Product map is bounded

Question

Let H be an infinite dimensional Hilbert space. Define the map $M: B(H)\otimes_{min} B(H) \mapsto B(H)$ by $M(A\otimes B)=AB$. How do I show that the map M is bounded? If I replace $B(H)$ by an arbitrary C* algebra, whether M is still bounded?

Answer 1

I don't think $M$ is bounded.

Let $H=\bigoplus_n \mathbb C^n$. For each $n$, consider the matrix units $\{E_{kj}{(n)}\}_{k,j=1}^n\subset M_n(\mathbb C)$.

Let $$X_n=\sum_{k,j=1}^nE_{kj}^{(n)}\otimes E_{jk}^{(n)}.$$ Then $X_n$ is selfadjoint, and $$ X_n^2=\sum_{k,j=1}^n\sum_{s,t=1}^nE_{kj}^{(n)}E_{st}^{(n)}\otimes E_{jk}^{(n)}E_{ts}^{(n)} =\sum_{k,j=1}^nE_{kk}^{(n)}\otimes E_{jj}^{(n)}=I_n. $$ Thus $$\tag1\|X_n\|^2=\|X_n^2\|=1.$$ On the other hand, $$\tag2 M(X_n)=\sum_{k,j=1}^nE_{kj}^{(n)}E_{jk}^{(n)}=n\,I_n. $$ Fix $m\in\mathbb N$. Define $X\in B(H)$ by $$X=\bigoplus_{n=1}^mX_n.$$ By $(1)$, $$\|X\|=1,$$ and by $(2)$, $$\|M(X)\|=m.$$ So $M$ is not bounded.