Bilinear map $m:B(H)\times H\to H$

Question

Let $m:B(H)\times H\to \mathbb C$ be bilinear form . Can we write it as $m(T,h)=\left<Th,x\right>$ for some $x\in H$?

Consider bilinear map $f:B(H)\times H\to H$ given by $f(T,h)=Th$. Now define $\phi:H\to \mathbb C$ as $$\phi(Th)=m(T,h)$$ this is a linear functional hence by Riesz representation $\phi(Th)=\left<Th,x\right>$ for some $x\in H$. Am I doing this right?

Answer 1

Even if $m$ is bounded, it doesn't have to be of the form you want. For instance choose $S\in B(H)$ to be trace-class, $y\in H$, and define $$ m(T,h)=\operatorname{Tr}(ST)\,\langle h,y\rangle. $$ If you had $m(T,h)=\langle Th,x\rangle$, then you would have $$ \operatorname{Tr}(ST)\,\langle h,y\rangle=\langle h,T^*x\rangle $$ for all $h$, so $T^*x=\operatorname{Tr}(ST)\,y$. But we can choose $T$ and $y$ so that $y$ is not in the image of $T^*$.

Answer 2

If $m$ just a continuous bilinear from on $B(H) \times H$ we cannot write in above form in general. If such an $x$ exists then we must have $m(T,h)=0$ whenever $Th=0$ but this is not guaranteed. For example we may have $m(T,h)=\langle T(S(h)), x_0 \rangle$ for some $S$ and some $x_0$ and there is no reason why $\langle T(h), x_0 \rangle$ implies $\langle T(S((h)), x_0 \rangle$