5. banach algebras

5. Banach algebras

In this section we discuss an important concept in Functional Analysis. Theresult presented here are needed in Section 7 as well as in Chapter V.

Definition. Let K be either R or C A normed algebra over K is an algebraA over K, which also carries a norm ‖ . ‖, which is sub-multiplicative, in the sensethat:

• ‖xy‖ ≤ ‖x‖ · ‖y‖, ∀x, y ∈ A(When there is no danger of confusion, the norm will be omitted from the notation.)If A is also a Banach space, then it will be called a Banach algebra.

Remarks 5.1. Suppose A is a normed algebra.A. For any element a ∈ A, one defines the left multiplication map La and the

right multiplication map Ra by

La : A 3 x 7−→ ax ∈ A,Ra : A 3 x 7−→ xa ∈ A.

It is a consequence of sub-multiplicativity that both La, Ra : A → A are linearcontinuous, and one has the inequalities ‖La‖ ≤ ‖a‖ and ‖Ra‖ ≤ ‖a‖, for all a ∈ A.

B. As a consequence of sub-multiplicativity, one has the inequality

(1) ‖xy − ab‖ ≤ ‖x‖ · ‖y − b‖+ ‖x− a‖ · ‖b‖, ∀ a, b, x, y ∈ A.In particular, when we equip the product space A×A with the product topology,the multiplication map M : A × A 3 (x, y) 7−→ xy ∈ A is continuous. Indeed, if(xn)∞n=1 ⊂ A is convergent to a ∈ A, and (yn)∞n=1 ⊂ A is convergent to b ∈ A, then(1) clearly gives limn→∞ ‖xnyn − ab‖ = 0, i.e. limn→∞ xnyn = ab.

C. If A is a normed algebra, then its completion A is a Banach algebra. Thepoint here is of course the fact that A can be equipped with a multiplication op-eration, which turns it into a normed algebra. This multiplication is defined asfollows. One starts with two elements x,y ∈ A, represented by two Cauchy se-quences (xn)∞n=1 ⊂ A and (yn)∞n=1 ⊂ A respectively, and using (1), one has

‖xmym − xnyn‖ ≤ ‖xm‖ · ‖ym − yn‖+ ‖xm − xn‖ · ‖yn‖, ∀m,n ≥ 1,

which clearly gives the fact that (xnyn)∞n=1 is a Cauchy sequence, hence it definessome element xy ∈ A. Of course, one has to show that this definition is independentof the choice of the two Cauchy sequences. This is pretty clear, for if one takes twoother Cauchy sequences (x′n)∞n=1 ⊂ A and (y′n)∞n=1 ⊂ A with

limn→∞

(x′n − xn) = limn→∞

(y′n − yn) = 0,

then again by (1) we have

‖x′ny′n − xnyn‖ ≤ ‖x′n‖ · ‖y′n − yn‖+ ‖x′n − xn‖ · ‖yn‖, ∀n ≥ 1,

so limn→∞(x′ny′n − xnyn) = 0. It is clear that the norm on A is sub-multiplicative.

Definition. If B is a Banach algebra, then for a subset M⊂ B we denote byalg(M) the subalgebra of B, generated by M, which consists all linear combinationsof (finite) products of elements in M. (This is the smallest subalgebra of B which

79

80 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS

contains M.) Its closure alg(M), in B (which can be treated as the completion),will be referred to as the Banach subalgebra of B, generated by M.

Exercise 1♦. Let B be a Banach algebra, and let M ⊂ B, be a commutingsubset, i.e.

xy = yx, ∀x, y ∈M.

Let A be the Banach subalgebra of B generated by M. Prove that A is a commu-tative algebra, i.e.

ab = ba, ∀ a, b ∈ A.

Examples 5.1. Let K be either R or C.A. Let X be a compact space. Then the Banach space (see Section 5)

CK(X) = f : X → K : f continuous ,is a Banach algebra over K, when equipped with the pointwise multiplication, andthe norm

‖f‖ = supx∈X

|f(x)|, f ∈ C(X).

B. More generally, if Ω is a topological space, then the Banach space

CKb (Ω) = f : Ω → K : f continuous, bounded

is a Banach algebra.C. If Ω is a locally compact space, then CK

c (Ω) is a normed algebra. Its com-pletion CK

0 (Ω) is a Banach algebra.D. Take X to be a Banach space over K, and define

L(X) = T : X → X : T linear and continous ,equipped with the pointwise vector space structure. The multiplication is the com-position, and the norm is:

‖T‖ = sup‖Tx‖ : x ∈ X, ‖x‖ ≤ 1, T ∈ L(X).

Then L(X) is a Banach algebra.

Definition. A normed algebra A is said to be unital, if it has a unit element1, i.e. 1 · x = x · 1 = x, ∀x ∈ A.

Remark 5.2. Suppose A is a unital normed algebra with unit 1 6= 0. Thenthere exists a new sub-multiplicative norm ‖ . ‖′, which is equivalent to ‖ . ‖, andsuch that ‖1‖′ = 1. Indeed, if ‖1‖ 6= 1, then by sub-multiplicativity we have‖1‖ ≤ ‖1‖2, which forces ‖1‖ > 1. We then define

‖a‖′ = ‖La‖, ∀ a ∈ A.On the one hand, by Remark 5.1.A, we know already that we have the inequality

‖a‖′ ≤ ‖a‖, ∀ a ∈ A.On the other hand, we also have

‖a‖ = ‖La1‖ ≤ ‖La‖ · ‖1‖ = ‖a‖′ · ‖1‖, ∀ a ∈ A.The new norm ‖ . ‖′ will be called the unital correction of ‖ . ‖.

The Banach algebras described in Examples 5.1.A, 5.1.B and 5.1.D are allunital. For A and B the unit is the constant function 1. For D the unit is theidentity map I : X → X. In these examples the norm is the correct one, in thesense that the unit has norm 1.

§5. Banach algebras 81

The following is a familiar construction in algebra, which is often used whendealing with non-unital algebras.

Definition. Suppose A is a K-algebra. Define the vector space space A =A×K, equipped with the operation

(2) (a, α) · (b, β) = (ab+ αb+ βa, αβ), a, b ∈ A, α, β ∈ K.

Then A is a unital K-algebra, with unit 1 = (0, 1). We identify A as a subalgebrain A via the map

ι : A 3 a 7−→ (a, 0) ∈ A.The pair (A, ι) is referred to as the unitization of A. Its key feature is the following:

• whenever B is a unital K-algebra, and Φ : A → B is an algebra homomor-phism, there exists a unique algebra homomorphism Φ : A → B, such thatΦ

∣∣A = Φ and Φ(1) = 1.

Explicitly, Φ(a, α) = Φ(a) + α1, ∀ (a, α) ∈ A.Remark that this construction makes sense even in the case when A already

has a unit. The unit 1 in A is then regarded as a “new” unit.

Remarks 5.3. In the normed algebra setting, it is natural to ask the followingquestion. Given a normed algebra A, does there exist a sub-multiplicative norm onthe unitized algebra A, which extends the given norm on A?

A. The answer to the above question is affirmative. One example is the norm‖ . ‖1 on A, defined by

‖(a, α)‖1 = ‖a‖+ |α|, ∀ (a, α) ∈ A.Notice that this norm has the properties

(i) ‖ι(a)‖1 = ‖a‖, ∀ a ∈ A;(ii) ‖1‖1 = 1.

A sub-multiplicative norm on A, with the properties (i) and (ii) above, will becalled a unitization norm.

B. In principle there exist many unitization norms on A. One way to constructsuch a norm is as follows (see Example 5.2 below for an illustration). Assumethere exists an isometric algebra homomorphism Φ : A → B, where B is a unitalnormed algebra (whose unit is denoted by 1B) with ‖1B‖ = 1, and such that1B 6∈ Φ(A). The second condition ensures that the homomorphism Φ : A → B,defined by Φ(a, α) = Φ(a) +α1B, ∀ a ∈ A, α ∈ K, is injective. With the help of thehomomorphism Φ one can then define a unitization norm ‖ . ‖′ on A by

‖x‖′ = ‖Φ(x)‖, ∀x ∈ A.

