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Oct 16, 2011 at 10:24 comment added Marc Palm Okay, excuse my false claim, I was overlooking that this holds for the subset $M^+(T)$ of positive Radon measure, and does not generalize to the complex linear span.
Oct 12, 2011 at 3:47 comment added Benjamin Hayes $X^{∗}$ in the weak∗ topology is a countable union of $\{\phi\in X^{*}:\|\phi\|\leq N\}$, which have empty weak∗ interior. Hence, if the weak∗ topology were metrizable, we get a contradiction to the Baire Category Theorem. Are you sure you don't mean the weak∗ topology on the state space of $C_{0}(X)?
Oct 12, 2011 at 3:42 comment added Benjamin Hayes I think $M(T)$ is not metrizable in the weak$^\ast$ topology, and in fact my claim that this fails for every infinite dimensional Banach space i also think is true. The rough outline of the proof I saw was this: 1. If $X^\ast$ is weak$^\ast$ metrizable, then a first countabliity at the origin argument implies that $X^\ast$ has a translation invariant metric given the weak$^\ast$ topology. 2. One can characterize completeness topologically for translation-invariant metrics, and see directly that if $X^\ast$ had a translation-invariant metric given the weak$^\ast$ topology it would be complete.
Oct 6, 2011 at 13:38 comment added Marc Palm Minor nitpick: Consider a locally compact Hausdorff space $T$. The $*$ topology on the dual of the $C^*$ algebra $C_0(T)$ is metrizable, if and only if $X$ is second countable. That is a theorem in Choquet's book on functional analysis. So your claim, that the first statement is never true in infinite dimensional situations, is false. Take e.g. $T$ being a circle.
May 4, 2011 at 15:00 history answered Benjamin Hayes CC BY-SA 3.0