6.4 Spectrum and Functional Calculus in C*-Algebras

Proof. We go straight to computation

applying this iteratively

By Proposition 6.2.6

$\square$

Proof. Let $u\in A_{u}$ then

hence $\norm{u}{}=\norm{u^*}{}=1$. By Proposition 6.4.3, $\norm{u}{sp}=\norm{u^*}{sp}=1$. Since $u^{*}=u\inv$ we have

so

Let $h\in A_{h}$ and set $u\define \exp(ih)$ then $u^{*}=\exp(-ih)$ so $u\in A_{u}$. By Proposition 6.2.11

which implies that $\spec{A}{h}\suf\R$$\square$

Proof. Since $A$ is abelian, every element $x\in A$ is normal so $\norm{x}{}=\norm{x}{sp}$ by Proposition 6.4.3. Hence $\scr{F}$ is isometric, showing that $A$ is semisimple. The image $\scr{F}(A)$ is complete therefore closed. Let $\om\in\OM(A)$ then for each $h\in A_{h}$

by Theorem 6.3.11 and Proposition 6.4.4. For every $x\in A$ we have $x=h+ik$ for some $h,k\in A_{h}$ and

since $*$-operation in $C_{\infty}(\OM(A))$ is complex conjugation, $\scr{F}:x\mto \ha{x}$ preserves the $*$-operation. Thus, $\scr{F}(A)$ is a self-adjoint, subalgebra of $C_{\infty}(\OM(A))$ that separates points of $\OM(A)$. By Stone-Weierstrass theorem, $\scr{F}(A)$ is dense in $C_{\infty}(\OM(A))$; hence $\scr{F}(A)=C_{\infty}(\OM)$ by closedness.$\square$

Proof. Since $\OM$ is LCH, it is a completely regular space so the map $\om\mto \ha{\om}$ is a homeomorphism of $\OM$ into $\OM(A)$. Let $\rho\in\OM(A)$ then $\rho$ is a linear functional on $C_{\infty}(\OM)$ such that $\rho\geq 0$ if $x\geq 0$ as a function. Hence there exists a positive Radon measure $\mu$ on $\OM$ such that $\rho(x)=\integral{\OM}{}x(\om)d\mu(\om)$ and $\mu(\OM)=1$. By multiplicativity of $\rho$

Hence for each $x\in C_{\infty}(\OM)$, $E_{x}\define\curl{\om\in\OM:x(\om)\neq \rho(x)}$ is a $\mu$-nullset. If $\supp{\mu}$ contains $\om_{1}\neq \om_{2}$ then $\exists x\in C_{\infty}(\OM)$ with $x(\om_{1})\neq x(\om_{2})$ since $C_{\infty}(\OM)$ separates points since $\OM$ is completely regular. By continuity $\exists U_{i}\ni \om_{i}$ open neighborhoods where $x|_{U_1}\no\equiv x|_{U_2}$ hence on $U_{1}\cup U_{2}$ we have $\abs{x(\om)-\rho(x)}^{2}>0$. Since $\mu$ is a positive Radon measure, $\mu(U_{i})>0$ thus

We conclude $\mu$ is concentrated at a single point $\om\in\OM$ so $\rho(x)=x(\om)$ and $\rho=\ha{\om}$.$\square$

Proof. Let $\curl{B_i}_{i\in I}$ be a collection of $C^{*}$-subalgebra of $A$ then $\caps{}{i\in I}B_{i}$ is closed topologically, under addition, multiplications and involution.$\square$

Proof. The normality of $x$ implies that $B$ is abelian. Since the polynomials of $x$ and $x^{*}$ form a dense subalgebra of $B$, the map $\psi:\om\in\OM(B)\mto\om(x)\in \spec{A}{x}$ is a homeomorphism. By Gelfand Representation 6.3.11. $\psi$ is well-defined and continuous. By Proposition 6.3.8. $\psi$ is bijective. By Propositionn 6.3.9, $\psi\inv$ is continuous. By Gelfand-Naimark 6.4.5, the composed map $\p:f\in C(\spec{A}{x})\mto\scr{F}\inv(f\circ\psi)\in B$ is the desired isomorphism.$\square$

Proof. If $\lam\nin\spec{A}{x}$ then $x-\lam$ is invertible in $B$ because $\lam-\indi{\spec{A}{x}}$ is invertible in $C(\spec{A}{x})$. Let $F$ be a holomorphic function on a neighborhood of $\spec{A}{x}$ and $F|_{\spec{A}{x}}=f$ then for any $\om\in\OM(B)$

where $C$ is a smooth simple closed curve enclosing $\spec{A}{x}$. Observe that

So we have for all $\om\in\OM(B)$

hence $F(x)=\p(f)$ and since $\p$ is a $*$-isomorphism we extend by density to $C(\spec{A}{x})$.$\square$

Proof. Let $\om_{0}$ be the homomorphism of $A_{I}$ with kernel $A$ then $\om_{0}(f(x))=f(\om_{0}(x))=f(0)=0$ so $f(x)\in \ker(\om_{0})=A$$\square$

Proof. By Proposition 6.4.4, Proposition 6.4.8 and Jordan decomposition in $\R$.$\square$

Proof. Since 1. follows from 2. by adjunction of an identity, it suffices to prove 2. In general, we have $\spec{A}{x}\suf \spec{B}{x}$.

Suppose that $x$ is self-adjoint. Let $\lam\nin\spec{A}{x}$ then by Proposition 6.4.4, $\spec{B}{x}\suf\R$ so to show that $\lam\nin\spec{B}{x}$ we may assume WLOG that $\lam\in\R$. For any $\e>0$, $\lam_{\e}=\lam+i\e\nin\spec{B}{x}$ so $(x-\lam_{\e})\inv\in B$. By continuity of inverse in $G(A)$, $(x-\lam_{\e})\inv\to (x-\lam)\inv\in G(A)$ as $\e\to 0$. Since $B$ is closed, $(x-\lam)\inv\in B$. Thus $\lam\nin\spec{B}{x}$ and $\spec{A}{x}=\spec{B}{x}$ for a self-adjoint $x\in B$.

Now if $x\in B$ is invertible in $A$ then $x^{*}x$ is invertible in $A$ so is in $B$ by self-adjointness of $x^{*}x$. Hence $x$ is left invertible in $B$. Similarly considering $xx^{*}$ shows that $x$ is right invertible in $B$ also. Therefore, $x$ is invertible in $B$. We apply this argument for $x-\lam,\lam\in\com$ to conclude 2.$\square$

Proof. Let $h\in A_{h}$ with $\norm{h}{}\leq 1$ and set $u=h+i(1-h^{2})^{1/2}$. By Proposition 6.4.8, $u$ is unitary and $h=\frac{1}{2}(u+u^{*})$. For general $x\in A$ we consider the real and imaginary parts of $x/\norm{x}{}$$\square$

Proof. By Stone-Weierstrass theorem, for any $\e>0$ there exists a polynomial $p$ of $\lam$ and $\cl{\lam}$ such that $\SUP{\lam\in K}\abs{p(\lam,\cl{\lam})-f(\lam)}<\e$. Since $p$ is a polynomial there exists $\de>0$ such that $\norm{p(x,x^*)-p(y,y^*)}{}<\e$ for $\norm{x-y}{}<\de$ and $\norm{x}{},\norm{y}{}\leq M$ for some constant $M>0$ that doesn’t depend on $x,y$. let $M=\SUP{\lam\in K}\abs{\lam}$. If $\norm{x-y}{}<\de$ and $x,y\in A_{K}$ then

$\square$