C.6 Adjoints for Operators on Hilbert Spacespeople.math.gatech.edu/~heil/7338/fall09/appendixc_2.pdfC.6 Adjoints for Operators on Hilbert Spaces 321 As a corollary, we obtain the following

C.6 Adjoints for Operators on Hilbert Spaces 317

Additional Problems

C.11. Let E ⊆ R be measurable. Given 1 ≤ p ≤ ∞ and a measurable weightfunction w : E → (0,∞), the weighted Lp space Lp

s(R) consists of all measur-able functions f : E → C for which ‖f‖p,w = ‖fw‖p < ∞. Prove that Lp

w(R)is a Banach space with respect to the norm ‖ · ‖p,w. Assuming Theorem C.38,

prove that if 1 ≤ p <∞ then Lpw(E)∗ ∼= Lp′

1/w(E).

C.6 Adjoints for Operators on Hilbert Spaces

If A is an m× n complex matrix and x · y is the ordinary dot product on Cd,then

Ax · y = x ·A∗y, x ∈ Cn, y ∈ Cm,

where A∗ = AT = AH is the Hermitian, or conjugate transpose, of A. As anapplication of the Riesz Representation Theorem, we will show that there isan analogue of the Hermitian matrix for linear operators on Hilbert spaces.In Section C.10 we will see that this extends further to operators on Banachspaces, but in that setting we need to appeal to the Hahn–Banach Theoremin order to construct the adjoint.

Throughout this section, we will let H and K denote Hilbert spaces.

C.6.1 Adjoints of Bounded Operators

Exercise C.40. Let L : H → K be a bounded linear operator.

(a) For each g ∈ K, define a functional µg : H → C by

〈f, µg〉 = 〈Lf, g〉, f ∈ H.

Note that, following Notation C.36, 〈f, µg〉 denotes the action of the func-tional µg on the vector f, while 〈Lf, g〉 denotes the inner product of thevectors Lf, g ∈ K. Show that µg ∈ H∗. Consequently, by the Riesz Rep-resentation Theorem, there exists a unique element g∗ ∈ H such that

〈f, µg〉 = 〈f, g∗〉, f ∈ H.

(b) Define L∗ : K → H by L∗g = g∗. Show that L∗ is a bounded linear map,that (L∗)∗ = L, and that ‖L∗‖ = ‖L‖.

We formalize this as a definition.

Definition C.41 (Adjoint). The adjoint of L ∈ B(H,K) is the unique op-erator L∗ : K → H that satisfies

〈Lf, g〉 = 〈f, L∗g〉, f ∈ H, g ∈ K.

318 C Functional Analysis and Operator Theory

When H = K, we use the following additional terminology.

Definition C.42 (Self-Adjoint and Normal Operators). Let L ∈ B(H)be given.

(a) We say that L is self-adjoint or Hermitian if L = L∗. Equivalently, L isself-adjoint if

∀ f, g ∈ H, 〈Lf, g〉 = 〈f, Lg〉.

(b) We say that L is normal if L commutes with its adjoint, i.e., if LL∗ = L∗L.

All self-adjoint operators are normal, but not all normal operators areself-adjoint (Problem C.19).

Remark C.43. Each complex m× n matrix A determines a linear map of Cn

to Cm. The adjoint A∗ of this map is determined by the conjugate transpose

of the matrix A. That is, identifying matrices with maps, we have A∗ = AT =AH. This conjugate transpose matrix AH is called the Hermitian of A. Formatrices it is customary to instead say that A is Hermitian if A = AH, insteadof saying that A is self-adjoint.

If L ∈ B(H,K), then L∗ is a linear map of H into K, but the mappingof B(H,K) to B(K,H) given by L 7→ L∗ is antilinear, as (αA + βB)∗ =αA∗+βB∗. Exercise C.40 shows that ‖L∗‖ = ‖L‖, and hence the map L 7→ L∗

is an isometry.We will need some facts about the relationship between invariant subspaces

and adjoints.

Definition C.44. Let A ∈ B(H) be given. A closed subspace M ⊆ H is saidto be invariant under A if A(M) ⊆M, where A(M) = {Af : f ∈M}.

Note that we do not require that A(M) be equal to M.The simplest example of an invariant subspace is M = span{f} where f

is an eigenvector of A.

Exercise C.45. Show that if a closed subspace M ⊆ H is invariant underA ∈ B(H), then M⊥ is invariant under A∗.

C.6.2 Adjoints of Unbounded Operators

Adjoints can also be defined for unbounded operators, although now we mustbe careful with domains. For example, consider the differentiation operatorDf = f ′. This operator is not defined on all of L2(R), but instead is denselydefined in the sense of Notation C.3. For example, D maps the dense subspaceS =

{f ∈ C1

0 (R) : f, f ′ ∈ L2(R)}

into L2(R). Although D : S → L2(R) isunbounded, if f, g ∈ S then we have fg ∈ C0(R), so integration by partsyields


〈Df, g〉 =

∫f ′(x) g(x) dx = −

∫f(x) g′(x) dx = −〈f,Dg〉.

Hence D∗ = −D as an operator mapping S into L2(R).The important point in the preceding example is that if g ∈ S is fixed,

then f 7→ 〈Df, g〉 is a bounded linear functional on S, even though D is anunbounded operator. The following exercise extends this to general operators.

Exercise C.46. Suppose that S is a dense subspace of H, and L : S → K isa linear, but not necessarily bounded, operator. Let

S∗ = {g ∈ K : f 7→ 〈Lf, g〉 is a bounded linear functional on S}.

Show that there is an operator L∗ : S∗ → H such that

〈Lf, g〉 = 〈f, L∗g〉, f ∈ S, g ∈ S∗.

If necessary, we can always restrict a densely defined operator to a smallerbut still dense domain. Given an operator L mapping some dense subspaceof H into H, if we can find some dense subspace S on which L is defined andsuch that 〈Lf, g〉 = 〈f, Lg〉 for f, g ∈ S, then we say that L is self-adjoint.

C.6.3 Bounded Self-Adjoint Operators on Hilbert Spaces

We now focus in more detail on bounded self-adjoint operators on Hilbertspaces, which have many useful properties and appear often throughout thisvolume. Some of the results that we describe next hold only for complexHilbert spaces; we will indicate the changes needed for real Hilbert spaces.Recall that, following the notational conventions laid out in Section A.1, wealways assume in this volume that a vector space is over the complex fieldunless specifically stated otherwise.

For complex Hilbert spaces, we have the following characterization of self-adjoint operators. This theorem does not have an analogue for real Hilbertspaces.

Theorem C.47. Let H be a (complex) Hilbert space, and let A ∈ B(H) begiven. Then:

A is self-adjoint ⇐⇒ 〈Af, f〉 ∈ R for all f ∈ H.

Proof. ⇐ . Assume that 〈Af, f〉 is real for all f. Choose any f, g ∈ H. Then

〈A(f + g), f + g〉 = 〈Af, f〉 + 〈Af, g〉 + 〈Ag, f〉 + 〈Ag, g〉.

Since 〈A(f + g), f + g〉, 〈Af, f〉, and 〈Ag, g〉 are all real, we conclude that〈Af, g〉 + 〈Ag, f〉 is real. Hence it equals its own complex conjugate:

〈Af, g〉 + 〈Ag, f〉 = 〈Af, g〉 + 〈Ag, f〉 = 〈g,Af〉 + 〈f,Ag〉. (C.11)


Similarly, after examining the equation

〈A(f + ig), f + ig〉 = 〈Af, f〉 − i〈Af, g〉 + i〈Ag, f〉 + 〈Ag, g〉,

we conclude that

〈Af, g〉 − 〈Ag, f〉 = −〈g,Af〉 + 〈f,Ag〉. (C.12)

Adding (C.11) and (C.12) together, we obtain

2〈Af, g〉 = 2〈f,Ag〉 = 2〈A∗f, g〉.

Since this is true for every f and g, we conclude that A = A∗. ⊓⊔

The next result gives an alternative formula for the operator norm of aself-adjoint operator. This theorem holds for both real and complex Hilbertspaces.

Theorem C.48. If A ∈ B(H) is self-adjoint, then

‖A‖ = sup‖f‖=1

|〈Af, f〉|.

Proof. Set M = sup‖f‖=1 |〈Af, f〉|. By Cauchy–Bunyakowski–Schwarz andthe definition of operator norm, it follows that M ≤ ‖A‖.

Choose any unit vectors f, g ∈ H. Then, by expanding the inner products,canceling terms, and using the fact that A = A∗, we see that

⟨A(f + g), f + g

⟩−

⟨A(f − g), f − g

⟩= 2 〈Af, g〉 + 2 〈Ag, f〉

= 2 〈Af, g〉 + 2 〈g,Af〉

= 4 Re〈Af, g〉.

Therefore, applying the definition of M and using the Parallelogram Law, wehave

4 Re〈Af, g〉 ≤ |〈A(f + g), f + g〉| + |〈A(f − g), f − g〉|

≤ M ‖f + g‖2 +M ‖f − g‖2

= 2M(‖f‖2 + ‖g‖2

)= 4M.

That is, Re〈Af, g〉 ≤ M for every choice of unit vectors f and g. Write|〈Af, g〉| = α 〈Af, g〉 where α ∈ C satisfies |α| = 1. Then αg is another unitvector, so

|〈Af, g〉| = α〈Af, g〉 = 〈Af, αg〉 ≤ M.

Hence‖Af‖ = sup

‖g‖=1

|〈Af, g〉| ≤ M,

and therefore ‖A‖ ≤M. ⊓⊔


As a corollary, we obtain the following useful fact for operators on a com-plex Hilbert space. This result also holds for operators on a real Hilbert spaceif we include the assumption A = A∗ as a hypothesis.

Exercise C.49. Let H be a (complex) Hilbert space. If A ∈ B(H) and〈Af, f〉 = 0 for every f, then A = 0.

C.6.4 Positive and Positive Definite Operators on Hilbert Spaces

Among the self-adjoint operators, we distinguish the special class of positiveoperators.

Definition C.50 (Positive and Positive Definite Operators). Let A ∈B(H) be given.

(a) We say that A is positive or nonnegative, denoted A ≥ 0, if 〈Af, f〉 ≥ 0for every f ∈ H.

(b) We say that A is positive definite or strictly positive, denoted A > 0, if〈Af, f〉 > 0 for every nonzero vector f ∈ H.

By Theorem C.47, since we are dealing with complex Hilbert spaces, allpositive and positive definite operators are self-adjoint. When dealing withreal Hilbert spaces, the assumption of self-adjointness should be added in thedefinition of positive and positive definite operators.

