MAT 579 Functional Analysis II Lecture Notes

MAT 579 Functional Analysis IILecture Notes

Professor: John Quigg

Semester: Spring 2020

Revised December 6, 2020

c©2020 Arizona State University School of Mathematical & Statistical Sciences

Contents

1 Banach algebras 2

2 The spectrum 7

3 Gelfand transform 13

4 Spectral permanence 17

5 Banach *-algebras 18

6 Commutative C∗-algebras 21

7 Functional calculus 28

8 Spectral theorem 31

9 Unitization 43

1

1 BANACH ALGEBRAS 2

Introduction

Most of these notes follow the books by Arveson, Conway, Dixmier, Rudin, Folland (partic-ularly Chapter 5), and Williams [1, 2, 3, 5, 4, 6].

This is the second semester of the two-semester sequence MAT 578–579 on functionalanalysis. The first semester covered the basics: Banach spaces and bounded linear maps,Hilbert spaces, the consequences of the Baire Category theorem, including the Open Mappingtheorem, the Closed Graph theorem, and the Principle of Uniform Boundedness, locallyconvex spaces and weak topologies, and the Krein-Milman theorem on extreme points.

This second semester focuses on spectral theory, culminating (in some sense) in theSpectral Theorem for normal operators on Hilbert space, one of the pinnacles of modernanalysis. It has been apparent for quite a while that the natural context for the developmentof spectral theory is C∗-algebras, and the build-up to that occupies almost half of the lecturenotes. The highlight along the way is the Gelfand transform, which is a vast generalizationof the Fourier transform.

Most of the time we will discuss spectral theory in the context of unital Banach algebras,i.e., those containing a multiplicative identity. At the end we briefly consider how to dealwith nonunital C∗-algebras, and the final exercises give a glimpse into the rich general theoryof C∗-algebras.

In the first semester we tried as much as possible to allow either the real or complexnumbers as the scalar field. But spectral theory demands complex numbers, so all vectorspaces throughout these notes will have scalar field C.

This semester the prerequisites include a few additions that we didn’t need in the firstsemester: a bit of (abstract) algebra, a bit of complex analysis, and somewhat more topology.Actually, we don’t need much algebra, mostly just a willingness to work with algebras overthe complex numbers. The main result we need from complex analysis is Liouville’s theo-rem, although we do need a version for functions taking values in a unital Banach algebra.Fortunately, the passage from complex-valued functions is painless. From topology, we needto work more seriously with locally compact Hausdorff spaces,

1 Banach algebras

Definition 1.1. An algebra is a vector space A equipped with an associative bilinear mapA× A→ A, called the multiplication and denoted by (a, b) 7→ ab.

A normed algebra is an algebra A that is also a normed space such that ‖ab‖ ≤ ‖a‖‖b‖for all a, b ∈ A. A Banach algebra is a normed algebra whose norm is complete.

1 BANACH ALGEBRAS 3

A unit, or identity, of an algebra A is a nonzero element 1 = 1A ∈ A such that 1a = a1 = afor all a ∈ A, and A is unital if it has a unit. A unit is unique if it exists. A unital normedalgebra is a normed algebra that is also a unital algebra in which ‖1A‖ = 1. The term unitalBanach algebra is self-explanatory.

Definition 1.2. A subalgebra of an algebra A is a vector subspace B that is closed undermultiplication in the sense that ab ∈ B for all a, b ∈ B. A unital subalgebra of a unitalalgebra A is a subalgebra B containing 1A.

For subsets B,C ⊆ A, we write

BC = bc : b ∈ B, c ∈ C,

so that the condition for B to be a subalgebra can be expressed as BB ⊆ B. A subalgebraof A is an algebra with the multiplication from A.

Every subalgebra of a normed algebra is itself a normed algebra. The closure of a subal-gebra of a normed algebra is a subalgebra. A subalgebra of a Banach algebra is a Banachalgebra if and only if it is closed, in which case we call it a Banach subalgebra.

Example 1.3. For any set X, the vector space CX of all functions from X to C is a unitalalgebra with pointwise multiplication

(fg)(x) = f(x)g(x).

The subspace `∞(X) of bounded functions is a Banach algebra with the sup norm

‖f‖∞ = supx∈X|f(x)|,

and is a unital subalgebra of CX .

Example 1.4. For any topological space X, the vector space C(X) of continuous functionsfrom X to C is a unital subalgebra of CX , and the subset Cb(X) of bounded continuousfunctions is a unital Banach subalgebra of `∞(X). If X is compact, then Cb(X) = C(X).

Example 1.5. For any locally compact Hausdorff space X, the set C0(X) of all continuousf that vanish at infinity in the sense that for every ε > 0 the set |f | ≥ ε is compact in Xis a Banach subalgebra of Cb(X). It is unital if and only if X is compact. The set Cc(X) ofcontinuous functions with compact support is a dense normed subalgebra that is completeif and only if X is compact.

Example 1.6. For any measure space (X,M, µ), the Banach space L∞(X,µ) is a unital Ba-nach algebra with the pointwise operations. Note that `∞(X) is the special case of countingmeasure.

Example 1.7. For any measurable space (X,M), the space B∞(X) of all bounded measur-able functions is a unital Banach subalgebra of `∞(X).

1 BANACH ALGEBRAS 4

Example 1.8. Let D be the closed unit disk in C. Then the disc algebra is the unitalBanach subalgebra of C(D) consisting of all continuous functions that are analytic in theinterior.

Example 1.9. Any vector space A can be made into an algebra with the trivial multiplicationab = 0 for all a, b ∈ A. Such an algebra is normed or Banach whenever it has these propertiesas a vector space, but is never unital.

Definition 1.10. Let A be an algebra and a, b ∈ A. Then a commutes with b if ab = ba.Note that a commutes with b if and only if b commutes with a, and we also say that a andb commute.

Definition 1.11. An algebra A is commutative if ab = ba for all a, b ∈ A.

Thus, A is commutative if and only if every pair of elements commute.

All the above examples are commutative. We will see many that are not.

Example 1.12. For any Banach space X, the vector space B(X) of bounded (linear) op-erators on X is a unital Banach algebra, where multiplication is composition. The setK(X) of compact operators on X is a Banach subalgebra, that is unital if and only if Xis finite-dimensional, in which case K(X) = B(X). Note that K(X), and hence B(X), isnoncommutative if dimX > 1 (because it contains the algebra of finite-rank operators).

Definition 1.13. If A and B are algebras, a linear map φ : A → B is a homomorphism ifφ(ab) = φ(a)φ(b) for all a, b ∈ A. An algebraic isomorphism from A to B is a bijective ho-momorphism, and A and B are algebraically isomorphic if there is an algebraic isomorphismφ : A → B. An algebraic automorphism on A is an algebraic isomorphism φ : A → A. Ahomomorphism φ : A→ B between unital algebras is unital if φ(1A) = 1B.

An algebraic isomorphism between unital algebras will automatically be unital.

For isomorphism of normed algebras, we want more:

Definition 1.14. If A and B are normed algebras, a homomorphism φ : A → B is anisomorphism if it is an algebraic isomorphism that is bounded with bounded inverse, and Aand B are isomorphic, written A ' B, if such a φ exists.

By the Open Mapping principle, if A and B are Banach algebras then every boundedalgebraic isomorphism φ : A→ B is an isomorphism of Banach algebras.

Example 1.15. For any n ∈ N, the set Mn of n×n matrices is a unital algebra with matrixmultiplication. It can be made into a normed algebra in lots of ways; for our purposes themost useful norm is the one it inherits under the algebraic isomorphism Mn ' B(Cn), wherethe norm on Cn is the one associated to the inner product 〈x, y〉 =

∑ni=1 xiyi.

1 BANACH ALGEBRAS 5

Example 1.16. If X is a Banach space of dimension n, then B(X) 'Mn (where of coursethe isomorphism depends upon a choice of basis for X).

Definition 1.17. An ideal of an algebra A is a subspace I ⊆ A such that IA ∪ AI ⊆ I.

An ideal is obviously a subalgebra. The closure of an ideal of a normed algebra is anideal. 0 and A are obviously ideals of A.

Definition 1.18. 0 is the zero ideal of A. An ideal I of A is proper if I 6= A. The algebraA is simple if there are no proper nonzero ideals.

Definition 1.19. A proper ideal I of A is maximal if for every ideal J of A with I ⊆ J wehave J = I or J = A.

Example 1.20. For any locally compact Hausdorff space X, C0(X) is an ideal of Cb(X)and Cc(X) is an ideal of C0(X).

Example 1.21. For any Banach space X, K(X) is an ideal of B(X), and the set of finite-rank operators is an ideal of K(X).

Example 1.22. The algebra Mn is simple.

Example 1.23. If φ : A → B is a homomorphism, then the kernel kerφ is an ideal. Moregenerally, if I is an ideal of B then φ−1(I) is an ideal of A.

Lemma 1.24. Let φ : A → B be a surjective homomorphism of algebras. Then the mapI 7→ φ−1(I) gives an inclusion-preserving bijection from the set of ideals of B to the set ofideals of A containing kerφ. Consequently, if B is nonzero, then B is simple if and only ifthe ideal kerφ is maximal.

Proposition 1.25. Let I be an ideal of an algebra A. Then the quotient vector space A/Ibecomes an algebra with (a+ I)(b+ I) = ab+ I, and the quotient map is a homomorphism.Moreover, if A is a normed algebra and I is closed then the quotient normed space A/I is anormed algebra, which is a Banach algebra if A is. Finally, if A is unital and I is proper.then A/I is unital and the quotient map is unital.

Proof. The proposed multiplication is well-defined since I is an ideal, and associativity isinherited from A. Suppose that A is a normed algebra and I is closed, and let u, v ∈ A/I.Then for any a ∈ u, b ∈ v we have ab ∈ uv, so

‖uv‖ ≤ ‖ab‖ ≤ ‖a‖‖b‖.

Letting ‖a‖ ‖u‖ and ‖b‖ ‖v‖ gives ‖uv‖ ≤ ‖u‖‖v‖. Thus A/I is a normed algebra.Since the quotient space is complete if A is, it is a Banach algebra if A is.

Suppose A is unital and I is proper. Then for all a ∈ A,

(1 + I)(a+ I) = 1a+ I = a+ I = a1 + I = (a+ I)(1 + I),

so 1 + I is a unit for A/I, and in particular the quotient map preserves units.

1 BANACH ALGEBRAS 6

Proposition 1.26. Let φ : A → B be a bounded homomorphism of Banach algebras. Thenthere is a unique homomorphism φ making the diagram

Aφ //

π

B

A/ kerφφ

;;

commute, where π is the quotient map. Moreover, φ is bounded, with ‖φ‖ = ‖φ‖.

Proof. For the first part, from linear algebra we know that there is a unique linear map φmaking the diagram commute, and moreover φ is injective. Since φ and π are homomor-phisms, a trivial computation verifies that φ is also a homomorphism.

For the second part, write I = kerπ, and note first that for every a ∈ A,

‖φ(a+ I)‖ = ‖φ(a)‖ ≤ ‖φ‖‖a‖.

Letting ‖a‖ ‖a+ I‖ gives ‖φ(a+ I)‖ ≤ ‖φ‖‖a+ I‖. It follows that ‖φ‖ ≤ ‖φ‖.

For the opposite inequality, since ‖π‖ = 1,

‖φ‖ = ‖φ π‖ ≤ ‖φ‖‖π‖ = ‖φ‖.

Exercises

1. In a normed algebra A, the multiplication is continuous from A× A to A.

2. Let A be an algebra, and let S ⊆ A.

(a) Prove that the intersection of any family of subalgebras of A is a subalgebra. You canassume that the family is nonempty.

(b) Use part (a) to help prove that there is a subalgebra B of A containing S that issmallest in the sense that B ⊆ C for any subalgebra C of A containing S. Then B is calledthe subalgebra generated by S.

(c) Prove that if a ∈ A then the subalgebra generated by a equals the set of all polynomialsin a with 0 constant term, i.e., elements of the form

∑ni=1 cia

i, where n ∈ N and ci ∈ C fori = 1, . . . , n. Moreover, if A is unital then the subalgebra generated by 1, a is the set ofall polynomials in a, and we sometimes call it the unital subalgebra generated by a.

(d) Prove that the intersection of any family of closed subalgebras of A is a subalgebra.

2 THE SPECTRUM 7

(e) Use the above to help prove that if A is a normed algebra then there is a smallestclosed subalgebra B of A containing S, and moreover B is the closure of the subalgebra ofA generated by S. This is called the closed subalgebra generated by S.

