This is the proof by Brezis on showing all compact operators have compact adjoints. I am a little bit lost in the last paragraph of the proof:
This is how I understood the last paragraph:
We know that $T^{**} \in K(E^{**}, F^{**})$. Therefore, we have $T^{**}(J_E(B_E)) \subseteq F^{**}$ to have compact closure, where $J_E: E \to E^{**}$ is the canonical injection such that $J_E(x)(f) = f(x)$ for all $f \in E^*$. Now we claim that $J_F(T(B_E)) = T^{**}(J_E(B_E))$, where $J_F$ is the canonical injection for the space $F$. To see this, let $g \in J_F(T(B_E)) \subseteq F^{**}$. Then for $f \in F^*$ and some $x \in B_E$, we have \begin{align*} \langle g, f \rangle &= \langle J_F(Tx), f \rangle \\ &= \langle f, Tx \rangle \\ &= \langle T^*f, x \rangle \\ &= \langle J_E x, T^*f \rangle \\ &= \langle T^{**}(J_E x), f \rangle \end{align*} where in the last line, we have $g \in T^{**}(J_E(B_E))$. The proof for the converse is similar. Now we know $J_F(F) \subseteq F^{**}$ is closed in $F^{**}$ as $J_F$ is continuous and $F$ is closed. Therefore, we now have:
$\overline{J_F(T(B_E))} = \overline{T^{**}(J_E(B_E))} \subseteq F^{**}$ is compact.
$J_F(F) \subseteq F^{**}$ is closed.
I am not seeing why 1 and 2 here would imply $\overline{T(B_E)} \subseteq F$ is compact. In particular, it seems like we do not have $F$ to be reflexive and hence $J_F$ does not necessarily have a continuous inverse, which would have mapped compact sets to compact sets and closed sets to closed sets. How do we reach the conclusion that Brezis concluded in the end?
Update After Reading the Comment:
$J_F: F \to F^{**}$ is an isometric injection implies that $J_F|_{J(F)}: F \to J(F)$ is an isometric isomorphism. Therefore, we have $$ J_F|_{J(F)}^{-1}(\overline{J_F(T(B_E))}) = \overline{T(B_E)} \subseteq J_F|_{J(F)}^{-1}(J_F(F)) = F. $$ Since $J_F|_{J(F)}^{-1}$ is continuous and $\overline{J_F(T(B_E))}$ is compact, we are done.
Second Update: We can also view the problem as such:
If we know $J_F \circ T: E \to F^{**}$ to be compact, then $T: E \to F$ is necessarily compact. Now since $J_F \circ T$ is compact, we have $\overline{J_F(T(B_E))} \subseteq F^{**}$ is compact. Now let $\{ Tx_n \}_n \subseteq T(B_E)$ be a sequence with $\| x_n \| \leq 1$. We show that $\{ Tx_n \}_n$ has a convergent subsequence. Note that we have for the sequence $\{ J_F(T(x_n)) \}_n$ to have a convergent subsequence $\{ J_F(T(x_{n_k})) \}_k$ as $J_F \circ T$ is compact. In particular, this implies that $\{ J_F(T(x_{n_k})) \}_k$ is Cauchy and hence $\{ T(x_{n_k}) \}_k$ is Cauchy. Since $F$ is a Banach space and $T(B_E)$ is closed, we have $\{ T(x_{n_k}) \}_k$ to be a convergent subsequence in $T(B_E)$. This shows that $\overline{T(B_E)} \subseteq F$ is compact. That is, we have $T$ is compact.
I am still, however, not seeing where the condition $J_F(F) \subseteq F^{**}$ is closed being used. I suppose that I did use $T(B_E)$ is closed in the above. Is this what Brezis meant though?