# Weakly compact operator on $c_0$ is compact

Show that if $T\in {\cal B}(c_0)$ and $T$ is weakly compact, then $T$ is compact.

My attempt: $T$ is weakly compact, so there is a reflexive space $X$ , and operators $A\in {\cal B}(X,c_0)$ and $B \in {\cal B}(c_0, X)$ such that $T=AB$. By reflexivity of $X$, operators $A,B$ are weakly compact. To show $B$ is compact suppose $\{x_n\}$ is a bounded sequence in $c_0$. There is a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that the sequence $\{Bx_{n_i}\}$ is weakly convergent. suppose $Bx_{n_i}\to y$ (weakly). Using reflexivity of $X$, $Bx_{n_i}\to y$ (norm), which shows $B$ is compact. Furthermore $T$ is compact.

I think my attempt is wrong, because I know on reflexive spaces , every operator is weakly compact, and using my attempt, every weakly compact operator is compact, while ${\cal B}(H)$ is a counterexample for it. I do not know where it were wrong. Please help me. Thanks in advance.

• Here's one way to prove it: The adjoint of $T$ is weakly compact (by Gantmacher's Theorem). Wealkly convergent sequences in $\ell_1$ are norm convergent ($\ell_1$ has the Schur property); so it follows that $T^*$ is compact. But then $T$ is compact by Schauder's Theorem ($T$ is compact if and only if $T^*$ is). – David Mitra Sep 18 '14 at 9:42
• In your proof, how does weak convergence in a reflexive space imply norm convergence? – David Mitra Sep 18 '14 at 9:47
• I think I made a mistake. Thanks for your answer – niki Sep 18 '14 at 14:02
• You can replace $c_0$ by any $C(K)$ with $K$ scattered. – Tomek Kania Sep 19 '14 at 15:33