(f) Prove that if A is a unital Banach algebra and a ∈ A then the Banach subalgebragenerated by 1, a is the closure of the set of all polynomials in a, and we sometimes call itthe unital Banach subalgebra generated by a.

(g) Show how to use the Stone-Weierstrass theorem to prove that C(T) is generated as aBanach algebra by z, z, where z is the inclusion map T → C.

3. Prove Lemma 1.24.

4. Prove that `1(Z) becomes a commutative unital Banach algebra with multiplication givenby convolution

(x ∗ y)n =∞∑

k=−∞

xkyn−k.

5. Prove that L1(R) becomes a commutative nonunital Banach algebra with convolution

(f ∗ g)(x) =

∫ ∞−∞

f(y)g(x− y) dy.

Hint: Fubini’s theorem is useful several times here.

6. Let µ be normalized arc-length measure on the unit circle T, i.e., arc length divided by 2π sothat µ(T) = 1 (a useful convention for compact groups), and write

∫T f(z) dµ(z) =

∫T f(z) dz.

Prove that L1(T) becomes a commutative nonunital Banach algebra with convolution

(f ∗ g)(z) =

∫Tf(w)g(w−1z) dw.

You may use without comment the property that µ is translation-invariant in the sense thatfor every Borel set E and every z ∈ T we have µ(zE) = µ(E), and the consequence that forevery f ∈ L1(T) and z ∈ T we have

∫T f(zw) dw =

∫T f(w) dw.

7. Let Aii∈S be a family of nonzero Banach algebras. Prove that the direct sum⊕∞

i∈S Ai ofbounded S-tuples with the sup norm is a Banach algebra that is unital if and only if everyAi is unital.

2 The spectrum

Definition 2.1. Let A be a unital algebra and a ∈ A. An inverse of a is an element b ∈ Asuch that ab = ba = 1, and a is invertible if it has an inverse. If a is invertible, its inverse isunique, and is denoted by a−1.

2 THE SPECTRUM 8

The last part of the above definition is justified by (2) of the following lemma:

Lemma 2.2. Let A be a unital algebra and a, b, c ∈ A.

(1) If ab = ca = 1, then b = c, so a is invertible and b is an inverse of a.

(2) If a is invertible then its inverse is unique.

(3) ab and ba are both invertible if and only if both a and b are invertible.

(4) If a and b commute then ab is invertible if and only if both a and b are invertible.

Example 2.3. Let S and T be respectively the forward and backward shifts on `2. ThenTS = 1 is invertible in B(`2), but neither S nor T is invertible.

Note that a proper ideal of a unital algebra cannot contain any invertible element.

Theorem 2.4. Let A be a unital Banach algebra, and let x ∈ A with ‖x‖ < 1. Then 1− xis invertible, and

(1− x)−1 =∞∑n=0

xn.

Moreover,

‖(1− x)−1‖ ≤ 1

1− ‖x‖(2.1)

‖(1− x)−1 − 1‖ ≤ ‖x‖1− ‖x‖

. (2.2)

Proof. The series∑∞

n=0 xn converges because it converges absolutely and A is complete. If

sn =∑n

k=0 xk, then

(1− x)sn = 1− xn+1 n→∞−−−→ 1,

and similarly sn(1− x)→ 1, so the sum∑∞

n=0 xn is an inverse of 1− x.

For the other part, we have

‖(1− x)−1‖ ≤∞∑n=0

‖xn‖ ≤∞∑n=0

‖x‖n =1

1− ‖x‖.

Similarly,

(1− x)−1 − 1 =∞∑n=1

xn = x∞∑n=0

xn,

so

‖(1− x)−1 − 1‖ =

∥∥∥∥∥x∞∑n=0

xn

∥∥∥∥∥ ≤ ‖x‖∥∥∥∥∥∞∑n=0

xn

∥∥∥∥∥ ≤ ‖x‖1− ‖x‖

.

2 THE SPECTRUM 9

Corollary 2.5. In a unital Banach algebra, the set Inv(A) of invertible elements is open,and the map a 7→ a−1 on Inv(A) is a homeomorphism.

Proof. By Theorem 2.4, the open ball B1(1) is contained in Inv(A). Let a ∈ Inv(A). Thenthe map b 7→ ab on Inv(A) is continuous, and is in fact a homeomorphism since its inverseis b 7→ a−1b. and so it maps B1(1) onto a neighborhood of a. Thus Inv(A) is open.

For the other part, the map a 7→ a−1 is self-inverse, so it suffices to show continuity. First,the inequality (2.2) implies that inversion is continuous at 1. Then if an → a in Inv(A) wehave a−1an → 1, so

a−1n a = (a−1an)−1 → 1,

and therefore a−1n → a−1.

Corollary 2.6. In a unital Banach algebra, the closure of every proper ideal is proper.

Proof. If I is a proper ideal in a unital Banach algebra A, then B1(1) is an open set disjointfrom I, and hence disjoint from I.

Definition 2.7. Let A be a unital algebra and a ∈ A. The spectrum of a is

σ(a) = λ ∈ C : a− λ1 is not invertible.

If A is a unital algebra and λ ∈ C, we frequently identify λ with the scalar multiple λ1A,so in particular we can write σ(a) = λ ∈ C : a− λ is not invertible.

Proposition 2.8. If A is a unital Banach algebra and a ∈ A, then σ(a) is compact and iscontained in the closed disk λ ∈ C : |λ| ≤ ‖a‖.

Proof. If |λ| > ‖a‖, then ‖λ−1a‖ < 1, and so

λ−1(a− λ) = λ−1a− 1 ∈ Inv(A),

and therefore a − λ ∈ Inv(A). Thus σ(a) is contained in the indicated disk, so to finish itsuffices to observe that the complement σ(a)c is open because it is the inverse image of theopen set Inv(A) under the continuous map

λ 7→ a− λ : C→ A.

We now need a bit of the theory of analytic functions with values in a Banach algebra,and we only need the unital case: let A be a unital Banach algebra, U ⊆ C be a connectedopen set, and f : U → A. Then f is analytic on U if for every z ∈ U the limit

f ′(z) = limw→z

f(w)− f(z)

w − xexists. f is entire if it is analytic on C. Most of the usual theory from complex analysiscarries over to this context. For example:

2 THE SPECTRUM 10

• The usual rules of differentiation:

∗ (f + g)′ = f ′ + g′;

∗ (cf)′ = cf ′ for c ∈ C;

∗ (fg)′ = f ′g + fg′.

• If f is analytic on a disk D = z ∈ C : |z − t| < r, then

f(z) =∞∑n=0

f (n)(t)

n!(z − t)n for all z ∈ D.

• Liouville’s theorem: If f is entire and bounded then it is constant.

Theorem 2.9 (Gelfand). If A is a unital Banach algebra and a ∈ A, then σ(a) 6= ∅.

Proof. Suppose σ(a) = ∅, so that a− λ ∈ Inv(A) for all λ ∈ C. Define f : C→ A by

f(λ) = (a− λ)−1.

If γ 6= λ then

(a− γ)−1 − (a− λ)−1 = (a− γ)−1((a− λ)− (a− γ)

)(a− λ)−1

= (γ − λ)(a− γ)−1(a− λ)−1.

Thus

f ′(λ) = limγ→λ

f(γ)− f(λ)

γ − λ= (a− λ)−2.

We see that f is differentiable on C, i.e., is entire.

When |λ| > ‖a‖ we have

‖(a− λ)−1‖ = |λ|−1∥∥(λ−1a− 1)−1

∥∥≤ |λ|−1

1− ‖λ−1a‖

=1

|λ| − ‖a‖|λ|→∞−−−−→ 0,

so

f(λ)|λ|→∞−−−−→ 0.

Therefore f is bounded. Thus f is constant by Liouville’s theorem. In this case the constantmust be 0, which is absurd because f(0) = a−1 is an invertible element of A.

We postponed the following definition until we were assured that the spectrum isnonempty, since it saves us wondering about the supremum of the empty set:

2 THE SPECTRUM 11

Definition 2.10. If A is a unital algebra and a ∈ A, then the spectral radius of a is

r(a) = supλ∈σ(a)

|λ|.

Example 2.11. If X is a compact Hausdorff space and f ∈ C(X), then σ(f) = f(X).

Example 2.12. If A ∈Mn, then σ(A) is the set of eigenvalues of A.

Definition 2.13. A division algebra is a unital algebra in which every nonzero element isinvertible.

Corollary 2.14 (Gelfand-Mazur). Every Banach division algebra is isomorphic to C.

Proof. Let A be a Banach division algebra, and define φ : C → A by φ(λ) = λ1. Then φ isan isomorphism of C onto the subalgebra C1 of A (note that it is bounded with boundedinverse since it is isometric). It suffices to show that φ is surjective. Let a ∈ A. By Gelfand’stheorem, we can choose λ ∈ σ(a). Then a− λ is not invertible, so we must have a− λ = 0,and therefore φ(λ) = a.

Theorem 2.15 (Spectral Radius formula). If A is a unital Banach algebra and a ∈ A, then

r(a) = limn→∞

‖an‖1/n.

Proof. We first note that for all λ ∈ σ(a) and n ∈ N we have

an − λn = (a− λ)q(a) = q(a)(a− λ)

for a suitable polynomial q. Since a− λ is not invertible, neither is an − λn, so λn ∈ σ(an),Thus |λn| ≤ ‖an‖, so

|λ| ≤ ‖an‖1/n.

It follows thatr(a) ≤ lim inf

n→∞‖an‖1/n.

To finish, it suffices to show that

lim supn→∞

‖an‖1/n ≤ r(a).

Without loss of generality a 6= 0.

Claim: for all λ ∈ C with |λ| < 1/r(a) (where we define 1/r(a) = ∞ if r(a) = 0), thesequence ((λa)n) is bounded in A. Define f on the (perhaps infinite) disk D = |λ| < 1/r(a)by

f(λ) = (1− λa)−1 = λ−1(λ−1 − a)−1.

2 THE SPECTRUM 12

By the proof of Theorem 2.9, f is analytic on the set λ−1 /∈ σ(a), in particular wheneverλ ∈ D. Thus f has a power series expansion on D. On the smaller disk |λ| < 1/‖a‖, byTheorem 2.4, we have

f(λ) =∞∑n=0

(λa)n =∞∑n=0

anλn.

By uniqueness of power series, for every λ ∈ D we must have

f(z) =∞∑n=0

anλn,

so the terms anλn are bounded, proving the claim.

So, given λ ∈ D, we can choose M > 0 such that for all n ∈ N we have∥∥λnan∥∥ ≤M,

so

‖an‖1/n ≤ M1/n

|λ|.

Thus

lim supn→∞

‖an‖1/n ≤ limn→∞

M1/n

|λ|=

1

|λ|.

Letting |λ| 1/r(a) giveslim supn→∞

‖an‖1/n ≤ r(a).

Exercises

1. Let A be a unital algebra, a ∈ A, and z ∈ C. Prove that

σ(a+ z) = σ(a) + z.

2. Prove that if X is a set and f ∈ `∞(X), then σ(f) = f(X).

3. Prove that if (X,M, µ) is a measure space and f ∈ L∞(X), then σ(f) is the essential rangeof f , which is defined as the complement in C of the set of z ∈ C for which there exists r > 0such that f−1(Br(z)) has measure 0.

4. Let A be a unital Banach algebra. Then a ∈ A is nilpotent if an = 0 for some n ∈ N. Usethe Spectral Radius formula to show that if a is nonzero and nilpotent then r(a) < ‖a‖.

3 GELFAND TRANSFORM 13

3 Gelfand transform

Proposition 3.1. In a commutative unital Banach algebra A,

(1) every maximal ideal of A is closed,

(2) every proper ideal of A is contained in a maximal ideal, and

(3) every noninvertible element is contained in a maximal ideal.

Proof. (1) Every proper ideal I is disjoint from the nonempty open set Inv(A), and henceso is I. Thus I is a proper ideal containing I, so we must have I = I by maximality.

(2) This is a routine Zorn’s lemma argument, using the elementary observation that theunion of a linearly ordered family Ijj∈S of proper ideals is an ideal, which must be properbecause 1 /∈ Ij for all j.

(3) If a is noninvertible, then the ideal aA is proper and contains a, so (3) follows from(2).

Corollary 3.2. Every commutative unital Banach algebra has at least one maximal ideal.

Definition 3.3. A character of a commutative Banach algebra A is a nonzero homomor-phism χ : A→ C. We write Ω = Ω(A) for the set of all characters of A.

Lemma 3.4. If A is a unital commutative Banach algebra and χ ∈ Ω, then χ(1) = 1.

Proof. We have χ(A) = C, and routine computations show that χ(1) is the unit of thealgebra χ(A).

