemma 3.3 (Characterization of Orthogonal Projections). Let E ∈ B(H) be a nonzero idempotent operator. Then the following statements are equivalent.

(a) E is a projection.

(b) E is the orthogonal projection of H onto range(E).

(c) kEk = 1.

(d) E is self-adjoint.

(e) E is normal.

(f) E is positive, i.e., hEf, fi ≥ 0 for every f ∈ H.

Proof. (e) ⇒ (a). Assume that E2 = E and E is normal. Then from Lemma 2.20 we know that kEfk = kEfk for every f ∈ H. Hence Ef = 0 if and only if Ef = 0, or in other words, ker(E) = ker(E). But we know from Theorem 2.26 that ker(E) = range(E). Hence we conclude that ker(E) = range(E), and therefore E is a projection.

The remaining implications are exercises. 

Definition 3.4(Orthogonal Direct Sum of Subspaces). Let{Mi}i∈I be a collection of closed subspaces of H such that Mi ⊥ Mj whenever i 6= j. Then the orthogonal direct sum of the Mi is the smallest closed subspace which contains every Mi. This space is

L

i∈I

Mi = span S

i∈I

Mi

 .

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Exercise 3.5. Suppose that M , N are closed subspaces of H such that M ⊥ N. Prove that M + N ={m + n : m ∈ M, n ∈ N} is a closed subspace of H, and that

M ⊕ N = M + N.

Show that every vector x∈ M ⊕ N can be written uniquely as x = m + n with m ∈ M and n∈ N.

Extend by induction to finite collections of closed, pairwise orthogonal subspaces. (Un- fortunately, the analogous statement is not true for infinite collections.)

Exercise 3.6. Show that if A∈ B(H, K) then H = ker(A) ⊕ range(A).

Definition 3.7. Let A∈ B(H) and M ≤ H.

(a) We say that M is invariant under A if A(M )⊆ M, where A(M ) = {Ax : x ∈ M}.

That is, M is invariant if x∈ M implies Ax ∈ M. Note that it need not be the case that A(M ) = M .

(b) We say that M is a reducing subspace for A if both M and M are invariant under A, i.e., A(M )⊆ M and A(M)⊆ M.

Proposition 3.8. Let A∈ B(H) and M ≤ H be given. Then the following statements are equivalent.

(a) M is invariant under A.

(b) P AP = AP , where P = PM is the orthogonal projection of H onto M .

Exercise 3.9. Define L : `2(Z)→ `2(Z) by

L(. . . , x−1, x0, x1, . . . ) = (. . . , , x0, x1, x2, . . . ),

where on the right-hand side the entry x1 sits in the 0th component position. That is, L slides each component one unit to the left (L is called a bilateral shift). Find a closed subspace of `2(Z) that is invariant but not reducing under L.

Exercise 3.10. Assume that M ≤ H is invariant under L ∈ B(H). Prove that M is invariant under L.

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4. Compact Operators

Definition 4.1 (Compact and Totally Bounded Sets). Let X be a Banach space, and let E ⊆ X be given.

(a) We say that E is compact if every open cover of E contains a finite subcover. That is, E is compact if whenever {Uα}α∈I is a collection of open sets whose union contains E, then there exist finitely many α1, . . . , αN such that E ⊆ Uα1 ∪ · · · ∪ UαN.

(b) We say that E is sequentially compact if every sequence {fn}n∈N of points of E contains a convergent subsequence {fnk}k∈N whose limit belongs to E.

(c) We say that E is totally bounded if for every ε > 0 there exist finitely many points f1, . . . , fN ∈ E such that

E ⊆ SN

k=1

B(fk, ε),

where B(fk, ε) is the open ball of radius ε centered at fk. That is, E is totally bounded if and only there exist finitely many points f1, . . . , fN ∈ E such that every element of E is within ε of some fk.

In finite dimensions, a set is compact if and only if it is closed and bounded. In infinite dimensions, all compact sets are closed and bounded, but the converse fails. Instead, we have the following characterization of compact sets. (this characterization actually holds in any complete metric space).

Theorem 4.2. Let E be a subset of a Banach space X. Then the following statements are equivalent.

(a) E is compact.

(b) E is sequentially compact.

(c) E is closed and totally bounded.

Proof. (b) ⇒ (a).2 Assume that E is sequentially compact. Our first step will be to prove the following claim, where the diameter of a set S is defined to be

diam(S) = sup{kf − gk : f, g ∈ S}.

Claim 1. For any open cover {Uα}α∈I of E, there exists a number δ > 0 (called a Lebesgue number for the cover) such that if S ⊆ E satisfies diam(S) < δ, then there is an α ∈ I such that S ⊆ Uα.

To prove the claim, suppose that {Uα}α∈I was an open cover of E such that no δ with the required property existed. Then for each n ∈ N, we could find a set Sn ⊆ E with diam(Sn) < n1 such that Sn is not contained in any Uα. Choose any fn ∈ Sn. Since E is sequentially compact, there must be a subsequence {fnk}k∈N that converges to an element of

2This proof is adapted from one given in J. R. Munkres, “Topology,” Second Edition, Prentice Hall, 2000.