Exercise C.51. Show that if A ∈ B(H,K), then A∗A ∈ B(H) and AA∗ ∈B(K) are both positive operators. Determine conditions on A that imply thatA∗A or AA∗ is positive definite.

Additional Problems

C.12. Let H1, H2, H3 be Hilbert spaces. Show that if A ∈ B(H1, H2) andB ∈ B(H2, H3), then (A∗)∗ = A and (BA)∗ = A∗B∗.

C.13. Show that if A ∈ B(H,K) is a topological isomorphism then A∗ ∈B(K,H) is also a topological isomorphism, and (A−1)∗ = (A∗)−1.

C.14. Show that if A ∈ B(H,K), then ‖A‖ = ‖A∗‖ = ‖A∗A‖1/2 = ‖AA∗‖1/2.

C.15. Show that A 7→ A∗ defines an involution on B(H) in the sense ofDefinition C.32.

Remark: In the language of operator theory, the fact that this involutionsatisfies ‖A∗A‖ = ‖A‖2 means that B(H) is an example of a C∗-algebra.

C.16. Given A ∈ B(H), show that A is normal if and only if ‖Af‖ = ‖A∗f‖for every f ∈ H.


C.17. Show that if A ∈ B(H,K) then the following statements hold.

(a) ker(A) = range(A∗)⊥.

(b) ker(A)⊥ = range(A∗).

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

C.18. Given A ∈ B(H), show that

ker(A) = ker(A∗A) and range(A∗A) = range(A∗).

Use this to show that if A is normal, then

ker(A) = ker(A∗) and range(A) = range(A∗).

C.19. Fix λ ∈ ℓ∞, and let Mλ be the multiplication operator defined inExercise C.14. Find M∗

λ , and show that Mλ is normal. Determine when Mλ

is self-adjoint, positive, or positive definite.

C.20. Fix φ ∈ L∞(R), and let Mφ be the multiplication operator defined inExercise C.15. Find M∗

φ , and show that Mφ is normal. Determine when Mφ

is self-adjoint, positive, or positive definite.

C.21. Given k ∈ L2(R2), let Lk be the integral operator with kernel k. Showthat the adjoint operator (Lk)∗ is the integral operator Lk∗ whose kernel is

k∗(x, y) = k(y, x). Characterize those kernels k ∈ L2(R2) corresponding toself-adjoint operators Lk.

C.22. Let L and R be the left- and right-shift operators from Problem C.2.Show that R = L∗, and conclude that L and R are not normal.

C.23. Let {en}n∈N be an orthonormal basis for a separable Hilbert space H.

Let T : H → ℓ2(N) be the analysis operator Tf = {〈f, en〉}n∈N. Find a formulafor the synthesis operator T ∗ : ℓ2(N) → H.

C.24. Let M be a closed subspace of H, and let P ∈ B(H) be given. Showthat P is the orthogonal projection of H onto M if and only if P 2 = P ,P ∗ = P, and range(P ) = M.

C.25. Suppose that A, B ∈ B(H) are self-adjoint. Show that ABA and BABare self-adjoint, but AB is self-adjoint if and only if AB = BA. Exhibit self-adjoint operators A, B that do not commute.

C.26. Let L : H → H be a self-adjoint operator on H, either bounded orunbounded and densely defined. Show that all eigenvalues of L are real, andthat eigenvectors of L corresponding to distinct eigenvalues are orthogonal.

C.27. Let L ∈ B(H) be normal. Show that if λ is an eigenvalue of L, thenλ is an eigenvalue of L∗, and the λ-eigenspace of L equals the λ-eigenspaceof L∗. Also show that eigenvectors of L corresponding to distinct eigenvaluesare orthogonal.

C.7 Compact Operators on Hilbert Spaces 323

C.28. Let A ∈ B(H) be given. Show that if A ∈ B(H) is a positive oper-ator, then all eigenvalues of A are real and nonnegative, and if A ∈ B(H)is a positive definite operator, then all eigenvalues of A are real and strictlypositive.

C.7 Compact Operators on Hilbert Spaces

In many senses, compact operators on a Hilbert space are the ones that aremost similar to linear operators on finite-dimensional spaces.

Throughout this section, H and K will denote Hilbert spaces.

C.7.1 Definition and Basic Properties

The closed unit ball makes frequent appearances in any discussion of compactoperators, so we introduce a notation for it.

Notation C.52. BallH will denote the closed unit ball in a Hilbert space H :

BallH ={f ∈ H : ‖f‖ ≤ 1

}.

By Problem A.25, BallH is compact if and only if H is finite-dimensional.If H and K are finite-dimensional and T : H → K is linear, then T willmap the closed unit ball in H to a closed ellipsoid in K, which is necessarilycompact. Even if K is infinite-dimensional, if H is finite-dimensional thenT (BallH) will be a “finite-dimensional ellipsoid” in K, and hence will still becompact. However, this need not be the case if H is infinite-dimensional. Forexample, if H is infinite-dimensional and I : H → H is the identity map, thenI(BallH) = BallH is not compact. In general, even if T : H → K is boundedand linear, T (BallH) need not be closed in K (see Problem C.32).

Definition C.53 (Compact Operators). A linear operator T : H → K iscompact 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).

The next exercise gives some useful reformulations of the definition ofcompact operator.

Exercise C.54. Given a linear operator T : H → K, show that the followingstatements are equivalent.

(a) T is compact.

(b) T (BallH) is totally bounded.


(c) If {fn}n∈N is a bounded sequence in H, then {Tfn}n∈N contains a con-vergent subsequence in K.

We show now that all compact operators are bounded.

Theorem C.55. B0(H,K) ⊆ B(H,K).

Proof. Assume that T : H → K is linear but unbounded. Then there existvectors fn ∈ H such that ‖fn‖ = 1 but ‖Tfn‖ ≥ n. Therefore every subse-quence of {Tfn}n∈N is unbounded, and hence cannot converge. Exercise C.54therefore implies that T is not compact. ⊓⊔

Now we consider the structure of the space of compact operators.

Theorem C.56 (Limits of Compact Operators). B0(H,K) is a closedsubspace of B(H,K). Specifically, if Tn : H → K are compact, T : H → K islinear and bounded, and ‖T − Tn‖ → 0, then T is compact.

Proof. Exercise: Show that B0(H,K) is a subspace of B(H,K).We must show that B0(H,K) is closed. Assume that Tn are compact op-

erators and that Tn → T in operator norm. By Exercise C.54, to show that Tis compact, it suffices to show that T (BallH) is a totally bounded subset of K.

Choose any ε > 0. Then there exists an n such that ‖T − Tn‖ < ε. SinceTn is compact, we know that Tn(BallH) is totally bounded. This implies thatthere exist finitely many points h1, . . . , hm ∈ BallH such that

Tn(BallH) ⊆m⋃

j=1

Bε(Tnhj). (C.13)

We will show that T (BallH) is totally bounded by showing that

T (BallH) ⊆m⋃

j=1

B3ε(Thj). (C.14)

Choose any element of T (BallH), i.e., any vector Tf with ‖f‖ ≤ 1. ThenTnf ∈ Tn(BallH), so by equation (C.13) there must be some j such that‖Tnf − Tnhj‖ < ε. Consequently,

‖Tf − Thj‖ ≤ ‖Tf − Tnf‖ + ‖Tnf − Tnhj‖ + ‖Tnhj − Thj‖

< ‖T − Tn‖ ‖f‖ + ε+ ‖Tn − T ‖ ‖hj‖

< ε · 1 + ε+ ε · 1 = 3ε.

Hence equation (C.14) follows, so T is compact. ⊓⊔

Compact operators are well-behaved with respect to compositions.

Exercise C.57. Let H1, H2, H3 be Hilbert spaces.


(a) Show that if A : H1 → H2 is bounded and linear and T : H2 → H3 iscompact, then TA : H1 → H3 is compact.

(b) Show that if T : H1 → H2 is compact and A : H2 → H3 is bounded andlinear, then AT : H1 → H3 is compact.

In particular, B0(H) is a closed, two-sided ideal in B(H) under compositionof operators.

It is possible for the product of two noncompact operators to be com-pact. An example is given in Exercise 2.52, which shows that the product ofa noncompact time-limiting operator with a noncompact frequency-limitingoperator can be compact.

C.7.2 Finite-Rank Operators

Recall that the rank of an operator T : H → K is the vector space dimensionof range(T ). We say that T is a finite-rank operator if range(T ) is finite-dimensional. We set

B00(H,K) ={T ∈ B(H,K) : T is finite-rank

},

and let B00(H) = B00(H,H).Problem C.8 shows that a linear, finite-rank operator need not be bounded

(that is why we include the assumption of boundedness in the definitionof B00(H,K) above). On the other hand, every finite-rank operator that isbounded must be compact.

Exercise C.58. Show that if T : H → K is bounded, linear, and has finiterank, then T is compact. Thus,

B00(H,K) ⊆ B0(H,K).

Combining the preceding exercise with Theorem C.56, we obtain a usefultechnique for showing that a given operator is compact. Specifically, if we aregiven T : H → K, and if we can find bounded linear finite-rank operators Tn

such that ‖T − Tn‖ → 0, then T must be compact. Moreover, the converse isalso true: Every compact operator T : H → K can be realized as the operatornorm limit of a sequence of bounded finite-rank operators.

Theorem C.59. B00(H,K) is a dense subspace of B0(H,K). In particular,if T : H → K is compact, then there exist bounded linear finite-rank operatorsTn : H → K such that Tn → T in operator norm.

Proof. Assume that T is compact, and set R = range(T ). If R is finite-dimensional, then T is finite-rank, and we are done. So, assume that Ris infinite-dimensional. Problem C.34 shows that R is separable. Therefore,


since R is closed, we can find a countable orthonormal basis {en}n∈N for R.For any f ∈ H we have Tf ∈ R, so

Tf =

∞∑

n=1

〈Tf, en〉 en, f ∈ H.

Define

TNf =

N∑

n=1

〈Tf, en〉 en, f ∈ H,

and note that TN = PNT where PN is the orthogonal projection ofK onto theclosed subspace span{e1, . . . , eN}. By definition, we have that TN convergesto T in the strong operator topology, i.e., TNf → Tf for every f. Our goal isto show that TN → T in operator norm.

Choose any ε > 0. Since T (BallH) is totally bounded, it is coveredby finitely many ε-balls centered at points in T (BallH). Hence, there existh1, . . . , hm ∈ BallH such that

T (BallH) ⊆m⋃

k=1

Bε(Thk).

Since limN→∞ ‖Thk − TNhk‖ = 0 for k = 1, . . . ,m, we can find an N0 suchthat

∀N > N0, ‖Thk − TNhk‖ < ε, k = 1, . . . ,m.

