# Is weak-* the strongest topology s.t. the unit ball is compact?

Let $B$ be a Banach space, $B^*$ be the dual of $B$ and $\mathcal B$ be the closed unit ball in $B^*$.

By the Banach-Alaoglu theorem, $\mathcal B$ is closed in the weak-* topology. My question is whether in general the weak-* topology on $B^*$ is the strongest topology such that this assertion holds?

Thank you

• Following the nice answer of @xyzzyz, I would restrict the question to topologies such that $B^*$ is a TVS. – the_lar Apr 1 '13 at 17:47
• The following facts may be relevant. The strongest linear topology on $E'$ which agrees with the weak star topology on the ball is locally convex and is, in fact, the topology of uniform convergence on the compacta of $E$. It coincides with the finest topology which agrees with the weak star topology on multiples of the unit ball. This is the essential content of the Banach-Dieudonné theorem. – jbc Apr 1 '13 at 17:57

For instance, introduce a new topology on $B^*$: a subset $U \subset B^*$ is open if and only if there's some set $V$ open in the weak-* topology such that $U \cap \mathcal{B} = V \cap \mathcal{B}$. This condition ensures that the induced topology on $\mathcal{B}$ is the same as topology induced from weak-* topology. Now, since these topologies are the same, $\mathcal{B}$ is compact in this new topology. The complement of the $\mathcal{B}$ in $B^*$ has a discrete topology, though.
• I like your answer, and indeed I felt that my question was a bit vague. But indeed, I was thinking of a topology that would at least make $B^*$ into a topological vector space. It would be even nicer if it was some "known" or otherwise nice topology. – the_lar Apr 1 '13 at 17:44