This blog post is intended to deliver a quick explanation of the
algebra of Borel measures on \(\mathbb{R}^n\). It will be broken into
pieces. All complex-valued complex Borel measures \(M(\mathbb{R}^n)\) clearly form a vector
space over \(\mathbb{C}\). The main
goal of this post is to show that this is a Banach space and also a
Banach algebra.

In fact, the \(\mathbb{R}^n\) case
can be generalised into any locally compact abelian group (see any
abstract harmonic analysis books), this is because what really matters
here is being locally compact and abelian. But at this moment we stick
to Euclidean spaces. Note since \(\mathbb{R}^n\) is \(\sigma\)-compact, all Borel measures are
regular.

To read this post you need to be familiar with some basic properties
of Banach algebra, complex Borel measures, and the most important,
Fubini's theorem.

(Gleason-Kahane-Żelazko) If \(\phi\) is a complex linear functional on a
unitary Banach algebra \(A\), such that
\(\phi(e)=1\) and \(\phi(x) \neq 0\) for every invertible \(x \in A\), then \[
\phi(xy)=\phi(x)\phi(y)
\] Namely, \(\phi\) is a complex
homomorphism.

Notations and remarks

Suppose \(A\) is a complex unitary
Banach algebra and \(\phi: A \to
\mathbb{C}\) is a linear functional which is not identically
\(0\) (for convenience), and if \[
\phi(xy)=\phi(x)\phi(y)
\] for all \(x \in A\) and \(y \in A\), then \(\phi\) is called a complex
homomorphism on \(A\). Note that a
unitary Banach algebra (with \(e\) as
multiplicative unit) is also a ring, so is \(\mathbb{C}\), we may say in this case \(\phi\) is a ring-homomorphism. For such
\(\phi\), we have an instant
proposition:

Proposition 0\(\phi(e)=1\) and \(\phi(x) \neq 0\) for every invertible \(x \in A\).

Proof. Since \(\phi(e)=\phi(ee)=\phi(e)\phi(e)\), we have
\(\phi(e)=0\) or \(\phi(e)=1\). If \(\phi(e)=0\) however, for any \(y \in A\), we have \(\phi(y)=\phi(ye)=\phi(y)\phi(e)=0\), which
is an excluded case. Hence \(\phi(e)=1\).

For invertible \(x \in A\), note
that \(\phi(xx^{-1})=\phi(x)\phi(x^{-1})=\phi(e)=1\).
This can't happen if \(\phi(x)=0\).
\(\square\)

The theorem reveals that Proposition \(0\) actually characterizes the complex
homomorphisms (ring-homomorphisms) among the linear functionals
(group-homomorphisms).

This theorem was proved by Andrew M. Gleason in 1967 and later
independently by J.-P. Kahane and W. Żelazko in 1968. Both of them
worked mainly on commutative Banach algebras, and the non-commutative
version, which focused on complex homomorphism, was by W. Żelazko. In
this post we will follow the third one.

Unfortunately, one cannot find an educational proof on the Internet
with ease, which may be the reason why I write this post and why you
read this.

Equivalences

Following definitions of Banach algebra and some logic manipulation,
we have several equivalences worth noting.

Subspace and ideal version

(Stated by Gleason) Let \(M\) be a linear subspace of codimension one
in a commutative Banach algebra \(A\)
having an identity. Suppose no element of \(M\) is invertible, then \(M\) is an ideal.

(Stated by Kahane and Żelazko) A subspace \(X \subset A\) of codimension \(1\) is a maximal ideal if and only if it
consists of non-invertible elements.

Spectrum version

(Stated by Kahane and Żelazko) Let \(A\) be a commutative complex Banach algebra
with unit element. Then a functional \(f \in
A^\ast\) is a multiplicative linear functional if and only if
\(f(x)=\sigma(x)\) holds for all \(x \in A\).

Here \(\sigma(x)\) denotes the
spectrum of \(x\).

The connection

Clearly any maximal ideal contains no invertible element (if so, then
it contains \(e\), then it's the ring
itself). So it suffices to show that it has codimension 1, and if it
consists of non-invertible elements. Also note that every maximal ideal
is the kernel of some complex homomorphism. For such a subspace \(X \subset A\), since \(e \notin X\), we may define \(\phi\) so that \(\phi(e)=1\), and \(\phi(x) \in \sigma(x)\) for all \(x \in A\). Note that \(\phi(e)=1\) holds if and only if \(\phi(x) \in \sigma(x)\). As we will show,
\(\phi\) has to be a complex
homomorphism.