C. If ‖ . ‖′ is any unitization norm on A, then one has the inequality

(3) ‖x‖′ ≤ ‖x‖1, ∀x ∈ A.This is pretty obvious, for if x = (a, α), then we can also write x = ι(a)+α1, whichby the properties (i) and (ii) gives

‖x‖′ ≤ ‖ι(a)‖′ + |α| · ‖1‖′ = ‖a‖+ |α| = ‖x‖1.Of course, (3) gives the fact that the identity map

Id : (A, ‖ . ‖1) → (A, ‖ . ‖′)


is continuous.D. If A is a Banach algebra, then any unitization norm on A is equivalent to

the norm ‖ . ‖1. This is a consequence of the Inverse Mapping Theorem, which givesthe continuity of the identity map

Id : (A, ‖ . ‖′) → (A, ‖ . ‖1).Since (A, ‖ . ‖1) is complete, it follows that any unitization norm turns A into aBanach algebra.

Exercise 2*. Consider the algebra C[t] of polynomials in one variable, and put

A = P ∈ C[t] : P (0) = 0.Equip A with the norm

‖P‖ = supt∈[1,2]

|P (t)|, P ∈ A,

so that if one defines for each P ∈ A the function fP : [1, 2] → C by fP (t) = P (t),t ∈ [1, 2], one gets an isometric algebra homomorphism

Φ : A 3 P 7−→ fP ∈ C([1, 2]).

Prove that the unitization norm ‖ . ‖′ on A, defined by Φ as in Remark 5.3.B, isnot equivalent to the norm ‖ . ‖1.

Example 5.2. Let Ω be a locally compact space Ω, and consider Banachalgebra CK

0 (Ω), identified as

CK0 (Ω) =

f ∈ CK(Ωα) : f(∞) = 0

,

where Ωα is the Alexandrov compactification of Ω. If we apply the constructionoutlined in Remark 5.3.B, for the inclusion Φ : CK

0 (Ω) → CK(Ωα), then this givesrise to a unitization norm, denoted ‖ . ‖st, defined as follows. If one starts with a

pair x = (f, λ) ∈ CK0 (Ω), then the element φx = Φ(f, λ) = Φ(f) + λ1 ∈ CK(Ωα) is

precisely the function defined by

φx(p) =f(p) + λ if p ∈ Ω

λ if p = ∞so that

‖(f, λ)‖st = maxp∈Ωα

|φx(p)| = max

maxp∈Ω

[f(p) + λ], |λ|.

The norm ‖ . ‖st is referred to as the standard norm. Of course, by construction

Φ : (CK0 (Ω), ‖ . ‖st) → CK(Ωα)

is an isometric algebra homomorphism. It turns out that Φ is in fact an isomor-phism. All we need is the surjectivity, which is pretty clear. Indeed, if one startswith some continuous function g : Ωα → K, then if we define f : Ω → K byf(p) = g(p) − g(∞), ∀ p ∈ Ω, then clearly f ∈ CK

0 (Ω), and moreover if we putλ = g(∞), then we clearly have g = Φ(f, λ).

Remarks 5.4. A. As we have seen in I.5, the construction of the compactspace Ωα makes sense also when Ω is already compact. At the Banach algebra levelthis corresponds to the unitization of an already unital Banach algebra.

B. In practice one would like to realize the unitization of algebras of the formCK

0 (Ω) in a more concrete way. This can be seen in the following situation. Assume


T is some compact Hausdorff space, and p is some point in T . Consider the openset Ω = T rp, which is a locally compact space, when equipped with the inducedtopology. Consider the space Ap =

f ∈ CK(T ) : f(p) = 0

. It is clear that Ap

is a closed subalgebra of CK(T ). Moreover, the restriction map gives rise to anisometric algebra isomorphism

R : Ap 3 f 7−→ f∣∣Ω∈ CK

0 (Ω).

Of course, if we denote the inverse of R by Φ, then we have an isometric algebrahomomorphism

Φ : CK0 (Ω) → CK(T ),

with RanΦ = Ap. Finally, if we consider the extension of Φ to an algebra homo-morphism

Φ : CK0 (Ω) → CK(T ),

defined byΦ(f, λ) = Φ(f) + λ1, ∀ f ∈ CK

0 (Ω), λ ∈ K,

then Φ is an isometric algebra isomorphism, when CK0 (Ω) is equipped with the

standard norm ‖ . ‖st.

An important role in Banach algebra theory is played by the invertible elements.The following technical result is very useful (note that the fact that we work in aBanach algebra is essential in the proof).

Lemma 5.1. Suppose A is a unital Banach algebra, and a ∈ A is an invertibleelement. Suppose x ∈ A is an element with

‖x− a‖ < 1‖a−1‖

.

Then x is also invertible, and

‖x−1 − a−1‖ ≤ ‖a−1‖2 · ‖x− a‖1− ‖a−1‖ · ‖x− a‖

.

Proof. Define the element y = 1− a−1x. We have

‖y‖ = ‖a−1(a− x)‖ ≤ ‖a−1‖ · ‖a− x‖ < 1.

Since we have‖yn‖ ≤ ‖y‖n, ∀n ≥ 1,

it follows that the series1 + y + y2 + y3 + . . .

is convergent to an element b ∈ A, and we have

‖1− b‖ ≤ ‖y‖+ ‖y‖2 + · · · ≤∞∑

n=1

(‖a−1‖ · ‖x− a‖)n =‖a−1‖ · ‖x− a‖

1− ‖a−1‖ · ‖x− a‖.

By the way b is defined, it is obvious that b(1 − y) = (1 − y)b = 1, so we get thefact that a−1x = 1 − y is invertible, and (a−1x)−1 = b. Because a is invertible, itfollows that x is invertible, with inverse x−1 = ba−1. We then have

‖x−1 − a−1‖ = ‖(1− b)a−1‖ ≤ ‖1− b‖ · ‖a−1‖ ≤ ‖a−1‖2 · ‖x− a‖1− ‖a−1‖ · ‖x− a‖

.


Corollary 5.1. If A is a unital Banach algebra, then the set

GL(A) =a ∈ A : a invertible

is open in A, and the map

GL(A) 3 a 7−→ a−1 ∈ GL(A)

is continuous.

Proof. Immediate from the above result.

Exercise 3. Suppose A is a Banach algebra and (an)∞n=1 is a sequence of in-vertible elements which converges to some element a ∈ A. Prove that the followingare equivalent:

(i) limn→∞ ‖a−1n ‖ = ∞;

(ii) the sequence(‖a−1

n ‖)n≥0

is unbounded;(iii) a is non-invertible.

Definition. Let A be a Banach algebra over K. A character of A is a nontrivial (i.e. not identically zero) algebra homomorphism γ : A → K, which meansthat:

• γ is linear;• γ is multiplicative, i.e. φ(ab) = φ(a)φ(b), ∀ a, b ∈ A.

Remark that, if A has unit 1, then the fact that γ is non-trivial forces the equalityγ(1) = 1.

We then define the space

Char(A) =γ : A → K : γ character

.

Proposition 5.1 (Automatic continuity of characters). Let A be a Banachalgebra over K, and let γ : A → K be a character. Then γ is continuous, and hasnorm ‖γ‖ ≤ 1.

Proof. We can assume that A is a unital Banach algebra, and ‖1‖ = 1. Ifthis is not the case, we extend γ to a character γ of A = A×K by

γ(a, α) = γ(a) + α, ∀ a ∈ A, α ∈ K.

In this case we have γ = γ ι, where ι : A 3 a 7−→ (a, 0) ∈ A is the inclusion map,so it will suffice to show that γ is continuous and has norm ‖γ‖ ≤ 1. (By Remark5.3.D, we know that A is a Banach algebra, when equipped with any unitizationnorm.)

In order to prove the desired statement, we need to prove that

(4) |γ(a)| ≤ ‖a‖, ∀ a ∈ A.

We argue by contradiction. Assume there exists a0 ∈ A with |γ(a0)| > ‖a0‖. Putλ = γ(a0), so that the element a1 = λ−1a0 has norm ‖a1‖ < 1. By Lemma 5.1,applied to a = 1 and x = 1 − a1, it follows that the element 1 − a1 = 1 − λ−1a0

is invertible. Since λ 6= 0, it follows that the element z = λ(1 − a1) = λ1 − a0 isinvertible. Notice however that by construction we have

γ(z) = λγ(1)− γ(a0) = λ− γ(a0) = 0.

Of course this forces 1 = γ(1) = γ(z−1z) = γ(z−1)γ(z) = 0, which is impossible.


Corollary 5.2. Let A be a unital Banach algebra. When equipped with thew∗-topology, the space Char(A) is compact.

