Let $H$ be a hilbert space (with inner product $(\cdot,\cdot)$) and suppose that $(S_{n})$ and $(T_{n})$ are sequences of bounded operators from $H$ to $H$ such that $(S_{n}x,y)\to(Sx,y)$ and $(T_{n}x,y)\to(Tx,y)$ for all $x,y\in H$. Is it then true that $(S_{n}T_{n}x,y)\to(STx,y)$ for all $x,y\in H$? I tried the following:
\begin{align*} |((S_{n}T_{n}-ST)x,y)|&\leq|((S_{n}T_{n}-ST_{n})x,y)|+|((ST_{n}-ST)x,y)|\\ &=|((S_{n}-S)(T_{n}x),y)|+|(T_{n}-T)(x),S^{*}y)|. \end{align*}
It is clear that $|(T_{n}-T)(x),S^{*}y)|\to0$, but I'm not sure about the convergence of $|((S_{n}-S)(T_{n}x),y)|$.
Any help would be greatly appreciated!
We will say that $S_n\to S$ in the weak operator topology when for all $x,y\in H$ it is $\langle S_nx,y\rangle\to\langle Sx,y\rangle$. Your question is: is multiplication sequentially continuous with respect to the weak operator topology? The answer is no.
For example, consider $H=\ell^2$ and for $x=(x_k)_{k=1}^\infty$ set $$S_nx= (x_n,x_{n+1},\dots)$$ and $$T_nx=(0,\dots,0,x_1,x_2,\dots)$$ where the number of $0$'s in the definition of $T_n$ is $n$. It is easy to verify that $(S_n),(T_n)\subset B(H)$ and also $S_n,T_n\to0$ in WOT. But on the other hand we have that $S_nT_n=\text{Id}_H$, so $S_nT_n\not\to 0$ in WOT.
It is true however that if $S_n\to S$ in WOT, then $S_nT\to ST$ and $TS_n\to TS$ in WOT. There are some more relative results, but multiplication is not continuous with respect to WOT in general.