Theorem 29.7.5 (Gleason-Kahane-Żelazko).label Let $A$ be a unital Banach algebra and $\phi \in A^{*}$, then the following are equivalent:
- (1)
$\phi$ is a multiplicative linear functional.
- (2)
$\phi(1) = 1$ and $\phi(G(A)) \subset \complex \setminus \bracs{0}$.
Proof, [Theorem I.4.5, Zhu93]. (1) $\Rightarrow$ (2): Proposition 29.7.2.
(2) $\Rightarrow$ (1): Let $x \in A$ and $\lambda \in \complex$ with $|\lambda| > [x]_{sp}$, then $\lambda - x \in G(A)$ and $\phi(\lambda - x) \ne 0$. Therefore $\phi(x) \subset \ol{B(0, [x]_{sp})}$, and $\norm{\phi}_{A^*}= 1$.
Let $x \in \ker \phi$ with $\norm{x}_{A} \le 1$, and let
then
- (a)
$f(0) = \phi(1) = 1$.
- (b)
$Df(0) = \phi(x) = 0$.
- (c)
Since $\norm{x}_{A} \le 1$, $|f| \le |\exp|$. As $\exp(\lambda x) \in G(A)$ for all $\lambda \in \complex$, $f(\lambda) \ne 0$ for all $\lambda \in \complex$.
By Proposition 27.6.6, $f = 1$, and $\phi(x^{2}) = 0$, so for any $x \in \ker\phi$, $x^{2} \in \ker\phi$ as well.
Now, for each $x, y \in A$, there exists $x_{0}, y_{0} \in \ker \phi$ such that $x = x_{0} + \phi(x)$ and $y = y_{0} + \phi(x)$. In which case,
In particular,
To conclude, let $x \in \ker\phi$, $y \in A$, then
so $xy + yx \in \ker\phi$ as well. Finally,
and $(xy - yx)^{2} \in \ker\phi$. Since
The commutator $xy - yx \in \ker \phi$ as well. Therefore $2xy = (xy + yx) - (xy - yx) \in \ker \phi$, $\ker \phi$ is an ideal, and $\phi$ is a homomorphism.$\square$
Post a Comment