Choose any f with ‖f‖ = 1 and any N > N0. Then Tf ∈ Bε(Thk) forsome k, i.e., ‖Tf − Thk‖ < ε. Therefore

‖TNf − TNhk‖ = ‖PNTf − PNThk‖ ≤ ‖PN‖ ‖Tf − Thk‖ < 1 · ε,

so

‖Tf − TNf‖ ≤ ‖Tf − Thk‖ + ‖Thk − TNhk‖ + ‖TNhk − TNf‖

< ε+ ε+ ε = 3ε.

This is true for every unit vector, so ‖T − TN‖ ≤ 3ε for all N > N0, and thus‖T − TN‖ → 0. ⊓⊔

The next exercise characterizes the finite-rank operators, and uses thischaracterization to show that compactness is preserved under adjoints.

Exercise C.60. (a) Show that T ∈ B(H,K) has finite rank if and only ifthere exist ϕ1, . . . , ϕN ∈ H and ψ1, . . . , ψN ∈ K such that

Tf =

N∑

k=1

〈f, ϕk〉ψk, f ∈ H. (C.15)

In case this holds, find T ∗ and show that T ∗ is also finite-rank.

(b) Show that if T ∈ B(H,K), then T is compact if and only if T ∗ is compact.

Following Notation C.18, an alternative way to write the finite-rank oper-ator given in equation (C.15) is T =

∑Nk=1 (ψk ⊗ ϕk).


C.7.3 Integral Operators with Square-Integrable Kernels

The next exercise shows that integral operators that have a square-integrablekernel (which we call the Hilbert–Schmidt integral operators) are compact. Wewill refine this yet further in Theorem C.79 below.

Exercise C.61 (Hilbert–Schmidt Integral Operators). Fix k ∈ L2(R2),and let Lk be the integral operator with kernel k.

(a) Let {emn}m,n∈N be an orthonormal basis for L2(R2) of the form emn =em ⊗ en, as constructed in Problem C.29. Define

kN =

N∑

m=1

N∑

n=1

〈k, emn〉 emn,

and show that the corresponding integral operator LkNis bounded and

finite-rank.

(b) Show that Lk −LkNis the integral operator with kernel k− kN , and that

‖Lk − LkN‖ ≤ ‖k − kN‖2. Conclude that Lk is a compact mapping of

L2(R) into itself.

Additional Problems

C.29. Let {en}n∈N be any orthonormal basis for L2(R). Define

emn(x, y) = (em ⊗ en)(x, y) = em(x) en(y), x, y ∈ R.

Show that {emn}m,n∈N is an orthonormal basis for L2(R2).

C.30. Suppose that T : H → K is compact and {en}n∈N is an orthonormalsequence in H. Show that Ten → 0.

C.31. LetH be an infinite-dimensional Hilbert space. Show that if T : H → Kis compact and injective, then T−1 : range(T ) → H is unbounded.

C.32. Fix λ ∈ ℓ∞, and let Mλ be the multiplication operator defined inExercise C.14. Show that Mλ is compact if and only if λ ∈ c0 (i.e., λn → 0as n→ ∞). Show that if λ ∈ c0 and λn 6= 0 for every n, then T (BallH) is notclosed in ℓ2.

C.33. Fix φ ∈ L∞(R), and let Mφ be the multiplication operator defined inExercise C.15, where we take p = 2. Show that Mφ is compact if and only ifφ = 0 a.e.

C.34. Prove that if T ∈ B0(H,K), then range(T ) is a separable subspaceof K.

C.35. Show that if T : H → H is compact and λ 6= 0 is an eigenvalue of T,then the corresponding eigenspace ker(T − λI) is finite-dimensional.


C.8 The Spectral Theorem for Compact Self-AdjointOperators

Throughout this section, H will denote a Hilbert space.The Spectral Theorem provides a fundamental decomposition for self-

adjoint operators on a Hilbert space. The version of the Spectral Theoremthat we will need in this volume asserts that if T is a compact self-adjointoperator on H, then there is an orthonormal basis for H that consists ofeigenvectors of T, and T has a simple representation with respect to this or-thonormal basis. This is only a basic version of the Spectral Theorem. Thetheorem generalizes to compact normal operators with little change, allowingeigenvalues to be complex instead of real. Much more deeply, however, thereis a Spectral Theorem for self-adjoint operators that are not compact, andeven for some that are unbounded. We refer to texts on operator theory forsuch extensions, e.g., see Conway [Con00].

C.8.1 Existence of an Eigenvalue

In the finite-dimensional setting, the first step in proving the Spectral The-orem is to note that every n × n matrix always has at least one (complex)eigenvalue. Unfortunately, this is not true in infinite dimensions, even for com-pact operators.

Exercise C.62. (a) The Volterra operator on L2[0, 1] is the integral operatorV f(x) =

∫ x

0 f(y) dy. Show that V is compact, not self-adjoint, and has noeigenvalues.

(b) Show that the multiplication operator M : L2[0, 1] → L2[0, 1] given byMf(x) = xf(x) is self-adjoint (in fact, positive definite), but is not compactand has no eigenvalues.

On the other hand, we will show that an operator that is both compactand self-adjoint must have an eigenvalue. In order to do this, we first need thefollowing sufficient condition for the existence of an eigenvalue of a compactoperator.

Lemma C.63. If T : H → H is compact and λ 6= 0, then

inf‖f‖=1

‖Tf − λf‖ = 0 =⇒ λ is an eigenvalue of T.

Proof. Suppose we can find unit vectors fn such that ‖Tfn−λfn‖ → 0. SinceT is compact, {Tfn}n∈N has a convergent subsequence, say Tfnk

→ g ∈ H.Then

λfnk=

(λfnk

− Tfnk

)+ Tfnk

→ 0 + g = g.

Since the fnkare unit vectors and λ 6= 0, we conclude that g 6= 0. Moreover,

since T is continuous it follows that λTfnk→ Tg. But we also know that

λTfnk→ λg, so we conclude that Tg = λg. ⊓⊔

C.8 The Spectral Theorem for Compact Self-Adjoint Operators 329

Now we show that compactness combined with self-adjointness implies theexistence of an eigenvalue. Since the eigenvalues of a self-adjoint operator arereal and are bounded in absolute value by the operator norm, we know thatany eigenvalue λ must lie in the range −‖T ‖ ≤ λ ≤ ‖T ‖.

Theorem C.64. If T : H → H is compact and self-adjoint, then either ‖T ‖or −‖T ‖ is an eigenvalue of T.

Proof. Since T is self-adjoint, we have ‖T ‖ = sup‖f‖=1 |〈Tf, f〉| by Theo-

rem C.48. Hence, there must exist unit vectors fn such that |〈Tfn, fn〉| → ‖T ‖.Since T is self-adjoint, every inner product 〈Tfn, fn〉 is real, so we can find asubsequence {〈Tgn, gn〉}n∈N that converges either to ‖T ‖ or to −‖T ‖. Let λbe either ‖T ‖ or −‖T ‖, as appropriate. Then we have ‖gn‖ = 1 for every nand 〈Tgn, gn〉 → λ. Hence, since both λ and 〈Tgn, gn〉 are real,

‖Tgn − λgn‖2 = ‖Tgn‖

2 − 2λ 〈Tgn, gn〉 + λ2 ‖gn‖2

≤ ‖T ‖2 ‖gn‖2 − 2λ 〈Tgn, gn〉 + λ2 ‖gn‖

2

= λ2 − 2λ 〈Tgn, gn〉 + λ2

→ λ2 − 2λ2 + λ2 = 0.

It therefore follows from Lemma C.63 that λ is an eigenvalue of T. ⊓⊔

C.8.2 The Spectral Theorem

We obtain the Spectral Theorem by iterating Theorem C.64. The index set Jin the following theorem is either J = {1, . . . , N} with N = dim(range(T ))if T has finite rank, or J = N otherwise.

Theorem C.65 (Spectral Theorem for Compact Self-Adjoint Oper-ators). Let T : H → H be compact and self-adjoint. Then, there exist:

(a) countably many nonzero real numbers {λn}n∈J , either finitely many orwith λn → 0 if infinitely many, and

(b) an orthonormal basis {en}n∈J of range(T ),

such thatTf =

∑

n∈J

λn 〈f, en〉 en, f ∈ H. (C.16)

Each λn is an eigenvalue of T, and each en is a corresponding eigenvector.

Proof. If T = 0 then we can take J = ∅, so assume that T is not the zero

operator. Since T is compact, Problem C.34 implies that range(T ) is separable,and hence has either a finite or a countably infinite orthonormal basis.

Define H1 = H and T1 = T. By Theorem C.64, T1 has an eigenvalue λ1

that satisfies |λ1| = ‖T1‖ > 0. Let e1 be a corresponding eigenvector, normal-ized so that ‖e1‖ = 1.


Let H2 = {e1}⊥ and let T2 = T |H2. If T2 = 0, then we stop at this point.

Otherwise, we continue as follows. Since span{e1} is invariant under T1, weknow from Exercise C.45 that H2 is invariant under T ∗

1 = T1.Exercise: Show that T2 : H2 → H2 is compact and self-adjoint.Therefore T2 has an eigenvalue λ2 such that |λ2| = ‖T2‖ > 0, and since T2

is a restriction of T1, we have |λ2| = ‖T2‖ ≤ ‖T1‖ = |λ1|. Let e2 be a corre-sponding unit eigenvector. By definition of H2, we have e2 ⊥ e1. Further, λ2

is an eigenvalue of T (not just T2), and e2 is the corresponding eigenvectorof T.

Now set H3 = {e1, e2}⊥ and T3 = T |H3. If T3 = 0, then we stop at this

point. Otherwise, we continue as before to construct an eigenvalue λ3 and uniteigenvector e3 (which will be orthogonal to both e1 and e2). As we continuein this process, there are two possibilities.

Case 1: TN+1 = 0 for some N . In this case, since HN+1 = {e1, . . . , eN}⊥,we have

H = span{e1, . . . , eN} ⊕HN+1.

Therefore, each f ∈ H can be written uniquely as

f =

N∑

n=1

〈f, en〉 en + vf

where vf ∈ HN+1. Since Tvf = TN+1vf = 0, we have

Tf =N∑

n=1

〈f, en〉Ten + Tvf =N∑

n=1

λn 〈f, en〉 en.

In this case T is finite-rank and has the required form.

Case 2: TN 6= 0 for any N . In this case we obtain countably many eigenval-ues λn and corresponding orthonormal eigenvectors en. Since T is compact,Problem C.30 implies that Ten → 0. Since Ten = λnen and ‖en‖ = 1, weconclude that λn → 0.

