Compact operator and dual space

Question

Let $A:Q\to Q^{*}$ be a bounded linear operator where $Q$ is a Banach space with dual $Q^{*}$. I have shown that for a sequence $\{\varphi_{n}\}$ in $Q$ with $\varphi_{n}\rightharpoonup 0$ it follows that $$\|A\varphi_{n}\|_{Q^{*}}\to 0.$$ Can I conclude that $A$ is compact?

Answer 1

Without further assumptions on $Q$ you cannot conclude that $A$ is compact.

Consider the natural embedding $A: Q = \ell^1 \hookrightarrow Q^* = \ell^\infty$ given by $Ax = x$.

Recall that $Q = \ell^1$ has the schur property which means that weakly convergent sequences in $\ell^1$ are norm convergent. Since, for $x \in \ell^1$, $\|x\|_{\ell^\infty} \leq \|x\|_{\ell^1}$ this means that $A$ trivially satisfies your property. However $A$ is not compact since if $e^i$ is the element of $\ell^1$ with $1$ in the $i$-th place and $0$ elsewhere then $\{e^i\}_{i \geq 1}$ is a bounded set in $\ell^1$ but $\{Ae^i\}_{i \geq 1}$ has no convergent subsequence in $\ell^\infty$.

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If you additionally assume that $Q$ is reflexive then the result is true.

In this case, every bounded sequence in $Q$ has a weakly convergent subsequence. Hence if $\varphi_n$ is a bounded sequence then there is a subsequence $\varphi_{n_k}$ such that $\varphi_{n_k} \rightharpoonup \varphi$. By applying your assumption to the sequence $\varphi_{n_k} - \varphi$ we see that $A\varphi_{n_k} \to A\varphi$ in the norm topology on $Q^*$. Hence $A\varphi_n$ has a norm convergent subsequence so $A$ is compact.