Proof. Since the unit ball (A∗)1 is compact, when equipped with the w∗-topology (see Section 3), all we need to show is the fact that Char(A) is w∗-closedin (A∗)1. Take a net (γλ)λ∈Λ ⊂ Char(A), and φ ∈ (A∗)1, with φ = w∗- limλ∈Λ γλ,and let us show that φ ∈ Char(A). By the definition of the w∗-topology, we knowthat

φ(a) = limλ∈Λ

γλ(a), ∀ a ∈ A.

This clearly implies the fact that φ is linear, multiplicative, as well as the fact thatφ is not identically zero (since φ(1) = limλ∈Λ γλ(1) = 1).

The above result can be generalized to include the non-unital case.

Proposition 5.2. Suppose A is a Banach algebra. Let A be the unitization ofA, and let ι : A → A denote the inclusion map. The correspondence

ι∗ : Char(A) 3 φ 7−→ φ ι ∈ (A∗)1has the following properties:

(i) ι∗ is injective;(ii) Ran ι∗ = Char(A) ∪ 0;(iii) when we equip A with any unitization norm, and the spaces Char(A) and

Char(A) ∪ 0 with the w∗-topology, the map

ι∗ : Char(A) → Char(A) ∪ 0

is a homeomorphism;(iv) in particular, Char(A) ∪ 0 is compact, in the w∗-topology.

Proof. (i). Let φ1, φ2 ∈ Char(A) be such that φ1 ι = φ2 ι, and let usshow that φ1 = φ2. Recall that A = A×K has unit 1 = (0, 1). Since any elementx = (a, λ) ∈ A can be written as

x = (a, 0) + λ(0, 1) = ι(a) + 1,

we get

φ1(x) = (φ1 ι)(a) + λφ1(1) = (φ1 ι)(a) + λ = (φ2 ι)(a) + λφ2(1) = φ2(x),

so we indeed have φ1 = φ2.(ii). It is obvious that, if φ ∈ Char(A), then ι∗(φ) = φ ι : A → K is an algebra

homomorphism. Then either ι∗(φ) = 0, or ι∗(φ) is a character on A. This provethe inclusion

Ran ι∗ ⊂ Char(A) ∪ 0.Conversely, if we start with some γ ∈ Char(A)∪ 0, then γ : A → K is an algebrahomomorphism (possibly zero), and using the key feature of A (see the definitionpreceding Remarks 5.3), there exists a (unique) algebra homomorphism γ : A → K,with γ(1) = 1, and γ ι = γ. Obviously the first propert says that γ is a characteron A, while the second property reads ι∗(γ) = γ.

(iii)-(iv). Using the fact that the character space Char(A) is compact, combinedwith (i) and (ii), we see that it suffices to show that ι∗ is continuous. But this factis clear.


Corollary 5.3. Let A be a Banach algebra. If the character space Char(A)is non-empty, then it is locally compact, when equipped with the w∗-topology.

Proof. PutK = Char(A)∪0 ⊂ (A∗)1, which is compact in the w∗-topology.Then Char(A) = K r 0 is open in K, so it is locally compact, when equippedwith the induced topology from K, which is precisely the w∗-topology.

Remark 5.5. Suppose A is a non-unital Banach algebra. It is possible for thecharacter space Char(A) to be empty (see Exercise ??). It is also possible for thecharacter space Char(A) to be non-empty and compact (see Exercise ?? below). Inthe event when Char(A) is non-compact, it is locally compact, and using the resultsfrom Section I.5, it follows that the compact space Char(A)∪0 is identified withChar(A)α - the Alexandrov compactification of Char(A) - with 0 identified withthe point at ∞.

In fact the topological (i.e. homeomorphic) identifications

(5) Char(A) ' Char(A) ∪ 0 ' Char(A)α

work also in the case when Char(A) is already compact, then only difference beingthat in this case the point at ∞ is isolated.

Using (5) we see that Char(A) always gets topologocally identified with anopen subset of the compact space Char(A). This identification is implemented bythe injective map

J : Char(A) 3 γ 7−→ γ ∈ Char(A),where for a character γ ∈ Char(A), one defines the character γ ∈ Char(A) by

γ(a, α) = γ(a) + α, ∀ a ∈ A, α ∈ K.

Exercise 4. Any Banach space X can be turned into a “trivial” Banach algebraby defining the product as

xy = 0, ∀x, y ∈ X.

Prove that, when equipped with this trivial Banach algebra structure, the characterspace is empty.

Exercise 5. Consider the vector space C2, equipped with the norm

‖(λ1, λ2)‖∞ = max|λ1|, |λ2|.Equipp C2 with the product

(λ1, λ2) · (µ1, µ2) = (λ1µ1, 0).

Prove that C2 is a commutative non-unital Banach algebra, whose character spaceis a singleton, hence compact.

Theorem 5.1. Let Ω be a locally compact space, and let K be either R or C.For each point p ∈ Ω, define the evaluation map

εp : CK0 (Ω) 3 f 7−→ f(p) ∈ K.

(i) For each p ∈ Ω, the map εp defines a character of the Banach algebraCK

0 (Ω).(ii) When we equipp the character space Char

(CK

0 (Ω))

with the w∗-toplogy,the correspondence

E : Ω 3 p 7−→ εp ∈ Char(CK

0 (Ω))

defines a homeomorphism.


Proof. (i). This part is obvious.(ii). We begin the proof with the following

Particular case: Assume Ω is compactWe first show that E is injective. We start with two points p, q ∈ Ω, p 6= q, andwe prove that E(p) 6= E(q). For this purpose we use Urysohn Lemma for locallycompact spaces to find a continuous function f : Ω → [0, 1] with f(p) = 1 andf(q) = 0. This means that

εp(f) = f(p) = 1 6= 0 = f(q) = εq(f)

so we indeed have E(p) 6= E(q).Next we show that E is continuous. This amounts to showing that, for a net

(pλ)λ∈Λ ⊂ Ω which converges to a point p ∈ Ω, we also have

w∗- limλ∈Λ

εpλ= εp.

This is however trivial, since the above condition reads

limλ∈Λ

f(pλ) = f(p), ∀ f ∈ CK(Ω).

To conclude the proof in our particular case, it suffices to show that E issurjective. (By the compactness assumption, this will force E to be a homeo-morphism). Start with some character γ : CK(Ω) → K, and consider the closedsubspace J = Ker γ.

Claim 1: For every finite subset F ⊂ J , the closed set

A(F) = p ∈ Ω : f(p) = 0, ∀ f ∈ F

is non-empty.Indeed, if we list the set F = f1, . . . , fn, then the function g = |f1|2+ · · ·+ |fn|2 =f1f1 + · · ·+ fnfn will satisfy

γ(g) = γ(f1)γ(f1) + · · ·+ γ(fn)γ(fn) = 0,

hence g ∈ G. Obviously, g cannot be invertible in CK(Ω), because otherwise we willhave 1 = γ(1) = γ(g−1)γ(g) = 0. In particular, this forces the set G = p ∈ Ω :g(p) = 0 to be non-empty. Then we are done since obviously G ⊂ A(F).

Using the above claim, combined with the fact that for any sequence F1, . . . ,Fk ∈Pfin(J ), we have

A(F1) ∩ · · · ∩A(Fk) = A(F1 ∪ · · · ∪ Fk) 6= ∅,

by compactness it follows that ⋂F∈Pfin(J )

A(F) 6= ∅.

Pick a point p in the above intersection, so that we have

(6) f(p) = 0, ∀ f ∈ J = Ker γ.

Claim 2: One has the equality γ = εp.Indeed, if we start with some arbitrary function f ∈ CK(Ω), and we set µ = γ(f),then the function g = µ1−f will satisfy γ(g) = µγ(1)−γ(f) = µ−γ(f) = 0, so using(6) we will have g(p) = 0, i.e. µ = f(p). In other words, we get γ(f) = f(p) = εp(f),and the claim is proven


Having proven property (ii) in the particular case when Ω is compact, we nowproceed with the proof in the case when Ω is non-compact. Consider the Alexandrovcompactification Ωα of Ω, and denote by

Eα : Ωα → Char(CK(Ωα)

)the corresponding E-map, which we know to be a homeomorphism, by the partic-ular case above.

We know (se Example 5.2) that CK(Ωα) is algebraically isomorphic to the

unitization CK0 (Ω). Using Proposition 5.2 (see also Remark 5.5) this isomporhism

gives rise to a homeomorphism

F : Char(CK(Ωα)

)→ Char

(CK

0 (Ω))∪ 0.

Explicitly, this homeomorphism is defined as follows. If we identify

CK0 (Ω) =

f ∈ CK(Ωα) : f(∞) = 0

,

as a (closed) subalgebra of CK(Ωα), then we have

F (γ) = γ∣∣CK

0 (Ω), ∀ γ ∈ Char

(CK(Ωα)

).