Let M = span{en}n∈N. Then {en}n∈N is an orthonormal basis for M, andH = M ⊕M⊥. Hence, each f ∈ H can be written uniquely as

f =

∞∑

n=1

〈f, en〉 en + vf

for some vf ∈M⊥. Therefore

Tf =

∞∑

n=1

〈f, en〉Ten + Tvf =

∞∑

n=1

λn 〈f, en〉 en + Tvf .

Note that since span{e1, . . . , eN} ⊆M, we have for any N > 0 that

C.8 The Spectral Theorem for Compact Self-Adjoint Operators 331

vf ∈ M⊥ ⊆ span{e1, . . . , eN}⊥ = HN .

Since T |HN= TN , we therefore have

‖Tvf‖ = ‖TNvf‖ ≤ ‖TN‖ ‖vf‖ = |λN | ‖vf‖ → 0 as N → ∞.

Consequently Tvf = 0, so T has the required form.

Exercise: Complete the proof by showing that, in either case, {en}n∈J is

an orthonormal basis for range(T ). ⊓⊔

Since the nonzero eigenvalues of a compact self-adjoint operator convergeto zero (if there are infinitely many), we usually order them in decreasingorder according to absolute value:

|λ1| ≥ |λ2| ≥ · · · .

The multiplicity of a given eigenvalue λ is the number of times it is repeatedin the list of eigenvalues given above, or, equivalently, it is the dimension ofthe λ-eigenspace ker(T−λI). The multiplicity of any given eigenvalue is finite.

The Spectral Theorem is stated above in terms of the nonzero eigenval-ues of T. The zero eigenspace is ker(T ), and the vectors in this space areeigenvectors of T for λ = 0. If H is separable, then ker(T ) is separable, soby combining an orthonormal basis for ker(T ) with the orthonormal basis for

range(T ) given by Theorem C.65, we obtain the following reformulation ofthe Spectral Theorem, which states that every compact, self-adjoint operatoron a separable Hilbert space is actually a multiplication operator Mλ of theform introduced in Exercise C.14.

Exercise C.66. LetH be a separable Hilbert space. Let J = {1, . . . ,dim(H)}if H is finite-dimensional, or J = N if H is infinite-dimensional. Given T ∈B(H), show that the following statements are equivalent.

(a) T is compact and self-adjoint.

(b) There exists a real sequence λ = {λn}n∈J , with λ ∈ c0 if J = N, and anorthonormal basis {en}n∈J for H such that

Tf = Mλf =∑

n∈J

λn 〈f, en〉 en, f ∈ H.

For the finite-dimensional setting, the Spectral Theorem corresponds tothe diagonalization of Hermitian matrices.

Exercise C.67. Let A be an n× n complex matrix, viewed as a mapping ofCn into itself. Show that A is a Hermitian matrix if and only if A = UΛU−1

where U is unitary and Λ is real and diagonal. Also show that, in this case,the diagonal entries of Λ are the eigenvalues of A, and the columns of U forman orthonormal basis for Cn consisting of eigenvectors of A.


As an application of the Spectral Theorem, we construct a square root ofa compact positive operator.

Exercise C.68. Assume that T : H → H is both compact and positive, andlet equation (C.16) be the representation of T given by the Spectral Theorem.Define

T 1/2f =∑

n∈J

λ1/2n 〈f, en〉 en, f ∈ H.

Show that T 1/2 is well-defined, compact, and positive, and that (T 1/2)2 =T 1/2T 1/2 = T.

Additional Problems

C.36. Define the spectral radius ρ(T ) of an m×m matrix T to be

ρ(T ) = max{|λ| : λ is an eigenvalue of T }.

Let A be an m×n complex matrix. Show directly that if we place the ℓ2 normon both Cn and Cm, then the operator norm of A as a mapping of Cn intoCm is ‖A‖ = ρ(A∗A)1/2. Compare this to Problem C.7.

C.9 Hilbert–Schmidt Operators

The Hilbert–Schmidt operators are a special subclass of the compact opera-tors on a Hilbert space. Although the definition of a Hilbert–Schmidt operatormakes sense when H is nonseparable (by using a complete orthonormal sys-tem instead of a countable orthonormal basis), in this volume we will onlyneed to consider Hilbert–Schmidt operators on separable spaces. Therefore,throughout this section, H will denote a separable Hilbert space.

C.9.1 Definition and Basic Properties

Definition C.69. An operator T ∈ B(H) is Hilbert–Schmidt if there existsan orthonormal basis {en}n∈N for H such that

∑∞n=1 ‖Ten‖2 <∞.

The next exercise shows that the choice of orthonormal basis is irrelevant.

Exercise C.70. Fix T ∈ B(H). Given an orthonormal basis E = {en}n∈N

for H, set

S(E) =

( ∞∑

n=1

‖Ten‖2

)1/2

.

Show that S(E) is independent of the choice of orthonormal basis E . That is,if S(E) is finite for one orthonormal basis then it is finite for all and alwaystakes the same value, and if S(E) is infinite for one orthonormal basis then itis infinite for all.

C.9 Hilbert–Schmidt Operators 333

Definition C.71. The space of Hilbert–Schmidt operators on H is

B2(H) ={T ∈ B(H) : T is Hilbert–Schmidt

}.

The Hilbert–Schmidt norm of T ∈ B2(H) is

‖T ‖HS = ‖T ‖B2=

( ∞∑

n=1

‖Ten‖2

)1/2

,

where {en}n∈N is any orthonormal basis for H.

The notation B2(H) emphasizes that the space of Hilbert–Schmidt oper-ators is the central (p = 2) space in the family of Schatten classes Bp(H)described below. We will use the notations ‖ · ‖HS and ‖ · ‖B2

interchangeably.Note that if we write

‖T ‖2B2

=

∞∑

n=1

‖Ten‖2 =

∞∑

n=1

∞∑

m=1

|〈Ten, em〉|2, (C.17)

then we see that the Hilbert–Schmidt norm is simply the ℓ2-norm of the matrixrepresentation of T with respect to the orthonormal basis {en}n∈N.

The fact that ‖·‖B2is indeed a norm on B2(H), and some of the other basic

properties of Hilbert–Schmidt operators, are developed in the next exercise.

Exercise C.72. Prove the following facts.

(a) ‖T ‖ ≤ ‖T ‖B2for all T ∈ B2(H).

(b) ‖ · ‖B2is a norm on B2(H), and B2(H) is complete in this norm.

(c) B2(H) is closed under adjoints, and ‖T ∗‖B2= ‖T ‖B2

for all T ∈ B2(H).

(d) If T ∈ B2(H) and A ∈ B(H), then AT , TA ∈ B2(H), and we have

‖AT ‖B2≤ ‖A‖ ‖T ‖B2

, ‖TA‖B2≤ ‖A‖ ‖T ‖B2

.

Consequently, B2(H) is a two-sided ideal in B(H).

(e) All finite-rank operators are Hilbert–Schmidt, i.e., B00(H) ⊆ B2(H).

(f) All Hilbert–Schmidt operators are compact, i.e., B2(H) ⊆ B0(H).

(g) The space B00(H) of finite-rank operators is dense in B2(H) with respectto both the operator norm and the Hilbert–Schmidt norm.

C.9.2 Singular Numbers and Schatten Classes

We will give an equivalent formulation of Hilbert–Schmidt operators in termsof their singular numbers. Singular numbers can be defined for any com-pact operator T : H → H. Although a compact operator need not have anyeigenvalues, the operator T ∗T is both compact and self-adjoint (in fact, pos-itive). Therefore, we can apply the Spectral Theorem to T ∗T to deduce that


there exists an orthonormal sequence {en}n∈J and corresponding real num-bers {µn}n∈J such that

T ∗Tf =∑

n∈J

µn 〈f, en〉 en, f ∈ H.

The µn are the nonzero eigenvalues of T ∗T, and since T ∗T is positive, theseµn are strictly positive. There are either finitely many nonzero eigenvalues ofT ∗T, or if infinitely many then we have µn → 0. The index set J is eitherJ = {1, . . . , N} with N = dim(range(T ∗T )) if T ∗T is finite-rank (which byProblem C.18 happens if and only if T is finite-rank), or J = N otherwise.

Definition C.73 (Singular Numbers). Let T : H → H be compact, andlet {µn}n∈J and {en}n∈J be as constructed above. Then the singular numbersor singular values of T are

sn(T ) = µ1/2n , n ∈ J,

taken in decreasing order:

s1(T ) ≥ s2(T ) ≥ · · · > 0.

The vectors en are corresponding singular vectors of T.

In particular, a bounded finite-rank operator has only finitely manynonzero singular numbers.

The singular numbers of a self-adjoint operator can be written directly interms of its eigenvalues.

Exercise C.74. Show that if T is compact and self-adjoint and {λn(T )}n∈J

are its nonzero eigenvalues taken in decreasing order, then sn(T ) = |λn(T )|.Consequently, if T is compact and positive then sn(T ) = λn(T ).

Exercise C.68 shows how to construct the square root of a compact positiveoperator. If T is an arbitrary compact operator, then T ∗T is positive andcompact, and we have a special name for its square root.

Definition C.75 (Absolute Value). If T : H → H is compact, then thepositive compact operator

|T | = (T ∗T )1/2

is the absolute value of T.

In this terminology, the singular numbers of a compact operator T are thenonzero eigenvalues of |T | = (T ∗T )1/2, so we have the relations

sn(T ) = λn(|T |) = λn(T ∗T )1/2.

In particular, since T ∗T is positive, its largest eigenvalue is its operator norm.Therefore,

s1(T ) = λ1(T∗T )1/2 = ‖T ∗T ‖1/2 = ‖T ‖, (C.18)

the final equality following from Problem C.14.


Remark C.76. It can be shown that every bounded operator A on H can bewritten in the form A = UP where P is positive and U is a partial isometry(meaning that ‖Uf‖ = ‖f‖ for all f ∈ (kerU)⊥), and this polar decompositionis unique if we impose the condition that ker(U) = ker(P ). In particular, ifA is compact then P coincides with the operator |A| defined above. Thus,absolute value can be defined for arbitrary bounded operators, although wewill only need it for compact operators in this volume.

Now we can reformulate the definition of Hilbert–Schmidt operators interms of singular numbers.

Exercise C.77. Let T : H → H be compact, and let s = {sn(T )}n∈J be thesingular numbers of T. Show that T is Hilbert–Schmidt if and only if s ∈ ℓ2(J),and, in this case,

‖T ‖B2= ‖s‖2 =

(∑

n∈J

sn(T )2)1/2

.

