6.2 Spectrum and Functional Calculus

Proof. Suppose $\lam\nin \spec{A}{xy}\cup\curl{0}$ then $(xy-\lam)\inv=u$ exists.

from which we conclude

which shows $yx-\lam$ is invertible and $\lam\nin \spec{A}{yx}\cup\curl{0}$. By symmetry, $\spec{A}{xy}\cup\curl{0}=\spec{A}{yx}\cup\curl{0}$. Lastly we have

noting that we are taking supremum over non-negative values.$\square$

Proof. By Lemma 6.1.7 we assume that $A$ is unital. Let $x\in A$ and consider the continuous map $f:\lam\in \com\mto x-\lam \in A$. The resolvent of $x$ is the inverse image $f\inv(G(A))$ thus is open by Proposition 6.1.17 so $\spec{A}{x}$ is closed. Suppose now $\abs{\lam}>\norm{x}{}$ then $\norm{\frac{x}{\lam}}{}<1$ so by Proposition 6.1.16 the series $\sums{\infty}{n=0}\braks{\frac{x}{\lam}}^{n}$ converges to $\braks{\frac{x}{\lam}-1}\inv$. Hence

and $\lam$ is in the resolvent of $x$. Therefore, $\spec{A}{x}$ is contained in the disk $\abs{\lam}\leq \norm{x}{}$. Hence it is compact.$\square$

Proof. Let $r(x)=\limitinf{n\to\infty}\norm{x^n}{}^{1/n}$ then the power series $-\sums{\infty}{n=0}\frac{x^{n}}{\lam^{n+1}}$ converges for $\abs{\lam}>r(x)$ and this is the inverse $(x-\lam)\inv$. Hence

Let $f(\lam)=(x-\lam)\inv$ for $\lam\nin\spec{A}{x}$. By Proposition 6.1.17 with $x\bij x-\lam$ and $x_{0}\bij x-\lam_{0}$ then if $\abs{(x-\lam)-(x-\lam_0)}<\norm{f(\lam_0)}{}\inv$ for $\lam_{0}\nin\spec{A}{x}$ and $\lam\in\com$ then $\lam\in B\parens{\lam_0,\norm{f(\lam_0)}{}\inv}\suf G(A)$ thus $\lam\nin \spec{A}{x}$ and

Hence $f(\lam)$ is a holomorphic function on the resolvent of $x$. The power series $\sums{\infty}{n=0}\frac{x^{n}}{\lam^{n+1}}$ is the Laurent expansion of $f(\lam)$ at $\infty$. Let $\p$ be a bounded linear functional of $A$ then

is a holomorphic function on the resolvent of $x$, where $\inner{a}{\p}$ denotes the value of $\p$ at $a\in A$. The function $f_{\p}(\lam)$ is holomorphic for $\abs{\lam}>\norm{x}{sp};$ hence its Laurent expansion $\sums{\infty}{n=0}\frac{\inner{x^n}{\p}}{\lam^{n+1}}$ converges for $\abs{\lam}>\norm{x}{sp}$. Hence we have

By uniform boundedness theorem, the sequence $\curl{\frac{x^{n}}{\lam^{n+1}}}$ is bounded for $\abs{\lam}>\norm{x}{sp}$ so that for any $\abs{\lam}>\norm{x}{sp}$, there exists $\al>0$ such that $\norm{x^n}{}\leq\al\abs{\lam}^{n+1}$, $n=1,2,...$. Therefore, we have

Thus

$\square$

Proof. Suppose $\spec{A}{x}=\emp$ for some $x\in A$. Using notation as in Proposition 6.2.6, the function $f_{\p}(\lam)$ is holomorphic everywhere; hence it is entire. But $\limit{\lam\to\infty}f_{\p}(\lam)=0$ so $f_{\p}$ is identically zero by Liouville’s Theorem, which means that $\inner{f(\lam)}{\p}\equiv 0$ for every $\p\in A^{*}$. Hence $f(\lam)\equiv 0$ contradicting $f(\lam)=(x-\lam)\inv$.$\square$

Proof. Let $x\in A$ then by Theorem 6.2.7, let $\lam\in \spec{A}{x}$. By definition $x-\lam$ is not invertible so $x=\lam$ by assumption.$\square$

Proof. For each $\p\in A^{*}$, we consider a continuous function $\lam\mto f(\lam)\inner{(\lam-x)\inv}{\p}\in\com$ on the curve $C$. Define a linear functional

then $F\in A^{**}$ since

where $l$ is the length of the curve $C$. Hence we may write

OTOH, the $A$-valued function $\lam\mto f(\lam)(\lam-x)\inv$ is continuous so the limit

exists, where $\curl{\lam_0,...,\lam_n,\lam_{n+1}=\lam_0}$ is a partition of the curve $C$. Then $\inner{y}{\p}=\inner{F}{\p}$ for all $\p\in A^{*}$. Hence $F\in A\suf A^{**}$. By Cauchy’s theorem for contour integrals, $F$ does not depend on the curve $C$. Therefore we denote $F$ by $f(x)$ and write

$\square$

Proof. Let $f,g$ be functions holomoprhic in neighborhoods $U_{f},U_{g}$ of $\spec{A}{x}$, respectively. Let $U=U_{f}\cap U_{g}$, and $C_{i}$ ($i=1,2$) be smooth simple closed curve in $U$ enclosing $\spec{A}{x}$ such that $C_{2}$ lies inside $C_{1}$ then

where the last equality is by partial fractions. Since $\frac{g(\mu)}{\lam-\mu}$ is holomorphic inside the curve $C_{2}$ if $\lam$ lies on $C_{1}$

Hence we get the map $f\mto f(x)$ is a homomorphism

Suppose now $f(\lam)\equiv 1,\lam\in\com$ then

where we pick $C=\curl{\lam:\abs{\lam}=\norm{x}{}+\e}$ for some $\e>0$. We have then

uniformly for every $\lam\in C$. Hence by residue theorem

Similarly if $f(\lam)=\lam,\fall \lam\in \com$ then $f(x)=x$.$\square$

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

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

By Proposition 6.2.10

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

$\square$

Proof. Let $x_{0}\in E_{U}$, the function $\lam\in U^{c}\mto (x_{0}-\lam)\inv\in A$ is continuous with

Let $\de=\INF{\lam\in U^c}\frac{1}{\norm{(x_0-\lam)\inv}{}}>0$. If $\norm{x-x_0}{}<\de$ then for any $\lam\in U^{c}$ we have

hence $(x-\lam)\inv$ exists by 6.1.17, so $\spec{A}{x}\suf U$ and $x\in E_{U}$ as desired.$\square$

Proof. By Spectral Mapping 6.2.11

Furthermore $\spec{A}{x}\suf\curl{\lam\in \com:-\pi<\Im\lam<\pi}$ hence

This shows that $\spec{A}{\exp x}\suf\text{dom(Log)}$ so $\log\in \holon{\spec{A}{\exp x}}{\com}$ and by Spectral Mapping 6.2.11 with $f=\exp, g=\log$ we have

$\square$