10. Commutative Banach algebras - Aarhus Universitet

10. Commutative Banach algebras
Klaus Thomsen matkt@imf.au.dk
Institut for Matematiske Fag
Det Naturvidenskabelige Fakultet
Aarhus Universitet
September 2005
Klaus Thomsen 10. Commutative Banach algebras

We read in W. Rudin: Functional Analysis
Chapter 10, 10.14, followed by the beginning of Chapter 11.
Klaus Thomsen 10. Commutative Banach algebras

The Gelfand-Mazur theorem
Let A and B be Banach algebras. An isometric isomorphism or just
an isomorphism between A and B is a linear bijection α : A → B
such that α(ab) = α(a)α(b) and α(a) = a for all a,b ∈ A.
We write A ≃ C in this case.
Theorem
(Theorem 10.14) Let A be a unital Banach algebra with the
property that every non-zero element is invertible. Then A ≃ C.
Proof.
Define α : C → A by α(λ) = λ1. Note that α is an algebra
homomorphism and α(λ) = λ1 = |λ|1 = |λ|. It remains
only to show that α is surjective. To this end consider an element
a ∈ A. By Theorem 10.13 there is an element λ in the spectrum of
a. Then λ1 − a is not invertible. Under the present assumption this
implies that a = λ1.
Klaus Thomsen 10. Commutative Banach algebras

Commutative Banach algebras - the definition
Let A be a Banach algebra. A is commutative or abelian when
ab = ba for all a,b ∈ A.
Let A be a commutative Banach algebra. An ideal in A is a
subspace J ⊆ A such that aj ∈ J when a ∈ A and j ∈ J.
Lemma
(Proposition 11.2) Let A be a commutative and unital Banach
algebra.
(a) The only ideal in A which contains an invertible element is A
itself.
(b) If J ⊆ A is an ideal then the closure J of J is also an ideal in A.
Klaus Thomsen 10. Commutative Banach algebras

Commutative Banach algebras - ideals
An ideal J ⊆ A is proper when J = A. A proper ideal J ⊆ A is
maximal when there is no proper ideal I ⊆ A such that J ⊆ I and
J = I.
Theorem
(Theorem 11.3) Let A be a commutative unital Banach algebra.
Then
(a) every proper ideal is contained in a maximal ideal in A, and
(b) every maximal ideal is closed.
Klaus Thomsen 10. Commutative Banach algebras

The proof of Theorem 11.3
Proof.
(a) : Let J be a proper ideal in A. The proper ideals in A which
contain J constitute a partially ordered set when ordered by
inclusion. By Hausdorff’s maximality theorem there is a maximal
totally ordered set Ω in this partially ordered set. Let I be the
union of the ideals in Ω.
It follows I is an ideal since Ω is totally ordered.
If 1 ∈ J there must be an ideal from Ω which contains 1. By (a) of
Lemma 2 this is not possible since the ideals in Ω are proper. This
shows that J is a proper ideal.
Note that J ⊆ I. By maximality of Ω there can not be a proper
ideal J ′ in A such that J ⊆ J ′ and J = J ′ , i.e. J is a maximal ideal.
Klaus Thomsen 10. Commutative Banach algebras

The proof of Theorem 11.3
Proof.
(b) Let J be a maximal ideal in A. Then J is an ideal in A by (b) of
Lemma 2.
If 1 ∈ J there is an element x ∈ J such that 1 − x < 1. But then
x = 1 − (1 − x) is invertible, which is not possible since J is a
proper ideal, cf. (a) of Lemma 2. Thus 1 /∈ J, proving that J is a
proper ideal.
Since J ⊆ J and J is maximal, we conclude that J = J, i.e. J is
closed.
Klaus Thomsen 10. Commutative Banach algebras

