Let $X,Y,Z$ be Banach Spaces. Let $T_n,T \subset BL(X,Y), S_n,S \in BL(Y,Z)$. Show that
a) $S_n \to S $(weakly) and $T_n \to T$ (strongly) implies $S_nT_n \to ST$ (weakly)
b) $S_n \to S $(uniformly) and $T_n \to T$ (weakly) implies $S_nT_n \to ST$ (weakly)
c)$S_n \to S $(weakly) and $T_n \to T$ (uniformly) doesn't necessarily imply $S_nT_n \to ST$ (strongly)
d)$S_n \to S $(strongly) and $T_n \to T$ (weakly) doesn't necessarily imply that $S_nT_n \to ST$ (weakly)
My try:
a) Since I need to show that for all $f \in BL(Z,K), f(S_nT_nx) \to f(STx)$, I add $f\circ S_n(Tx)$ and subtract it to get $$|f(S_nT_nx) - f(STx)|\le ||f||||S_n||||T_nx-Tx||+||f\circ S_n(Tx)-f\circ S(Tx)||$$
For a fixed $x$, Let $E=\{S_nx: n \ge 1\}$. Now $f(E)$ is bounded for all $f\in BL(Z,K)$. B Resonance theorem $E$ is bounded. Hence by Uniform Boundedness Principle $\sup_{n}||S_n|| \lt \infty$. Hence the first thing goes to zero as $T_nx \to Tx $ (strongly) and the second thing too by the weak convergence of $S_n$ to $S$.
b) $$||(f\circ S_n)(T_n(x)-(fos)(Tx)||\le ||f||||S_n-S||||T_n(x)||+||(f\circ S)(T_n(x)-T(x))||$$
Again I will use Resonance Theorem to conclude that $T_n(x)$ is bounded for each $x$ and Uniform Boundedness Principle to conclude that $\sup_{n}||T_n|| \lt \infty$ and that would make the first term to go to $0$. The second term goes to $0$ because $f\circ S \in BL(Y,K)$ and weak convergence of $T_n$.
Is this alright??
I am unable to find counterexamples for (c) and (d).