10. Commutative Banach algebras - Aarhus Universitet
10. Commutative Banach algebras - Aarhus Universitet
10. Commutative Banach algebras - Aarhus Universitet
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<strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong><br />
Klaus Thomsen matkt@imf.au.dk<br />
Institut for Matematiske Fag<br />
Det Naturvidenskabelige Fakultet<br />
<strong>Aarhus</strong> <strong>Universitet</strong><br />
September 2005<br />
Klaus Thomsen <strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong>
We read in W. Rudin: Functional Analysis<br />
Chapter 10, <strong>10.</strong>14, followed by the beginning of Chapter 11.<br />
Klaus Thomsen <strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong>
The Gelfand-Mazur theorem<br />
Let A and B be <strong>Banach</strong> <strong>algebras</strong>. An isometric isomorphism or just<br />
an isomorphism between A and B is a linear bijection α : A → B<br />
such that α(ab) = α(a)α(b) and α(a) = a for all a,b ∈ A.<br />
We write A ≃ C in this case.<br />
Theorem<br />
(Theorem <strong>10.</strong>14) Let A be a unital <strong>Banach</strong> algebra with the<br />
property that every non-zero element is invertible. Then A ≃ C.<br />
Proof.<br />
Define α : C → A by α(λ) = λ1. Note that α is an algebra<br />
homomorphism and α(λ) = λ1 = |λ|1 = |λ|. It remains<br />
only to show that α is surjective. To this end consider an element<br />
a ∈ A. By Theorem <strong>10.</strong>13 there is an element λ in the spectrum of<br />
a. Then λ1 − a is not invertible. Under the present assumption this<br />
implies that a = λ1.<br />
Klaus Thomsen <strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong>
<strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong> - the definition<br />
Let A be a <strong>Banach</strong> algebra. A is commutative or abelian when<br />
ab = ba for all a,b ∈ A.<br />
Let A be a commutative <strong>Banach</strong> algebra. An ideal in A is a<br />
subspace J ⊆ A such that aj ∈ J when a ∈ A and j ∈ J.<br />
Lemma<br />
(Proposition 11.2) Let A be a commutative and unital <strong>Banach</strong><br />
algebra.<br />
(a) The only ideal in A which contains an invertible element is A<br />
itself.<br />
(b) If J ⊆ A is an ideal then the closure J of J is also an ideal in A.<br />
Klaus Thomsen <strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong>
<strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong> - ideals<br />
An ideal J ⊆ A is proper when J = A. A proper ideal J ⊆ A is<br />
maximal when there is no proper ideal I ⊆ A such that J ⊆ I and<br />
J = I.<br />
Theorem<br />
(Theorem 11.3) Let A be a commutative unital <strong>Banach</strong> algebra.<br />
Then<br />
(a) every proper ideal is contained in a maximal ideal in A, and<br />
(b) every maximal ideal is closed.<br />
Klaus Thomsen <strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong>
The proof of Theorem 11.3<br />
Proof.<br />
(a) : Let J be a proper ideal in A. The proper ideals in A which<br />
contain J constitute a partially ordered set when ordered by<br />
inclusion. By Hausdorff’s maximality theorem there is a maximal<br />
totally ordered set Ω in this partially ordered set. Let I be the<br />
union of the ideals in Ω.<br />
It follows I is an ideal since Ω is totally ordered.<br />
If 1 ∈ J there must be an ideal from Ω which contains 1. By (a) of<br />
Lemma 2 this is not possible since the ideals in Ω are proper. This<br />
shows that J is a proper ideal.<br />
Note that J ⊆ I. By maximality of Ω there can not be a proper<br />
ideal J ′ in A such that J ⊆ J ′ and J = J ′ , i.e. J is a maximal ideal.<br />
Klaus Thomsen <strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong>
The proof of Theorem 11.3<br />
Proof.<br />
(b) Let J be a maximal ideal in A. Then J is an ideal in A by (b) of<br />
Lemma 2.<br />
