Let $X$ be a Banach space, $x_0 \in X$ and $\varphi_0 \in X^*.$ Define $T : X^* \longrightarrow X^*$ by $\ T(\psi) = \psi(x_0) \varphi_0,\ \psi \in X^*.$ Prove that $T$ is compact.
What I need to show is that $\overline {T(A)}$ is compact for every bounded subset $A \subseteq X^*$ i.e. for any sequence $\{y_n \}_{n \geq 1}$ in $\overline {T(A)}$ $\exists$ a subsequence $\{y_{n_k} \}_{k \geq 1}$ of the sequence $\{y_n \}_{n \geq 1}$ which is convergent. What I observe is that $T(A)$ is bounded because since $A$ was taken to be bounded there exists $M \gt 0$ such that $\|\psi\|_{\text {op}} \leq M, \ $ for every $\psi \in A.$ But then $\|T(\psi) \|_{\text {op}} \leq M \|x_0\| \|\varphi_0\|_{\text {op}},$ for every $\psi \in A,$ proving that $T(A)$ is bounded. Since closure of a bounded set is also bounded it follows that $\overline {T(A)}$ is bounded. Since $X$ is a Banach space so is the dual space $X^*$ and $\overline {T(A)},$ being a closed subset of the Banach space $X^*,$ is also a Banach space. If we can show that $\overline {T(A)}$ is totally bounded then we are through. Can anybody please help me in this regard?
Thanks in advance.
Any finite rank operator is compact.
In this case the range of $T$ is contained in the one dimensional subspace spanned by $\varphi_0$. Any closed and bounded set in a finite dimensional space is compact. So total boundedness is automatic from boundeness.