I read that if $A:E\to E$ is a bounded linear operator where $E$ is a Banach space and $A^{\ast}$ is a compact operator, then $A$ is a compact operator.
I know that the converse is true (th. 4 here), but I cannot prove the lemma above. I have tried to apply Arzelà's theorem, similarly to how it is done in the quoted th. 4, but I cannot get anything...
Thank you so much for any help!!!
Let $J\colon E\hookrightarrow E^{\ast\ast}$ be the canonical embedding. For $x\in E$ and $\lambda \in E^\ast$, we have
$$A^{\ast\ast}(Jx)(\lambda) = (Jx \circ A^\ast)(\lambda) = Jx(A^\ast\lambda) = (A^\ast\lambda)(x) = (\lambda\circ A)(x) = \lambda(Ax) = J(Ax)(\lambda),$$
so $A^{\ast\ast}\circ J = J\circ A$. Looking at the commutative diagram
$$\begin{matrix}\;\;\; E & \xrightarrow{A} &\mspace{-11mu} E\\ J\downarrow & & \downarrow J \\ \;\;\;\;E^{\ast\ast} & \xrightarrow[A^{\ast\ast}]{} & \mspace{-3mu} E^{\ast\ast} \end{matrix}$$
and keeping in mind that $J(E)$ is a closed subspace of $E^{\ast\ast}$, and of course the compactness of $A^{\ast\ast}$, can you see that there is a compact subset $K$ of $E$ such that $A(B_E) \subset K$, where $B_X$ denotes the (closed) unit ball of the space $X$?