If 1 ∈ J there is an element x ∈ J such that 1 − x < 1. But then<br />
x = 1 − (1 − x) is invertible, which is not possible since J is a<br />
proper ideal, cf. (a) of Lemma 2. Thus 1 /∈ J, proving that J is a<br />
proper ideal.<br />
Since J ⊆ J and J is maximal, we conclude that J = J, i.e. J is<br />
closed.<br />
Klaus Thomsen <strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong>
Quotients<br />
Let X be a <strong>Banach</strong> space and J ⊆ X a closed subspace. Then the<br />
quotient space X/J comes equipped with the norm<br />
x + J = inf {x − j : j ∈ J} ,<br />
called the quotient norm.<br />
The reader should check that this is indeed a norm on the quotient<br />
space. E.g the triangle inequality is established as follows:<br />
When x,y ∈ X and j,j ′ ∈ J,<br />
x + y + J ≤ x + y − j − j ′ ≤ x − j + y − j ′ .<br />
By taking the infimum over all j ′ ∈ J we conclude that<br />
x + y + J ≤ x − j + y + J;<br />
then we take the infimum over all j ∈ J to conclude that<br />
x + y + J ≤ x + J + y + J.<br />
In the following we will often write q(x) = x + J and consider<br />
q : X → X/J as a linear surjection - the quotient map. Note that<br />
q(x) ≤ x for all x ∈ X.<br />
Klaus Thomsen <strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong>
Quotients<br />
Lemma<br />
Let X be a <strong>Banach</strong> space and J ⊆ X a closed subspace. Then X/J<br />
is a <strong>Banach</strong> space in the quotient norm.<br />
Proof.<br />
Let {yn} ∞<br />
n=1 be a Cauchy sequence in X/J. For each n there is an<br />
Nn ∈ N such that yi − yj < 2−n−1 when i,j ≥ Nn. We arrange<br />
that N1 < N2 < N3 < ....<br />
Set zj = yNj and note that zj − zj+1 < 2−j for all j = 1,2,3,... .<br />
and note that<br />
Choose any x ′ j ∈ X such that q(x′ j ) = yj. Set x1 = x ′ 1<br />
x1 − x ′ 2 + J < 2−1 . It follows that there is a j2 ∈ J such that<br />
x1 − x ′ 2 − j2 < 2 −1 . Set x2 = x ′ 2 + j2 and note that q (x2) = z2<br />
and that x1 − x2 < 2 −1 .<br />
Note that x2 − x ′ 3 + J < 2−2 . It follows that there is a j3 ∈ J<br />
such that x2 − x ′ 3 − j3 < 2 −2 . Set x3 = x ′ 3 + j3 and note that<br />
q (x3) = z3 and that x2 − x3 < 2 −2 .<br />
Klaus Thomsen <strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong>
Quotients<br />
Proof.<br />
We continue by induction to construct a sequence {xj} ∞ j=1<br />
such that<br />
a) q (xj) = zj = yNj ,<br />
b) xj − xj+1 < 2 −j ,<br />
in X<br />
for all j.<br />
It follows from b) that xm − xn ≤<br />
xm − xm−1 + xm−1 − xm−2 + · · · + xn+1 − xn ≤ m−1<br />
j=n 2−j .<br />
Thus {xj} is a Cauchy sequence in X and x = limj→∞ xj exists in<br />
X because X is complete.<br />
We finish the proof by checking that y = q(x) is the limit {yn} in<br />
X/J:<br />
Klaus Thomsen <strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong>
Quotients<br />
Proof.<br />
First we observe that y − zj = q(x) − q(xj) ≤ x − xj,<br />
proving that that y = limj→∞ zj in X/J.<br />
Let ǫ > 0. Since {yn} is Cauchy there is an M ∈ N such that<br />
yn − ym < ǫ when n,m ≥ M. In particular, zj − ym < ǫ for all<br />
m > M and all sufficiently large j.<br />
By letting j tend to infinity we deduce that y − ym < ǫ for all<br />
m > M.<br />
Klaus Thomsen <strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong>
Quotients<br />
Let A be a <strong>Banach</strong> algebra - not necessarily abelian. A closed<br />
two-sided ideal in A is a closed subspace J ⊆ A such that<br />
aj ∈ J, ja ∈ J when j ∈ J and a ∈ A.<br />
Lemma<br />
Let A be a <strong>Banach</strong> algebra and J ⊆ A a closed two-sided ideal.<br />
Then A/J is a <strong>Banach</strong> algebra in the quotient norm, and the<br />
quotient map q : A → A/J is an algebra homomorphism.<br />
Proof.<br />
Except for some algebraic trivialities - which we omit here - it<br />
remains only to prove the submultiplicativity of the norm in A/J,<br />
