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Jan 11, 2021 at 19:07 vote accept Ruy
Jan 11, 2021 at 8:36 comment added Robert Furber Side note: These questions can bump into set-theoretic undecidability. Consider $\ell^\infty(\mathbb{R})$ (the bounded, not required to be measurable, functions $\mathbb{R} \rightarrow \mathbb{R}$) as the dual space of $\ell^1(\mathbb{R})$. There is a sequentially weak-* continuous functional that is not weak-* continuous (i.e. a point evaluation) if and only if there is an atomless real-valued measurable cardinal (i.e. $(\mathbb{R}, \mathcal{P}(\mathbb{R}))$ admits a probability measure vanishing on singletons).
Jan 11, 2021 at 7:57 comment added Robert Furber Related: mathoverflow.net/q/284276/61785
Jan 11, 2021 at 6:35 answer added Nate Eldredge timeline score: 7
Jan 10, 2021 at 23:59 comment added Ruy @Mikael, here is what I was able to do based on your comment: math.stackexchange.com/questions/3977058/… Thanks again!
Jan 10, 2021 at 20:03 comment added Mikael de la Salle Indeed, I had missed this Corollary. Note however that separability of $E$ is assumed, and used essentially in the proof.
Jan 10, 2021 at 19:58 comment added Ruy @Mikael, thanks for pointing it out. In fact the following Corollary in Conway's book, namely V.12.8, is exactly what I am asking!
Jan 10, 2021 at 19:42 comment added Mikael de la Salle This is at least true if $E$ is separable. Indeed, in that case, by the Krein-Smulian theorem, a convex subset of $E'$ is $\sigma(E',E)$-closed if and only if it is $\sigma(E',E)$ sequentially closed. See Corollary V.12.7 in Conway's course in functional analysis. Apply this to the kernel of $\varphi$.
Jan 10, 2021 at 18:09 history asked Ruy CC BY-SA 4.0