More generally, the Schatten class Bp(H) is the space of compact operatorsthat have ℓp singular numbers.

Definition C.78 (Schatten Class). Given 1 ≤ p < ∞, the Schatten classBp(H) consists of all compact operators T : H → H whose correspondingsingular numbers {sn(T )}n∈J satisfy

‖T ‖Bp=

(∑

n∈J

sn(T )p

)1/p

< ∞.

Note that since a bounded finite-rank operator has only finitely manysingular numbers, we have B00(H) ⊆ Bp(H), for each 1 ≤ p < ∞. In fact, itcan be shown that the finite-rank operators are dense in Bp(H) for each p.Since the ℓ∞-norm of the singular values is

‖s‖∞ = supn∈J

sn(T ) = s1(T ) = ‖T ‖,

we usually define the p = ∞ Schatten class to be B∞(H) = B0(H), the spaceof all compact operators on H.

As in the discussion leading up to Exercise C.66, it is often convenient toextend the singular vectors so that we obtain an orthonormal basis for all ofH.The singular vectors {en}n∈J as we have defined them form an orthonormal

basis for range(T ). Since T ∗T is self-adjoint, we have

range(T )⊥

= ker(T ∗T ).

Hence we can form an orthonormal basis for all of H by combining {en}n∈N

with an orthonormal basis {fm}m∈I for ker(T ∗T ). We can regard the vectors


fm as being singular vectors of T corresponding to zero singular numbers.Each vector f ∈ H can be written

f =∑

n∈J

〈f, en〉 en +∑

n∈I

〈f, fm〉 fm,

and we have

T ∗Tf =∑

n∈J

〈f, en〉T∗Ten +

∑

n∈I

〈f, fm〉T ∗Tfm

=∑

n∈J

sn(T )2 〈f, en〉 en.

C.9.3 Hilbert–Schmidt Integral Operators

We now specialize to the case of integral operators on L2(R). Theorem C.19showed that if k ∈ L2(R2), then the corresponding integral operator Lk is abounded operator on L2(R), with operator norm ‖Lk‖ ≤ ‖k‖2. Exercise C.61further showed that Lk is compact. The next result shows that Lk is actuallyHilbert–Schmidt when k ∈ L2(R2), and further that all Hilbert–Schmidt op-erators on L2(R) can be written as integral operators with square-integrablekernels.

Theorem C.79 (Hilbert–Schmidt Kernel Theorem).

(a) If k ∈ L2(R2), then the integral operator Lk with kernel k is Hilbert–Schmidt, and

‖Lk‖B2= ‖k‖2.

(b) If T is a Hilbert–Schmidt operator on L2(R), then there exists k ∈ L2(R2)such that T = Lk.

(c) The mapping k 7→ Lk is an isometric isomorphism of L2(R2) ontoB2(L

2(R)).

Proof. (a) Assume that k ∈ L2(R2), and choose any orthonormal basis{en}n∈N for L2(R). Problem C.29 shows that if we set

emn(x, y) = (em ⊗ en)(x, y) = em(x) en(y),

then {emn}m,n∈N is an orthonormal basis for L2(R2). Since k · emn ∈ L1(R2),Fubini’s Theorem allows us to interchange integrals in the following calcula-tion:

〈k, emn〉 =

∫∫k(x, y) em(x) en(y) dx dy

=

∫ (∫k(x, y) en(y) dy

)em(x) dx

=

∫Lken(x) em(x) dx = 〈Lken, em〉.


Therefore,

‖Lk‖2B2

=

∞∑

n=1

‖Lken‖2 =

∞∑

n=1

( ∞∑

m=1

|〈Lken, em〉|2)

=∞∑

m=1

∞∑

n=1

|〈k, emn〉|2 = ‖k‖2

2 < ∞,

so Lk is Hilbert–Schmidt and the mapping k 7→ Lk is isometric.

(b) Suppose that T is Hilbert–Schmidt on L2(R). Choose any orthonormalbasis {en}n∈N for L2(R), and let emn be as above. Equation (C.17) shows that{〈Tem, en〉}m,n∈N ∈ ℓ2(N2). Therefore, the series

k =

∞∑

m=1

∞∑

n=1

〈Ten, em〉 emn

converges unconditionally in L2(R2). Hence k ∈ L2(R2), so the integral oper-ator Lk is Hilbert–Schmidt. Since part (a) shows that k 7→ Lk is an isometry,we conclude that

Lk =∞∑

m=1

∞∑

n=1

〈Ten, em〉Lemn, (C.19)

with unconditional convergence of this series in Hilbert–Schmidt norm. Sincethe Hilbert–Schmidt norm dominates the operator norm, the series in equa-tion (C.19) also converges unconditionally in operator norm. Hence, givenf ∈ L2(R), we have that

Lkf =∞∑

m=1

∞∑

n=1

〈Ten, em〉Lemnf,

where the series converges unconditionally in L2(R). Now, Example C.17shows that Lemn

is the rank one operator Lemnf = 〈f, en〉 em. Therefore,

using the unconditionality of the convergence to reorder the summations, wehave

Lkf =

∞∑

m=1

∞∑

n=1

〈Ten, em〉〈f, en〉 em

=

∞∑

n=1

〈f, en〉

( ∞∑

m=1

〈Ten, em〉 em

)

=

∞∑

n=1

〈f, en〉Ten

= T

( ∞∑

n=1

〈f, en〉 en

)= Tf,


where the continuity of T is used to obtain the penultimate equality. HenceT = Lk. ⊓⊔

Thus B2(L2(R)) is isometrically isomorphic to L2(R2), and therefore in-

herits a Hilbert space structure from L2(R2). In terms of kernels, the innerproduct on B2(L

2(R)) is〈Lk, Lh〉 = 〈k, h〉.

B2(L2(R)) is a separable Hilbert space, for if {en}n∈N is any orthonormal basis

for L2(R) and we set emn = em ⊗ en, then {Lemn}m,n∈N is an orthonormal

basis for B2(L2(R)). By Example C.17, each operator Lemn

is a bounded,rank one operator whose range is span{em}. As discussed in Notation C.18,we often identify this operator with the function em ⊗ en, and write

Lemn= em ⊗ en.

Thus {em ⊗ en}m,n∈N is an orthonormal basis for B2(L2(R)) consisting of

bounded rank one operators, and equation (C.19) is the orthonormal basisrepresentation of a Hilbert–Schmidt operator as a superposition of rank oneoperators, which we can also write as

T =

∞∑

m=1

∞∑

n=1

〈Ten, em〉 (em ⊗ en).

This series converges both in Hilbert–Schmidt norm and in operator norm.

C.9.4 Trace-Class Operators

Definition C.80 (Trace-Class Operators). A compact operator T ∈ B(H)is trace-class if its singular numbers {sn(T )}n∈J satisfy

∑

n

sn(T ) < ∞.

The space of trace-class operators on H is the Schatten class B1(H).

Since the largest singular number of a compact operator T is s1(T ) = ‖T ‖,the operator norm is dominated by the trace-class norm, or indeed by anySchatten-class norm:

‖T ‖ ≤ ‖T ‖Bp, 1 ≤ p ≤ ∞.

The next exercise provides an equivalent definition of the trace-class normreminiscent of Definition C.71.

Exercise C.81. Suppose that T : H → H is compact, and prove the followingfacts.

C.10 The Hahn–Banach Theorem 339

(a) The quantity∑∞

n=1

⟨|T |en, en

⟩is independent of the choice of orthonor-

mal basis {en}n∈N.

(b) T ∈ B1(H) if and only if∑∞

n=1

⟨|T |en, en

⟩< ∞ for some orthonormal

basis {en}n∈N.

(c) If T ∈ B1(H), then ‖T ‖B1=

∑∞n=1

⟨|T |en, en

⟩.

(d) T ∈ B1(H) if and only if |T |1/2 ∈ B2(H).

The trace class B1(H) is a Banach space with respect to the norm ‖ · ‖B1,

and is also a two-sided ideal in B(H). All finite-rank operators are trace-class,and every trace-class operator is the limit in trace-class norm of a sequenceof finite-rank operators.

The trace of a trace-class operator T ∈ B1(H) is

trace(T ) =∞∑

n=1

⟨Ten, en

⟩,

where {en}n∈N is an orthonormal basis for H. This series converges and isindependent of the choice of orthonormal basis. Thus the trace is the sum ofthe diagonal elements of the matrix representation of T with respect to anorthonormal basis.

The trace induces an inner product on the space of Hilbert–Schmidt op-erators. Specifically, if A, B ∈ B2(H), then AB∗ is trace-class, and

〈A,B〉 = trace(AB∗)

defines an inner product on B2(H) whose induced norm is exactly ‖ · ‖B2.

Additional Problems

C.37. Given g, h ∈ H and 1 ≤ p <∞, show that the Schatten-class norm ofg ⊗ h is ‖g ⊗ h‖Bp

= ‖g‖ ‖h‖, where (g ⊗ h)(f) = 〈f, h〉 g.

C.10 The Hahn–Banach Theorem

The Hahn–Banach Theorem is a powerful result dealing with extension oflinear functionals defined on subspaces of a space.

C.10.1 Abstract Statement of the Hahn–Banach Theorem

We concentrate in this volume on complex vector spaces. For such spaces, theabstract version of the Hahn–Banach Theorem is as follows.


Theorem C.82 (Hahn–Banach Theorem). Let X be a complex vectorspace and let ρ be a seminorm on X. If M is a subspace of X and λ : M → C

is a linear functional on M satisfying

|〈f, λ〉| ≤ ρ(f), f ∈M,

then there exists a linear functional Λ: X → C such that

Λ|M = λ and |〈f,Λ〉| ≤ ρ(f), f ∈ X.

For real vector spaces, the dominating seminorm ρ can be replaced bya sublinear function q : X → R that satisfies q(f + g) ≤ q(f) + q(g) andq(cf) = q(f) for all f, g ∈ X and scalars c ≥ 0.

The proof of the Hahn–Banach Theorem takes some preparation, andtherefore we will omit it. The proof for arbitrary vector spaces is not con-structive, as it relies on the Axiom of Choice in the form of Zorn’s Lemma.The Hahn–Banach Theorem is intimately tied to convexity, and there are gen-eralizations to more abstract settings such as locally convex topological vectorspaces.

C.10.2 Implications of the Hahn–Banach Theorem

In practice, it is usually not the Hahn–Banach Theorem itself but rather itscorollaries that are applied. Therefore we will concentrate on the implicationsof the Hahn–Banach Theorem. Since these corollaries are so important, wheninvoking any one of them it is customary to write “by the Hahn–BanachTheorem” instead of “by a corollary to the Hahn–Banach Theorem.”

