a Positive Bilinear Map is Necessarily Bounded

Suppose $A$ and $B$ are Banach spaces, $C$ is a normed space, and $\sigma:A\times B\to C$ is a bilinear form. For each $a\in A$, denote the map $b\mapsto \sigma(a,b)$ by $\hat{a}:B\to C$; and for each $b\in B$ denote the map $a\mapsto \sigma(a,b)$ by $\hat{b}:A\to C$. **Lemma**. The bilinear form $\sigma$ is bounded is equivalent to $\hat{a}$ and $\hat{b}$ are both bounded for all $a\in A$ and $b\in B$. *Proof*. Since $\hat{b}$ is a bounded map on $A$ for all elment $b$ in the closed unit ball of $B$, $\sup_{a\in A_1}\|\hat{b}(a)\|\leq \|\hat{b}\|$, i.e., $\sup_{a\in A_1}\|\hat{a}(b)\|\leq \|\hat{b}\|$. By PUB, the family $\{\hat{a}|a\in A_1\}$ of bounded maps is bounded uniformly, that is, $\sup_{a\in A_1}\|\hat a\|\leq M$ for some $M>0$ . Hence, $\sup_{a\in A_1, b\in B_1}\|\sigma(a, b)\|=\sup_{a\in A_1, b\in B_1}\|\hat{a}(b)\|\leq M$. **Application**. Let $A$, $B$ and $C$ be C*-algebras.$\newcommand{\tensor}{\otimes}$ By a *positive* bilinear form $\sigma$ from $A\times B$ to $C$, we mean $\sigma(a,b)\geq 0$ for all positive elements $a$ in $A$ and $b$ in $B$, or equivalently, $\hat{a}$ and $\hat{b}$ are positive linear maps on $B$ and $A$ respectively. Since every positive linear map between two C*-algebra is bounded, by the above lemma, >a positive bilinear form in C*-algebras must be bounded. If $A$ and $B$ are C*-algebras, $\gamma$ is a C*-norm on $A\tensor B$, then the map $A\times B\to A\tensor_\gamma B$ is a positive bilinear form and thus is bounded, that is, $\gamma(a,b)\leq M\|a\|\|b\|$ for some positive number $M$.

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 $C(X)$

$\newcommand{\M}{\operatorname{Max}}\newcommand{\C}{\mathbb{C}}$**Theorem**. If $X$ is a compact Hausdorff space, and $\M=\{\text{non-zero algebra homomorphism of }C(X) \text{to } \C\}$ is the maximal ideal space of $C(X)$, then there is a homeomorphism of $X$ onto $\M$. *Proof*. For each $x\in X$, define \begin{equation*} ev_x(f)=f(x), f\in C(X), \end{equation*} we have 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.