Let us consider now the composition

E = F Eα : Ωα → Char(CK

0 (Ω))∪ 0,

which is a homeomorphism. Remark that, since we obviously have E(∞) = 0, itfollows that

E∣∣Ω

: Ω → Char(CK

0 (Ω))

is also a homeomorphism. (We identify Ω with the open set Ωα r ∞ ⊂ Ωα.) Butnow we are done, because one clearly has the equality E

∣∣Ω

= E.

Having brought the w∗-topology into the picture, the following result is a wel-come addition to the theory.

Proposition 5.3. Let A be a Banach algebra, with the property that its char-acter space Char(A) is non-empty. Equip Char(A) with the w∗-toplogy, so that itbecomes a locally compact space.

(i) For any element a ∈ A, the map

a : Char(A) 3 γ 7−→ γ(a) ∈ C

is continuous.(ii) If Char(A) is non-compact, then for any element a ∈ A, the continuous

function a : Char(A) → C has limit 0 at ∞.(iii) The correspondence

ΓA : A 3 a 7−→ a ∈ C0

(Char(A)

)is an algebra homomorphism. If A is unital, with unit 1, then 1 is theconstant function 1.

(iv) The map ΓA : A → C0

(Char(A)

)is continuous, with ‖ΓA‖ ≤ 1.

Proof. (i). The continuity is trivial, by the way the w∗-topology is defined.(ii). If Char(A) is non-compact, then we know (see Remark 5.7, and Section I.5)

that the Alexandrov compactification of Char(A) is identified with Char(A)∪ 0,with the zero map 0 playing the role of the point at ∞. The proof of the desired


statement amount to showing that whenever (γλ)λ∈Λ ⊂ Char(A) is a net withw∗- limλ∈Λ γλ = 0, we have limλ∈Λ a(γλ) = 0, and this is trivial.

(iii). In order to check that ΓA is an algebra homomorphism, we need to checkthat

• a+ b = a+ b, ∀ a, b ∈ A;• αa = αa, ∀ a ∈ A, α ∈ C;• ab = ab, ∀ a, b ∈ A.

All these conditions are trivially verified. For example the first condition is checkedpointwise:

a+ b(γ) = γ(a+ b) = γ(a) + γ(b) = a(γ) + b(γ), ∀ γ ∈ Char(A).

Likewise, if A is unital, then we have

1(γ) = γ(1) = 1, ∀ γ ∈ Char(A).

(iv). By the definition of the norm on C0

(Char(A)

), we have

‖a‖ = max|a(γ)| : γ ∈ Char(A)

= max

|γ(a)| : γ ∈ Char(A)

.

By Corollary 5.3 we know that |γ(a)| ≤ ‖a‖, ∀ γ ∈ Char(A), and then the aboveequalities immediately prove the estimate ‖a‖ ≤ ‖a‖.

Definition. Let A be a Banach algebra, with non-empty character spaceChar(A). The continuous algebra homomorphism

ΓA : A → C0

(Char(A)

),

defined above, is called the Gelfand correspondence. For an element a ∈ A, thefunction ΓA(a) = a ∈ C0

(Char(A)

)is called the Gelfand transform of a.

As with ring theory, an important role in the study of algebras is played byideals. Since the algebras we deal with are not necessarily commutative, there arethree types of ideals one can consider, as explained below.

Definitions. Let A be some algebra over a field K.A. A subset J ⊂ A is called a left ideal in A, if

• J is a linear subspace of A;• ax ∈ J , ∀ a ∈ A, x ∈ J .

B. A subset J ⊂ A is called a right ideal in A, if• J is a linear subspace of A;• xa ∈ J , ∀ a ∈ A, x ∈ J .

C. A subset J ⊂ A is called a two sided ideal in A, if J is both a left and aright ideal in A.

Of course, if A is commutative, these three notions coincide.In the study of (Banach) algebras, one deals almost exclusively with two sided

ideals. The reason is the followingFact: If J is a two sided ideal in an algebra A, then the quotient space A/J

carries a unique algebra structure, which makes the quotient map

Q : A 3 a 7−→ [a] ∈ A/Jan algebra homomorphism. Moreover, if A has a unit 1, and J ( A, thenthe algebra A/J has unit 1 = Q(1).

In the Banach algebra setting, one has the following properties.


Theorem 5.2. Let A be a Banach algebra.

A. If J is a left (or right, or two sided) ideal in A, then so is its closure J .B. If A is a unital Banach algebra, and J ( A is a left (or right, or two

sided) ideal, then J ( A.C. If J is a closed two sided ideal in A, then when equipped with the quotient

norm, the algebra A/J is a Banach algebra. Moreover, if A is unital, andJ ( A, then A/J is a unital Banach algebra.

Proof. A. Assume J is a left ideal. Clearly J is a linear subspace, so we onlyneed to check the second condition. Fix a ∈ A and x ∈ J , and let us prove thatax ∈ J . Since x ∈ J , there exists a sequence (xn)∞n=1 ⊂ J , with x = limn→∞ xn.Then the continuity of the left multiplication map La : A → A gives

ax = Lax = limn→∞

Laxn = limn→∞

axn.

Since axn ∈ J , ∀n ≥ 1, this forces ax ∈ J .In the case when J is a right ideal, the proof is identical. In the case when J

is two sided, we use the above cases.B. Assume J is a left ideal. We argue by contradiction. Suppose J = A. In

particular it follows that J contains 1, the unit in A. Since the set GL(A) of allinvertible elements is open, and contains 1, the fact that 1 belongs to J gives thefact that the intersection GL(A) ∩ J is non-empty. In particular, this means thatJ contains some invertible element x. Of course, if we define a = x−1 ∈ A, we haveJ 3 ax = 1. Since J contains 1, it will contain all elements of A, which contradictsthe strict inclusion J ( A.

In the case when J is a right ideal, the proof is identical. In the case when Jis two sided, we use the above cases.

C. Let us prove the first assertion. We know (see Section 2) that the quotientspace A/J is a Banach space, so the only thing we need to prove is the secondcondition in the definition. Start with two elements v,w ∈ A/J , and let us provethe inequality

(7) ‖vw‖A/J ≤ ‖v‖A/J · ‖w‖A/J .

Use the notations from Section 1. For each ε > 0, we choose aε ∈ v and bε ∈ w,such that

‖aε‖ ≤ ‖v‖A/J + ε and ‖bε‖ ≤ ‖w‖A/J + ε.

Since aεbε ∈ vw, we immediately get

‖vw‖A/J ≤ ‖aεbε‖ ≤ ‖aε‖ · ‖bε‖ ≤(‖v‖A/J + ε

)·(‖v‖A/J + ε

).

Since the inequality ‖vw‖A/J ≤(‖v‖A/J + ε

)·(‖v‖A/J + ε

)holds for all ε > 0,

it will clearly force (7).

Definition. Let A be an algebra over a field K. A maximal left (or right,or two sided) ideal in A is a left (or right, or two sided) ideal M ( A, with theproperty:

• there is no left (or right, or two sided) ideal J in A, with

M ( J ( A.


Remarks 5.6. A. Let A be an algebra over a field K, and let M be a two sidedideal in A with the property that the quotient algebra A/M is one dimensional.Then M is a maximal ideal. Indeed, if M were not maximal, we could find anothertwo sided ideal J with

M ( J ( A.As in the proof of Proposition 5.1, this would give rise to an injective, but notsurjective linear map

J /M→A/M,

which is impossible, since the target space is one dimensional.B. If A is an algebra over a field K, and if A has a unit, then for every left (or

right, or two sided) ideal J ( A, there exists at least one maximal (or right, ortwo sided) ideal M with J ⊂M. This follows from Zorn’s Lemma, applied to thecollection

C =N ⊂ A : N left (or right, or two sided) ideal, J ⊂ N and 1 6∈ N

,

ordered with inclusion.C. Suppose A is a Banach algebra, which has at least one character. By the

remark A above, it follows that the collection

Max(A) =M⊂ A : M maximal two sided ideal in A

is non-empty, and one has a correspondence

(8) Char(A) 3 γ 7−→ Ker γ ∈ Max(A).

