## 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 $C(X)$


*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.