Our first corollary states that any bounded linear functional λ on a sub-space M of a normed space X has an extension to the entire space whoseoperator norm on X equals the operator norm of λ on M. This is easy toprove when the space is a Hilbert space, but it is far from obvious that suchan extension should be be possible on non-inner product spaces. Note that wedo not require the subspace M to be closed.

Corollary C.83 (Hahn–Banach). Let X be a normed space and M a sub-space of X. If λ ∈M∗, then there exists Λ ∈ X∗ such that

Λ|M = λ and ‖Λ‖ = ‖λ‖.

Proof. Set ρ(f) = ‖λ‖ ‖f‖ for f ∈ X. Note that ρ is defined on all of X, andis a seminorm on X (in fact, it is a norm if λ 6= 0). Further,

∀ f ∈M, |〈f, λ〉| ≤ ‖f‖ ‖λ‖ = ρ(f).

Hence, Theorem C.82 implies that there exists a linear functional Λ: X → C

such that Λ|M = λ (which implies ‖Λ‖ ≥ ‖λ‖) and

∀ f ∈ X, |〈f,Λ〉| ≤ ρ(f) = ‖λ‖ ‖f‖,

which implies that ‖Λ‖ ≤ ‖λ‖. ⊓⊔


Given a normed space X and given µ ∈ X∗, the operator norm of µ is

‖µ‖ = supf∈X, ‖f‖=1

|〈f, µ〉|.

Thus, we obtain the operator norm of µ by “looking back” at its action on X.The next corollary provides a complementary viewpoint: The norm of f ∈ Xcan be obtained by “looking forward” to its action on X∗.

Corollary C.84 (Hahn–Banach). If X is a normed space and f ∈ X, then

‖f‖ = supµ∈X∗, ‖µ‖=1

|〈f, µ〉|. (C.20)

Further, the supremum is achieved.

Proof. Fix f ∈ X, and let α denote the supremum on the right-hand side ofequation (C.20). Since |〈f, µ〉| ≤ ‖f‖ ‖µ‖, we have α ≤ ‖f‖.

Let M = span{x}, and define λ : M → C by 〈cf, λ〉 = c ‖f‖. Then λ ∈M∗

and ‖λ‖ = 1. Corollary C.83 therefore implies that there exists some Λ ∈ X∗

with Λ|M = λ and ‖Λ‖ = ‖λ‖ = 1. In particular, since f ∈ M, we haveα ≥ |〈f,Λ〉| = |〈f, λ〉| = ‖f‖, and therefore the supremum in equation (C.20)is achieved. ⊓⊔

Now we can give one of the most powerful and often-used implicationsof the Hahn–Banach Theorem. It states that we can find a bounded linearfunctional that separates a point from a closed subspace of a normed space.This is easy to prove constructively for the case of a Hilbert space, but it isquite amazing that we can do this in arbitrary normed spaces.

Corollary C.85 (Hahn–Banach). LetX be a normed linear space. Supposethat:

(a) M is a closed subspace of X,

(b) f0 ∈ X\M, and

(c) d = dist(f0,M) = inf{‖f0 −m‖ : m ∈M

}.

Then there exists µ ∈ X∗ such that

〈f0, µ〉 = 1, µ|M = 0, and ‖µ‖ =1

d.

Proof. Note that d > 0 since M is closed. Define M1 = span{M,x0}. Sincef0 /∈ M, each f ∈ M1 can be written uniquely as f = mf + tff0 for somemf ∈ M and tf ∈ C. Define λ : M1 → C by 〈f, λ〉 = tf . Then λ is linear,λ|M = 0, and 〈f0, λ〉 = 1.

If f ∈M1 and tf 6= 0, then we have mf/tf ∈M, so

‖f‖ = ‖tff0 +mf‖ = |tf |∥∥∥f0 −

(−mf

tf

)∥∥∥ ≥ |tf | d.


If tf = 0 (so f ∈ M), this is still true. Hence, |〈f, λ〉| = |tf | ≤ ‖f‖/d for allf ∈M1. Therefore ‖λ‖ ≤ 1/d.

On the other hand, there exist mn ∈M such that ‖f0−mn‖ → d. Since λvanishes on M, we therefore have

1 = 〈f0, λ〉 = 〈f0 −mn, λ〉 ≤ ‖f0 −mn‖ ‖λ‖ → d ‖λ‖.

Therefore ‖λ‖ ≥ 1/d.Applying Corollary C.83, there exists a µ ∈ X∗ such that µ|M1

= λ and‖µ‖ = ‖λ‖. This functional µ has all of the required properties. ⊓⊔

C.10.3 Orthogonal Complements in Normed Spaces

We can generalize the notion of orthogonal complements of subsets of Hilbertspaces to normed spaces by allowing the orthogonal complement of S ⊆ X tobe a subset of X∗ rather than X.

Definition C.86 (Orthogonal Complement). If S is a subset of a normedspace X, then its orthogonal complement S⊥ is the subset of X∗ defined by

S⊥ ={µ ∈ X∗ : 〈f, µ〉 = 0 for all f ∈ S

}.

Exercise C.87. Show that S⊥ is a closed subspace of X∗.

The next exercise gives a useful characterization of complete sequences ina normed space. The analogous characterization for the case of Hilbert spacesappeared as Exercise A.101.

Exercise C.88. Let S be a subset of a normed space X. Prove that

S is complete ⇐⇒ S⊥ = {0}.

Consequently, a subspace M of X is dense in X if and only if M⊥ = {0}.

C.10.4 X∗∗ and Reflexivity

Let X be a normed linear space. Following our notational conventions, iff ∈ X and µ ∈ X∗ then we write 〈f, µ〉 to denote the action of µ on f, andwe take this notation to be linear in f and antilinear in µ. Holding µ fixed,the mapping f 7→ 〈f, µ〉 is a bounded linear functional on X. On the otherhand, if we hold f fixed, then µ 7→ 〈f, µ〉 is an antilinear functional on X∗.

We can make a linear mapping by considering µ 7→ 〈f, µ〉 instead. Thus, each

element f ∈ X determines a linear functional f : X∗ → C by the formula

〈µ, f 〉 = 〈f, µ〉 for µ ∈ X∗. If this linear functional is bounded, then it is abounded linear functional on X∗ and hence belongs to the dual space of X∗,

which we denote by X∗∗. The next result shows that f is indeed bounded and

has operator norm ‖f ‖ = ‖f‖. This gives us a a natural identification of Xwith a subset of X∗∗.


Theorem C.89. Let X be a normed space. Given f ∈ X, define f : X∗ → C

by

〈µ, f 〉 = 〈f, µ〉, µ ∈ X∗.

Then f is a bounded linear functional on X∗, with operator norm ‖f ‖ = ‖f‖.Consequently, the mapping

T : X → X∗∗

f 7→ f ,

is a linear isometry of X into X∗∗.

Proof. By definition of the operator norm, we have

‖f ‖ = supµ∈X∗, ‖µ‖=1

|〈µ, f 〉|.

On the other hand, by the Hahn–Banach Theorem in the form of Corol-lary C.84, we have

‖f‖ = supµ∈X∗, ‖µ‖=1

|〈f, µ〉|.

Since |〈f, µ〉| = |〈µ, f 〉|, the result follows. ⊓⊔

Definition C.90 (Natural Embedding of X into X∗∗). Let X be anormed space.

(a) The mapping T : f 7→ f defined in Theorem C.89 is called the naturalembedding or the canonical embedding of X into X∗∗.

(b) If the natural embedding of X into X∗∗ is surjective, then we say that Xis reflexive.

Remark C.91. We emphasize that in order for X to be called reflexive, thenatural embedding must be a surjective isometry. There exist Banach spacesXsuch that X ∼= X∗∗ even though X is not reflexive [Jam51].

Exercise C.92. Let E ⊆ Rd be Lebesgue measurable. Show that ℓp andLp(E) are reflexive for each 1 < p <∞.

C.10.5 Adjoints of Operators on Banach Spaces

We saw in Section C.6 that if H, K are Hilbert spaces and A ∈ B(H,K),then there exists an adjoint operator A∗ ∈ B(K,H) uniquely defined by thecondition

∀ f ∈ H, ∀ g ∈ H, 〈Af, g〉K = 〈f,A∗g〉H . (C.21)

Now we will consider the case whereX, Y are Banach spaces and A ∈ B(X,Y ).We will see that, as a consequence of the Hahn–Banach Theorem, there ex-ists a unique adjoint A∗ ∈ B(Y ∗, X∗) defined by an equation analogous toequation (C.21). However, while we have A : X → Y, the adjoint will be amap A∗ : Y ∗ → X∗. In particular, unlike the Hilbert space case, we cannotconsider compositions of A with A∗.


Exercise C.93. Let X, Y be Banach spaces, and fix A ∈ B(X,Y ).

(a) Fix µ ∈ Y ∗. Define A∗µ : X → C to be the functional whose rule is

〈f,A∗µ〉 = 〈Af, µ〉, f ∈ X.

Show that A∗µ ∈ X∗.

(b) Show that A∗ : µ 7→ A∗µ is bounded and linear, and ‖A∗‖ = ‖A‖.

(c) Prove that A∗ is the unique operator from Y ∗ to X∗ that satisfies

∀ f ∈ X, ∀µ ∈ Y ∗, 〈Af, µ〉 = 〈f,A∗µ〉. (C.22)

Definition C.94 (Adjoint). Let X, Y be Banach spaces. The adjoint of A ∈B(X,Y ) is the unique operator A∗ : Y ∗ → X∗ that satisfies equation (C.22).

Remark C.95. Our use of a sesquilinear form notation for linear functionalsallows us to create a definition of the adjoint that directly generalizes thedefinition of the adjoint of an operator on Hilbert spaces. In other words, ifH and K are Hilbert spaces and A ∈ B(H,K) then there is no differencebetween the adjoint A∗ defined in Section C.6, where we considering H andK to be Hilbert spaces, and the adjoint A∗ defined above, where we considerH and K to be Banach spaces. If we instead used a bilinear form notation,there would be a distinction between these two adjoints.

Additional Problems

C.38. Let X be a normed space. Show that if X∗ is separable then X isseparable, but the converse can fail.

C.39. Let X, Y be Banach spaces. Show that if T : X → Y is a topologicalisomorphism, then the adjoint T ∗ : Y ∗ → X∗ is a topological isomorphism,and similarly if T is an isometric isomorphism then so is T ∗.

C.11 The Baire Category Theorem

It is not possible to write the Euclidean plane R2 as the union of countablymany straight lines. This is a special case of the Baire Category Theorem,which states that a complete metric space cannot be written as a countableunion of “nowhere dense” sets.

