### Category: functional analysis

## Borel Functional Caculus on Orthogonal Hermitian Elements

Let \(A\) be a C*-algebra, \(X\) be a compact subset of \(\mathbb{C}\), and \(c\) be a hermitian element in \(A\) such that \(\sigma\left( c \right) \subset X\). The restriction of a function on \(X\) to \(\sigma\left( c \right)\) is a *-homomorphism from \(C\left( X \right)\) to \(C\left( {\sigma\left( c \right)} \right)\), the Gelfand representation of \(C\left( {\sigma\left( c \right)} \right)\) is a *-isomorphism between \(C\left( {\sigma\left( c \right)} \right)\) and the unital C*-algebra, say \(C^{\ast}\left( c \right)\) generated by \(a\), the inclusion of \(C^{\ast}\left( c \right)\) into \(A\) and the atomic representation \(\left( {\pi_{a},H_{a}} \right)\) of \(A\) are both *-homomorphism. Hence the composition of them defines a *-homomorphism \(H\) from \(C\left( X \right)\) to \(B\left( H_{a} \right)\). Let \(\widetilde{H}\) be the extension of \(H\) to \(C\left( X \right)^{\ast \ast}\), then \(\widetilde{H}\left( f \right)\) coincides with \(f|_{\sigma{(c)}}\left( c \right)\) for each continuous function on \(X\). Since \(f_{i} \in C\left( X \right)\) \(\sigma\left( {C\left( X \right)^{\ast \ast},C\left( X \right)^{\ast}} \right)\) converges to a Borel measurable bounded function \(f\) implies that \(f_{i}|_{\sigma{(c)}}\) \(\sigma\left( {C\left( {\sigma\left( c \right)} \right)^{\ast \ast},C\left( {\sigma\left( c \right)} \right)^{\ast}} \right)\) converges to \(f|_{\sigma{(c)}}\), \(\widetilde{H}\left( f \right)\) coincides with \(f|_{\sigma{(c)}}\left( c \right)\) for each Borel measurable bounded function on \(X\). Hence there is no ambiguity to write \(f\left( c \right)\) for \(\widetilde{H}\left( f \right)\).

Let \(a\) and \(b\) whose spectrums are both contained in \(X\) be two hermitian elements such that \(ab = 0\). If the spectrum of \(a + b\) is also contained in \(X\), then for each bounded Borel measurable function \(f\) on \(X\),

\[f\left( {a + b} \right) = f\left( a \right) + f\left( b \right).\]This can be verified for polynomials, continuous functions and bounded Borel measurable functions on \(X\) in turn.

## A compact Hausdorff space $X$ is homeomorphic to the maximal ideal space of $C(X)$

**Claim.**\begin{equation*} H:X\to \M, x\to ev_x \end{equation*} is a homeomorphism. (a)Obviously, $H$ is well-defined. (b)By Urysohn's lemma, $H$ is an injection. (c)If $\{x_\lambda\}$ is a net in $X$ converges to $x\in X$, then $f(x_\lambda)\to f(x)$ for all $f\in C(X)$, i.e., $H(x_\lambda)\xrightarrow{weak*}H(x)$. Hence $H$ is continuous. (d)Let $I$ be a none-trivial ideal of $C(X)$ and denote all the zero points of $f\in C(X)$ by $Z(f)$. For any finite continuous functions $f_i\in I, i=1,2,\cdots, n$, $$f:=\bar{f_1}f_1+\bar{f_2}f_2+\cdots \bar{f_n}f_n\in I,$$ therefore $ f$ is not invertible, and thus $Z(f)\neq\emptyset$. Since $$Z(f_1)\cap Z(f_2)\cap\cdots\cap Z(f_n)\supset Z(\bar{f_1}f_1+\bar{f_2}f_2+\cdots \bar{f_n}f_n),$$ $Z(f_1)\cap Z(f_2)\cap\cdots\cap Z(f_n)\neq\emptyset.$ Hence we have $$\bigcap_{f\in I}Z(f)\neq\emptyset.$$ Note that $Z(f)$ is closed for every continuous function and that $X$ is compact. In particular, take $I=\ker\tau, \tau\in \M$. Suppose $x\in \bigcap_{f\in I}Z(f)$, then $\ker\tau=I\subset \{f\in C(X)|f(x)=0\}=\ker ev_x$, so $\tau=ev_x$. Therefore, $H$ is a surjection. (e) Since $X$ is compact and $\M$ is Hausdorff, $H$ is a homeomorphism.