Definition 3.5. We write M = M(A) for the set of maximal ideals of a commutativeBanach algebra A.

Proposition 3.6. If A is a commutative unital Banach algebra, then the map χ 7→ kerφgives a bijection Ω→M.

Proof. Let χ ∈ Ω. Since χ is nonzero, we have χ(A) = C. Thus kerχ is a proper ideal, whichmust be maximal since C is simple. To see that χ 7→ kerχ is injective, let χ, τ ∈ Ω, andsuppose that kerφ = ker τ . Then there exists c ∈ C such that χ = cτ . Since χ(1) = τ(1) = 1,we must have c = 1, so χ = τ .

For the surjectivity, let I ∈ M. Then A/I is a simple commutative unital Banachalgebra. But then every nonzero element of A/I is invertible, so there is an isomorphismθ : A/I → C by the Gelfand-Mazur theorem. Letting π : A→ A/I be the quotient map, wehave a surjective homomorphism χ = θ π : A → C with kerχ = I. Since χ 6= 0, we haveχ ∈ Ω.


Theorem 3.7. If A is a commutative unital Banach algebra and a ∈ A, then

σ(a) = χ(a) : χ ∈ Ω.

Proof. If λ ∈ σ(a), then a− λ is noninvertible, and hence is contained in a maximal ideal I,so by Proposition 3.6 we can choose χ ∈ Ω such that a− λ ∈ kerχ, and then χ(a) = λ.

On the other hand, if χ ∈ Ω and λ = χ(a), then a − λ ∈ kerχ. Since kerχ is a properideal, a− λ is noninvertible, so λ ∈ σ(a).

Corollary 3.8. Let A be a commutative unital Banach algebra and χ ∈ Ω. Then χ isbounded, with ‖χ‖ = 1.

Proof. For all a ∈ A we have|χ(a)| ≤ r(a) ≤ ‖a‖,

so ‖χ‖ ≤ 1. Since χ(1A) = 1 and ‖1A‖ = 1, we must have ‖χ‖ = 1.

Definition 3.9. If A is a commutative unital Banach algebra, we give Ω the relative weak*topology from A∗, and call it the spectrum of A.

Remark 3.10. Another popular notation for Ω(A) is A.

Corollary 3.11. If A is a commutative unital Banach algebra, then the spectrum Ω of A iscompact Hausdorff.

Proof. Ω is Hausdorff because the weak* topology is. By Corollary 3.8 and Alaoglu’s theo-rem, to see that Ω is compact it suffices to observe that it is weak* closed since the propertiesχ(1) = 1 and χ(ab) = χ(a)χ(b) are obviously preserved by weak* limits.

Definition 3.12. Let A be a commutative unital Banach algebra. The Gelfand transformof an element a ∈ A is the function a : Ω→ C defined by

a(χ) = χ(a).

Also, the Gelfand transform of A is the map Γ defined by

Γ(a) = a for a ∈ A.

Theorem 3.13. For a commutative unital Banach algebra the Gelfand transform Γ is abounded unital homomorphism A→ C(Ω), with ‖Γ‖ = 1, and

‖Γ(a)‖ = r(a) for all a ∈ A.

Corollary 3.14. If A is a commutative unital Banach algebra and a ∈ A, then a is invertibleif and only if 0 is not in the range of a.

It’s frequently instructive to examine the special case C(X):


Example 3.15. Let X be a compact Hausdorff space, and let A = C(X). For each x ∈ Xdefine ωx ∈ Ω by

ωx(f) = f(x).

Then the map x 7→ ωx : X → Ω is a homeomorphism.

To see this, note straightaway that x 7→ ωx is a continuous injection since A is a set ofcontinuous functions that separates points. Since X is compact and Ω is Hausdorff, we canfinish by showing that every character of A is of the form ωx. Let χ ∈ Ω, and put I = kerχ.It suffices to show that there exists x ∈ X such that

I = kerωx.

Suppose not. Then, since I is maximal, for every x ∈ X we have I 6⊆ kerωx, so we canchoose fx ∈ I such that fx(x) 6= 0. Then by continuity we can choose an open set Ux ⊆ Xsuch that fx(y) 6= 0 for all y ∈ Ux. By compactness, we can choose a finite set S ⊆ X suchthat X =

⋃x∈S Ux. Then the function

f =∑x∈S

fxfx ∈ C(X)

does not take the value 0, and so is invertible in A. Since I is an ideal we have f ∈ I. Butthen I cannot be proper, which is a contradiction.

Example 3.16. Let A = `1(Z), which we have seen is a commutative unital Banach algebraunder convolution. For each n ∈ Z let δn ∈ A be the characteristic function of n. Notethat δ0 is the unit of A, and for all n ∈ Z we have (δ1)n = δn (and in particular δ−1 = (δ1)−1).Each f ∈ A can be expressed as an absolutely convergent series

f =∑n∈Z

f(n)δn.

For z ∈ T define ωz : A→ C by

ωz(f) =∑n∈Z

f(n)zn

(which obviously converges). It is an exercise that ωz is a character. For f ∈ A defineγ(f) : T → C by γ(f)(z) = ωz(f). Then γ(f) ∈ C(T) since it is a uniform limit of thecontinuous functions z 7→

∑|n|≤k f(n)zn for k ∈ N. Note that

ωz(δ1) = z,

so the mapz 7→ ωz : T→ Ω

is injective. It is also continuous, by continuity of each function γ(f). We will now showthat it is surjective: let χ ∈ Ω. Put

z = χ(δ1).


Then|z| ≤ ‖δ1‖ = 1.

But also ∣∣z−1∣∣ =

∣∣χ(δ1)−1∣∣ = |χ(δ−1)| ≤ 1,

so we must have z ∈ T. We check that in fact χ = ωz: for all f ∈ A we have

χ(f) =∑n∈Z

f(n)χ(δn) =∑n∈Z

f(n)zn = ωz(f).

Thus the map z 7→ ωz is a continuous bijection of T onto Ω, and hence is a homeomorphismsince T is compact and Ω is Hausdorff. Moreover, we may now identify the map γ : A→ C(T)

as the Gelfand transform Γ of A, and we may write f = γ(f).

Note that for all f ∈ A and n ∈ N,∫Tf(z)z−n dz =

∑k∈Z

∫Tf(k)zkz−n dz = f(n).

It follows that Γ(A) coincides with the set of functions in C(T) whose Fourier series isabsolutely convergent. Wiener proved that if g ∈ C(T) has absolutely convergent Fourierseries and is never 0, then 1/g also has absolutely convergent Fourier series. Wiener’s proofused a lot of hard classical analysis. But Gelfand gave the following “soft” proof: we haveg = f for some f ∈ A, and f is never 0, so f is invertible in A. Then Γ(f−1) is themultiplicative inverse 1/g of g. The image Γ(A) ⊆ C(T) is sometimes called the Wieneralgebra.

Proposition 3.17. Let A be a commutative unital Banach algebra generated by 1, a. Thenthe map

a : Ω→ σ(a) given by χ 7→ χ(a)

is a homeomorphism.

Proof. The map is continuous and surjective, and is injective since if χ, τ ∈ Ω and χ(a) =τ(a) then χ = τ since they agree on a generating set. Since Ω is compact and σ(a) isHausdorff, we are done.

Example 3.18. Let A be the disk algebra of continuous functions on the closed unit diskD that are analytic on the interior D. It is an exercise in complex analysis that A isgenerated by 1, z. Thus by Proposition 3.17 Ω is homeomorphic to σ(z). We will showthat σ(z) = D. First, for all λ ∈ D the function z − λ is clearly not invertible in A, soλ ∈ σ(z). On the other hand, for all λ ∈ σ(z) we have

|λ| ≤ ‖z‖ = 1,

so λ ∈ D. Note that in this case the Gelfand transform of A is the inclusion A → C(D).

4 SPECTRAL PERMANENCE 17

Exercises

1. Prove Corollary 3.2.

2. Prove Theorem 3.13.

3. Prove Corollary 3.14.

4. Prove that if A is a commutative unital Banach algebra then Ω 6= ∅.

5. Let A and B be commutative unital Banach algebras, and let φ : A→ B be a unital homo-morphism (not assumed bounded). Assume that B is semisimple, i.e., the intersection of allmaximal ideals of B is zero.

(a) Prove that φ is bounded. Hint: Closed Graph theorem.

(b) Prove that every algebraic automorphism of B is a Banach-algebra isomorphism.

4 Spectral permanence

If A is a unital Banach algebra, we will frequently write the spectrum of an element a ∈ Aas σA(a) if there may be other Banach algebras around.

Definition 4.1. Let K ⊆ C be compact and nonempty. A hole of K is a bounded componentof Kc.

Theorem 4.2 (Spectral Permanence theorem). Let A be a unital Banach algebra, B a unitalBanach subalgebra, and b ∈ B. Then σB(b) is the union of σA(b) and some (possibly none)of its holes.

Proof. First, any element of B that is invertible in B is invertible in A, so σA(b) ⊆ σB(b).

Claim: ∂σB(b) ⊆ σA(b). To see this, let (λn) be a sequence in σB(b)c converging toλ ∈ ∂σB(b). Suppose that λ /∈ σA(b). Then, in A, (b − λn) is a sequence of invertibleelements converging to the invertible element b− λ. Since inversion is continuous, we have

(b− λn)−1 → (b− λ)−1.

Since (b− λn)−1 ∈ B for all n, we have

(b− λ)−1 ∈ B = B,

so λ /∈ σB(b), a contradiction.

5 BANACH *-ALGEBRAS 18

Now let U be a component of σA(b)c, and put

V = U \ σB(b) and W = U ∩ σB(b).

Then U = V tW , and V is obviously open. On the other hand, if λ ∈ W , then we cannothave λ ∈ ∂σB(b), since

λ ∈ U ⊆ σA(b)c.

Thus λ ∈ σB(b), and so we see that

W = U ∩ σB(b),

which is therefore open. Since U is connected, either V or W is empty. Therefore U is eithercontained in σB(b) or disjoint from σB(b). Since σB(b) is bounded, it cannot contain theunbounded component of σA(b)c. Thus, by the above we see that σB(b) is disjoint from theunbounded component of σA(b)c. Therefore, σB(b) \ σA(b) is a union of holes of σA(b).

Example 4.3. Let A be the disk algebra, and define a unital homomorphism φ : A→ C(T)by φ(f) = f |T. By the Maximum Modulus principle, φ is isometric. Thus φ is an isometricisomorphism of A onto a unital Banach subalgebra B of C(T). Put b = φ(z). Since isometricisomorphisms preserve spectra, for every a ∈ A we have

σA(a) = σB(φ(a)).

Since A is generated by 1, z, by Proposition 3.17 and Example 3.18 we see that

σA(z) = D = |λ| ≤ 1.

So, we must also haveσB(b) = D.

But it follows from Theorem 3.7 and Example 3.15 that

σC(T)(b) = T.

Thus in this case the spectrum of the element b gets strictly larger when we pass from C(T)to the unital Banach subalgebra B. Note that, in accordance with the Spectral Permanencetheorem, the spectrum relative to the subalgebra B is gotten by adding the hole |λ| < 1.

5 Banach *-algebras

Definition 5.1. An involution on an algebra A is a map a 7→ a∗ on A such that for alla, b ∈ A and λ ∈ C:

(1) (a+ b)∗ = a∗ + b∗;


(2) (λa)∗ = λa∗;

(3) (ab)∗ = b∗a∗;

(4) a∗∗ = a.

A *-algebra is an algebra equipped with an involution.

A normed *-algebra is a normed algebra that equipped with an involution in which‖a∗‖ = ‖a‖ for all a. A Banach *-algebra is a complete normed *-algebra.

Definition 5.2. A *-subalgebra of a *-algebra A is a subalgebra B of A that is closed underinvolution in the sense that a∗ ∈ B for all a ∈ B.

Caution: if A is a *-algebra and S ⊆ A, we will not write S∗ for the set a∗ : a ∈ S,because of the possible confusion with the dual of a normed space.

Every *-subalgebra of a normed *-algebra is itself a *-algebra. The closure of a *-subalgebra of a normed *-algebra is a *-subalgebra. A *-subalgebra of a Banach *-algebra isa Banach *-algebra if and only if it is closed, in which case we call it a Banach *-subalgebra.

Example 5.3. For any Hilbert space H, B(H) is a Banach *-algebra with the usual operatoradjoint:

〈Tx, y〉 = 〈x, T ∗y〉 for all T ∈ B(H), x, y ∈ H.

K(X) is a (typically nonunital) Banach *-subalgebra.

Example 5.4. For any n ∈ N, Mn is a Banach *-algebra under conjugate transpose: ifa = (aij) ∈Mn then a∗ is the matrix with ij-entry aji.