Tools to prove the theorem

Lemma 0 Suppose \(A\) is a unitary Banach algebra, \(x \in A\), \(\lVert x \rVert<1\), then \(e-x\) is invertible.

This lemma can be found in any functional analysis book introducing
Banach algebra.

Lemma 1 Suppose \(f\) is an entire function of one complex
variable, \(f(0)=1\), \(f'(0)=0\), and \[
0<|f(\lambda)| \leq e^{|\lambda|}
\] for all complex \(\lambda\),
then \(f(\lambda)=1\) for all \(\lambda \in \mathbb{C}\).

Note that there is an entire function \(g\) such that \(f=\exp(g)\). It can be shown that \(g=0\). Indeed, if we put \[
h_r(\lambda) = \frac{r^2g(\lambda)}{\lambda^2[2r-g(\lambda)]}
\] then we see \(h_r\) is
holomorphic in the open disk centred at \(0\) with radius \(2r\). Besides, \(|h_r(\lambda)| \leq 1\) if \(|\lambda|=r\). By the maximum modulus
theorem, we have \[
|h_r(\lambda)| \leq 1
\] whenever \(|\lambda| \leq
r\). Fix \(\lambda\) and let
\(r \to \infty\), by definition of
\(h_r(\lambda)\), we must have \(g(\lambda)=0\).

Jordan homomorphism

A mapping \(\phi\) from one ring
\(R\) to another ring \(R'\) is said to be a Jordan
homomorphism from \(R\) to
\(R'\) if \[
\phi(a+b)=\phi(a)+\phi(b)
\] and \[
\phi(ab+ba)=\phi(a)\phi(b)+\phi(b)\phi(a).
\] It's of course clear that every homomorphism is Jordan. Note
if \(R'\) is not of characteristic
\(2\), the second identity is
equivalent to \[
\phi(a^2)=\phi(a)^2.
\]To show the equivalence, one let \(b=a\) in the first case and puts \(a+b\) in place of \(a\) in the second case.

Since in this case \(R=A\) and \(R'=\mathbb{C}\), the latter of which is
commutative, we also write \[
\phi(ab+ba)=2\phi(a)\phi(b).
\] As we will show, the \(\phi\)
in the theorem is a Jordan homomorphism.

The proof

We will follow an unusual approach. By keep 'downgrading' the goal,
one will see this algebraic problem be transformed into a pure analysis
problem neatly.

To begin with, let \(N\) be the
kernel of \(\phi\).

Step
1 - It suffices to prove that \(\phi\)
is a Jordan homomorphism

If \(\phi\) is a complex
homomorphism, it is immediate that \(\phi\) is a Jordan homomorphism.
Conversely, if \(\phi\) is Jordan, we
have \[
\phi(xy+yx) =2\phi(x)\phi(y).
\] If \(x\in N\), the right hand
becomes \(0\), and therefore \[
xy+yx \in N \quad \text{if } x \in N, y \in A.
\] Consider the identity \[
(xy-yx)^2+(xy+yx)^2=2[x(yxy)+(yxy)x]
\]

Therefore \[
\begin{aligned}
\phi((xy-yx)^2+(xy+yx)^2)&=\phi((xy-yx)^2)+\phi((xy+yx)^2) \\
&=\phi(xy-yx)^2+\phi(xy+yx)^2 \\
&= \phi(xy-yx)^2 \\
&=2\phi[x(yxy)+(yxy)x] \\
&=0
\end{aligned}
\] Since \(x \in N\) and \(yxy \in A\), we see \(x(yxy)+(yxy)x \in N\). Therefore \(\phi(xy-yx)=0\) and \[
xy-yx \in N
\] if \(x \in N\) and \(y \in A\). Further we see \[
xy-yx+xy+yx=2xy \in N \quad \text {and}\quad xy+yx-xy+yx = 2yx \in N,
\] which implies that \(N\) is
an ideal. This may remind you of this classic diagram (we will not use
it since it is additive though):

For \(x,y \in A\), we have \(x \in \phi(x)e+N\) and \(y \in \phi(y)e+N\). As a result, \(xy \in \phi(x)\phi(y)e+N\), and therefore
\[
\phi(xy)=\phi(x)\phi(y)+0.
\]

Step 2 - It
suffices to prove that \(\phi(a^2)=0\)
if \(\phi(a)=0\).

