(Dis-)continuity of the multiplication in the space $\mathcal{B}(\mathcal{H})$ of bounded linear operators on a Hilbert space

Question

Let $\mathcal{H}$ be an infinite-dimensional Hilbert space. Let $\mathcal{B}(\mathcal{H})$ be the space of bounded linear operators on $\mathcal{H}$. Let $\mathcal{B}(\mathcal{H})_1$ be the ball of radius $1$ in $\mathcal{B}(\mathcal{H})$.

I would like to study the continuity of the multiplication $(T,S) \mapsto TS$ in the product of the SOT topologies to the SOT topology.

Using the characterization of continuity in terms of convergence of nets, we can show the continuity of this map if we restrict its domain to $\mathcal{B}(\mathcal{H})_1 \times \mathcal{B}(\mathcal{H})_1$.

For example, this is done here https://math.stackexchange.com/a/1468506/862585

My question is the following. I know that this map isn't SOT$\times$SOT-SOT continuous when we unrestrict the domain to $\mathcal{B}(\mathcal{H}) \times \mathcal{B}(\mathcal{H})$. How can I show that it isn't? Is the map continuous in the WOT$\times$WOT-WOT sense?

Answer 1

Fix an orthonormal basis $\{e_n\}$. It is shown here that $0$ belongs to the weak-closure of the set $\{\sqrt{n}\,e_n:\ n\}$. This means that there exists a net $\{n_j\}$ of natural numbers such that $$\sqrt{n_j}\,e_{n_j}\xrightarrow{\rm weak}0.$$ Now define operators $T_j\in B(H)$ by $$ T_jx=\sqrt{n_j}\langle x,e_{n_j}\rangle\,e_{n_j}. $$ Then $$ \|T_jx\| %=\langle \sqrt{n_j}\langle x,e_{n_j}\rangle\,e_{n_j},\sqrt{n_j}\langle x,e_{n_j}\rangle\,e_{n_j}\rangle^{1/2} =|\langle x,\sqrt{n_j}\,e_{n_j}\rangle|\to0. $$ So $T_j\to0$ sot. Meanwhile, $$ \|T_j^2x\|=\|n_j\langle x,e_{n_j}\rangle\,e_{n_j}\|=n_j\,|\langle x,e_{n_j}\rangle|. $$ Let $x=\sum_n\frac1n\,e_n$. Then $\langle n\,e_n,x\rangle=1$ for all $n$. So $V=\{y\in H:\ |\langle y,x\rangle|<1\}$ is a weak-open neighbourhood of $0$ that does not contain $ne_n$ for any $n$. In particular $n_j\,|\langle x,e_{n_j}\rangle|$ cannot converge to $0$, which means that $T_j^2x$ does not converge to $0$. A priori this does not show that $T_j^2$ does not converge, but it precludes continuity of the product because $T_j\xrightarrow{\rm sot}0$ and $T_j^2$ does not.

As for the wot, the same example works. But the continuity of the product fails even in the unit ball. Because it is not too hard to show that every $T\in B(H)$ with $\|T\|\leq1$ is a wot-limit of projections. If the multiplication map were continuous, a limit of idempotents would be an idempotent. But there exists a net $\{P_j\}$ of projections such that $P_j\xrightarrow{\rm wot}\frac12\,I$, for instance.