Definition C.96 (Nowhere Dense Sets). Let X be a metric space, andlet E ⊆ X be given.

(a) E is nowhere dense or rare if X \ E is dense in X.

(b) E is meager or first category if it can be written as a countable union ofnowhere dense sets.

C.11 The Baire Category Theorem 345

(c) E is nonmeager or second category if it is not meager.

Exercise C.97. Given a nonempty subset E of a metric space X, show thatthe following statements are equivalent.

(a) E is nowhere dense.

(b) E contains no nonempty open subsets.

The Cantor set is an example of a nowhere dense subset of [0, 1]. The set ofrationals Q is not a nowhere dense subset of R, but it is meager in R. Further,Q with the metric d(x, y) = |x − y| is an incomplete metric space that is ameager subset of itself.

Now we can prove the Baire Category Theorem, which states that no com-plete metric space can be a meager subset of itself. Unlike the Hahn–BanachTheorem, the driving property here is completeness rather than convexity.

Theorem C.98 (Baire Category Theorem). A nonempty complete met-ric space is nonmeager in itself. Consequently, if

X =∞⋃

n=1En

where each En is a closed subset of X, then at least one En contains anonempty open subset.

Proof. Suppose that X = ∪En where each En is nowhere dense. Then Un =

X \ En is dense and open for each n.Choose f1 ∈ U1 and let r1 > 0 be such that B1 = Br1

(f1) ⊆ U1. Thensince U2 is dense, we can find an f2 ∈ U2 ∩ B1. Since U2 and B1 are bothopen, there exists some r2 > 0 such that B2 = Br2

(f2) ⊆ U2 ∩ B1. Withoutloss of generality, we can take r2 small enough that we have both r2 < r1/2

and B2 ⊆ B1.Continuing in this way we obtain points fn ∈ Un and ballsBn = Brn

(fn) ⊆Un such that

rn <rn−1

2and Bn ⊆ Bn−1.

In particular, rn → 0, and the balls Bn are nested.Fix ε > 0, and let N be large enough that rN < ε/2. If m, n > N, then

we have fm, fn ∈ BN . Hence d(fm, fn) < 2rN < ε. Thus {fn}n∈N is Cauchy,and since X is complete, there exists some f ∈ X such that fn → f.

Now fix any N > 0. Then, since the Bn are nested, we have fn ∈ BN+1

for all n > N. As fn → f, this implies that f ∈ BN+1 ⊆ BN . This is true forevery N, so

f ∈∞⋂

n=1

Bn ⊆∞⋂

n=1

Un =∞⋂

n=1

(X \ En).

But then f /∈ ∪En, which is a contradiction. ⊓⊔


The next exercise applies the Baire Category Theorem to show that Hamelbases for Banach spaces are generally too unweildy to be of much use.

Exercise C.99. Let X be an infinite-dimensional Banach space, and provethe following.

(a) Any Hamel basis for X must be uncountable.

(b) Any infinite-dimensional subspace of X that has a countable Hamelbasis is a meager subset of X, and cannot be a closed subspace of X.

(c) Cc(R) is a meager, dense subspace of C0(R) that does not have acountable Hamel basis.

C.12 The Uniform Boundedness Principle

The Uniform Boundedness Principle states that a family of bounded linearoperators on a Banach space that are uniformly bounded at each individualpoint must actually be uniformly bounded in operator norm.

Theorem C.100 (Uniform Boundedness Principle). Let X be a Ba-nach space and Y a normed space. If {Ai}i∈I is any collection of operatorsin B(X,Y ) such that

∀ f ∈ X, supi∈I

‖Aif‖ < ∞,

thensupi∈I

‖Ai‖ < ∞.

Proof. Set

En ={f ∈ X : sup

i∈I‖Aif‖ ≤ n

}.

Then X = ∪En by hypothesis, and since each Ai is continuous it followsthat En is closed. Consequently, the Baire Category Theorem implies thatsome En must contain an open ball, say Br(f0) ⊆ En.

Given an arbitrary f ∈ X, if we set g = f0 + sf with s = r2‖f‖ then we

have g ∈ Br(f0) ⊆ En, and therefore

‖Aif‖ =∥∥∥Ai

(g − f0s

)∥∥∥ ≤1

s

(‖Aig‖ + ‖Aif0‖

)≤

2 ‖f‖

r2n =

4n

r‖f‖.

Consequently, ‖Ai‖ ≤ 4n/r, which is a constant independent of i. ⊓⊔

The following special case of the Uniform Bounded Principle is often use-ful (sometimes the names “Uniform Boundedness Principle” and “Banach–Steinhaus Theorem” are used interchangeably).

Exercise C.101 (Banach–Steinhaus Theorem). Let X, Y be Banachspaces. If An ∈ B(X,Y ) for each n ∈ N and Af = limn→∞ Anf exists foreach f ∈ X, then A ∈ B(X,Y ) and ‖A‖ ≤ sup ‖An‖ <∞.

C.13 The Open Mapping Theorem 347

C.13 The Open Mapping Theorem

Recall that a function is continuous if the inverse image of any open set isopen. It is often important to consider direct images of open sets as well.

Definition C.102 (Open Mapping). Let X, Y be topological spaces. Afunction U : X → Y is an open mapping if

U is open in X =⇒ A(U) is open in Y.

In general, a continuous function need not be an open mapping, but theOpen Mapping Theorem asserts that any continuous linear surjection of oneBanach space onto another must do so. The key to the proof is the followinglemma. For clarity, we will write BX

r (f) and BYr (g) to distinguish open balls

in X from open balls in Y.

Lemma C.103. Let X, Y be Banach spaces and fix A ∈ B(X,Y ). If A(BX1 (0))

contains an open ball in Y, then A(BX1 (0)) contains an open ball BY

r (0) forsome r > 0.

Proof. Suppose that A(BX1 (0)) contains an open ball BY

s (g). It follows that

BYr (0) ⊆ A(BX

1 (0)) (C.23)

where r = s/2. We will show that BYr/2(0) ⊆ A(BX

1 (0)). To see this, sup-

pose h ∈ BYr/2(0), i.e., ‖h‖ < r/2. Rescaling equation (C.23), we have

h ∈ A(BX1/2(0)), so there exists some f1 ∈ X with ‖f1‖ < 1/2 such that

‖h−Af1‖ < r/4. Then h−Af1 ∈ BYr/4(0) ⊆ A(BX

1/4(0)), so there exists some

f2 ∈ X with ‖f2‖ < 1/4 such that ‖(h − Af1) − Af2‖ < r/8. Continuing inthis way, we obtain vectors fn ∈ X with ‖fn‖ < 2−n such that

‖h−Agn‖ <r

2n+1,

where gn =∑n

k=1 fk. Hence Agn → h. However, {gn}n∈N is Cauchy in X, sogn → g for some g ∈ X. Since A is continuous, it follows that h = Ag. Since‖g‖ < 1, we therefore have h ∈ A(BX

1 (0)). ⊓⊔

Theorem C.104 (Open Mapping Theorem). If X, Y are Banach spacesand A : X → Y is a continuous linear surjection, then A is an open mapping.

Proof. Since A is surjective we have

Y =∞⋃

k=1

A(BXk (0)).

The Baire Category Theorem therefore implies that some set A(BXk (0)) must

contain an open ball. Therefore, by Lemma C.103, there is some r > 0 suchthat


BYr (0) ⊆ A(BX

1 (0)). (C.24)

Now suppose that U ⊆ X is open and y ∈ A(U). Then y = Ax forsome x ∈ U, so BX

s (x) ⊆ U for some s > 0. Rescaling equation (C.24),BY

t (0) ⊆ A(BXs (0)) for some t > 0. Therefore

BYt (y) = BY

t (0) +Ax ⊆ A(BXs (0) + x) = A(BX

s (x)) ⊆ A(U),

so A(U) is open. ⊓⊔

Specializing to the case of bijections yields the following result.

Theorem C.105 (Inverse Mapping Theorem). If X, Y are Banachspaces and A : X → Y is a continuous linear bijection, then A−1 : Y → Xis continuous. Consequently A is a topological isomorphism.

The next exercise is a typical application of the Inverse Mapping Theorem.

Exercise C.106. Suppose X is a vector space that is complete with respectto two norms ‖ · ‖ and ||| · |||. If there exists C > 0 such that ‖f‖ ≤ C |||f ||| forall f ∈ X, then ‖ · ‖ and ||| · ||| are equivalent norms on X.

Additional Problems

C.40. Let X and Y be Banach spaces. Show that A ∈ B(X,Y ) is surjectiveif and only if range(A) is not meager in Y.

C.14 The Closed Graph Theorem

The Closed Graph Theorem provides a convenient means of testing whethera linear operator on Banach spaces is continuous.

Theorem C.107 (Closed Graph Theorem). Let X and Y be Banachspaces. If A : X → Y is linear, then the following statements are equivalent.

(a) A is continuous.

(b) graph(A) ={(f,Af) : f ∈ X

}is a closed subset of X × Y.

(c) If fn → f in X and Afn → g in Y, then g = Af.

Proof. (c) ⇒ (a). Assume that statement (c) holds. For clarity, write ‖ · ‖X

and ‖ · ‖Y for the norms on X and Y, and define

|||f ||| = ‖f‖X + ‖Af‖Y , f ∈ X.

Exercise: Show that ||| · ||| is a norm on X, and that X is complete with respectto this norm.

Since we have ‖f‖X ≤ |||f ||| for f ∈ X and X is complete with respect toboth norms, it follows from Exercise C.106 that there exists a constant C > 0such that |||f ||| ≤ C ‖f‖X for f ∈ X. Consequently, ‖Af‖Y ≤ C ‖f‖X , so A isbounded. ⊓⊔

C.15 Schauder Bases 349

C.15 Schauder Bases

Schauder bases were introduced and briefly discussed in Section A.10. In thissection we derive some additional properties of Schauder bases and relatedsystems.

C.15.1 Continuity of the Coefficient Functionals

Recall from Definition A.80 that a sequence F = {fk}k∈N is a Schauder basisfor a Banach space X if every f ∈ X can be written as

f =∞∑

k=1

αk(f) fk (C.25)

for a unique choice of scalars αk(f), where the series converges in the normof X.

With k fixed, αk is a mapping of X to scalars, i.e., it is a functional on X.Further, it follows immediately from the definition that αk is linear. We call{αk}k∈N the associated family of coefficient functionals. With j fixed, we have

∞∑

k=1

δjk fk = fj =

∞∑

k=1

ak(fj) fk,

so by uniqueness we must have ak(fj) = δjk. We therefore say that {fk}k∈N

and {αk}k∈N are biorthogonal sequences.Despite the fact that the definition of Schauder basis does not require the

functionals αk to be bounded, we will shortly see the surprising result thateach αk must be continuous, and hence belongs to X∗. In order to prove this,for each N ∈ N define the partial sum operator SN : X → X by

SNf =

N∑

k=1

αk(f) fk, f ∈ X.

