# 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!

• this property suprises me... my first guess: that might have something to do with the product topology. And also with the spaces: with $X=Y=Z=\mathbb{R}$ and $\mathcal{L}\left(\mathbb{R},\mathbb{R}\right)=\mathbb{R}$ i think it is continuous or am I wrong? – Max Feb 23 '14 at 12:38
• I've found this question in many books and is very often left as exercise... other times the answer is simply "Take the shift operator." But I can't figure why... – Benzio Feb 23 '14 at 12:41
• sorry I forgot to write the adjective "jointly". Now it's edited – Benzio Feb 23 '14 at 12:44
• @Max For finite-dimensional spaces, all Hausdorff vector space topologies coincide, and all multilinear maps are continuous. One needs infinite-dimensional spaces for it to be discontinuous. – Daniel Fischer Feb 23 '14 at 12:46
• Please, let me "see" it! – Benzio Feb 23 '14 at 12:49

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 $0$.

• Could you explain more about why the multiplication is jointly continuous on bounded sets? I tried to show myself but it seems much trickier than expected... I cannot figure out at all where the boundedness comes in.... – Keith Oct 3 '19 at 5:25
• @Keith Take a bounded net $(A_i,B_i)$ converging to $(A,B)$. Then, for any $x$, you have $\Vert A_iB_i x-ABx\Vert \leq \Vert A_i\Vert \Vert B_ix-Bx\Vert + \Vert A_i Bx-ABx\Vert\leq M\Vert B_ix-Bx\Vert +\Vert (A_i-A) Bx\Vert$ for some constant $M$; so $A_iB_ix\to ABx$ for all $x$. – Etienne Oct 4 '19 at 14:23