Proposition 6.2.11: Spectral Mapping Theorem

Proposition 6.2.11 (Spectral Mapping Theorem).label Let $A$ be a unital Banach algebra and fix $x\in A$. If $f\in \holo{\spec{A}{x}}{\com}$ then

\begin{align*}\spec{A}{f(x)}=f\parens{\spec{A}{x}}\end{align*}

Furthermore, if $g\in \holo{f(\spec{A}{x})}{\com}$ then

\begin{align*}(g\circ f)(x)=g(f(x))\end{align*}

Proof. If $\mu\nin f\parens{\spec{A}{x}}$ then $h(\lam)=\frac{1}{f(\lam)-\mu}\in\holo{\cali{N}\parens{\spec{A}{x}}}{\com}$. By Proposition 6.2.10

\begin{align*}(f(x)-\mu)h(x)=h(x)(f(x)-\mu)=(h(f-\mu))(x)=1\end{align*}

Hence $\mu\in\com\del\spec{A}{f(x)}$. Conversely if $\mu\in f(\spec{A}{x})$ then $\mu=f(\lam_{0})$ for some $\lam_{0}\in \spec{A}{x}$. Since $f\in\holo{\cali{N}\parens{\spec{A}{x}}}{\com}$, $\exists h\in\holon{\spec{A}{x}}{\com}$ such that

\begin{align*}f(\lam)-\mu=(\lam-\lam_{0})h(\lam)\end{align*}

By Proposition 6.2.10

\begin{align*}f(x)-\mu=(x-\lam_{0})h(x)=h(x)(x-\lam_{0})\end{align*}

Since $x-\lam_{0}$ is not invertible, neither is $y-\mu$ thus $\mu\in \spec{A}{f(x)}$

Lastly pick smooth simple closed curves $C_{1}$ and $C_{2}$ such that $C_{1}$ encloses $f\parens{\spec{A}{x}}$ and is contained in the domain of $g$, $C_{2}$ encloses preimage of $C_{1}$ under $f$ and is contained in the domain of $f$. We compute

\begin{align*}(g\circ f)(x)&=\frac{1}{2\pi i}\integral{C_1}{}(g\circ f)(\lam)(\lam-x)\inv d\lam\\&=-\frac{1}{4\pi^{2}}\integral{C_1}{}\parens{\integral{C_2}{}g(\mu)(\mu-f(\lam))\inv d\mu}(\lam-x)\inv d\lam\\&=-\frac{1}{4\pi^{2}}\integral{C_2}{}g(\mu)\parens{\integral{C_1}{}(\mu-f(\lam))\inv(\lam-x)\inv d\lam}d\mu\\&=\frac{1}{2\pi i}\integral{C_2}{}g(\mu)(\mu-f(x))\inv d\mu\\&=g(f(x))\end{align*}

$\square$

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Proposition 6.2.10: Spectrum Evaluation Homomorphism

The Eigenspace

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The Eigenspace

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