The 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):

Ring Homomorphism

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
I. N. Herstein, Jordan Homomorphisms