Example 5.5. For any set X, CX is a *-algebra with conjugation:

f ∗(x) = f(x).

We write f = f ∗. `∞(X) is a *-subalgebra (but we don’t call it a Banach *-subalgebrabecause CX is typically not a Banach *-algebra).

Example 5.6. For any topological space X, Cb(X) is a Banach *-subalgebra of `∞(X).

Example 5.7. For any locally compact Hausdorff space X, C0(X) is a Banach *-subalgebraof Cb(X), and Cc(X) is a *-subalgebra of C0(X).

Example 5.8. For any measure space (X,M, µ), L∞(X) is a Banach *-algebra with con-jugation.

Example 5.9. For any measurable space (X,M), B∞(X) is a Banach *-subalgebra of`∞(X). Note: if µ is a measure on the σ-algebraM, then B∞(X) is in general not a Banach*-subalgebra of L∞(X,M, µ), because for f ∈ B∞(X) the uniform norm could be strictlylarger than the ∞-norm.


Example 5.10. `1(Z) is a Banach *-algebra with

f ∗(n) = f(−n).

Similarly, L1(R) is a Banach *-algebra with

f ∗(x) = f(−x),

and L1(T) is a Banach *-algebra with

f ∗(z) = f(z−1).

Example 5.11. The disk algebra is a Banach *-algebra with

f ∗(z) = f(z).

Definition 5.12. A *-homomorphism φ : A → B between *-algebras is a homomorphismsuch that

φ(a∗) = φ(a)∗ for all a ∈ A.

The terms unital *-homomorphism, algebraic *-homomorphism, algebraic *-isomorphism,and algebraic *-automorphism are self-explanatory.

Definition 5.13. A *-isomorphism between normed *-algebras is an algebraic*-isomorphism that is bounded with bounded inverse.

Example 5.14. If H is a finite-dimensional Hilbert space and eini=1 is an orthonormalbasis, then the associated isomorphism B(H) 'Mn is an isometric *-isomorphism of Banachalgebras.

Example 5.15. If (X,M, µ) is a measure space, then the identity map on B∞(X) gives riseto a surjective unital *-homomorphism to L∞(µ) that is in general not injective (because anonzero measurable function could be 0 µ-a.e.)

Definition 5.16. A *-ideal of a *-algebra is an ideal that is also a *-subalgebra. A *-algebrais simple if it has no proper nonzero *-ideals.

Example 5.17. For any locally compact Hausdorff space X and any closed subset S, theset

f ∈ C0(X) : f(x) = 0 for all x ∈ Sis a closed *-ideal of C0(X).

We will not need to officially record a separate concept of maximal *-ideal.

Lemma 5.18. If φ : A → B is a surjective *-homomorphism, then the map I 7→ φ−1(I)gives a bijection from the set of *-ideals of B to the set of *-ideals of A containing kerφ.

Proposition 5.19. If I is a *-ideal of a *-algebra A, then the quotient A/I is a *-algebrawith (a+ I)∗ = a∗ + I, and the quotient map is a *-homomorphism.

6 COMMUTATIVE C∗-ALGEBRAS 21

Exercises



3. (a) Define and prove the existence of the *-subalgebra of a *-algebra generated by a subsetS, and the Banach *-subalgebra of a Banach *-algebra generated by S.

(b) Prove that C(T) is generated as a Banach *-algebra by z.

4. Prove that a direct sum⊕∞

i∈S Ai of Banach *-algebras is a Banach *-algebra in a naturalway.

6 Commutative C∗-algebras

Definition 6.1. A C∗-algebra is a Banach *-algebra A such that

‖a∗a‖ = ‖a‖2 for all a ∈ A. (6.1)

(6.1) is called the C∗-condition, and we also say that ‖ · ‖ is a C∗-norm.

A C∗-norm automatically satisfies one of the axioms of a normed *-algebra:

Lemma 6.2. Let A be a *-algebra that is also a normed algebra, and assume that the normsatisfies the C∗-condition (6.1). Then

‖a∗‖ = ‖a‖ for all a ∈ A.

Proof. Let a ∈ A, and without loss of generality assume that a 6= 0.

‖a‖2 = ‖a∗a‖ ≤ ‖a∗‖‖a‖,

so ‖a‖ ≤ ‖a∗‖. Then replacing a by a∗ gives the opposite inequality ‖a∗‖ ≤ ‖a‖.

Example 6.3. If A is a C∗-algebra and B is a closed *-subalgebra, then B is a C∗-algebra,and we call it a C∗-subalgebra of A. If S ⊆ A, we write C∗(S) for the closed *-algebra of Agenerated by S.

Example 6.4. For any Hilbert space H, B(H) is a C∗-algebra.


One of the cornerstones of the theory is the Gelfand-Naimark theorem, which says thatevery C∗-algebra is isometrically *-isomorphic to a C∗-subalgebra of B(H) for some H.

Of course, B(H) is noncommutative if dimH > 1. In this section our primary interest iscommutative C∗-algebras, the most common of which are recorded in the following example:

Example 6.5. For any set X, `∞(X) is a C∗-algebra, and so are Cb(X) if X is a topologicalspace, C0(X) if X is a locally compact Hausdorff space, L∞(X,µ) for a measure space(X,M, µ), and B∞(X) for a measurable space (X,M).

Remark 6.6. We will show in the Gelfand-Naimark Theorem 6.20 that every unital com-mutative C∗-algebra is of the form C(X) for an essentially unique compact Hausdorff space.This will allow quite simple proofs of many results about (not necessarily commutative)C∗-algebras, but until we get there we have to use more abstract techniques.

Definition 6.7. Let A be a C∗-algebra and a ∈ A. Then a is:

(1) self-adjoint if a = a∗;

(2) normal if aa∗ = a∗a;

(3) an isometry if (A is unital and) a∗a = 1;

(4) unitary if (A is unital and) a−1 = a∗;

(5) a projection if a = a2 = a∗.

Example 6.8. If H is a Hilbert space and T ∈ B(H), then T is

(1) self-adjoint if and only if

〈Tx, y〉 = 〈x, Ty〉 for all x, y ∈ H,

(2) normal if and only if

〈Tx, Ty〉 = 〈T ∗x, T ∗y〉 for all x, y ∈ H,

(3) an isometry if and only if

〈Tx, Ty〉 = 〈x, y〉 for all x, y ∈ H,

(4) unitary if and only if it is a surjective isometry, and

(5) a projection if and only if it is the orthogonal projection of H onto the range T (H).

Example 6.9. If X is a locally compact Hausdorff space and f ∈ C0(X), then f is

(1) self-adjoint if and only if f(X) ⊆ R,


(2) automatically normal,

(3) unitary if and only if f(X) ⊆ T (of course, this requires X compact), and

(4) a projection if and only if f(X) ⊆ 0, 1.

Lemma 6.10. Let A be a unital C∗-algebra and a ∈ A.

(1) If a is invertible, then so is a∗, with (a∗)−1 = (a−1)∗.

(2) If a is unitary then ‖a‖ = 1.

Lemma 6.11. Let A be a C∗-algebra and a ∈ A. Then a = b + ic for unique self-adjointb, c, namely

b = Re a =a+ a∗

2, and c = Im a =

a− a∗

2i.

Lemma 6.12. Let φ : A→ B be a *-homomorphism between C∗-algebras and a ∈ A.

(1) If a is self-adjoint, normal, or a projection then so is φ(a).

(2) If A and B are unital and φ is also unital, then φ(a) is unitary if a is.

(3) If A and B are unital and φ is also unital, then σ(φ(a)) ⊆ σ(a).

Theorem 6.13. If A is a unital C∗-algebra and a ∈ A is self-adjoint, then r(a) = ‖a‖.

Proof. Since a = a∗, we have ‖a2‖ = ‖a‖2. By induction, ‖a2n‖ = ‖a‖2n for all n ∈ N.Therefore by the Spectral Radius formula

r(a) = lim ‖ak‖1/k = lim ‖a2n‖1/2n = ‖a‖.

Corollary 6.14. If A is a unital Banach *-algebra and B is a unital C∗-algebra, then everyunital *-homomorphism φ : A→ B is norm-decreasing.

Proof. We must show that if a ∈ A then ‖φ(a)‖ ≤ ‖a‖, and this follows from the computation

‖φ(a)‖2 = ‖φ(a)∗φ(a)‖= ‖φ(a∗)φ(a)‖= ‖φ(a∗a)‖= r(φ(a∗a))

≤ r(a∗a) (because σ(φ(a∗a)) ⊆ σ(a∗a))

≤ ‖a∗a‖≤ ‖a‖2.


Definition 6.15. If A is a unital C∗-algebra and a ∈ A, we define

ea =∞∑n=0

an

n!,

where a0 = 1 by convention.

This makes sense since the series converges absolutely.

Lemma 6.16. Let A be a unital C∗-algebra and a, b ∈ A.

(1) If a and b commute thenea+b = eaeb.

(2) (ea)∗ = ea∗.

(3) If a is self-adjoint then eia is unitary.

Proposition 6.17. If A is a commutative unital C∗-algebra and u ∈ A is unitary, thenσ(u) ⊆ T.

Proof. It suffices to show that if χ ∈ Ω then χ(u) ∈ T. We have

|χ(u)| ≤ ‖u‖ = 1.

On the other hand, u−1 = u∗ is also unitary, so

1

|χ(u)|= |χ(u)−1| = |χ(u−1)| ≤ 1.

Therefore we must have |χ(u)| = 1.

Corollary 6.18. Let A be a commutative unital C∗-algebra. Then every character of A isa *-homomorphism.

Proof. Let χ ∈ Ω. We must show that χ(a∗) = χ(a) for all a ∈ A, and since a = b + ic forself-adjoint b, c, it suffices to show that if a is self-adjoint then χ(a) ∈ R. Define a unitaryu ∈ A by

u = eia =∞∑n=0

in

n!an,

Then by linearity and continuity

χ(u) =∞∑n=0

in

n!χ(a)n = eiχ(a).

Thus, eiχ(a) ∈ T by Corollary 6.17, and therefore χ(a) ∈ R.


Corollary 6.19. Let A be a unital C∗-algebra and a ∈ A be self-adjoint. Then σ(a) ⊆ R.

Proof. Let B = C∗(1, a) Then B is commutative because a is self-adjoint. By Corollary 6.18,χ(a) ∈ R for all χ ∈ Ω(B), so σB(a) ⊆ R. Since σA(a) ⊆ σB(a), we are done.

Theorem 6.20 (Gelfand-Naimark). If A is a commutative unital C∗-algebra then theGelfand transform Γ: A→ C(Ω) is an isometric *-isomorphism.

Proof. It follows from Corollary 6.18 that Γ is a *-homomorphism. Thus Γ(A) is a self-adjoint subalgebra of C(Ω) that separates points and contains the constants. By the Stone-Weierstrass theorem, Γ(A) is dense in C(Ω). We will show that Γ is isometric, from whichit will follow that Γ(A) is closed, and therefore must be all of C(Ω).

Let a ∈ A. Then‖Γ(a)‖2 = ‖Γ(a)∗Γ(a)‖ = ‖Γ(a∗a)‖

and similarly for ‖a‖2, so without loss of generality a is self-adjoint. Then

‖Γ(a)‖ = r(a) = ‖a‖

by Theorem 6.13.

Actually, there are two results called the Gelfand-Naimark theorem (and the other onewas mentioned earlier), so we have to keep track from the context which one is intended.

There is a generalization of the Gelfand-Naimark theorem to the nonunital case (andC0(Ω)), but since for the present we are primarily interested in spectral theory we don’tneed this now.

For C∗-algebras the Spectral Permanence theorem takes a quite simple form:

Corollary 6.21. Let A be a unital C∗-algebra and B a unital C∗-subalgebra. Then for alla ∈ B we have

σB(a) = σA(a).

Proof. First assume that a is self-adjoint. Then by Corollary 6.19

σA(a) ⊆ σB(a) ⊆ R.

Thus σA(a) has no holes, and so by the Spectral Permanence theorem σB(a) = σA(a).

For the general case, it suffices to show that if a is invertible in A then it is invertiblein B. Let a be invertible in A, so we have b ∈ A such that ab = ba = 1. Then a∗ is alsoinvertible in A, and hence so is a∗a. By the first part of the proof, a∗a is also invertible inB. Since the inverse of a∗a in A is bb∗, we see that bb∗ ∈ B. Then

b = b1 = b(ab)∗ = (bb∗)a∗ ∈ B.


