Closure of a set is compact in the bidual

Question

Let $F$ be a Banach space, $J:F\to F^{**}$ the canonical embedding. Let $B\subset F$ be such that $\overline{J(B)}$ is compact. Can I conclude that $\overline{B}$ is compact? I need this to show that $T^{*}:F^{*}\to E^{*}$ compact implies $T:E\to F$ compact. More specifically, $\overline{J(T(B_E))} = \overline{T^{**}(J(B_E))}$ compact implies $\overline{(T(B_E))}$ compact.

Answer 1

Every sequence $(x_n)$ in $B = J(B)$ has a convergent subsequence $(x_{n_k})$ that converges in norm to a $L\in F^{**}$, that is $$\lim_{k\to\infty} \|x_{n_k} - L\|\to 0.$$ Then, $(x_{n_k})$ is norm-Cauchy in $F$, so converges since $F$ is norm-complete. Thus, $L\in F$. Since $\overline{B}$ is norm-closed and $\{x_{n_k}:k\in\mathbb{N}\}\subseteq B\subseteq \overline{B}$, then $L\in\overline{B}$.

To summarize; every sequence in $B$ has a norm-convergent subsequence with a limit in $\overline{B}$. Thus, $\overline{B}$ is norm-compact.

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Note/Extra: The same argument doesn't hold for weak topology. Let $\overline{B}^{w}$ be the closure of $B$ in the weak topology of $F$ and $\overline{J(B)}^{w*}$ be the closure of $J(B)$ in the weak$^{*}$-topology of $F^{**}$. $\overline{J(B)}^{w*}$ can be weak$^{*}$-compact in $F^{**}$ while $\overline{B}^{w}$ is not weakly compact.