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 $M$.

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