Exercises



3. Prove that if A is a C∗-algebra that has a unit 1A, then 1A is self-adjoint and has norm 1.

4. Prove that if A is a unital C∗-algebra and a ∈ A is normal then ‖a‖ = r(a).

5. Prove there is at most one norm on a unital *-algebra making it a C∗-algebra.

6. Let A be a unital C∗-algebra and a ∈ A. Prove that if a is normal and σ(a) ⊆ Bε(1) then‖a− 1‖ < ε.

7. Let A be a unital C∗-algebra. Prove that if A contains a normal element with disconnectedspectrum then A contains a projection different from 0 and 1.

8. Let X and Y be topological spaces.

(a) Prove that if h : X → Y is a continuous map, then the map h∗ : Cb(Y )→ Cb(X) definedby

h∗(f) = f h

is a unital *-homomorphism. (Note: here the notation h∗ has nothing to do with an involutionin an algebra.)

(b) Prove that if k : Y → Z is another continuous map, then

(k h)∗ = h∗ k∗.

(c) For the rest of this problem assume that X, Y are compact Hausdorff. Prove that thehomomorphism h∗ of part (a) is injective if and only if h is surjective.

(d) Prove that h∗ is surjective if and only if h is injective.

(e) Prove that h is an isomorphism if and only if h is a homeomorphism.

(f) Prove that if φ : C(Y ) → C(X) is a unital homomorphism (not assumed to be *-preserving) then there is a unique continuous map h : X → Y such that φ = h∗. Con-sequently, φ is automatically a *-homomorphism.

9. Let A be a unital C∗-algebra, and let u ∈ A be invertible. Define φ : A → A by φ(a) =uau−1.


(a) Prove that φ is a unital automorphism.

(b) Given an example where φ is not *-preserving.

(c) Prove that if u is unitary then φ is *-preserving.

10. Let A and B commutative unital C∗-algebras and φ : A→ B a unital homomorphism. Provethat φ is actually a *-homomorphism. Hint: Problem 8 and the Gelfand-Naimark theorem.Instructor comment : Alternatively, a proof can be constructed based upon Corollary 6.18.

11. (a) Prove that a direct sum⊕∞

i∈S Ai of C∗-algebras is a C∗-algebra.

(b) Then do the same with the c0-direct sum, defined as the set of all a in⊕∞

i∈S Ai for whichthe function i 7→ ‖ai‖ is in c0(S).

12. Let A and B be commutative unital C∗-algebras. Describe (with complete justifications, asusual) the spectrum of the direct sum A⊕B.

13. Let X be a compact Hausdorff space.

(a) For each subset S ⊆ X let

k(S) = f ∈ C(X) : f(x) = 0 for all x ∈ S.

Prove that k(S) is a closed ideal of C(X).

(b) For each closed ideal I of C(X) let

h(I) = x ∈ X : f(x) = 0 for all f ∈ I.

Prove that h(I) is a closed subset of X.

(c) Prove that if S ⊆ X thenh(k(S)) = S.

(d) Prove that the assignment S 7→ k(S) restricts to a bijection from the family of all closedsubsets of X to the set of all closed ideals of C(X).

14. Let X be a locally compact Hausdorff space, let A = Cb(X), and let Ω = Ω(A). Defineω : X → Ω by

ωx(f) = f(x) for x ∈ X, f ∈ A.

(a) Prove that ω is a homeomorphism of X onto a dense open subset of Ω.

7 FUNCTIONAL CALCULUS 28

(b) Use the above and the results of this section to prove that if h : X → Y is a continuous

map into a compact Hausdorff space Y , then there is a unique continuous map h : Ω → Ysuch that h = h ω.

Instructor comment : Thus Ω is a version of the Stone-Cech compactification βX. Abstractly,X is usually identified with its image in βX, and is characterized by the universal propertythat every continuous map of X into a compact Hausdorff space has a unique continuousextension to βX. In particular, every bounded continuous function on X has a uniquecontinuous extension to βX.

7 Functional calculus

We begin with a C∗-version of Theorem 3.17.

Theorem 7.1. Let A be a unital commutative C∗-algebra generated by a. Then the Gelfandtransform a : Ω→ σ(a) is a homeomorphism.

Note: here we mean that A is generated as a unital C∗-algebra by a, and it so happensthat A is commutative, i.e., a is normal.

Proof. By Theorem 3.7 we know that the continuous function a maps Ω onto σ(a). Itremains to show that a is injective. Let χ, τ ∈ Ω, and assume that χ(a) = τ(a). Sinceχ(1) = τ(1) = 1, and since A is generated as a Banach algebra by 1, a, a∗, it suffices toobserve that we also have χ(a∗) = τ(a∗) because χ and τ are *-homomorphisms.

Theorem 7.2 (Functional calculus). Let A be a unital C∗-algebra, and let a ∈ A be nor-mal. Then there is a unique unital *-homomorphism φ : C(σ(a)) → A such that φ(z) = a.Moreover, φ is an isometric isomorphism onto C∗(1, a).

Proof. For this proof let B = C∗(1, a). Let Γ: B → C(Ω(B)) be the Gelfand trans-form. By the commutative Gelfand-Naimark theorem, Γ is a *-isomorphism. By The-orem 7.1, a : Ω(B) → σ(a) is a homeomorphism. Then by Problem 8 in Section 6 themap a∗ : C(σ(a)) → C(Ω(B)) is a *-isomorphism. Therefore there is a unital isometric*-isomorphism φ making the diagram

A C(σ(a))

a∗'

φ

'

BΓ

' //?

OO

C(Ω(B))

Clearly φ(z) = a, and it only remains to show the uniqueness, i.e., if also τ : C(σ(a))→ B is aunital *-homomorphism with τ(z) = a then φ = τ . But since φ and τ are *-homomorphismswe have φ(z) = τ(z), so because C(σ(a)) is generated as a unital C∗-algebra by z we aredone.


Definition 7.3. The homomorphism φ : C(σ(a))→ A from Theorem 7.2 is called the func-tional calculus at a, and for f ∈ C(σ(a)) we write

f(a) = φ(f).

Note that the above notation f(a) is consistent with p(a) for a polynomial p.

It’s instructive to examine the functional calculus when A = C(X):

Proposition 7.4. Let X be a compact Hausdorff space, g ∈ C(X), and f ∈ C(g(X)). Then

f(g) = f g. (7.1)

Proof. Note that this makes sense since g(X) = σ(g). Identifying the spectrum of C(X)with X, the Gelfand transform of C(X) is the identity automorphism. The left-hand sideof (7.1) uses the functional calculus at g, which is the unique unital *-homomorphism fromC(g(X)) to C(X) taking z to g. The map f 7→ f g from C(g(X)) to C(X) is a unital*-homomorphism taking z to g, so the result follows from uniqueness.

Homomorphisms are compatible with the functional calculus:

Theorem 7.5. Let π : A → B be a unital *-homomorphism of unital C∗-algebras, and leta ∈ A be normal. Then

π(f(a)) = f(π(a)) for all f ∈ C(σ(a)).

Proof. Note that π(a) is normal in B, so it has a functional calculus. Since σ(π(a)) ⊆ σ(a),every continuous function on σ(a) restricts to a continuous function on σ(π(a)). Both f 7→π(f(a)) and f 7→ f(π(a)) are unital *-homomorphisms C(σ(a)) → B taking z to π(a), sothey must be equal because z generates C(σ(a)) as a unital C∗-algebra.

Corollary 7.6. Let Γ be the Gelfand transform of a commutative unital C∗-algebra A. Thenfor all a ∈ A and f ∈ C(σ(a)),

Γ(f(a)) = f (Γ(a)).

Note the careful placement of parentheses in the above displayed equation. We could havewritten f(Γ(a)), because σ(Γ(a)) = σ(a), but it would have perhaps been a bit confusing.

Proof. Apply Theorem 7.5 with π = Γ.

Corollary 7.7. Let A be a unital commutative C∗-algebra generated by a, and let f ∈C(σ(a)). Then

χ(f(a)) = f(χ(a)) for all χ ∈ Ω(A).


Proof. Apply Theorem 7.5 with π = χ.

Theorem 7.8 (Spectral Mapping). Let a be a normal element of a unital C∗-algebra A, andlet f ∈ C(σ(a)). Then

σ(f(a)) = f(σ(a)),

and if g ∈ C(σ(f(a))) then(g f)(a) = g(f(a)).

Proof. Let B = C∗(1, a), and keep the notation from the proof of the commutative Gelfand-Naimark theorem. Since φ : C(σ(a))→ B is an isomorphism, it preserves spectra. Thus

σ(f(a)) = σ(φ(f)) = σ(f) = f(σ(a)).

For the other part, by Theorem 7.5 applied to the *-homomorphism f 7→ f(a), we haveg(f)(a) = g(f(a)), and g(f) = g f by Proposition 7.4.

Corollary 7.9. Let φ : A → B be an injective unital *-homomorphism between unital C∗-algebras. Then φ is isometric, and φ(A) is a C∗-subalgebra of B.

Proof. For the first part, by Corollary 6.14 it suffices to show that for all a ∈ A we have‖φ(a)‖ ≥ ‖a‖, and as usual it suffices to do this for self-adjoint a. For this, it is enoughto show that σ(a) = σ(φ(a)). If not, then we can choose a nonzero f ∈ C(σ(a)) such thatf = 0 on σ(φ(a)). Then f(φ(a)) = 0. By Theorem 7.5 we have f(φ(a)) = φ(f(a)). But thenf(a) ∈ kerφ = 0, contradicting injectivity of the functional calculus. Now the other partfollows immediately: φ(A) is a complete, and hence closed, *-subalgebra of B.

Remark 7.10. Alternatively, in the above corollary, once we had reduced to self-adjoint a,we could have replaced A and B by the commutative unital C∗-subalgebras C∗(1, a) andC∗(1, φ(a)), respectively, then by the Gelfand-Naimark theorem it would be enough to proveit for φ : C(X)→ C(Y ) with X, Y compact Hausdorff, and we could have applied Section 6Problem 8. However, although this might seem conceptually more elementary, it would nothave saved any writing.

Exercises

1. Prove that a normal element a of a unital C∗-algebra A is:

(a) self-adjoint if and only if σ(a) ⊆ R;

(b) unitary if and only if σ(a) ⊆ T;

(c) a projection if and only if σ(a) ⊆ 0, 1.

8 SPECTRAL THEOREM 31

2. Let A be a unital C∗-algebra and a ∈ A be normal. Prove that there exists λ ∈ σ(a) suchthat |λ| = ‖a‖.

3. Let A be a unital C∗-algebra. Then a ∈ A is called positive, written a ≥ 0, if a is self-adjointand σ(a) ⊆ [0,∞). Prove that if a ≥ 0 then a has a unique positive square root, i.e., there isa unique b ≥ 0 in A such that b2 = a. Hint: any such b commutes with a, and B = C∗(1, a, b)is a unital commutative C∗-subalgebra of A containing a. Then B contains C = C∗(1, a),and hence contains the unique positive square root c of a in C, and it can be shown fromthis that c = b.

Instructor comment : The unique square root of a is denoted by a1/2 =√a.

4. Let A be a unital C∗-algebra, and let a ∈ A be normal and invertible. Then a has a uniquepolar decomposition a = up, where u is unitary and p ≥ 0. Hint: show that p = (a∗a)1/2 isinvertible.

5. Let A be a unital C∗-algebra, and let a ∈ A be normal. We now have two definitions ofea: one as a power series, and the other using the functional calculus. Prove that these areconsistent.

8 Spectral theorem

Definition 8.1. A spectral measure on a compact Hausdorff space X is a map P from theBorel σ-algebra of X to the set of projections on a Hilbert space H such that

(1) P (∅) = 0 and P (X) = 1;

(2) P (E ∩ F ) = P (E)P (F ) for all Borel E,F ⊆ X;

(3) for all x, y ∈ H the map Px,y defined by Px,y(E) = 〈P (E)x, y〉 is a regular complexBorel measure on X.

A spectral measure is sometimes called a resolution of the identity, and is sometimesconsidered in more general measurable spaces, but we will have no use for this.

Remark 8.2. Of course, the Hilbert space H is in some sense attached to the spectralmeasure. However, many (but not all) authors do not mentionH explicitly in the terminologyor the notation for P , and I’ll continue this tradition.

Some authors formulate condition (3) above in terms of countable additivity of the mapP itself, and this is justified by:


Lemma 8.3. Let P be a map from the Borel sets on X to the projections on H satisfying(1)–(2) in Definition 8.1. Then P is a spectral measure if and only if:

(3)′ for every sequence (En) of pairwise disjoint Borel subsets of X,

P

(∞⋃n=1

En

)=∞∑n=1

P (En),

where the series converges in the weak operator topology, i.e.,⟨∞∑n=1

P (En)x, y

⟩=∞∑n=1

〈P (En)x, y〉 for all x, y ∈ H.

