Let $X,Y$ be Banach spaces and let $u: X \to Y$ be a linear operator. Let $u^\ast: Y^\ast \to X^\ast$ denote its transpose and assume that $u^\ast$ is compact. I am trying to prove that $u$ is compact. Could somebody please help me by checking my proof?
My proof so far:
Let $u^\ast$ be compact. It is a known theorem that if $u$ is compact then $u^\ast$ is compact. We apply this theorem to $u^\ast$ to get that $u^{\ast \ast}$ is compact.
Next note that the map $\varphi: X \to X^{\ast\ast}$ defined by $x \mapsto e_x$ where $e_x$ is the evaluation map $f \mapsto f(x)$ is injective. Similarly, $\psi : Y \to Y^{\ast \ast}$ is injective. The maps $\varphi : X \to \varphi (X)$ and $\psi : Y \to \psi (Y)$ therefore define bijections. It is easy to verify that they are linear homomorphisms. We can therefore identify $X$ with $\varphi (X)$ and $Y$ with $\psi (Y)$. Also, $u = \psi \circ u^{\ast \ast}\circ \varphi$. Therefore $u$ is compact if and only if $u^{\ast \ast }$ is which was what we wanted to show.
Would somebody please help me by showing me how to write whatever I am missing? I can't do it.
$u=\psi\circ u^{\star\star}\circ \phi$ is not correct because $\psi : Y\rightarrow Y^{\star\star}$. Try writing $\psi\circ u=u^{\star\star}\circ\phi$.
Assume $u^{\star}$ is compact, and show $u^{\star}$ is compact:
To do this, assume $\{ x_{n}\} \subset X$ is a bounded sequence, and show that there is subsequence $\{ x_{n_{k}}\}$ such that $\{ u(x_{n_{k}})\}$ converges. So assume $\{ x_{n}\}$ is a bounded sequence in $X$. $u^{\star\star}$ is compact because $u^{\star}$ is compact. Then there is a convergent subsequence of $\{ u^{\star\star}\circ\phi(x_{n})\}$ because $\{ \phi(x_{n})\}$ is a bounded sequence and $u^{\star\star}$ is compact. So, some $\{ u^{\star\star}\circ\phi(x_{n_{k}})\}$ converges, which means this sequence is a Cauchy sequence, which guarantees that $\{\psi\circ u(x_{n_{k}})\}$ is a Cauchy sequence. But $\psi$ is isometric. So $\{ u(x_{n_{k}})\}$ is a Cauchy sequence and, therefore, converges because $Y$ is complete.