Again, if \(\phi\) is Jordan, we
have \(\phi(x^2)=\phi(x)^2\) for all
\(x \in A\). Conversely, if \(\phi(a^2)=0\) for all \(a \in N\), we may write \(x\) by \[
x=\phi(x)e+a
\] where \(a \in N\) for all
\(x \in A\). Therefore \[
\begin{aligned}
\phi(x^2)&=\phi((\phi(x)e+a)^2)=\phi(x)^2+2\phi(x)\phi(a)+\phi(a)^2=\phi(x)^2,
\end{aligned}
\] which also shows that \(\phi\) is Jordan.

Step
3 - It suffices to show that the following function is constant

Fix \(a \in N\), assume \(\lVert a \rVert = 1\) without loss of
generality, and define \[
f(\lambda)=\sum_{n=0}^{\infty}\frac{\phi(a^n)}{n!}\lambda^n
\] for all complex \(\lambda\).
If this function is constant (lemma 1), we immediately have \(f''(0)=\phi(a^2)=0\). This is
purely a complex analysis problem however.

Step
4 - It suffices to describe the behaviour of an entire function

Note in the definition of \(f\), we
have \[
\lvert \phi(a^n) \rvert \leq \lVert \phi \rVert \lVert a^n \rVert \leq
\lVert \phi \rVert \lVert a \rVert^n=\lVert \phi \rVert.
\] So we expect the norm of \(\phi\) to be finite, which ensures that
\(f\) is entire. By reductio ad
absurdum, if \(\lVert e-a \rVert <
1\) for \(a \in N\), by lemma 0,
we have \(e-e+a=a\) to be invertible,
which is impossible. Hence \(\lVert e-a \rVert
\geq 1\) for all \(a \in N\). On
the other hand, for \(\lambda \in
\mathbb{C}\), we have the following inequality: \[
\begin{aligned}
\lVert \lambda e-a \rVert = \lambda\lVert e-\lambda^{-1}a \rVert
&\geq|\lambda| \\
&= |\phi(\lambda e)-\phi(a)| \\
&= |\phi(\lambda e-a)|
\end{aligned}
\] Therefore \(\phi\) is
continuous with norm less than \(1\). The continuity of \(\phi\) is not assumed at the beginning but
proved here.

For \(f\) we have some immediate
facts. Since each coefficient in the series of \(f\) has finite norm, \(f\) is entire with \(f'(0)=\phi(a)=0\). Also, since \(\phi\) has norm \(1\), we also have \[
|f(\lambda)|=\left|\sum_{n=0}^{\infty}\frac{\phi(a^n)}{n!}\lambda^n\right|
\leq \sum_{n=0}^{\infty}\frac{|\lambda^n|}{n!}=e^{|\lambda|}.
\] All we need in the end is to show that \(f(\lambda) \neq 0\) for all \(\lambda \in \mathbb{C}\).

The series \[
E(\lambda)=\exp(a\lambda)=\sum_{n=0}^{\infty}\frac{(\lambda a)^n}{n!}
\] converges since \(\lVert a
\rVert=1\). The continuity of \(\phi\) shows now \[
f(\lambda)=\phi(E(\lambda)).
\] Note \[
E(-\lambda)E(\lambda)=\left(\sum_{n=0}^{\infty}\frac{(-\lambda
a)^n}{n!}\right)\left(\sum_{n=0}^{\infty}\frac{(\lambda
a)^n}{n!}\right)=e.
\] Hence \(E(\lambda)\)is invertible for all \(\lambda \in
C\), hence \(f(\lambda)=\phi(E(\lambda)) \neq 0\). By
lemma 1, \(f(\lambda)=1\) is constant.
The proof is completed by reversing the steps. \(\square\)

References / Further reading

Walter Rudin, Real and Complex Analysis

Walter Rudin, Functional Analysis

Andrew M. Gleason, A Characterization of Maximal
Ideals

J.-P. Kahane and W. Żelazko, A Characterization of Maximal
Ideals in Commutative Banach Algebras

W. Żelazko A Characterization of Multiplicative linear
functionals in Complex Banach Algebras