In fact, condition (3)′ also implies condition (2).

Note that in Definition 8.1 we don’t impose any relationship between the norms of thevectors x, y and the measure Px,y. However:

Lemma 8.4. If P is a spectral measure on X and x, y ∈ H, then

‖Px,y‖ ≤ ‖x‖‖y‖.

Proof. Let E1, . . . , En be a Borel partition of X. Then

n∑i=1

|Px,y(Ei)| =n∑1

|〈P (Ei)x, y〉|

=n∑1

|〈P (Ei)x, P (Ei)y〉|

≤n∑1

‖P (Ei)x‖‖P (Ei)y‖

≤

(n∑1

‖P (Ei)x‖2

)1/2( n∑1

‖P (Ei)y‖2

)1/2

= ‖x‖‖y‖,

by the Pythagorean theorem, since x =∑n

1 P (Ei)x is a sum of pairwise orthogonal vectors,and similarly for y.

For the next proof we need to know a bit about sesquilinear forms. A sesquilinear formon a Hilbert space H is a map

ζ : H ×H → C


that is linear in the first variable and conjugate-linear in the second. A sesquilinear form ζis bounded if there exists M ≥ 0 such that∣∣ζ(x, y)

∣∣ ≤M‖x‖‖y‖ for all x, y ∈ H.

Every sesquilinear form ζ satisfies a version of the polarization identity :

ζ(x, y) =1

4

3∑k=0

ikζ(x+ iky, x+ iky) for all x, y ∈ H.

Thus ζ is determined by its values at the ordered pairs (x, x). The essential fact we need,which can be proved similarly to the existence of adjoint operators T ∗, is:

Lemma 8.5. For every bounded sesquilinear form ζ on H there is a unique operator T ∈B(H) such that

〈Tx, y〉 = ζ(x, y) for all x, y ∈ H.

We use Lemma 8.5 to define integration with respect to a spectral measure:

Theorem 8.6. Let P be a spectral measure on a compact Hausdorff space X. Then forevery f ∈ B∞(X) there is a unique operator

∫f dP ∈ B(H) such that⟨(∫

f dP

)x, y

⟩=

∫f dPx,y for all x, y ∈ H.

Proof. Let f ∈ B∞(X). The map ζ : H ×H → C defined by

ζ(x, y) =

∫f dPx,y for x, y ∈ H

is a sesquilinear form, which is bounded because∣∣ζ(x, y)∣∣ ≤ ∫ |f | d|Px,y| ≤ ‖f‖‖Px,y‖ ≤ ‖f‖‖x‖‖y‖,

and so the result follows from Lemma 8.5.

In Theorem 8.6, note that∫χE dP = P (E) for every Borel E ⊆ X.

Theorem 8.7. If P is a spectral measure on a compact Hausdorff space X, then the mapφ : B∞(X)→ B(H) defined by

φ(f) =

∫f dP

is a unital *-homomorphism.


Proof. Routine computations show that φ is linear, and φ is bounded since for all x, y ∈ Hthe computations in the proof of Lemma 8.6 imply that

|〈φ(f)x, y〉| ≤ ‖f‖‖x‖‖y‖.

Thus, to show that φ preserves products and adjoints, it suffices to check on the densesubspace of simple functions. Then routine computations with linear combinations showthat it is enough to check for characteristic functions. So, let E,F ⊆ X be Borel. Then

φ(χE)φ(χF ) = P (E)P (F ) = P (E ∩ F ) = φ(χE∩F ) = φ(χEχF )

andφ(χE)∗ = P (E)∗ = P (E) = φ(χE) = φ(χE).

Finally, φ is unital:

φ(1) =

∫1 dP =

∫χX dP = P (X) = 1.

Theorem 8.8. Let X be a compact Hausdorff space and φ : C(X) → B(H) a unital *-homomorphism. Then there is a unique spectral measure P on X such that

φ(f) =

∫f dP for f ∈ C(X).

Moreover, S ∈ B(H) commutes with φ(f) for all f ∈ C(X) if and only if S commutes withP (E) for every Borel set E ⊆ X.

Proof. This proof takes a while, because there are lots of things to check. For all x, y ∈ Hdefine a linear functional ωx,y on C(X) by

ωx,y(f) = 〈φ(f)x, y〉.

Then ‖ωx,y‖ ≤ ‖x‖‖y‖, so in particular ωx,y ∈ C(X)∗, and hence there is a unique µx,y ∈M(X) such that

ωx,y(f) =

∫f dµx,y for all f ∈ C(X).

Moreover,‖µx,y‖ = ‖ωx,y‖ ≤ ‖x‖‖y‖.

Let f ∈ B∞(X), and define ζ : H ×H → C by

ζ(x, y) =

∫f dµx,y.

Claim: ζ is a bounded sesquilinear form. To see this, let λ ∈ C, x, y, z ∈ H. then for allg ∈ C(X) we have ∫

g dµλx+z,y = 〈φ(g)(λx+ z), y〉


= λ〈φ(g)x, y〉+ 〈φ(g)z, y〉

= λ

∫g dµx,y +

∫g dµz,y

=

∫g d(λµx,y + µz,y),

and soµλx+z,y = λµx,y + µz,y.

Thus

ζ(λx+ z, y) =

∫f dµλx+z,y

=

∫f d(λµx,y + µz,y)

= λ

∫f dµx,y +

∫f dµz,y

= λζ(x, y) + ζ(z, y).

Similarly,ζ(x, λy + z) = λζ(x, y) + ζ(x, z),

the only difference being that we factor λ out of the second variable of the inner product,giving λ. Thus ζ is sesquilinear. For the boundedness, let x, y ∈ H. Then

|ζ(x, y)| ≤∫|f | dµx,y ≤ ‖f‖‖µx,y‖,

so now it suffices to show ‖µx,y‖ ≤ ‖x‖‖y‖: for all g ∈ C(X),∣∣∣∣∫ g dµx,y

∣∣∣∣ = |〈φ(g)x, y〉|

≤ ‖φ(g)‖‖x‖‖y‖≤ ‖g‖‖x‖‖y‖,

so

‖µx,y‖ = supg∈C(X)‖g‖≤1

∣∣∣∣∫ g dµx,y

∣∣∣∣ ≤ ‖x‖‖y‖.We have verified the claim. Thus there is a unique operator ψ(f) ∈ B(H) such that

ζ(x, y) = 〈ψ(f)x, y〉 for all x, y ∈ H.

Moreover, if f ∈ C(X) then

〈ψ(f)x, y〉 =

∫f dµx,y = ωx,y(f) = 〈φ(f)x, y〉 for all x, y ∈ H,


and so ψ : B∞(X)→ B(H) extends φ. Clearly ψ is linear, because for all x, y ∈ H the mapf 7→

∫f dµx,y is linear.

If f ∈ C(X) is self-adjoint, then φ(f) is self-adjoint, so for every x ∈ H the number∫f dµx,x = 〈φ(f)x, x〉

is real, and hence the measure µx,x is real. Thus, for all self-adjoint f ∈ B∞(X) we have

〈ψ(f)x, x〉 =

∫f dµx,x ∈ R for all x ∈ H,

so ψ(f) is self-adjoint. It follows that ψ(f) = ψ(f)∗ for all f ∈ B∞(X).

We will show multiplicativity of ψ in a few steps. First, for all f, g ∈ C(X) and x, y ∈ H,∫fg dµx,y = 〈φ(fg)x, y〉

= 〈φ(f)φ(g)x, y〉

=

∫f dµφ(g)x,y,

and sogµx,y = µφ(g)x,y.

Next, for all f ∈ B∞(X), g ∈ C(X), and x, y ∈ H,∫gf dµx,y =

∫fg dµx,y

=

∫f dµφ(g)x,y

= 〈ψ(f)φ(g)x, y〉= 〈φ(g)x, ψ(f)∗y〉= 〈φ(g)x, ψ(f)y〉

=

∫g dµx,ψ(f)y,

sofµx,y = µx,ψ(f)y.

Thus, for all f, g ∈ B∞(X), x, y ∈ H,

〈ψ(fg)x, y〉 =

∫fg dµx,y

=

∫gf dµx,y


=

∫g dµx,ψ(f)y

= 〈ψ(g)x, ψ(f)y〉= 〈ψ(f)ψ(g)x, y〉,

and thereforeψ(fg) = ψ(f)ψ(g).

We have shown that ψ : B∞(X) → B(H) is a *-homomorphism extending φ, and hence isunital.

For each Borel set E ⊆ X define

P (E) = ψ(χE).

We will show that P is a spectral measure. Each P (E) is a projection because χE is. Wehave

P (∅) = ψ(χ∅) = ψ(0) = 0,

andP (X) = ψ(χX) = ψ(1) = 1.

If E and F are Borel subsets of X, then

P (E ∩ F ) = ψ(χE∩F )

= ψ(χEχF )

= ψ(χE)ψ(χF )

= P (E)P (F ).

Finally for every Borel E ⊆ X and x, y ∈ H,

〈P (E)x, y〉 = 〈ψ(χE)x, y〉

=

∫χE dµx,y

= µx,y(E),

so E 7→ 〈P (E)x, y〉 is the regular complex Borel measure µx,y.

For the uniqueness, suppose that P ′ is any spectral measure on X such that

φ(f) =

∫f dP ′ for all f ∈ C(X).

Then for all f ∈ C(X) and x, y ∈ X,∫f dP ′x,y = 〈φ(f)x, y〉 =

∫f dPx,y,


so P ′x,y = Px,y, and hence for all Borel E ⊆ X we have

〈P ′(E)x, y〉 =

∫χE dP

′x,y =

∫χE dPx,y = 〈P (E)x, y〉,

and therefore P ′(E) = P (E).

We turn to the commutativity clause. Let S ∈ B(H), and suppose that S commuteswith φ(f) for all f ∈ C(X). Then for all f ∈ C(X) and x, y ∈ H,∫

f dPSx,y = 〈φ(f)Sx, y〉

= 〈Sφ(f)x, y〉= 〈φ(f)x, S∗y〉

=

∫f dPx,S∗y,

so PSx,y = Px,S∗y. Thus for every Borel E ⊆ X and x, y ∈ H we have

〈P (E)Sx, y〉 =

∫χE dPSx,y

=

∫χE dPx,S∗y

= 〈P (E)x, S∗y〉= 〈SP (E)x, y〉,

and so P (E)S = SP (E). Conversely, assuming P (E)S = SP (E) for all Borel E, essentiallyreverse the above steps to see that for all x, y ∈ H we have PSx,y = Px,S∗y, so for everyf ∈ C(X) we get 〈φ(f)Sx, y〉 = 〈Sφ(f)x, y〉, and therefore φ(f)S = Sφ(f).

Note that in the above proof we did not need to show explicitly that ψ is bounded —as soon as we know it’s a unital *-homomorphism between unital C∗-algebras then we knowautomatically that it’s bounded (and in fact has norm one).

The Spectral theorem is a special case:

Theorem 8.9 (Spectral theorem). Let T ∈ B(H) be normal. Then there is a unique spectralmeasure P on σ(T ) such that T =

∫z dP . Moreover, S ∈ B(H) commutes with T if and

only if S commutes with P (E) for every Borel set E ⊆ σ(T ).

Proof. Apply Theorem 8.8 to the continuous functional calculus for T . The only subtletyis the new fact that if S commutes with T then it also commutes with T ∗ (and hence withφ(f) for all f ∈ C(σ(T ))). This is a consequence of Fuglede’s theorem below.

Theorem 8.10 (Fuglede’s theorem). Let A be a unital C∗-algebra and a, b ∈ A. If a isnormal and b commutes with a, then b also commutes with a∗.


Proof. An induction argument shows that b commutes with every polynomial in a, and hencewith eiza for all z ∈ C. Thus

eizab = beiza for all z ∈ C.

Define an entire function f : C→ A by

f(z) = e−iza∗beiza

∗.

Thenf(z) = e−iza

∗e−izabeizaeiza

∗= e−i(za

∗+za)bei(za+za∗).

Since za∗ + za is self-adjoint, we have unitaries u(z) and v(z) such that

f(z) = u(z)bv(z) for all z ∈ C,

and so f is bounded. Since f is entire, by Liouville’s theorem it is constant, so

0 = f ′(0) = −ia∗b+ iba∗,

therefore a∗b = ba∗.

Definition 8.11. In the Spectral theorem, the values of the spectral measure P of T arecalled spectral projections of T , and the unital *-homomorphism

f 7→∫f dP : B∞(X)→ B(H)

is called the Borel functional calculus at T . If f ∈ B∞(σ(T )) we write

f(T ) =

∫f dP.