i.e. we must show that q(x)q(y) ≤ q(x) q(y) for all<br />
x,y ∈ A.<br />
Let j,j ′ ∈ J and note that q(x)q(y) = xy + J ≤<br />
xy − xj ′ − jy + jj ′ = (x − j)(y − j ′ ) ≤ x − j y − j ′ .<br />
The desired inequality follows by taking the infimum over j and j ′ .<br />
Klaus Thomsen <strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong>
Theorem 11.5<br />
Let A be a <strong>Banach</strong> algebra. Recall that a complex homomorphism<br />
or character on A is a non-zero homomorphism A → C.<br />
Theorem<br />
Let A be a unital commutative <strong>Banach</strong> algebra and ∆ the set of<br />
characters on A.<br />
a) Every maximal ideal in A is the kernel of an element of ∆.<br />
b) Let ω ∈ ∆. The the kernel of ω is a maximal ideal in A.<br />
c) An element a ∈ A is invertible if and only if ω(a) = 0 for all<br />
ω ∈ ∆.<br />
d) An element of A is invertible if and only it is not contained in<br />
a proper ideal of A.<br />
e) Let x ∈ A. Then<br />
σ(x) = {ω(x) : ω ∈ ∆}.<br />
Klaus Thomsen <strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong>
Theorem 11.5<br />
Proof.<br />
a) Let J be a maximal ideal in A. Then J is closed by b) of<br />
Theorem 11.3. Hence J is a closed two-sided ideal in A, and we<br />
conclude that A/J is a <strong>Banach</strong> algebra by Lemma 5.<br />
We assert that every non-zero element of A/J is invertible. To see<br />
this, let z ∈ A/J be a non-zero element. Then<br />
z (A/J) = {zξ : ξ ∈ A/J} is an ideal in A/J, and it follows that<br />
q −1 (z (A/J))<br />
is an ideal in A.<br />
Note that J ⊆ q −1 (z (A/J)) and that J = q −1 (z (A/J)). Since J<br />
is maximal ideal, q −1 (z (A/J)) can then not be a proper ideal in A,<br />
i.e. A = q −1 (z (A/J)). In particular, 1 = q(1) ∈ z (A/J), proving<br />
that z is invertible.<br />
Klaus Thomsen <strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong>
Theorem 11.5<br />
Proof.<br />
It follows that A/J = C1 by the Gelfand-Mazur theorem. Define<br />
ω : A → C such that ω(a)1 = q(a), where q : A → A/J is the<br />
quotient map. Then ω ∈ ∆ and J = ker ω.<br />
b): Let µ ∈ ∆. Then ker µ is clearly (!) an ideal in A. Let I ⊆ A be<br />
a proper ideal such that ker µ ⊆ I and ker µ = I.<br />
Let x ∈ I be an element such that µ(x) = 0. Then λµ(x) = 1 for<br />
some λ ∈ C, and it follows that µ (x − µ(x)1) = 0. Hence<br />
x − µ(x)1 ∈ ker µ ⊆ I, and we deduce that µ(x)1 ∈ I.<br />
This contradicts the properness of I, and we conclude that ker µ is<br />
a maximal ideal.<br />
Klaus Thomsen <strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong>
Theorem 11.5<br />
Proof.<br />
c) If x ∈ A is invertible, xx −1 = 1 and it follows from Proposition<br />
<strong>10.</strong>6 that 1 = µ(1) = µ(x)µ x −1 , and we conclude that µ(x) = 0.<br />
Conversely, if x is not invertible, the set xA = {xa : a ∈ A} is a<br />
proper ideal in A. By (a) of Theorem 11.3 it is then contained in a<br />
maximal ideal in A, and it follows from (a), proved above, that it is<br />
contained in the kernel of an element ω ∈ ∆.<br />
d) : If x ∈ A is invertible, it is not an element of any proper ideal<br />
by (a) of Proposition 11.2 in Rudin.<br />
Conversely, if x lies in no proper ideal, it does, in particular, not lie<br />
in any maximal ideal. Hence ω(x) = 0 for all ω ∈ ∆ by (a) and<br />
thus x is invertible by c).<br />
Klaus Thomsen <strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong>
Theorem 11.5<br />
Proof.<br />
(e) : If λ ∈ σ(x), λ1 − x is not invertible, and hence<br />
µ (λ1 − x) = 0 for some µ ∈ ∆. It follows that<br />
λ ∈ {ω(x) : ω ∈ ∆} since µ(1) = 1 by Proposition <strong>10.</strong>6.<br />
If λ = ω(x) for some ω ∈ ∆, we conclude from (c) that λ1 − x is<br />
not invertible (since ω(λ1 − x) = 0), and it follows that λ ∈ σ(x).<br />
Klaus Thomsen <strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong>
The Gelfand transform - Theorem 11.9<br />
