Let $\mathcal A,\mathcal B,\mathcal C$ be von Neumann algebras. Let $F:\mathcal A\otimes\mathcal B\to\mathcal C$ be a bounded linear map. Assume $F(a\otimes b)=0$ for all $a,b\in\mathcal A,\mathcal B$. Does this imply $F=0$?
And does the answer change if we require $\mathcal A,\mathcal B,\mathcal C$ to be sets of all bounded operators?
Here the tensor product $\mathcal A\otimes\mathcal B$ is the tensor product of von Neumann algebras, i.e., the smallest von Neumann algebra containing all $a\otimes b$.
What I know:
- If we define $\mathcal A\otimes\mathcal B$ as the spatial tensor product (i.e., the completion of the algebraic tensor product w.r.t. the operator norm), then the answer is yes because $F$ is $0$ on the algebraic tensor product and continuous w.r.t. the operator norm.
- If we restrict $F$ to be WOT-continuous, then it also holds (if I am not mistaken) because the smallest von Neumann algebra is the WOT-closure of the algebraic tensor product. Thus any $x\in \mathcal A\otimes\mathcal B$ is a WOT-limit of $x_i$ in the algebraic tensor product. Thus $F(x)$ is a WOT-limit of $F(x_i)=0$. Thus $F(x)=0$. Hence $F=0$.
- In the context of von Neumann algebras, normal maps are relevant (i.e., weak$^*$-continuous positive maps). Maybe for weak$^*$-continuous maps the above holds? (Without "positive".) That would be a useful fallback for me, but I don't see the proof.