Let $H$ be a separable, infinite-dimensional complex Hilbert space. How do we show that multiplication: $$ \mathcal{L}(H) \times \mathcal{L}(H) \rightarrow \mathcal{L}(H), \quad(A, B) \mapsto A B $$ is not continuous with respect to the strong operator topology
So I want to clarify what it means for multiplication to be continuous in the first place. So if we have a sequence of operators $A_n \to A$ and $B_n \to B$, then we should have that $lim A_nB_n \to AB$. So being not continuous, means this does not hold correct? Since we are working on a separable hilbert space, we can assume it is $l^2$. The only non-trivial linear bounded operator I know of on $l^2$ is just multiplication by a $l^\infty$ sequence. Now we are suppose to find a counterexample in the strong operator topology, which means we need two sequence of operators and some $x\in l^2$ where $||(A_nB_n - AB)x||$ does not converge to zero. Am I interpreting this problem correctly?