109 lines
4.5 KiB
TeX
109 lines
4.5 KiB
TeX
\section{Multiplicative Functionals}
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\label{section:multiplicative-functional}
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\begin{definition}[Multiplicative Functional]
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\label{definition:multiplicative-functional}
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Let $A$ be a unital Banach algebra and $\phi \in A^*$, then $\phi$ is \textbf{multiplicative} if $\phi \ne 0$ and for each $x, y \in A$, $\phi(xy) = \phi(x)\phi(y)$.
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\end{definition}
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\begin{proposition}
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\label{proposition:multiplicative-unit}
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Let $A$ be a unital Banach algebra and $\phi \in A^*$ be a multiplicative functional, then
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\begin{enumerate}
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\item For each $x \in A$, $|\phi(x)| \le [x]_{sp} \le \norm{x}_A$.
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\item $\norm{\phi}_{A^*} = 1$.
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\item $\phi(G(A)) \subset \complex \setminus \bracs{0}$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(3): For each $x \in G(A)$, $1 = \phi(xx^{-1}) = \phi(x)\phi(x^{-1}) \ne 0$.
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(1): By (3), for every $\lambda \in \complex$ with $|\lambda| > [x]_{sp}$, $\lambda x \in G(A)$ and $\lambda - \phi(x) \ne 0$. Therefore $\phi(x) \in \ol{B(0, [x]_{sp})}$.
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(2): For each $\lambda \in \complex$, $\phi(\lambda 1) = \lambda$, so $\norm{\phi}_{A^*} \le 1$.
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\end{proof}
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\begin{definition}[Space of Multiplicative Linear Functionals]
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\label{definition:multiplicative-linear-functional-space}
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Let $A$ be a unital Banach algebra, then $\Omega(A)$ is the \textbf{space of multiplicative linear functionals}, which is a compact Hausdorff space under the weak-* topology.
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\end{definition}
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\begin{proof}
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By the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}.
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\end{proof}
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\begin{proposition}
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\label{proposition:commutative-maximal-ideal-space}
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Let $A$ be a commutative unital Banach algebra, then the mapping
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\[
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\Omega(A) \to \cm(A) \quad \phi \mapsto \ker(\phi)
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\]
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is a bijection.
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\end{proposition}
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\begin{proof}
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For each $\phi \in \cm(A)$, $\ker(\phi)$ is an ideal of codimension $1$, and must be maximal.
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On the other hand, for each $I \in \Omega(A)$, the quotient $A/I$ is a field, so by the \hyperref[Gelfand-Mazur Theorem]{theorem:gelfand-mazur}, it is isomorphic to $\complex$. Thus the canonical projection $A \to A/I \iso \complex$ induces a multiplicative functional on $A$.
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\end{proof}
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\begin{theorem}[Gleason-Kahane-Żelazko]
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\label{theorem:gkz}
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Let $A$ be a unital Banach algebra and $\phi \in A^*$, then the following are equivalent:
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\begin{enumerate}
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\item $\phi$ is a multiplicative linear functional.
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\item $\phi(1) = 1$ and $\phi(G(A)) \subset \complex \setminus \bracs{0}$.
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\end{enumerate}
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\end{theorem}
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\begin{proof}[Proof, {{\cite[Theorem I.4.5]{Zhu}}}. ]
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(1) $\Rightarrow$ (2): \autoref{proposition:multiplicative-unit}.
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(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 $\norm{\phi}_{A^*} = 1$.
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Let $x \in \ker \phi$ with $\norm{x}_A \le 1$, and let
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\[
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f: \complex \to \complex \quad \lambda \mapsto \phi[\exp(\lambda x)] = \sum_{n = 0}^\infty \frac{\phi[(\lambda x)^n]}{n!}
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\]
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then
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\begin{enumerate}[label=(\alph*)]
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\item $f(0) = \phi(1) = 1$.
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\item $Df(0) = \phi(x) = 0$.
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\item 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 $\lambda \in \complex$.
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\end{enumerate}
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By \autoref{proposition:entire-constant}, $f = 1$, and $\phi(x^2) = 0$, so for any $x \in \ker\phi$, $x^2 \in \ker\phi$ as well.
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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,
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\begin{align*}
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\phi(xy) &= \phi(x_0y_0 + \phi(x)y_0 + \phi(y)x_0 + \phi(x)\phi(y)) \\
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&= \phi(x_0y_0) + \phi(x)\phi(y)
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\end{align*}
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In particular,
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\[
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\phi(x^2) = \phi(x_0^2) + \phi(x)^2 = \phi(x)^2
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\]
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To conclude, let $x \in \ker\phi$, $y \in A$, then
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\begin{align*}
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\phi((x + y)^2) &= (\phi(x + y))^2 = \phi(x)^2 + 2\phi(x)(y) + \phi(y)^2 \\
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\phi(xy + yx) &= 2\phi(x)\phi(y)
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\end{align*}
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so $xy + yx \in \ker\phi$ as well. Finally,
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\begin{align*}
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(xy + yx)^2 + (xy - yx)^2 &= 2(xyxy + yxyx) \\
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&= 2[x(yxy) + y(xyx)] \in \ker \phi
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\end{align*}
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and $(xy - yx)^2 \in \ker\phi$ as well, and
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\[
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(\phi(xy - yx))^2 = \phi((xy - yx)^2) = 0
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\]
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Therefore $2xy = (xy + yx) - (xy - yx) \in \ker \phi$, $\ker \phi$ is an ideal, and $\phi$ is a homomorphism.
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\end{proof}
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