By definition, for each f ∈ X we have that SNf → f as N → ∞. Clearly,range(SN ) ⊆ span{f1, . . . , fN}, and by biorthogonality we actually have that

range(SN ) = span{f1, . . . , fN}.

Each SN is a linear mapping on X, but we do not yet know whether it iscontinuous. This will be shown in the next exercise. The main idea is to showthat

|||f ||| = supN

‖SNf‖, f ∈ X.

defines an equivalent norm for X.


Exercise C.108. Let {fk}k∈N be a Schauder basis for a Banach space X.Using the notations given above, prove the following facts.

(a) Show that ||| · ||| is a norm on X, and ‖f‖ ≤ |||f ||| for all f ∈ X.

(b) Suppose that {gn}n∈N is a Cauchy sequence in X with respect to ||| · |||.With N fixed, show that {SNgn}n∈N is Cauchy with respect to ‖ · ‖.Let hN be such that ‖hN − SNgn‖ → 0 as n → ∞, and observe thathN ∈ span{f1, . . . , fN}.

(c) Show that

limn→∞

(supN

‖hN − SNgn‖)

= 0,

and use this to show that {hN}N∈N is Cauchy with respect to ‖ · ‖. Letg ∈ X be the element such that ‖g − hN‖ → 0.

(d) Show that SN (hN+1) = hN , and use this to show that hN =∑N

k=1 ckfk

where ck is independent of N.

(e) Show that g =∑∞

k=1 ckfk, and hence hN = SNg. Use this to show that|||g − gn||| → 0, and conclude that X is complete with respect to ||| · |||.

(f) Show that ‖ · ‖ and ||| · ||| are equivalent norms on X, and use this to showthat

C = supN

‖SN‖ < ∞.

The number C is called the basis constant for {fk}k∈N.

(g) Show that 1 ≤ ‖αk‖ ‖fk‖ ≤ 2C for each k ∈ N.

Notation C.109. Since the coefficient functionals associated with a Schauderbasis are continuous, we adopt our standard functional notation and write〈f, αk〉 instead of αk(f), with the understanding that this functional notationis linear in the first variable but antilinear in the second. More details on thisfunctional notation are given in Notation C.36.

C.15.2 Minimal Sequences

As we have seen, if {fk}k∈N is a Schauder basis with coefficient functionals{αk}k∈N, then we have the biorthogonality condition 〈fj , αk〉 = δjk. Unfortu-nately, the existence of a biorthogonal sequence is not sufficient by itself toguarantee that we have a Schauder basis. So, let us spend a little time con-sidering sequences that possess a biorthogonal sequence, and then see whatextra conditions we need to impose in order to have a Schauder basis.

The next exercise will recast the existence of a biorthogonal sequence interms of a type of “generalized linear independence” property called minimal-ity.

Definition C.110 (Minimal and Biorthogonal Sequences). Let {fk}k∈N

be a sequence in a Banach space X.

C.15 Schauder Bases 351

(a) {fk}k∈N is minimal if no vector fj lies in the closed span of the other fk:

∀ j ∈ N, fj /∈ span({fk}k 6=j

).

(b) A sequence {µk}k∈N in X∗ is biorthogonal to {fk}n∈N if 〈fj , µk〉 = δjk.

(c) A sequence that is both minimal and complete is said to be exact.

Compare the following characterization of minimal sequences in Banachspaces to the analogous result for Hilbert spaces given in Problem A.28.Whereas the proof for Hilbert spaces relies on the existence of an orthog-onal complements, the key for generic Banach spaces is the Hahn–BanachTheorem.

Exercise C.111. Let {fk}n∈N be a sequence in a Banach spaceX. Prove thatthe following statements are equivalent.

(a) {fk}k∈N is minimal.

(b) There exists a sequence {µk}k∈N in X∗ biorthogonal to {fk}k∈N.

Additionally, prove that the following statements are equivalent.

(a’) {fk}k∈N is exact (both minimal and complete).

(b’) There exists a unique sequence {µk}k∈N in X∗ biorthogonal to {fk}k∈N.

Every Schauder basis is minimal since it possesses a biorthogonal sequence,and it is complete by definition. In finite dimensions, a sequence {x1, . . . , xn}is minimal if and only if it is linearly independent, and in this case it is a basisfor its span. These simple facts do not extend to infinite sequences. Instead,there are several “shades of grey” in the meaning of independence in infinite-dimensional spaces, as illustrated in the following exercise and examples (seeProblem A.27) for the definition of an ω-independent sequence).

Exercise C.112. Let {fk}k∈N be a sequence in a Banach space X, and provethe following implications:

Schauder basis =⇒ minimal =⇒ ω-independent =⇒finitely

independent.

The following examples show that none of the converse implications holdin general, even in a Hilbert space.

Example C.113. (a) Problem 1.62 shows that {xe2πinx}n6=0 is both minimaland complete in L2[0, 1], but is not a Schauder basis for L2[0, 1].

(b) Let {en}n∈N be an orthonormal basis for a Hilbert spaceH. Set f1 = e1,and for n > 2 define fn = e1 + (en/n). Exercise: {fn}n∈N is ω-independentand complete, but is not minimal.

(c) Problem A.27 constructs an example of a sequence in ℓ2 that is finitelylinearly independent and complete, but is not ω-independent.


C.15.3 A Characterization of Schauder Bases

Now we will determine exactly what extra property an exact system mustpossess in order to be a Schauder basis.

Suppose that {fk}k∈N is a minimal sequence in a Banach space X. Then,by definition, there exists a sequence {αk}k∈N in X∗ that is biorthogonalto {fk}k∈N . Therefore, even though we do not know whether {fk}k∈N is aSchauder basis, we can define partial sum operators

SNf =

N∑

k=1

〈f, αk〉 fk, f ∈ X,

and we do have that each SN is a bounded operator on X. If we have thatSNf → f, then we will have f =

∑∞k=1 〈f, αk〉 fk, and the minimality as-

sumption will ensure that this representation is unique. The question is underwhat conditions we can be sure that the partial sums will converge, and thisis answered in the next exercise.

Exercise C.114. Let {fk}k∈N be a minimal sequence in a Banach space X,and let {αk}k∈N be its biorthogonal sequence in X∗. Prove that the followingstatements are equivalent.

(a) {fk}k∈N is a Schauder basis for X.

(b) SNf → f for each f ∈ X.

(c) {fk}k∈N is complete, and for each f ∈ X we have sup ‖SNf‖ <∞.

(d) {fk}k∈N is complete, and sup ‖SN‖ <∞.

Thus, a Schauder basis is precisely an exact sequence whose basis constantC = sup ‖SN‖ is finite.

C.15.4 Unconditional Bases

An unconditional basis for a Banach space X is a Schauder basis {fk}k∈N

which has the property that the representations f =∑∞

k=1 〈f, αk〉 fk convergeunconditionally for each f ∈ X. Not every Schauder basis is unconditional. Forexample, as is discussed in Section 1.11, while {e2πinx}n∈Z is an orthonormaland hence unconditional basis for L2[0, 1], if 1 < p <∞ and p 6= 2 then it is aSchauder basis for Lp[0, 1] that is not unconditional. A Hilbert space exampleof a Schauder basis that is not unconditional is

{∣∣x− 12

∣∣1/4e2πinx

}n∈Z

in L2[0, 1] [Bab48] (see also the discussion in [Sin70, pp. 351–354]). In contrast,to these systems, the Haar system

{χ[0,1]

}∪

{2n/2ψ(2nx− k)

}n≥0, k=0,...,2n−1

,

where ψ = χ[0,1/2)−χ[1/2,1), forms an unconditional basis for Lp[0, 1] for each1 < p <∞ (compare Problem B.29 for the case p = 2).

C.16 Weak and Weak* Convergence 353

C.16 Weak and Weak* Convergence

In this section we will discuss weak and weak* convergence of sequences. Adetailed discussion of the weak and weak* topologies and the correspondingconvergence criterion for nets is given in Section E.6. For now, we take thefollowing as our definition of weak or weak* convergence of sequences.

Definition C.115. Let X be a normed linear space.

(a) A sequence {fn}n∈N in X converges weakly to f ∈ X, denoted fnw→ f, if

∀µ ∈ X∗, limn→∞

〈fn, µ〉 = 〈f, µ〉.

(b) A sequence {µn}n∈N in X∗ converges weak* to µ ∈ X∗, denoted µnw*−→µ,

if∀ f ∈ X, lim

n→∞〈f, µn〉 = 〈f, µ〉.

Thus, weak* convergence is just “pointwise convergence” of the opera-tors µn. Weak* convergence is only defined for sequences that lie in a dualspace X∗. Given a sequence {µn}n∈N in X∗, we can consider three typesof convergence: strong (norm), weak, and weak* convergence. By definition,these are:

µn → µ ⇐⇒ limn→∞

‖µ− µn‖ = 0,

µnw→µ ⇐⇒ ∀T ∈ X∗∗, lim

n→∞〈µn, T 〉 = 〈µ, T 〉,

µnw*−→µ ⇐⇒ ∀x ∈ X, lim

n→∞〈x, µn〉 = 〈x, µ〉.

Exercise C.116. Let X be a normed space and suppose that fn, f ∈ X andµn, µ ∈ X∗. Show that

fn → f =⇒ fnw→ f

andµn → µ =⇒ µn

w→µ =⇒ µn

w*−→µ.

If X is reflexive, then weak and weak* convergence coincide. In par-ticular, this is the case for Hilbert spaces. In any finite-dimensional space,strong and weak convergence coincide, but they are distinct in any infinite-dimensional space. For example, if {en}n∈N is an orthonormal sequence in aHilbert space H, then it follows from Bessel’s Inequality that 〈en, f〉 → 0 for

each f ∈ H and hence enw→ 0, but ‖en − 0‖ = 1→/ 0.

Here are some properties of weak and weak* convergent sequences. Whilethe fact that norm-convergent sequences are bounded is trivial, the corre-sponding fact for weak and weak* convergent sequences is much more subtleand is most easily proved by applying the Uniform Boundedness Principle.

Documents

C.6 Adjoints for Operators on Hilbert Spacespeople.math.gatech.edu/~heil/7338/fall09/appendixc_2.pdfC.6 Adjoints for Operators on Hilbert Spaces 321 As a corollary, we obtain the following