Quotients
Let X be a Banach space and J ⊆ X a closed subspace. Then the
quotient space X/J comes equipped with the norm
x + J = inf {x − j : j ∈ J} ,
called the quotient norm.
The reader should check that this is indeed a norm on the quotient
space. E.g the triangle inequality is established as follows:
When x,y ∈ X and j,j ′ ∈ J,
x + y + J ≤ x + y − j − j ′ ≤ x − j + y − j ′ .
By taking the infimum over all j ′ ∈ J we conclude that
x + y + J ≤ x − j + y + J;
then we take the infimum over all j ∈ J to conclude that
x + y + J ≤ x + J + y + J.
In the following we will often write q(x) = x + J and consider
q : X → X/J as a linear surjection - the quotient map. Note that
q(x) ≤ x for all x ∈ X.
Klaus Thomsen 10. Commutative Banach algebras

Quotients
Lemma
Let X be a Banach space and J ⊆ X a closed subspace. Then X/J
is a Banach space in the quotient norm.
Proof.
Let {yn} ∞
n=1 be a Cauchy sequence in X/J. For each n there is an
Nn ∈ N such that yi − yj < 2−n−1 when i,j ≥ Nn. We arrange
that N1 < N2 < N3 < ....
Set zj = yNj and note that zj − zj+1 < 2−j for all j = 1,2,3,... .
and note that
Choose any x ′ j ∈ X such that q(x′ j ) = yj. Set x1 = x ′ 1
x1 − x ′ 2 + J < 2−1 . It follows that there is a j2 ∈ J such that
x1 − x ′ 2 − j2 < 2 −1 . Set x2 = x ′ 2 + j2 and note that q (x2) = z2
and that x1 − x2 < 2 −1 .
Note that x2 − x ′ 3 + J < 2−2 . It follows that there is a j3 ∈ J
such that x2 − x ′ 3 − j3 < 2 −2 . Set x3 = x ′ 3 + j3 and note that
q (x3) = z3 and that x2 − x3 < 2 −2 .
Klaus Thomsen 10. Commutative Banach algebras

Quotients
Proof.
We continue by induction to construct a sequence {xj} ∞ j=1
such that
a) q (xj) = zj = yNj ,
b) xj − xj+1 < 2 −j ,
in X
for all j.
It follows from b) that xm − xn ≤
xm − xm−1 + xm−1 − xm−2 + · · · + xn+1 − xn ≤ m−1
j=n 2−j .
Thus {xj} is a Cauchy sequence in X and x = limj→∞ xj exists in
X because X is complete.
We finish the proof by checking that y = q(x) is the limit {yn} in
X/J:
Klaus Thomsen 10. Commutative Banach algebras

Quotients
Proof.
First we observe that y − zj = q(x) − q(xj) ≤ x − xj,
proving that that y = limj→∞ zj in X/J.
Let ǫ > 0. Since {yn} is Cauchy there is an M ∈ N such that
yn − ym < ǫ when n,m ≥ M. In particular, zj − ym < ǫ for all
m > M and all sufficiently large j.
By letting j tend to infinity we deduce that y − ym < ǫ for all
m > M.
Klaus Thomsen 10. Commutative Banach algebras

Quotients
Let A be a Banach algebra - not necessarily abelian. A closed
two-sided ideal in A is a closed subspace J ⊆ A such that
aj ∈ J, ja ∈ J when j ∈ J and a ∈ A.
Lemma
Let A be a Banach algebra and J ⊆ A a closed two-sided ideal.
Then A/J is a Banach algebra in the quotient norm, and the
quotient map q : A → A/J is an algebra homomorphism.
Proof.
Except for some algebraic trivialities - which we omit here - it
remains only to prove the submultiplicativity of the norm in A/J,
i.e. we must show that q(x)q(y) ≤ q(x) q(y) for all
x,y ∈ A.
Let j,j ′ ∈ J and note that q(x)q(y) = xy + J ≤
xy − xj ′ − jy + jj ′ = (x − j)(y − j ′ ) ≤ x − j y − j ′ .
The desired inequality follows by taking the infimum over j and j ′ .
Klaus Thomsen 10. Commutative Banach algebras