This will be referred to as the character correspondence. Remark that this corre-spondence is injective. Indeed, if γ1 and γ2 are characters on A, with Ker γ1 =Ker γ2 = M, and if we take a ∈ A with γ1(a) = 1 (which is possible, since γ1 isnon-zero), then on the one hand, we also have γ2(a) 6= 0 (since a 6∈ M). On therother hand, since γ1(a2 − a) = γ1(a)2 − γ1(a) = 0, it follows that a ∈ M, whichin turn forces γ2(a)2 − γ2(a) = γ2(a2 − a) = 0, and since γ2(a) 6= 0, it follows thatγ2(a) = 1 = γ1(a). Finally, if we start with an arbitrary element x ∈ A, then

γ1

(γ1(x)a− x

)= γ1(x)γ1(a)− γ1(x) = 0,

which gives the fact that γ1(x)a − x belongs to M, and in particular we get thefact that

0 = γ2

(γ1(x)a− x

)= γ1(x)γ2(a)− γ2(x) = γ1(x)− γ2(x).

D. If A is a unital Banach algebra, then every maximal left (or right, or twosided) ideal is closed. Indeed, if M is a maximal left (or right, or two sided) idealin M, then by Theorem 5.1 we have the inclusions

M⊂M ( A,

with the closure M again a left (or right, or two sided) ideal. The maximality ofM then forces the equality M = M.

We will return to the study of the relationship between characters and maximalideals later in this section.

Exercise 6. Let n ≥ 2 be an integer. Equipp the space Cn with any norm,and take A = L(Cn). Prove that the only two sided ideals of A are 0 and A.Conclude that the Banach algebra A has no characters.


We now introduce an important tool in the analysis in Banach algebras.

Definition. Let A be a complex unital Banach algebra (the term “complex”means that the algebra is over C). For an element a ∈ A, the set

SpecA(a) = λ ∈ C : λ1− a is not invertible in A

is called the spectrum of a relative to A.

The following result collects some of the properties of the spectrum.

Theorem 5.3. Let A be a complex unital Banach algebra.(i) For any a ∈ A, the set SpecA(a) is compact and non-empty.(ii) If A and B are unital Banach algebras, and Φ : A → B is a unital homo-

morphism of algebras, i.e. Φ(1) = 1, then for every element a ∈ A onehas the inclusion

SpecB(Φ(a)) ⊂ SpecA(a).

Proof. (i) Let us show first that SpecA(a) is closed. We prove this indirectly,by showing that its complement is open. By definition, we have

C r SpecA(a) = λ ∈ C : λ1− a invertible in A.

If we define the mapϕ : C 3 λ 7−→ λ1− a ∈ A,

then we have the equality

C r SpecA(a) = ϕ−1(GL(A)

),

and then the fact that C r SpecA(a) is open is a consequence of the (obvious)continuity of ϕ, combined with the fact that GL(A) is open.

Next we prove that SpecA(a) is bounded. For the remainder of the proof, we aregoing to assume that the norm satisfies ‖1‖ = 1. (If not, we replace it with its unitalcorrection, which does not change anything, as far as the statements are concerned).Observe that if λ ∈ C satisfies |λ| > ‖a‖, then the element x(λ) = λ1−a will satisfythe inequality

‖x(λ)− λ1‖ = ‖a‖ < |λ| = ‖(λ1)−1‖−1,

so by Lemma 5.1, it follows that x(λ) is invertible. This means that λ 6∈ SpecA(a).The above argument proves that

SpecA(a) ⊂ λ ∈ C : |λ| ≤ ‖a‖,

so indeed SpecA(a) is bounded.We now prove that the set SpecA(a) is non-empty. We argue by contradiction.

Assume SpecA(a) = ∅. Define the function F : C → A by F (ζ) = (ζ1− a)−1. Onthe one hand, F is continuous.

On the other hand, if |ζ| > 2‖a‖, then ‖ζ−1a‖ < 12 , so by Lemma 5.1 we have

‖(1− ζ−1a)−1 − 1‖ ≤ ‖ζ−1a‖1− ‖ζ−1a‖

=1

1− ‖ζ−1a‖− 1 ≤ 1

1− 12

− 1 = 1,

which gives ‖(1− ζ−1a)−1‖ ≤ 2, thus proving the inequality

‖(ζ1− a)−1‖ = ‖ζ−1(1− ζ−1a)−1‖ ≤ 2|ζ|.


Among other things, this proves

(9) limζ→∞

‖F (ζ)‖ = 0.

Finally, it is clear that

F (λ)− F (ζ) = (λ1− a)−1 − (ζ1− a)−1 = (λ1− a)−1[1− (λ1− a)(ζ1− a)−1

]=

= (λ1− a)−1[(ζ1− a)− (λ1− a)

](ζ1− a)−1 = −(λ− ζ)F (λ)F (ζ),

so by the continuity of F , we get

(10) limλ→ζ

1λ− ζ

[F (λ)− F (ζ)] = −F (ζ)F (ζ) = −(ζ1− a)−2, ∀ ζ.

For each φ ∈ A∗ (the topological dual of A) we consider the map Fφ = φ F : C →C. By (10) it is clear that Fφ is holomorphic (being complex differentiable). Sinceby (9) we also have limζ→∞ Fφ(ζ) = 0, using Liouville’s Theorem, it follows thatFφ is constant zero. Fix now ζ ∈ C. The fact that

φ(F (ζ)

)= Fφ(ζ) = 0, ∀φ ∈ A∗,

combined with the Hahn-Banach Theorem (see Appendix D), forces F (ζ) = 0,which is clearly impossible.

(ii) This inclusion is an obvious consequence of the implication

λ1− a invertible in A =⇒ λ1− Φ(a) invertible in B.

Definition. Suppose A is a complex unital Banach algebra. For an elementa ∈ A, the number

radA(a) = max|λ| : λ ∈ SpecA(a)

is called the spectral radius of a in A. Recall that, during the proof of the Theorem,when we showed that SpecA(a) is bounded, we actually proved the inequality:

(11) radA(a) ≤ ‖a‖, ∀ a ∈ A

(In fact we showed the inequality radA(a) ≤ ‖a‖′, where ‖ . ‖′ is the unital correctionof ‖ . ‖. The inequality (11) follows then from the inequality ‖a‖′ ≤ ‖a‖.)

Remarks 5.7. Suppose B is a complex unital Banach algebra, and A is aclosed subalgebra of A, which contains the unit of B

A. As a particular case of property (ii) above, we see that for every a ∈ A onehas the inclusion

SpecA(a) ⊃ SpecB(a).

In other words, “the spectrum increases as the algebra gets smaller.” There areexamples when the inclusion is strict, and such examples will be discussed in thenext section.

B. For an element a ∈ A, the following are equivalent:

(i) SpecA(a) = SpecB(a);(ii) A ⊃

(λ1− a)−1 : λ ∈ C r SpecB(a)

.


Indeed, if the equality (i) holds, then one has the equality

C r SpecB(a) = C r SpecA(a),

which gives in particular the fact that for every λ ∈ CrSpecB(a), the element λ1−ais also invertible in A, so its inverse (λ1 − a)−1 must belong to A. Conversely, ifone has the inclusion (ii), then for every λ ∈ C r SpecB(a) it follows that λ1 − ais in fact invertible in A, which means that we have the inclusion C r SpecB(a) ⊂CrSpecA(a). In other words, we get SpecA(a) ⊂ SpecB(a), which by part A aboveforces the equality (i).

The following exercise clarifies how much the spectrum can increase.

Exercise 7. (Shilov’s boundary property). Let B be a complex unital Banachalgebra, and let A be a closed subalgebra of A, which contains the unit of B. Forthen for every element a ∈ B one has the inclusion

∂ SpecB(a) ⊂ SpecA(a).

(Note that the boundary ∂ is taken in C. For example ∂[0, 1] = [0, 1].)Hint: Use Exercise 5.

Exercise 8. Let B be a complex unital Banach algebra, and let A be a closedsubalgebra of A, which contains the unit of B. Prove that if a ∈ A has its A-spectrum a perfect set, i.e. ∂ SpecA(a) = SpecA(a), then one has the equalitySpecA(a) = SpecB(a).

Exercise 9. Prove the following continuity result for the spectrum. Let A be acomplex unital Banach algebra, and let (an)∞n=1 ⊂ A be a sequence which convergesto some a ∈ A. If D ⊂ C is some open set, with D ⊃ SpecA(a), then there existssome integer nD ≥ 1, such that

SpecA(an) ⊂ D, ∀n ≥ nD.

