It is well-known that if $V, W$ are Banach spaces, $T : V \to W$ a compact operator and $(x_n)$ a sequence in $V$ that converges for the weak topology towards some $x \in V$, then the sequence $(Tx_n)$ converges (strongly) to $Tx$ in $W$.
In order to prove that, one would observe that for all $\phi \in W'$, one has $T^*\{ \phi \} \in V'$, therefore $\phi (Tx_n) = T^* \{ \phi \} (x_n) \to \phi(Tx)$ and thus $(Tx_n)$ is weakly convergent in $W$. Thus any subsequence $(Tx_{n_k})$ that converges strongly in $W$ will necessarily (by uniqueness of the weak limit) converge to $Tx$.
Now, if $(\ell_n) \in V'$ is a sequence of continuous linear functionals that converges to some $\ell \in V'$ for the weak-* topology on $V'$, and $S : V' \to W$ is a compact operator, I was wondering whether this would be sufficient to entail the strong convergence of $(S\ell_n)$ in $W$. I don't think this would be the case, since one would fail to establish the weak convergence of $(S\ell_n)$ as mentioned above (one could only ensure that $\phi(S\ell_n) \to \phi(S\ell)$ if $T^* \{ \phi \}$ is an evaluation map, which need not be the case unless $V$ is reflexive).
Nonetheless, I could not manage to find a counterexample. I was thinking maybe one could be found in the case of $V = c_0$, the space of sequences converging to zero (thus $V' = \ell^1, V'' = \ell^\infty$), however I have not been able to find it. Any ideas?
Best Answer
This is not true even if $T$ has rank $1$. Consider the canonical sequence $(e_n)$ in $\ell^{1}=(c_0)^{*}$. This sequence converges to $0$ in weak* topology. Let $T(a_n)=\sum (-1)^{n} a_n$. This is bounded operator of rank $1$, hence compact. But $Te_n=(-1)^{n}$ which does not converge.