Strongest topology makes unit ball compact Let $X$ be a Banach space and $X^*$ be its dual. Let $\mathbb{B}^*$ be the closed unit ball in $X^*$. The Banach-Alaoglu Theorem asserts that $\mathbb{B}^*$ is compact in the topology $\sigma(X^*, X)$. My question is whether there exists a topology $\tau$ on $X^*$, $\tau\ne\sigma(X^*,X)$ such that $\sigma(X^*,X)\subset\tau\subset\sigma(X^*, X^{**})$ or $\sigma(X^*,X)\subset\tau\subset\tau(\|.\|)$ and $\mathbb{B}^*$ is compact in the topology $\tau$.
Thank you for all kind help.
 A: There does not. This follows from the following well-known theorem.

Theorem. Let $\langle X,\tau\rangle$ be any compact Hausdorff space. Suppose that $\tau'$ and $\tau''$ are topologies on $X$ such that $\tau'\subsetneqq\tau\subsetneqq\tau''$. Then $\langle X,\tau'\rangle$ is compact but not Hausdorff, and $\langle X,\tau''\rangle$ is Hausdorff but not compact. That is, a compact Hausdorff topology is maximal compact and minimal Hausdorff.

The positive parts of the theorem are obvious. To see that $\langle X,\tau''\rangle$ is not compact, let $K\subseteq X$ be $\tau''$-closed but not $\tau$-closed. Compact subsets of Hausdorff spaces are closed, so $K$ is not $\tau$-compact. But then clearly $K$ is not $\tau''$-compact either, and since $K$ is $\tau''$-closed, it follows that $\langle X,\tau''\rangle$ is not compact.
To see that $\langle X,\tau'\rangle$ is not Hausdorff, let $K\subseteq X$ be $\tau$-closed but not $\tau'$-closed. $K$ is $\tau$-compact, so it’s $\tau'$-compact. If $\langle X,\tau'\rangle$ were Hausdorff, $K$ would therefore by $\tau'$-closed. Since it’s not, $\langle X,\tau'\rangle$ is not Hausdorff.