Examples 5.3. A. Let n ≥ 1 be an integer. Equip Cn with any norm (it turnsout that any two norms on Cn are equivalent) so that Cn becomes a Banach space,by Exercise 1 in Section 2. If we work in the unital Banach algebra L(Cn), thenfor an element T ∈ L(Cn), its spectrum is

SpecL(Cn)(T ) =λ ∈ C : λ eigenvalue for T

.

In other words, for a complex number λ, the condition λ ∈ SpecL(Cn)(T ) is equiv-alent to the existence of a vector ξ ∈ Cn with ‖ξ‖ = 1, such that Tξ = λξ.

B. Let X be some compact Hausdorff space. Work in the Banach algebra C(X).Then for an element f ∈ C(X), one has the equality

SpecC(X)(f) = Ran f.

This equality is based on the observation that a function g ∈ C(X) is invertible inC(X), if and only if g(p) 6= 0, ∀ p ∈ X.

Remark 5.8. Suppose A is a unital Banach algebra. For each element a ∈ A,the following are equivalent

(i) a is invertible in A;(ii) the left multiplication La is invertible in L(A).


Indeed, the implication (i) ⇒ (ii) is trivial, since the correspondence

Λ : A 3 a 7−→ La ∈ L(A)

is a unital algebra homomorphism. Conversely, if La is invertible in L(A), and ifwe define b = (La)−1(1) ∈ A, then on the one hand, we have

(12) ab = a · (La)−1(1) = La

[(La)−1(1)

]= 1.

On the other hand, using this equality, we also have

La(ba) = aba = a = La(1),

and then the fact that La is invertible will force ba = 1, which combined with (12)gives the invertibility of a (and a−1 = b).

Comment. The above remark shows, among other things, that when A is acomplex unital Banach algebra, one has the equality

SpecA(a) = SpecL(A)(La), ∀ a ∈ A.In other words, the problem of computing the spectrum of an element in a complexunital Banach algebra is equivalent to the computation of the spectrum of a linearcontinuous map on a complex Banach space. In connection with example A, onemay think that this is equivalent with an eigenvalue problem. As example B sug-gests, this is not the correct point of view. For example if we work with the Banachalgebra A = C([0, 1]) and with the function f(t) = t, t ∈ [0, 1], then the linearcontinuous map Lf ∈ L(A) has no eigenvalues, although we have the equality

SpecL(A)(Lf ) = SpecA(f) = Ran f = [0, 1].

The following exercise describes a weaker notion of eigenvalues.

Exercise 10*. Let X be a complex Banach space, and let T ∈ L(X). We calla complex number λ an approximate eigenvalue for T , if there exists a sequence(xn)∞n=1 ⊂ X with

• ‖xn‖ = 1, ∀n ≥ 1;• limn→∞ ‖Txn − λxn‖ = 0.

Prove the following.A. If λ is an approximate eigenvalue for T , then λ ∈ SpecL(X)(T ).B. If λ ∈ ∂ SpecL(X)(T ), then λ is an approximate eigenvalue for T .C. Analyze the example given in the comment above, and for each λ ∈ [0, 1]

indicate how a sequence of functions (gn)∞n=1 ⊂ C([0, 1]) can be con-structed, such that ‖gn‖ = 1, ∀n ≥ 1, and limn→∞ ‖fgn − λgn‖ = 0.

Hint: For part B use Exercise 5.

The following discussion deals with the notion of Functional Calculus. Thegeneral theme of functional calculus is to make sense of expressions like f(a), wherea is an element in a Banach algebra A, and f is some kind of function f : C → C.

Example 5.4. Fix some element a in a complex unital Banach algebra A.Suppose f : C → C is a polynomial function, that is

f(ζ) = α0 + α1ζ + · · ·+ αnζn, ∀ ζ ∈ C,

where α0, α1, . . . , αn ∈ C are some constants. We define the element

f(a) = α01 + α1a+ · · ·+ αnan.


If we denote by Pol(C) the algebra of polynomial functions, then the correspondence

Pola : Pol(C) 3 f 7−→ f(a) ∈ Ais called the polynomial functional calculus on a.

The exercise below summarizes the properties of polynomial functional calculus.

Exercise 11♦. With the notations above, prove the following.A. The polynomial functional calculus defines a unital algebra homomor-

phism Pola : Pol(C) → A.B. For every polynomial function f , one has the equality

SpecA(f(a)

)= f

(SpecA(a)

).

The above equality is referred to as the Spectral Mapping Formula.Hint: For part B. prove the two inclusions separately. To get the inclusion “⊂” start with some

λ ∈ SpecA(f(a)

), consider the polynomial g(ζ) = λ− f(ζ), and use the Fundamental Theorem of

algebra to find µ1, . . . , µn ∈ C such that

g(ζ) = (−1)nαn(µ1 − ζ) · · · (µn − ζ), ∀ ζ ∈ CUse part A to conclude that

λ1− f(a) = (−1)nαn(µ11− a) · · · (µn1− a),

and conclude that for one of the µ’s, say µk, the element µk1−a is not invertible, so µk ∈ SpecA(a).

To prove the inclusion “⊃” start with some µ ∈ SpecA(a), put λ = f(µ) and consider thepolynomial function g(ζ) = λ− f(ζ), which factors as

g(ζ) = (µ− ζ)h(ζ),

for some other polynomial h(ζ). Use part A again, to conclude that

λ1− f(a) = (µ1− a)h(a),

anc conclude that λ1− f(a) is not invertible, hence λ ∈ SpecA(f(a)

).

Example 5.5. The following is an extension of polynomial functional calculus.Fix some element a in a complex unital Banach algebra A. Suppose f : C → C isan entire function, that is f is holomorphic, which means that it it defined by apower series

(13) f(ζ) =∞∑

n=1

αnζn, ∀ ζ ∈ C,

where (αn)∞n=0 ⊂ C are some constants. Remark that, since f is entire, the powerseries has infinite radius of convergence, i.e.

∞∑n=0

|αn|ρn <∞, ∀ ρ ≥ 0.

In particular (use the convention a0 = 1), we have∞∑

n=0

‖αnan‖ ≤

∞∑n=0

|αn| · ‖a‖n <∞,

which in particular gives the fact that the sequence of partial sums (bk)∞k=1 ⊂ A,defined by bk =

∑kn=0 αna

n, has a limit which we denote by f(a). If we denote byEnt(C) the algebra of polynomial functions, then the correspondence

Enta : Ent(C) 3 f 7−→ f(a) ∈ Ais called the entire functional calculus on a.


The exercise below summarizes the properties of entire functional calculus.

Exercise 12♦. With the notations above, prove the following.A. The entire functional calculus defines a unital algebra homomorphism

Enta : Ent(C) → A.B. For every entire function f , one has Spectral Mapping Formula:

SpecA(f(a)

)= f

(SpecA(a)

).

Hint: For part B. follow the hints in the preceding exercise, with suitable modification. Instead

of the Fundamental Theorem of Algebra, use the fact that for an entire function f , its range f(C)

is either a point, or C. The factoring property used in the proof of the inclusion “⊂” still holds

with entire functions.

Exercise 13*. Let A be a complex unital Banach algebra. Prove the SpectralRadius Formulas:

radA(a) = limn→∞

‖an‖1/n = inf‖an‖1/n : n ≥ 1

.

Hints: This is a very difficult problem! It requires several results from Complex Analysis. Followthe steps below to carry on the proof. We can assume that a 6= 0 (otherwise everything is trivial).Define the numbers

L = lim supn→∞

‖an‖1/n and M = inf‖an‖1/n : n ≥ 1

.

It is obvious that L ≥ M . Denote the spectral radius radA(a) simply by R The proof will bedivided in two parts, by showing the inequalities: (a) R ≥ L; (b) R ≤ M .

To prove (a) we define D to be the open disk in C, of radius 1/R, centered at 0. (In thecase when R = 0, we put D = C.) Consider the map F : D → A defined by F (ζ) = (1 − ζa)−1.Remark first that if we consider D0 to be the open disk of radius 1/‖a‖, centered at 0, then we

have the inclusion D0 ⊂ D. Moreover, since ‖ζa‖ < 1, ∀ ζ ∈ D0, by (the proof of) Lemma 5.1 wehave the equality

F (ζ) =∞∑

n=0

ζnan, ∀ ζ ∈ D0.

The key idea is to think F as a “holomorphic A-valued function.” To avoid any problems withmaking sense of this notion, we use the following trick. For every linear continuous map φ : A → C,we define Fφ : D → C as the composition Fφ = φ F . It is easy to show that Fφ is holomorphic.

On the one hand, we have

Fφ(ζ) =∞∑

n=0

φ(an)ζn, ∀ ζ ∈ D0.