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E, say fnk → a ∈ E. But we must have a ∈ Uα for some α, and since Uα is open there must exist some ε > 0 such that B(a, ε)⊆ Uα. Now choose k large enough that we have both

1 nk

< ε

2 and ka − fnkk < ε 2.

The first inequality above implies that diam(Snk) < ε2. Therefore, using this and second inequality, we have Snk ⊆ B(a, ε) ⊆ Uα, which is a contradiction. Therefore the claim is proved.

Next, we will prove the following claim.

Claim 2. For any ε > 0, there exist finitely many f1, . . . , fN ∈ E such that E ⊆ SN

k=1

B(fk, ε).

To prove this claim, assume that there is an ε > 0 such that E cannot be covered by finitely many ε-balls centered at points of E. Choose any f1 ∈ E. Since E cannot be covered by a single ε-ball, we have E 6⊆ B(f1, ε). Hence there exists f2 ∈ E \ B(f1, ε), i.e., f2 ∈ E and kf2− f1k ≥ ε. But E cannot be covered by two ε-balls, so there must exist an f3 ∈ E \ B(f1, ε)∪ B(f2, ε). In particular, we have kf3− f1k, kf3− f2k ≥ ε. Continuing in this way we obtain a sequence of points{fn}n∈N in E which has no convergent subsequence, which is a contradiction. Hence the claim is proved.

Finally, we show that E is compact. Let {Uα}α∈I be any open cover of E. Let δ be the Lebesgue number given by Claim 1, and set ε = δ3. By Claim 2, there exists a covering of E by finitely many ε-balls. Each ball has diameter smaller than δ, so by Claim 1 is contained in some Uα. Thus we find finitely many Uα that cover E.

(c)⇒ (b). Assume that E is closed and totally bounded, and let {fn}n∈N be any sequence of points in E. Since E is covered by finitely many balls of radius 12, one of those balls must contain infinitely many fn, say {fn(1)}n∈N. Then we have

∀ m, n ∈ N, kfm(1)− fn(1)k < 1.

Since E is covered by finitely many balls of radius 14, we can find a subsequence{fn(2)}n∈N of {fn(1)}n∈N such that

∀ m, n ∈ N, kfm(1)− fn(1)k < 1 2.

By induction we keep constructing subsequences {fn(k)}n∈N such that kfm(k)− fn(k)k < k1 for all m, n∈ N.

Now consider the “diagonal subsequence” {fn(n)}n∈N. Given ε > 0, let N be large enough that N1 < ε. If m ≥ n > N, then fm(m) is one element of the sequence {fk(n)}k∈N, say fm(m)= fk(n). Then

kfm(m)− fn(n)k = kfk(n)− fn(n)k < 1 n < ε.

Thus {fn(n)}n∈N is Cauchy and hence converges. Since E is closed, it must converge to some element of E.

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(a) ⇒ (c). Exercise.  Exercise 4.3. Show that if E is a totally bounded subset of a Banach space X, then its closure E is compact. A set whose closure is compact is said to be precompact.

Notation 4.4. We let BallH denote the closed unit sphere in H, i.e., BallH = Ball(H) = {f ∈ H : kfk ≤ 1}.

Exercise 4.5. Prove that if H is infinite-dimensional, then BallH is not compact.

Definition 4.6 (Compact Operators). Let H, K be Hilbert spaces. A linear operator T : H → K is compact if T (BallH) has compact closure in K. We define

B0(H, K) = {T : H → K : T is compact}, and set B0(H) = B0(H, H).

By definition, a compact operator is linear, and we will see that all compact operators are bounded. Thus it will turn out thatB0(H, K)⊆ B(H, K). In fact, we will see that B0(H, K) is a closed subspace ofB(H, K).

The following result gives some useful reformulations of the definition of compact operator.

Proposition 4.7 (Characterizations of Compact Operators). Let T : H → K be linear.

Then the following statements are equivalent.

(a) T is compact.

(b) T (BallH) is totally bounded.

(c) If{fn}n∈N is a bounded sequence in H, then{T fn}n∈N contains a convergent subse- quence.

Proof. (a) ⇔ (b). This follows from Theorem 4.2 and Exercise 4.3.

(a) ⇒ (c). Suppose that T is compact and that {fn}n∈N is a bounded sequence in H.

By rescaling the sequence (i.e., multiplying by an appropriate scalar), we may assume that fn∈ BallH for every n. Therefore T fn∈ T (BallH)⊆ T (BallH). Since T (BallH) is compact, it follows from Theorem 4.2 that {T fn}n∈N contains a subsequence which converges to an element of T (BallH).

(c) ⇒ (a). Exercise. 

Proposition 4.8. If T : H → K is compact, then it is bounded. That is, B0(H, K) ⊆ B(H, K).

Proof. Assume that T : H → K is linear but unbounded. Then there exist vectors fn ∈ H such thatkfnk = 1 but kT fnk ≥ n. Therefore every subsequence of {T fn}n∈Nis unbounded, and hence cannot converge. Therefore T is not compact by Proposition 4.7. 

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