Thus,

〈f(T )x, x〉 =

∫〈f dPx,y〉.

To avoid confusion, the functional calculus in Definition 7.3 is sometimes called the contin-uous functional calculus.

Remark 8.12. The spectral measure P of a normal operator T is sometimes called theresolution of the identity for T , although in the past this was sometimes reserved for self-adjoint T , and then the spectral measure was defined for all Borel subsets of R. Moreover,spectral measures for normal T are sometimes defined for all Borel subsets of C. We will nothave any use for these terms or conventions.


Exercises

1. Prove that T ∈ B(H) is self-adjoint if and only if 〈Tx, x〉 ∈ R for all x ∈ H. Hint:polarization.

2. Let T ∈ B(H) be normal, with spectral measure P , and let U ⊆ σ(T ) be nonempty andrelatively open. Prove that

Px,x(U) > 0 for all nonzero x ∈ χU(T )H.

3. Prove that if P is the spectral measure of a normal T ∈ B(H), then P (U) 6= 0 for everynonempty relatively open subset U ⊆ σ(T ).

4. Let T ∈ B(H). Prove that 〈Tx, x〉 ≥ 0 for all x ∈ H if and only if T is self-adjoint andσ(T ) ⊆ [0,∞). Instructor comment : In this case T is called positive, written T ≥ 0. Thus,this usage of the term “positive” is consistent with Problem 3 in Section 7, i.e., T is positiveas an element of the C∗-algebra B(H) as in that earlier problem if and only if 〈Tx, x〉 ≥ 0for all x ∈ H.

5. Prove that T ∈ B(H) is normal if and only if ‖Tx‖ = ‖T ∗x‖ for all x ∈ H.

6. Prove that if T ∈ B(H) is normal then kerT = kerT ∗ and TH = T ∗H.

Instructor comment : There was a typo in this problem: it should be TH = T ∗H insteadof TH = T ∗H. If T (and hence T ∗) doesn’t have closed range, the question of whetherTH = T ∗H is delicate (at least as far as I know). Also, this problem is quite elementary(in particular, does not relate directly to the Spectral Theorem), and has been moved muchearlier in the exercises than it originally was.

7. Prove that T ∈ B(H) is an isometry if and only if ‖Tx‖ = ‖x‖ for all x ∈ H.

Instructor comment : Thus, this use of the term isometry is consistent with the more generalconcept for operators on a Banach space.

8. Let T ∈ B(H) be normal. Then T has a polar decomposition T = UP , where U is unitary,P ≥ 0, and T, U, P commute. Hint: consider suitable Borel functions on σ(T ).

9. T ∈ B(H) is called a partial isometry if T ∗T is a projection.

(a) Prove that the following are equivalent:

(1) T is a partial isometry;

(2) TT ∗T = T ;

(3) T ∗ is a partial isometry;


(4) T ∗TT ∗ = T ∗;

(5) T is isometric on (kerT )⊥.

(b) Instructor comment : Originally, there were typos: T ∗T and TT ∗ should have been T ∗THand TT ∗H, respectively.

Let T be a partial isometry. Then T ∗TH is called the initial subspace of T , and TT ∗H iscalled the final subspace of T . Prove:

(1) the range of T is the final subspace of T ;

(2) the range of T ∗ is the initial subspace of T .

10. For every T ∈ B(H) there is a unique polar decomposition T = UP with P ≥ 0 and U apartial isometry with initial subspace PH. Hint: take P = (T ∗T )1/2, and show that it makessense to define U : PH → TH by UPx = Tx.

11. Prove that if T ∈ B(H) and T ∗T is compact then T is compact.



14. Let H be a separable infinite-dimensional Hilbert space, and let T ∈ B(H) be diagonal withrespect to an orthonormal basis en. Describe in precise terms the spectral measure P forT .

15. Let X be a compact Hausdorff space, and let µ be a (positive) regular Borel measure on X.For f ∈ L∞(X,µ), recall that the multiplication operator Mf on L2(X,µ) is defined by

(Mfξ) = fξ for ξ ∈ L2(µ).

Prove that the assignment E 7→MχE

from Borel sets to operators is a spectral measure.

16. Let µ be a regular Borel measure on C with compact support K. Define T ∈ L2(C, µ) by(Tξ)(z) = zξ(z). Prove:

(a) T is normal.

(b) σ(T ) = K.

(c) For every f ∈ B∞(C) define Mf ∈ B(L2(µ)) by Mfξ = fξ. Then f(T ) = Mf .

17. Let T ∈ B(H) be normal, with spectral measure P , and f ∈ B∞(σ(T )). In the follow-ing parts, the phrase “a.e. z ∈ σ(T )” will mean “Px,y-a.e. z ∈ σ(T ), for all x, y ∈ H”.Prove:


(a) f(T ) is self-adjoint if and only if f(z) ∈ R for a.e. z ∈ σ(T ).

(b) f(T ) is unitary if and only if f(z) ∈ T for a.e. z ∈ σ(T ).

(c) f(T ) is a projection if and only if f(z) ∈ 0, 1 for a.e. z ∈ σ(T ).

18. Let T ∈ B(H) be normal and g ∈ C(C). Prove that

(g f)(T ) = g(f(T )) for all f ∈ B∞(σ(T )).

Hint: first show it when g is a polynomial in z and z, then use the Stone-Weierstrass theorem.

19. Prove that for every unitary U ∈ B(H) there is a self-adjoint T ∈ B(H) such that U = eiT .Hint: consider the inverse of the function

t 7→ eit : [0, 2π)→ T,

and apply the preceding problem.

20. Let T ∈ B(H) be normal, with spectral measure P . Let U be a nonempty relatively opensubset of σ(T ) such that 0 /∈ U . Prove that P (U)H ⊆ TH. (Note: TH = Tx : x ∈ H isshorthand for the range of T , and similarly for P (U).) Hint: consider the function

g(z) =

(1/z)χU(z) if z 6= 0

0 if z = 0.

21. Prove that if T ∈ B(H) is normal, then every isolated point in σ(T ) is an eigenvalue.

22. Let T ∈ B(H) be normal, with spectral measure P . Prove that λ is an eigenvalue of T ifand only if P (λ) 6= 0, in which case P (λ)H is the associated eigenspace.

23. Let T ∈ B(H) be normal, with spectral measure P . For every n ∈ N define Pn = P (|z| <1/n). Prove that limn→∞ ‖TPn‖ = 0.

24. Let T ∈ B(H) be normal. Prove that TH is closed if and only if 0 is not a limit point ofσ(T ). Hint: show that TH = TK ⊆ K, where K = (kerT )⊥.

25. Let T ∈ B(H) be normal. Prove that

‖T‖ = sup|〈Tx, x〉| : ‖x‖ ≤ 1.

Hint: Without loss of generality T 6= 0. Choose λ ∈ σ(T ) such that |λ| = ‖T‖. For ε > 0let Uε = Bε(λ), and use the functions

fε = (z − λ)χUε .

to help show that there exists a unit vector x ∈ H such that

|〈Tx, x〉 − λ| ≤ ε.

9 UNITIZATION 43

26. Let T ∈ B(H) be normal, with spectral measure P . Prove that T is compact if and only iffor every ε > 0 the spectral projection Pε = P (|z| > ε) is finite rank.

27. (Spectral Theorem for compact normal operators) Let T ∈ B(H) be normal and compact,with spectral measure P . Prove that σ(T ) is either finite or a sequence converging to 0, andEλ = P (λ) is finite-rank for all nonzero λ ∈ σ(T ). Moreover, prove that

T =∑

λ∈σ(T )

λEλ,

where the series converges in the operator norm.

9 Unitization

We’ve developed spectral theory using unital algebras. But many Banach algebras come“out of the box” without a unit, and it’s convenient to be able to pass (perhaps temporarily)to the unital case. It’s all quite routine, but it takes a disappointing amount of fussinessto get all the details right. We start with a very general context, then quickly specialize toC∗-algebras.

First suppose that A is any algebra. Let A = A ⊕ C as a vector space, and give it themultiplication

(a, λ)(b, µ) = (ab+ λb+ µa, λµ).

Then A is a unital algebra, with unit (0, 1), containing an isomorphic copy A⊕ 0 of A asa maximal ideal.

If A is a *-algebra, we give A the involution

(a, λ)∗ = (a∗, λ).

If A is a normed algebra, we can make A into a normed algebra by

‖(a, λ)‖ = ‖a‖+ |λ|.

If A is a normed *-algebra, then so is A, and if A is complete then so is A.

Definition 9.1. A is the unitization of A.

Lemma 9.2. If A is a unital Banach algebra, then A is isomorphic to the Banach-algebradirect sum A⊕ C via the map

(a, λ) 7→ (a+ λ1A, λ).

If A has an involution then the above isomorphism is *-preserving.

9 UNITIZATION 44

Remark 9.3. In fact, we made a choice here: some authors would define A = A if A alreadyhas a unit.

In practice we simplify the notation, identifying A with its image in A, and writing a+λ1,or just a+ λ, instead of (a, λ).

Unitizations are useful in many ways. For example, we can define the spectrum in thenonunital case:

Definition 9.4. If A is a nonunital Banach algebra and a ∈ A, we define the spectrum of aas the spectrum of a in A, and keep the same notation σ(a).

Remark 9.5. Note that if A is a nonunital Banach algebra then 0 ∈ σ(a) for all a ∈ A.

Remark 9.6. The following two propositions are parallel, and each is useful. The secondone is an almost immediate corollary of the first, but I chose to give an independent proofbecause I wanted each to stand on its own.

Proposition 9.7. Let π : A → B be a homomorphism of Banach algebras, with B unital.Then π extends uniquely to a unital homomorphism π : A → B. If A and B are Banach*-algebras then π is a *-homomorphism. If A is nonunital and π is injective, then π is alsoinjective.

Proof. Defineπ(a+ λ) = π(a) + λ1B.

Routine computations show that π is a unital homomorphism extending π. It is obviousthat the extension is unique. Then the case of *-algebras is also routine.

Finally, assume that A is nonunital and π is injective. Suppose that π(a+ λ) = 0. Thenπ(a) = −λ1, so because π(A) is nonunital (being isomorphic to A), we must have λ = 0,and hence a = 0 because π is injective, and therefore a+ λ = 0.

Proposition 9.8. If π : A → B is a homomorphism of Banach algebras, then there is aunique unital homomorphism π making the diagram

Aπ // _

B _

Aπ// B

commute, where the vertical arrows are inclusion maps. If A and B are Banach *-algebrasthen π is a *-homomorphism. If A is nonunital and π is injective, then π is also injective.

Proof. Everything except the last assertion is completely routine: define

π(a+ λ) = π(a) + λ,

9 UNITIZATION 45

and check everything (again, uniqueness is obvious).

For the last part, assume that A is nonunital and π is injective. Suppose that π(a+λ) = 0.Then π(a) = −λ1, so because π(A) is nonunital (being isomorphic to A), we must have λ = 0,and hence a = 0 because π is injective, and therefore a+ λ = 0.

Remark 9.9. As it happens, in these notes we are only really interested in the C∗-case.This forces us to do a bit more work to get the right norm.1

Theorem 9.10. For every C∗-algebra A there is a unique C∗-norm on A.

Proof. We argue differently in the unital case: if A is unital just give A the C∗-norm of theC∗-algebra direct sum A⊕ C via the isomorphism of Lemma 9.2.

Now suppose A is nonunital. For each a ∈ A define La ∈ B(A) (the algebra of boundedoperators on A) by

La(b) = ab for b ∈ A.

Then L : A → B(A) is an algebraic homomorphism. We show that L is isometric: for alla ∈ A we have

‖La(b)‖ = ‖ab‖ ≤ ‖a‖‖b‖ for all b ∈ A,

so ‖La‖ ≤ ‖a‖. For the opposite inequality ‖La‖ ≥ ‖a‖, just note that ‖La(a∗)‖ = ‖a‖‖a∗‖.Thus L(A) is a Banach subalgebra of B(A). Enlarge it to a unital subalgebra C by

C = L(A) + C1,

where 1 denotes the identity operator on A. Then C is closed in B(A) because L(A) is closedand C1 is finite-dimensional.

Extend L to a unital homomorphism L : A → B(A) as in Proposition 9.7. Since L isinjective, so is L, by Proposition 9.7, and hence L is an isomorphism onto C.

Define an operation * on C by

(La + λ1)∗ = La∗ + λ1.

Routine computations show that this is an involution, and then by construction L : A→ Cis a *-isomorphism.