Let A be a commutative <strong>Banach</strong> algebra with unit. For any<br />
character ω : A → C of A,we know from (c) of Theorem <strong>10.</strong>7 that<br />
x < 1 ⇒ |ω(a)| < 1.<br />
This shows that every character is continuous and has norm ≤ 1.<br />
(Since ω(1) = 1 by Proposition <strong>10.</strong>6 we have in fact that ω = 1).<br />
Thus<br />
∆ ⊆ {τ ∈ A ∗ : τ ≤ 1} .<br />
In particular, ∆ comes equipped with a natural topology - namely<br />
the weak*-topology inherited from A ∗ .<br />
By Theorem 11.5 there is a bijective correpsondance between the<br />
character space ∆ and the set of maximal ideals of A. That’s why<br />
Rudin calls it the maximal ideal space of A. We will here call it the<br />
character space.<br />
Now, every a ∈ A defines a continuous function â : ∆ → C such<br />
that<br />
â(ω) = ω(a). (1)<br />
Klaus Thomsen <strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong>
The Gelfand transform - Theorem 11.9<br />
The map<br />
is called the Gelfand transform.<br />
Theorem<br />
A ∋ a ↦→ â ∈ C(∆)<br />
a) The character space ∆ is compact in the weak*-topology.<br />
b) The Gelfand transform is a homomorphism from A onto a<br />
subalgebra  of C(∆) whose kernel is the radical of A, i.e. the<br />
intersection of the maximal ideals in A.<br />
c) For each a ∈ A,<br />
max {|â(t)| : t ∈ ∆} = ρ(a) ≤ a.<br />
Klaus Thomsen <strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong>
The Gelfand transform - the proof of Theorem 11.9<br />
Proof.<br />
a): Since ∆ is contained in the unit ball of A ∗ , and the latter is<br />
compact in the weak*-topology by <strong>Banach</strong>-Alaoglu theorem<br />
(Theorem 3.15), it suffices to show that ∆ is closed in the<br />
weak*-topology.<br />
Let therefore ω ∈ A ∗ be an element of the closure ∆ of ∆ in A ∗ .<br />
We want to show that ω ∈ ∆.<br />
There is a δ > 0 such that |st − ω(a)ω(b)| < ǫ when |s − ω(a)| < δ<br />
and |t − ω(b)| < δ.<br />
Let a,b ∈ A, and let ǫ > 0. Since U =<br />
{ν ∈ A ∗ : |ν(a) − ω(a)| < δ, |ν(b) − ω(b)| < δ, |ν(ab) − ω(ab)| < ǫ}<br />
is an open neighborhood of ω in the weak*-topology of A ∗ there is<br />
a character ξ ∈ ∆ contained in U because ω ∈ ∆.<br />
Klaus Thomsen <strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong>
The Gelfand transform - the proof of Theorem 11.9<br />
Proof.<br />
Since ξ(ab) = ξ(a)ξ(b) and ξ ∈ U it follows that<br />
|ω(a)ω(b) − ω(ab)| < 2ǫ.<br />
Since a,b ∈ A and ǫ > 0 were arbitrary we conclude that<br />
ω(a)ω(b) = ω(ab) for all a,b ∈ A.<br />
In order to conclude that ω ∈ ∆ it remains only to show that<br />
ω = 0. But {ν ∈ A ∗ : |ν(1) − ω(1)| < 1} is an open neighborhood<br />
of ω in the weak*-topology of A ∗ and hence it contains an element<br />
ξ from ∆.<br />
Since ξ(1) = 1 we conclude that ω(1) = 0, and the proof of a) is<br />
complete.<br />
Klaus Thomsen <strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong>
The Gelfand transform - the proof of Theorem 11.9<br />
Proof.<br />
b): It should be obvious that the Gelfand transform is a<br />
homomorphism:<br />
a +<br />
λb(t) = t(a+λb) = a(t)+λt(b) = â(t)+λˆ <br />
b(t) = â + λˆ <br />
b (t)<br />
and<br />
ab(t) = t(ab) = t(a)t(b) = a(t) b(t)<br />
The kernel of the Gelfand transform is {a ∈ A : t(a) = 0 ∀t ∈ ∆}.<br />
By Theorem 11.5 this is the same as the set of elements of A that<br />
lie in every maximal ideal of A.<br />
Klaus Thomsen <strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong>
The Gelfand transform - the proof of Theorem 11.9<br />
Proof.<br />
c): By e) of Theorem 11.5 σ(a) = {â(t) : t ∈ ∆} so we see that<br />
sup {|â(t)| : t ∈ ∆} = sup {|λ| : λ ∈ σ(a)} .<br />
Thus a∞ ≤ ρ(a). By Theorem <strong>10.</strong>13 ρ(a) = infn an 1<br />
n. But<br />
a n 1<br />
n ≤ (a n ) 1<br />
n = a.<br />
Klaus Thomsen <strong>10.</strong> <strong>Commutative</strong> <strong>Banach</strong> <strong>algebras</strong>