On the other hand, Fφ is defined on the whole open disk D. This means that the radius of D isno greater than the radius of convergence of the above power series. This gives the inequality

1

R≤

1

lim supn→∞ |φ(an)|1/n,

thus proving the inequality

R ≥ lim supn→∞

|φ(an)|1/n, ∀φ ∈ A∗.

Analyze carefully this inequality, using for instance the equality

‖a‖ = maxφ∈(A∗)1

|φ(a)|,

to conclude that one actually has R ≥ lim supn→∞ ‖(an)‖1/n.To prove (b) fix some λ ∈ SpecA(a), with R = |λ|. Argue that, for every n ≥ 1, one has

λn ∈ SpecA(an), so

Rn = |λn| ≤ radA(an) ≤ ‖an‖,

which forces R ≤ ‖an‖1/n.


The following explains how the computation of the spectrum can be reducedto the commutative case.

Remark 5.9. Let B be a complex unital Banach algebra, and let a ∈ B bean arbitrary element. Then there exists a closed commutative subalgebra A whichcontains the unit of B, such that

SpecA(a) = SpecB(a).

Indeed if we take

M =1, a

∪

(λ1− a)−1 : λ ∈ C r SpecB(a)

,

then on the one hand, it is clear that M is commuting, hence by Remark 5.1, theclosed subalgebra A = alg(M) is commutative. On the other hand, the equalitySpecA(a) = SpecB(a) is immediate by Remark 5.2.B.

Comment. The above observation indicates that, although most of the inter-esting Banach algebras are non-commutative, the study of commutative ones shouldbe relevant. It turns out that a “commutative result” also has applications to thegeneral case.

What makes the commutative theory particularly manageable is the followingresult from Commutative Algebra.

Lemma 5.2. Let A be a commutative algebra with unit, over a field K. If x ∈ Ais a non-invertible element, then there exists at least one maximal ideal M of A,which contains x.

Proof. Since we work in a commutative algebra, we use only the term “ideal.”Consider the space

Jx = ax : a ∈ A.It is obvious that Jx is an ideal, and moreover, since x is non-invertible, it followsthat Jx 63 1. Use then Remark 5.2.B.

In the Banach algebra setting, one has the following remarkable result.

Theorem 5.4. Let A be a commutative complex unital Banach algebra.A. The character correspondence

Char(A) 3 γ 7−→ Ker γ ∈ Max(A)

is bijective.B. If x ∈ A is a non-invertible element, then there exists at least one char-

acter γ ∈ Char(A), such that γ(x) = 0.

Proof. The key step in proving the theorem is contained in the followingClaim: If M is a maximal ideal in A, then the quotient algebra A/M is one

dimensional.To prove this, we fix a maximal ideal M. On the one hand, using Remark 5.2.D itfollows thatM is closed. Denote for simplicity the quotient algebra A/M by B, andlet Q : A → B denote the quotient map. By Theorem 5.1.C the quotient algebraB, when equipped with the quotient norm, becomes a unital Banach algebra, withunit 1 = Q(1). Of course, B is commutative. Fix for the moment an arbitraryelement v ∈ B. Choose some number λ ∈ SpecB(v). (Here we use in an essentialway the fact that B is a unital Banach algebra, so by Theorem 5.3 the spectrum is


non-empty.) Let us examine the element w = λ1−v ∈ B. Since w is non-invertible,the set

Jw = bw : b ∈ Bis an ideal in B, which does not contain the unit 1. Then it is obvious that the setN = Q−1(Jw) is an ideal in A, which does not contain the unit 1. Since we clearlyhave

N ⊃ KerQ = M,

the maximality of M will force the equality N = M. In particular, by the surjec-tivity of Q, this forces Jw = 0, i.e. w = 0, so we get the equality v = λ1. Thisway we have shown that every element v ∈ B is a scalar multiple of the unit 1, soB is indeed one dimensional.

Having proven the Claim, we now proceed with the proof of the Theorem. Wealready know that the character correspondence is injective (Remark 5.2.C). Toprove the surjectivity, we start with a maximal ideal M, so using the Claim (andthe notations used in its proof) it follows that the quotient algebra B = A/Mis a one dimensional commutative unital Banach algebra. Obviously B is thenisomorphic, as an algebra, to C, the isomorphism being

ω : C 3 λ 7−→ λ1 ∈ B,where 1 denotes the unit of B. The composition ω−1 Q : A → C is then acharacter, which clearly has Ker γ = KerQ = M.

Part B follows immediately from Lemma 5.3, combined with part A.

Corollary 5.4. If A is a complex commutative unital Banach algebra, thenfor every element a ∈ A, one has the equality

SpecA(a) =γ(a) : γ ∈ Char(A)

.

Proof. Fix a ∈ A, and denote for simplicity the right hand side of the aboveequality by S. On the one hand, if λ ∈ S, there exists a character γ ∈ Char(A)with γ(a) = λ. Since γ(1) = 1, this forces γ(λ1 − a) = 0, which means that theelement λ1 − a belongs to the maximal ideal Ker γ. Since Ker γ ( A, it followsthat Ker γ contains no invertible element, so in particular λ1− a is non invertible,hence λ ∈ SpecA(a).

Conversely, if λ ∈ SpecA(a), then λ1 − a is non invertible, so by the aboveresult, there exists some character γ with γ(λ1− a) = 0, i.e. γ(a) = λ.

The above result can be equivalently reformulated in terms of the Gelfandtransform.

Corollary 5.5. If A is a complex commutative unital Banach algebra, thenfor every element a ∈ A, one has the equality

SpecA(a) = SpecC(Char(A))(a) = Ran a.

In particular one has the equality

radA(a) = ‖a‖.

Proof. Immediate from the definition of the Gelfand transform. The secondequality is clear, since

radA(a) = max|λ| : λ ∈ SpecA(a)

= max

|λ| : λ ∈ Ran a

=

= max|a(γ)| : γ ∈ Char(A)

= ‖a‖.


Example 5.6. Let B be a complex unital Banach algebra. For an elementa ∈ B, we consider the commutative unital Banach subalgebra A constructed inRemark 5.7, that is A = alg(M, where

M =1, a

∪

(λ1− a)−1 : λ ∈ C r SpecB(a)

.

We already know that we have the equality SpecA(a) = SpecB(a). Let us considerthe map

(14) Fa : Char(A) 3 γ 7−→ γ(a) ∈ C.

The map Fa is obviously continuous.Remark that Fa is injective. Indeed, if γ1, γ2 ∈ Char(A) satisfy γ1(a) = γ2(a),

then it is clear that we also have

γ1

((λ1− a)−1

)=

1λ− γ1(a)

=1

λ− γ2(a)= γ2

((λ1− a)−1

), ∀λ ∈ C r SpecB(a),

which means that γ1

∣∣M = γ2

∣∣M. Then it follows that γ1

∣∣alg(M)

= γ2

∣∣alg(M)

, andby continuity, we must have γ1 = γ2

Finally, by the properties of the Gelfand correspondence, we also know thatRanFa = SpecA(a).

Combining all these facts, it follows that the map (14) establishes a homeomor-phism

Fa : Char(A) ∼−→ SpecA(a) = SpecB(a).

Exercise 14. Let A be a unital Banach algebra, and let a, b ∈ A be commutingelements, i.e. satisfying ab = ba. Prove the following inclusions

(i) SpecA(a+ b) ⊂ SpecA(a) + SpecA(b);(ii) SpecA(ab) ⊂ SpecA(a) · SpecA(b).

Hint: Construct a closed commutative subalgebra U ⊂ A, with 1, a, b ⊂ U , and such that

SpecU (x) = SpecA(x), ∀x ∈ a, b, a + b, ab.

Then work in U and use Gelfand transforms.

The exercise below discusses a very interesting example. We shall ellaborateon it later in Chapter VII.

Exercise 15*. Consider the Banach space `1(Z).A. Prove that, for functions f, g ∈ `1(Z), and an integer k, the function

h : Z → C defined by h(n) = f(n)g(k − n), ∀n ∈ Z, is summable.B. Use part A to define, for f, g ∈ `1(Z) the function f × g : Z → C by

(f × g)(k) =∑n∈Z

f(n)g(k − n), ∀ k ∈ Z.

Prove that f ×g is summable, hence it defines an element in `1(Z). More-over, prove that

‖f × g‖1 ≤ ‖f‖1 · ‖g‖1, ∀ f, g ∈ `1(Z).