It now suffices to show that C is a C∗-algebra, and for this it suffices to show that ifT = La + λ1 then ‖T‖2 ≤ ‖T ∗T‖ in B(A): for all unit vectors b ∈ A we have

‖T (b)‖2 = ‖(ab+ λb)‖2

= ‖(ab+ λb)∗(ab+ λb)‖= ‖b∗T ∗T (b)‖

1Nevertheless, it’s useful to know that unitization can be performed for arbitrary Banach algebras.

9 UNITIZATION 46

≤ ‖(T ∗T )(b)‖≤ ‖T ∗T‖,

so ‖T‖2 ≤ ‖T ∗T‖.

The uniqueness follows from Section 6 Problem 5.

Remark 9.11. When A is a C∗-algebra we always give the unitization A the C∗-norm fromTheorem 9.10.

Remark 9.12. It is easy — and useful — to extend various concepts from the unital caseto arbitrary C∗-algebras; for example, in any C∗-algebra A we can define an element a to bepositive if it is self-adjoint and σ(a) ⊆ [0,∞).

Corollary 9.13. Let π : A → B be a *-homomorphism of C∗-algebras. Then ‖π‖ ≤ 1.Moreover, if π is injective then it is isometric and π(A) is a C∗-subalgebra of B,

Proof. Replacing B by B if necessary, without loss of generality B is unital. If A is alsounital, we are done, by Corollaries 6.14 and 9.13. Otherwise, extend to π : A→ B. Then

‖π‖ ≤ ‖π‖ ≤ 1.

Now suppose that π is injective. Then π is also injective by Proposition 9.7, and so π, andhence also π, is isometric, and hence the *-subalgebra π(A) of B is complete and thereforea C∗-subalgebra.

We define characters of a nonunital commutative C∗-algebra A just as in the unital case,namely nonzero homomorphisms into C, and we will again write Ω = Ω(A) for the set of allcharacters of A.

In the following corollary we exclude the case A = 0 because, besides the fact that itis trivial and does not require a theorem, it would violate the noncompactness of Ω(A) —indeed, the spectrum of the zero C∗-algebra is the empty set.

Corollary 9.14. Let A be a nonzero nonunital commutative C∗-algebra. Then every char-acter χ of A extends uniquely to a character χ of A. Consequently, every character of A is*-preserving. There is exactly one character of A that does not arise as the extension of acharacter of A, namely the character ω0 whose kernel is the maximal ideal A.

It follows that Ω = Ω(A) is contained in the unit ball of A∗, and the spectrum Ω(A)can be embedded in the unit ball of A∗ by sending the special character ω0 to the 0 vector.Then:

Corollary 9.15. With the same hypotheses as Corollary 9.14, the spectrum Ω(A) is locallycompact Hausdorff in the relative weak* toplogy of A∗. Moreover, the one-point compactifi-cation of Ω(A) is Ω(A).

9 UNITIZATION 47

Proof. We know that Ω(A) is compact, and hence the subset Ω(A) obtained by removing a

single point is locally compact Hausdorff. Moreover, the compact space Ω(A) is the one-pointcompactification of the subset Ω(A).

We will show later that in fact Ω(A) is contained in the unit sphere of A∗. Before this,we introduce the Gelfand transform Γ = ΓA of A just as in the unital case: if a ∈ A wedefine a = Γ(a) : Ω(A)→ C by

a(χ) = χ(a).

Then (still assuming A is nonunital)

σ(a) \ 0 = a(χ) : χ ∈ Ω(A).

Corollary 9.16. We have a commutative diagram

AΓA // _

C0(Ω(A)) _

AΓA

// C(Ω(A)).

The following is a complete characterization of commutative C∗-algebras, and includes(but of course is proved using) the unital version:

Corollary 9.17 (Gelfand-Naimark theorem). If A is a commutative C∗-algebra with spec-trum Ω, the Gelfand transform is an isometric C∗-isomorphism of A onto C0(Ω).

Corollary 9.18. A commutative C∗-algebra is unital if and only it its spectrum is compact.

As in the unital case, the continuous functional calculus is an almost immediate corollaryof the Gelfand-Naimark Theorem:

Corollary 9.19 (Nonunital Functional Calculus). Let A be a nonunital C∗-algebra, let a ∈ Abe normal, and let

B = f ∈ C(σ(a)) : f(0) = 0.Then there is a unique *-homomorphism φ : B → A such that φ(z) = a. Moreover, φ is anisometric isomorphism onto C∗(a).

Proof. The unital continuous functional calculus gives a unique *-homomorphismψ : C(σ(a))→ A such that ψ(z) = a. Consider the commutative C∗-subalgebra

C = C∗(a)

of A. We have (see Problem 1)

C = C∗(1, a).

The canonical homeomorphismΩ(C)→ σ(a)

takes the character with kernel C to 0, so it maps Ω(C) homeomorphically onto σ(a) \ 0,and the theorem then follows from the Gelfand-Naimark theorem.

9 UNITIZATION 48

As before, we write f(a) = φ(a).

Now we can finally show:

Corollary 9.20. With the hypotheses of Corollary 9.14, we have ‖χ‖ = 1 for all χ ∈ Ω(A).

Proof. We already know that ‖χ‖ ≤ 1, so we only have to show the opposite inequality. Bythe Gelfand-Naimark Theorem, we can assume without loss of generality that A = C0(X)for a locally compact Hausdorff space X, and then we can find the unique x ∈ X such that

χ(f) = f(x) for f ∈ C0(X).

By the locally compact Hausdorff version of Urysohn’s Lemma, there exists f ∈ C0(X) suchthat ‖f‖ = 1 and f(x) = 1. Therefore ‖χ‖ = 1.

Exercises

1. Let A be a C∗-algebra, let B be a unital C∗-algebra, and let π : A→ B be a *-isomorphismonto an ideal of B of codimension one. Prove that there is a unique *-isomorphism θ : A→ Bsuch that θ(a) = π(a) for all a ∈ A.

Instructor comment : In this situation, we frequently identify A with its image π(A), andrefer to B as the unitization of A. There is no danger of confusion, due to the uniqueness ofθ; in other words, we consider the unitization of A to be any unital C∗-algebra B with theproperties stated in this exercise.

2. Prove that positive elements of arbitrary C∗-algebras have unique positive square roots.Note: we define positive exactly as in the unital case: self-adjoint with nonnegative spectrum.

3. Let X be a nonempty compact Hausdorff space, pick a point t ∈ X, and let Y = X \ t.Identify C0(Y ) with the set of all functions f ∈ C(X) vanishing at t. Prove that (in thesense of Problem 1) C(X) is the unitization of C0(Y ).

Instructor comment : Thus if Y is any locally compact Hausdorff space then the spectrumof the unitization of C0(Y ) is a version of the one-point compactification of Y .

4. Let H be an infinite-dimensional Hilbert space. Prove that the unitization of K(H) (in thesense of Problem 1) is the subalgebra

K(H) + C1 = T + λ1 : T ∈ K(H), λ ∈ C

of B(H).

9 UNITIZATION 49

5. Let A be a C∗-algebra. An approximate identity for A is a net (ei) of positive elements ofnorm at most 1 such that ei − ej ≥ 0 whenever i ≥ j, and

aei → a for all a ∈ A.

In each part below, prove that the given net is an approximate identity:

(a) Let H be a separable infinite-dimensional Hilbert space with orthonormal basis un.For each n ∈ N let en be the orthogonal projection of H onto spanuk : k ≤ n. Then (en)is an approximate identity for K(H).

(b) Let X be a locally compact Hausdorff space. Consider the partially ordered set

F = φ ∈ Cc(X) : 0 ≤ φ ≤ 1,

with the usual pointwise ordering. Then F , indexed by itself, can be regarded as a net, andthen it is an approximate identity for C0(X).

(c) In part (b), if there exists g ∈ C0(X) that vanishes nowhere, with 0 ≤ g ≤ 1, then(g1/n)∞n=1 is an approximate identity for C0(X).

6. Let A be a nonunital C∗-algebra. In this problem you will prove that A has “local approxi-mate identities”, i.e., for all a ∈ A there is a sequence (en) of positive elements of A of normat most 1 such that aen → a in A.

(a) First suppose that a is self-adjoint. Define a sequence of functions fn : R→ R by

fn(x) =nx2

1 + nx2,

and let en = fn(a). Prove that aen → a in A by justifying the following assertions:

• First, we can work in A, and then we have a− aen = a(1− en).

• Next,

‖a(1− en)‖2 ≤ ‖a2(1− en)‖ ≤ 1

n.

(b) For the general case, replace en by fn(a∗a). Then

‖a(1− en)‖2 ≤ ‖a∗a(1− en)‖ → 0.

Instructor comment : Actually, A has an approximate identity in the sense of Problem 5, butthis is harder to prove, and we will not need it in these notes.

7. Let X be a locally compact Hausdorff space, and let A = C0(X). Let C be the family ofclosed subsets of X, and let I be the family of closed ideals of A. Define maps φ : C → Iand σ : I → C by

φ(S) = f ∈ A : f(x) = 0 for all x ∈ Sσ(I) = x ∈ X : f(x) = 0 for all f ∈ I.

Prove the following:

9 UNITIZATION 50

(a) σ φ = idC.

(b) φ σ = idI .

(c) For all S, T ∈ C, if S ⊆ T then φ(S) ⊇ φ(T ).

(d) For all I, J ∈ I, if I ⊆ J then σ(I) ⊇ σ(J).

(e) Let S ∈ C, and put U = X \ S. Then the map

f 7→ f |U : φ(S)→ C0(U)

is a *-isomorphism.

(f) For all S ∈ C, the mapf 7→ f |S : A→ C0(S)

is a surjective *-homomorphism with kernel φ(S). Hint: for the surjectivity, use an appro-priate version of the Tietze Extension Theorem.

8. Let I be a closed ideal in a C∗-algebra A.

(a) Prove that I is automatically self-adjoint, i.e., I is a *-ideal. Hint: for a ∈ I, use asequence (en) as in Problem 6 and consider adjoints. Note: you will need the following fact,which you must prove: en ∈ I for each n ∈ N. This is a bit subtle, because you do not yetknow that I is a C∗-algebra!

(b) Prove that the quotient norm on the Banach *-algebra A/I satisfies the C∗-condition.Hint: fill in the details in the following outline:

• For x = a+ I ∈ A/I, it suffices to show ‖x‖2 ≤ ‖x∗x‖.• Let E be the set of self-adjoint elements of I with spectrum in [0, 1]. Then ‖x‖ =

infe∈E ‖a − ae‖, since first of all ‖x‖ ≤ infe∈E ‖a − ae‖ because aE ⊆ I, and for theopposite inequality, if we fix d ∈ I then, with (en) chosen as in Problem 6 for the elementd, we have

‖a+ d‖ ≥ ‖(a+ d)(1− en)‖ = ‖(a− aen) + (d− den)‖ for all n,

and so

‖a+ d‖ ≥ lim infn→∞

‖a− aen‖ ≥ infe∈E‖a− ae‖,

and it follows that‖x‖ ≥ inf

e∈E‖a− ae‖.

• Now we have

‖x‖2 = infe∈E‖(1− e)a∗a(1− e)‖

≤ infe∈E‖a∗a(1− e)‖

= ‖x∗x‖.

REFERENCES 51

(c) Now let B be another C∗-algebra and π : A → B a *-homomorphism with kernel I.Prove that π(A) is a C∗-subalgebra of B that is isometrically isomorphic to A/I. Hint:factor π to make the diagram

A π //

Q

B

A/Iψ

==

commute, where Q is the quotient map, and show that ψ is isometric.

Instructor comment : This problem gives a peek into the general theory of C∗-algebras, andin particular shows how well-behaved they are.

References

[1] W. Arveson. A short course on spectral theory, volume 209 of Graduate Texts in Mathe-matics. Springer-Verlag, New York, 2002.

[2] J. B. Conway. A course in functional analysis, volume 96 of Graduate Texts in Mathe-matics. Springer-Verlag, New York, second edition, 1990.

[3] J. Dixmier. C∗-algebras. North-Holland, 1977.

[4] G. B. Folland. Real analysis. Pure and Applied Mathematics (New York). John Wiley& Sons, Inc., New York, second edition, 1999.

[5] W. Rudin. Functional analysis. International Series in Pure and Applied Mathematics.McGraw-Hill, Inc., New York, second edition, 1991.

[6] Dana P. Williams. A (Very) Short Course on C∗-Algebras. https://math.dartmouth.

edu/~dana/bookspapers/cstar.pdf (9/23/2019). Dartmouth College.

https://math.dartmouth.edu/~dana/bookspapers/cstar.pdf

https://math.dartmouth.edu/~dana/bookspapers/cstar.pdf

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MAT 579 Functional Analysis II Lecture Notes

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