29.7 Multiplicative Functionals
Definition 29.7.1 (Multiplicative Functional).label Let $A$ be a unital Banach algebra and $\phi \in A^{*}$, then $\phi$ is multiplicative if $\phi \ne 0$ and for each $x, y \in A$, $\phi(xy) = \phi(x)\phi(y)$.
Proposition 29.7.2.label Let $A$ be a unital Banach algebra and $\phi \in A^{*}$ be a multiplicative functional, then
- (1)
For each $x \in A$, $|\phi(x)| \le [x]_{sp}\le \norm{x}_{A}$.
- (2)
$\norm{\phi}_{A^*}= 1$.
- (3)
$\phi(G(A)) \subset \complex \setminus \bracs{0}$.
Proof. (3): For each $x \in G(A)$, $1 = \phi(xx^{-1}) = \phi(x)\phi(x^{-1}) \ne 0$.
(1): By (3), for every $\lambda \in \complex$ with $|\lambda| > [x]_{sp}$, $\lambda x \in G(A)$ and $\lambda - \phi(x) \ne 0$. Therefore $\phi(x) \in \ol{B(0, [x]_{sp})}$.
(2): For each $\lambda \in \complex$, $\phi(\lambda 1) = \lambda$, so $\norm{\phi}_{A^*}\le 1$.$\square$
Definition 29.7.3 (Space of Multiplicative Linear Functionals).label Let $A$ be a unital Banach algebra, then $\Omega(A)$ is the space of multiplicative linear functionals, which is a compact Hausdorff space under the weak-* topology.
Proof. By the Banach-Alaoglu Theorem.$\square$
Proposition 29.7.4.label Let $A$ be a commutative unital Banach algebra, then the mapping
is a bijection.
Proof. For each $\phi \in \cm(A)$, $\ker(\phi)$ is an ideal of codimension $1$, and must be maximal.
On the other hand, for each $I \in \Omega(A)$, the quotient $A/I$ is a field, so by the Gelfand-Mazur Theorem, it is isomorphic to $\complex$. Thus the canonical projection $A \to A/I \iso \complex$ induces a multiplicative functional on $A$.$\square$
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.$\square$
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