C. Prove that the operation × is associative, and distributive with respectto addition. Using part B, conclude that, when equipped with × as themultiplication, `1(Z) becomes a Banach algebra. Moreover, prove that`1(Z) is a unital commutative algebra. The unit is the function e : Z → Cdefined by e(n) = δn0.


D. Consider the unit circle T = ζ ∈ C : |ζ| = 1. For each ζ ∈ T, and eachf ∈ `1(Z), consider the function fζ : Z → C defined by

fζ(n) = ζnf(n), ∀n ∈ Z.

Prove that fζ is summable. Moreover, for a fixed ζ ∈ T, the correspon-dence

γζ : `1(Z) 3 f 7−→∑n∈Z

fζ(n) ∈ C

defines a character on `1(Z).E. Prove that the map

T 3 ζ 7−→ γζ ∈ Char(`1(Z)

)is a homeomorphism.

The above results state that after identifying the character space Char(`1(Z)

)with T, the Gelfand correspondence gets identified with an algebra homomorphism

`1(Z) 3 f 7−→ f ∈ C(T),

defined by

f(ζ) =∑n∈Z

f(n)ζn, ∀ f ∈ `1(Z).

For f ∈ `1(Z) the function f is called the Fourier transform of f . The multiplication× is called the convolution product, and the Banach algebra `1(Z) defined above iscalled the Fourier algebra of Z.

The machinery developed in this section will be employed in the study of somespecial types of Banach algebras, as we shall see later on in this chapter (see Section??). In preparation for this, we conclude this section with a discussion on somespecial type of Banach algebras.

Definitions. Let A be an algebra over C. A map A 3 a 7−→ a? ∈ A is calledan involution, if it satisfies the following conditions

• (a+ b)? = a? + b?, ∀ a, b ∈ A;• (ζa)? = ζa?, ∀ a ∈ A, ζ ∈ C;• (ab)? = b?a?, ∀ a, b ∈ A;• (a?)? = a, ∀ a ∈ A.

(Notice that if A has unit 1, then the above conditions force 1? = 1.) In this setting,the system (A, ?) is called an involutive algebra (or ?-algebra).

Given involutive algebras (A, ?) and (B, ?), a map Φ : A → B is called a?-homomorphism, if

• Φ is an algebra homomorphism;• Φ(a?) = Φ(a)?, ∀ a ∈ A.

A system (A, ?) is called an involutive normed algebra, if A is a normed algebra,and ? is an involution on A, which is isometric, in the sense that

‖a?‖ = ‖a‖, ∀ a ∈ A.

If, in addition to these features, A is complete as a normed vector space, then A iscalled an involutive Banach algebra.


Remarks 5.10. A. If (A, ?) is an involutive normed algebra, then its com-pletion A carries a natural involution, defined as the unique extension of of theinvolution on A (which is isometric) by continuity. When equipped with this invo-lution, A becomes an involutive Banach algebra.

B. If A is an involutive algebra, then the unitized algebra A carries a uniqueinvolution, defined as

(a, α)? = (a?, α), ∀ a ∈ A, α ∈ C.

C. Remark also that, if A is an involutive normed algebra, then the unitizationnorm ‖ . ‖1 on A, defined in Remark 5.3.A by

‖(a, α)‖1 = ‖a‖+ |α|, ∀ a ∈ A, α ∈ C,

turns the unitized algebra A into an involutive normed algebra. A unitization normon A, which makes the involution isometric, will be called a unitization ?-norm.

D. In general, there may exist many unitization ?-norms on A, but if A is aninvolutive Banach algebra, then any such norm is equivalent to ‖ . ‖1.

Examples 5.7. A. If Ω is a locally compact space, then Cb(Ω) is an invo-lutive Banach algebra, with the complex conjugation f 7−→ f is the involution.The Banach algebra C0(Ω) is also an involutive Banach algebra, with the sameinvolution. These involutions are called the standard involutions. Unless otherwisestated, these are the only involutions used on these algebras.

B. The Fourier algebra `1(Z) becomes an involutive Banach algebra, with theinvolution `1(Z) 3 f 7−→ f? ∈ `1(Z) defined by

f?(n) = f(−n), ∀n ∈ Z.

Definitions. If (A, ?) is a Banach algebra, then the dual space A∗ is equippedwith a map

A∗ 3 φ 7−→ φ? ∈ A∗,defined by

φ?(a) = φ(a?), ∀ a ∈ A, φ ∈ A∗.The properties of this map are similar to the ones of the involution

• (φ+ ψ)? = φ? + ψ?, ∀φ, ψ ∈ A∗;• (ζφ)? = ζφ?, ∀φ ∈ A∗, ζ ∈ C;• (φ?)? = φ, ∀φ ∈ A∗.

With this terminology, we say that a character γ ∈ Char(A) is involutive (or a?-character), if γ? = γ, which means that

γ(a?) = γ(a), ∀ a ∈ A.

(This means that γ : A → C is a ?-homomorphism, when C is equipped with thecomplex conjugation as its involution.) The set of ?-characters will be denotedby Char?(A, ?) (or simply Char?(A), when there is no danger of confusion on theinvolution on A).

Remark 5.11. If (A, ?) is an involutive Banach algebra, then Char?(A) isw∗-closed in Char(A). If one considers the restriction map

R : C0

(Char(A)

)3 f 7−→ f

∣∣Char?(A)

∈ C0

(Char?(A)

),


then we can construct a new algebra homomorphism

Γ(A,?) : A 3 a 7−→ a = R(a) ∈ C0

(Char?(A)

),

which is referred to as the Gelfand ?-correspondence. The key feature of this mapis the fact that it defines a ?-homomorphism. For an element a ∈ A, we call thecontinuous function a the Gelfand ?-transform of a.

Comment. In general, the Gelfand ?-transform is not sensitive to the spec-trum. This is due to the fact that, in general, the inclusion

Char?(A) ⊂ Char(A)

may be strict (see the exercise below). One can then talk about “nice” involutiveBanach algebras, as being the ones for which one has the equality

(15) Char?(A) = Char(A).

For “nice” involutive Banach algebras, the Gelfand correspondence and the Gelfand?-correspondence coincide.

Remark 5.12. Let (A, ?) be an involutive unital Banach algebra. For anyelement a ∈ A one has the equality

SpecA(a?) = λ : λ ∈ SpecA(a).

This is due to the fact that the involution ? satisfies the equivalence

x ∈ GL(A) ⇔ x? ∈ GL(A).

Proposition 5.4. Let (A, ?) be an involutive unital commutative Banach al-gebra. The following conditions are equivalent.

(i) (A, ?) is “nice”, in the sense that one has the equality

Char?(A) = Char(A).

(ii) Every element a ∈ A with a? = a, has real spectrum, i.e. SpecA(a) ⊂ R.

Proof. (i) ⇒ (ii). Assume (i), let a ∈ A be an element with a? = a, andlet λ ∈ SpecA(a). Using the properties of the Gelfand transform, there existsγ ∈ Char(A), such that λ = γ(a). Since every character is involutive, we have

λ = γ(a) = γ(a?) = γ(a) = λ,

so λ ∈ R.(ii) ⇒ (i). Assume (ii), and let us prove that every character γ ∈ Char(A) is

involutive, i.e

(16) γ(a?) = γ(a), ∀ a ∈ A.

To prove this fact, we start with some a ∈ A and we define the elements a=12 (a+a?)

and a2 = 12i (a− a?). It is pretty obvious that ak = a?

k, k = 1, 2, and we also havea = a1 + ia2 and a? = a1 − ia2, so that

γ(a) = γ(a1) + iγ(a2) and γ(a?) = γ(a1)− iγ(a2).

But now the equality (16) is clear, since γ(ak) ∈ SpecA(ak) ⊂ R, k = 1, 2.


Exercise 16. Let T be the unit circle in C. Consider the Banach algebra C(T),and equip it with a non-standard involution, defined by

f?(ζ) = f(ζ), ∀ ζ ∈ T, f ∈ C(T).

Prove that(C(T), ?

)is still an involutive Banach algebra. Identify the character

space Char(C(T)

)with T, as in Exercise ??. Prove that, under this identification,

one has the equalityChar?

(C(T), ?

)= −1, 1.

Exercise 17. Let Ω be a locally compact space. Consider the involutive Banachalgebra C0(Ω) (this means that it comes equipped with the standard involution).Prove that C0(Ω) is “nice” in the sense defined above.

Exercise 18. Prove that the Fourier algebra `1(Z), equipped with the involutiondefined in Example 5.5.B, is “nice.”

5. banach algebras

Documents