Let $H,K$ are Hilbert spaces, I want to show there is an isometric linear correspondence between bounded sesquilinear forms $S(H,K)$ and bounded linear operators $B(H,K)$. ( $\Phi: B(H,K)\to S(H,K)$ such that $\Phi(t)(h\otimes k) = (th,k))$.
I have two questions.
1- I have to show that for every bounded linear operator $t\in B(H, K)$, $||t||=\sup\{|(th,k) : ||h||,||k||\leq 1\}$, while I can not show it.
2- About $\Phi$ is surjective , suppose $A\in S(H,K)$. By Riesz representation, there is a unique vector $\eta_1\otimes \eta_2$ such that $A(\xi_1\otimes \xi_2)= (\xi_1\otimes \xi_2 , \eta_1\otimes\eta_2)$. Now $t=\eta_2\otimes \eta_1$ is an operator in $B(H,K)$. But I can not show that $A(\xi_1,\xi_2)=(t\xi_1,\xi_2)$ .
Please help me understand them. Thanks in advance.
You have $$|(th, k)|\leq\|t\|\,\|h\|\,\|k\|\leq\|t\|. $$ On the other hand, if you take $ k=th/\|th\|$, then $(th, k)=\|th\|$. As we can choose $ h $ unital with $\|th\|$ arbitrarily close to $\|t\|$, thd other inequality is established.
You sentence "By Riesz..." is not correct as stated. What you want to do is, for $\xi\in H $, to consider the map $\eta\longmapsto \overline{A (\xi,\eta)} $. By the boundedness of $ A $ this map is a bounded functional on $ K $. By the Riesz Representation Theorem there exists a vector $ t\xi\in K $ such that $ \overline{A (\xi,\eta)}=(\eta,t\xi) $ for all $ \eta\in K $. That is, $ A (\xi,\eta)=(t\xi,\eta) $. Now you check that $\xi\longmapsto t\xi $ defines a bounded linear operator.