Proving that composition is continuous in the strong operator topology when the domain is bounded Let $X$, $U$, $W$ be banach spaces and $L(X,U)$ be the set of bounded linear maps from $X$ to $U$ with the operator norm.
Then The composition of linear maps is a mapping from $L(X,U) \times L(U,W)$ to $L(X,W)$.
Now let $A$ and $B$ be bounded sets in each of $L(X,U)$ and $L(U,W)$. Then I have to show that the composition is a continuous mapping from $A \times B$ to $L(X,W)$ in the strong operator topology.
I know that the form of the subbasis for open sets in strong operator topology, which is in the wikipedia. However I cannot proceed how to show that the mapping is continuous. Especially I cannot figure out where the boundedness condition on the domain comes in....Could anyone please help me?
 A: In fact, we only need to assume that $B$ is bounded. I will prove that composition is continuous from $ L(X,U) \times B $ to $L(X,W)$ for $B$ bounded. I will denote the composition map by $$\Phi: L(X,U) \times B \to L(X,W).$$
Note that a basis for the strong operator topology on $L(Y,Z)$ is given by sets of the form
$$U((T_i)_{i=1}^n, (x_i)_{i=1}^n, \varepsilon) = \{T : \|Tx_i - T_i x_i\| < \varepsilon \text{ for } i = 1, \dots, n\}.$$
Let $O$ be an open set in $\operatorname{Ran}\Phi \subseteq L(X,W)$. We want to show that $\Phi^{-1}(O)$ is open in $L(X,U) \times B$ so pick $(S,T) \in \Phi^{-1}(O)$.
Since $TS = \Phi(S,T) \in O$ there is an open neighbourhood of $TS$, $O' \subseteq O$ of the form 
$$O' = \{T'S' : \|T'S' x_i - TSx_i\|< \varepsilon \text{ for } i = 1, \dots n\}.$$ Note here that since $B$ is bounded there exists a $C$ such that $\|T'\| \leq C$ for each $T' \in B$.
We then have for $(S',T') \in A \times L(U,W)$,
\begin{align}
\|T'S'x_i - TS x_i\| &\leq \|T'\| \|S' x_i - S x_i\| + \|(T' - T)S x_i\|
\\& \leq C \| S' x_i - S_ix_i\| + \|(T' - T)S x_i\|
\end{align}
This means that $$O'' = \{(S',T') \in A \times L(X,W): \|(T'-T)Sx_i\| < \frac{\varepsilon}{2} , \|S'x_i - Sx_i\| < \frac{\varepsilon}{2C} \text{ for } i = 1, \dots, n \}$$
is a product of basic open sets for the relevant strong operator topologies such that 
$\Phi(O'') \subseteq O' \subseteq O$. This means that $O''$ is an open neighbourhood of $(S,T)$ contained in $\Phi^{-1}(O)$ which proves that $\Phi^{-1}(O)$ is open and so $\Phi$ is continuous as desired.
