Multiplication operator is not jointly continuous in strong topology How can I show that multiplication operator ($M:\mathcal{L}(X,Y) \times \mathcal{L}(Y,Z) \rightarrow \mathcal{L}(X,Z)$; $M(A,B)=AB)$ is not jointly continuous in strong topology?
I have to show that if I take an open set $O$ in $\mathcal{L}(X,Z)$ (with norm topology) then $M^{-1}O$ is not always an open set in $\mathcal{L}(X,Y) \times \mathcal{L}(Y,Z)$ with the strong topology. Right? But How!?
Thank you!
 A: Let us show that if $Y$ is infinite-dimensional, then multiplication is not jointly continuous on $\mathcal L(X,Y)\times \mathcal L(Y,Z)$ with respect to the strong operator topology.
Choose any non-zero $x_0\in X$. It is enough to show that the set 
$\mathcal M:=\{ (A,B);\; \Vert BAx_0\Vert <1\} $
is not an $SOT\times SOT$-neighbourhood of $(0,0)$ in $\mathcal L(X,Y)\times \mathcal L(Y,Z)$. Equivalently, let us show that for any neighbourhood $\mathcal U$ of $(0,0)$, one can find $(A,B)\in\mathcal U$ such that $\Vert BA x_0\Vert\geq 1$.
Choose $\varepsilon >0$ and finite sets $E\subset X$ and $F\subset Y$ such that
$$\Bigl(\Vert Au\Vert<\varepsilon\;\hbox{for all $u\in F$ and}\; \Vert Bv\Vert<\varepsilon\;\hbox{for all $v\in F$}\Bigr)\implies (A,B)\in\mathcal U\, . $$
Since $\dim(Y)=\infty$, one can find an operator $A\in\mathcal L(X,Y)$ such that $y_0:=Ax_0\not\in \hbox{span}(F)$. Moreover, multiplying $A$ by a suitable constant, we may assume that $\Vert A\Vert$ is arbitrarily small, so that in particular $\Vert Au\Vert<\varepsilon$ for all $u\in E$. 
Next, since $y_0\not\in \hbox{span}(F)$, one can find an operator $B\in\mathcal L(Y,Z)$ such that $B\equiv 0$ on $F$ and $By_0\neq 0$; and multiplying $B$ by a suitable constant we may assume that $\Vert By_0\Vert=1$.
By the definition of $A$ and $B$, we then have $(A,B)\in\mathcal U$ and $\Vert BAx_0\Vert=\Vert By_0\Vert=1$, which concludes the proof.
Note however that multiplication is jointly continuous on bounded sets. (This is not difficult to prove). Hence, by the Uniform Boundedness Principle (assuming that $X,Y,Z$ are Banach spaces) it is not possible to find two sequences $(A_n)$ and $(B_n)$ such that $A_n\xrightarrow{SOT} 0$ and $B_n\xrightarrow{SOT} 0$ but $B_nA_n$ does not tend to $0$.
A: Here is a natural example in the case of a Hilbert space.
if $H$ is an infinite-dimensional Hilbert space with orthonormal basis $\{e_n\}$, it is not hard to show that there is a net $\{n_\alpha\}$ of positive integers such that $\{\sqrt{n_\alpha}\,e_{n_\alpha}\}_\alpha$ converges weakly to $0$.
Let $T_\alpha\in B(H)$ be the rank-one operator
$$
T_\alpha\,x=\sqrt{n_\alpha}\,\langle x,e_{n_\alpha}\rangle\,e_{n_\alpha}.
$$
Then
$$
\|T_\alpha\,x\|=|\sqrt{n_\alpha}\,\langle x,e_{n_\alpha}\rangle|\to0.
$$
That is, $T_\alpha\xrightarrow{\rm sot}0$. We will see, however, that $\{T_j^2\}$ does not converges sot to $0$.
We easily see that
$$
\|T_\alpha^2\,x\|=|{n_\alpha}\,\langle x,e_{n_\alpha}\rangle|.
$$
If $T_\alpha^2\xrightarrow{\rm sot}0$, this would imply that $0$ is in the weak closure of $\{ne_n\}$, which is a contradiction. To see the contradiction, consider the weak neighbourhood of $0$ given by $V=\{y:\ |\langle x,y\rangle|<1\}$, where
$$
y=\sum_n\frac1n\,e_n. 
$$
Then
$$
\langle ne_n,y\rangle=1,\qquad\qquad n\in\mathbb N, 
$$
and so $V\cap \overline{\{ne_n\}}^{w}=\emptyset$, implying that $0\not\in\overline{\{ne_n\}}^{w}$.