Theorem 11.5
Let A be a Banach algebra. Recall that a complex homomorphism
or character on A is a non-zero homomorphism A → C.
Theorem
Let A be a unital commutative Banach algebra and ∆ the set of
characters on A.
a) Every maximal ideal in A is the kernel of an element of ∆.
b) Let ω ∈ ∆. The the kernel of ω is a maximal ideal in A.
c) An element a ∈ A is invertible if and only if ω(a) = 0 for all
ω ∈ ∆.
d) An element of A is invertible if and only it is not contained in
a proper ideal of A.
e) Let x ∈ A. Then
σ(x) = {ω(x) : ω ∈ ∆}.
Klaus Thomsen 10. Commutative Banach algebras

Theorem 11.5
Proof.
a) Let J be a maximal ideal in A. Then J is closed by b) of
Theorem 11.3. Hence J is a closed two-sided ideal in A, and we
conclude that A/J is a Banach algebra by Lemma 5.
We assert that every non-zero element of A/J is invertible. To see
this, let z ∈ A/J be a non-zero element. Then
z (A/J) = {zξ : ξ ∈ A/J} is an ideal in A/J, and it follows that
q −1 (z (A/J))
is an ideal in A.
Note that J ⊆ q −1 (z (A/J)) and that J = q −1 (z (A/J)). Since J
is maximal ideal, q −1 (z (A/J)) can then not be a proper ideal in A,
i.e. A = q −1 (z (A/J)). In particular, 1 = q(1) ∈ z (A/J), proving
that z is invertible.
Klaus Thomsen 10. Commutative Banach algebras

Theorem 11.5
Proof.
It follows that A/J = C1 by the Gelfand-Mazur theorem. Define
ω : A → C such that ω(a)1 = q(a), where q : A → A/J is the
quotient map. Then ω ∈ ∆ and J = ker ω.
b): Let µ ∈ ∆. Then ker µ is clearly (!) an ideal in A. Let I ⊆ A be
a proper ideal such that ker µ ⊆ I and ker µ = I.
Let x ∈ I be an element such that µ(x) = 0. Then λµ(x) = 1 for
some λ ∈ C, and it follows that µ (x − µ(x)1) = 0. Hence
x − µ(x)1 ∈ ker µ ⊆ I, and we deduce that µ(x)1 ∈ I.
This contradicts the properness of I, and we conclude that ker µ is
a maximal ideal.
Klaus Thomsen 10. Commutative Banach algebras

Theorem 11.5
Proof.
c) If x ∈ A is invertible, xx −1 = 1 and it follows from Proposition
10.6 that 1 = µ(1) = µ(x)µ x −1 , and we conclude that µ(x) = 0.
Conversely, if x is not invertible, the set xA = {xa : a ∈ A} is a
proper ideal in A. By (a) of Theorem 11.3 it is then contained in a
maximal ideal in A, and it follows from (a), proved above, that it is
contained in the kernel of an element ω ∈ ∆.
d) : If x ∈ A is invertible, it is not an element of any proper ideal
by (a) of Proposition 11.2 in Rudin.
Conversely, if x lies in no proper ideal, it does, in particular, not lie
in any maximal ideal. Hence ω(x) = 0 for all ω ∈ ∆ by (a) and
thus x is invertible by c).
Klaus Thomsen 10. Commutative Banach algebras

Theorem 11.5
Proof.
(e) : If λ ∈ σ(x), λ1 − x is not invertible, and hence
µ (λ1 − x) = 0 for some µ ∈ ∆. It follows that
λ ∈ {ω(x) : ω ∈ ∆} since µ(1) = 1 by Proposition 10.6.
If λ = ω(x) for some ω ∈ ∆, we conclude from (c) that λ1 − x is
not invertible (since ω(λ1 − x) = 0), and it follows that λ ∈ σ(x).
Klaus Thomsen 10. Commutative Banach algebras

The Gelfand transform - Theorem 11.9
Let A be a commutative Banach algebra with unit. For any
character ω : A → C of A,we know from (c) of Theorem 10.7 that
x < 1 ⇒ |ω(a)| < 1.
This shows that every character is continuous and has norm ≤ 1.
(Since ω(1) = 1 by Proposition 10.6 we have in fact that ω = 1).
Thus
∆ ⊆ {τ ∈ A ∗ : τ ≤ 1} .
In particular, ∆ comes equipped with a natural topology - namely
the weak*-topology inherited from A ∗ .
By Theorem 11.5 there is a bijective correpsondance between the
character space ∆ and the set of maximal ideals of A. That’s why
Rudin calls it the maximal ideal space of A. We will here call it the
character space.
Now, every a ∈ A defines a continuous function â : ∆ → C such
that
â(ω) = ω(a). (1)
Klaus Thomsen 10. Commutative Banach algebras

The Gelfand transform - Theorem 11.9
The map
is called the Gelfand transform.
Theorem
A ∋ a ↦→ â ∈ C(∆)
a) The character space ∆ is compact in the weak*-topology.
b) The Gelfand transform is a homomorphism from A onto a
subalgebra  of C(∆) whose kernel is the radical of A, i.e. the
intersection of the maximal ideals in A.
c) For each a ∈ A,
max {|â(t)| : t ∈ ∆} = ρ(a) ≤ a.
Klaus Thomsen 10. Commutative Banach algebras

The Gelfand transform - the proof of Theorem 11.9
Proof.
a): Since ∆ is contained in the unit ball of A ∗ , and the latter is
compact in the weak*-topology by Banach-Alaoglu theorem
(Theorem 3.15), it suffices to show that ∆ is closed in the
weak*-topology.
Let therefore ω ∈ A ∗ be an element of the closure ∆ of ∆ in A ∗ .
We want to show that ω ∈ ∆.
There is a δ > 0 such that |st − ω(a)ω(b)| < ǫ when |s − ω(a)| < δ
and |t − ω(b)| < δ.
Let a,b ∈ A, and let ǫ > 0. Since U =
{ν ∈ A ∗ : |ν(a) − ω(a)| < δ, |ν(b) − ω(b)| < δ, |ν(ab) − ω(ab)| < ǫ}
is an open neighborhood of ω in the weak*-topology of A ∗ there is
a character ξ ∈ ∆ contained in U because ω ∈ ∆.
Klaus Thomsen 10. Commutative Banach algebras

The Gelfand transform - the proof of Theorem 11.9
Proof.
Since ξ(ab) = ξ(a)ξ(b) and ξ ∈ U it follows that
|ω(a)ω(b) − ω(ab)| < 2ǫ.
Since a,b ∈ A and ǫ > 0 were arbitrary we conclude that
ω(a)ω(b) = ω(ab) for all a,b ∈ A.
In order to conclude that ω ∈ ∆ it remains only to show that
ω = 0. But {ν ∈ A ∗ : |ν(1) − ω(1)| < 1} is an open neighborhood
of ω in the weak*-topology of A ∗ and hence it contains an element
ξ from ∆.
Since ξ(1) = 1 we conclude that ω(1) = 0, and the proof of a) is
complete.
Klaus Thomsen 10. Commutative Banach algebras

The Gelfand transform - the proof of Theorem 11.9
Proof.
b): It should be obvious that the Gelfand transform is a
homomorphism:
a +
λb(t) = t(a+λb) = a(t)+λt(b) = â(t)+λˆ
b(t) = â + λˆ
b (t)
and
ab(t) = t(ab) = t(a)t(b) = a(t) b(t)
The kernel of the Gelfand transform is {a ∈ A : t(a) = 0 ∀t ∈ ∆}.
By Theorem 11.5 this is the same as the set of elements of A that
lie in every maximal ideal of A.
Klaus Thomsen 10. Commutative Banach algebras

The Gelfand transform - the proof of Theorem 11.9
Proof.
c): By e) of Theorem 11.5 σ(a) = {â(t) : t ∈ ∆} so we see that
sup {|â(t)| : t ∈ ∆} = sup {|λ| : λ ∈ σ(a)} .
Thus a∞ ≤ ρ(a). By Theorem 10.13 ρ(a) = infn an 1
n. But
a n 1
n ≤ (a n ) 1
n = a.
Klaus Thomsen 10. Commutative Banach algebras