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Given a normed vector space $E$, its dual space $E^*$ has a standard norm itself. However, we can define the weak$^*$ topology $\sigma(E^*,E)$ over $E^*$ to be the minimal one in which the evaluation (linear) maps are continuous. Here, for a vector $x \in E$, its evaluation map $\mathrm{ev}_x : E^*\to E$ is computed by $\mathrm{ev}_x(f) = f(x)$. If $\overline{E}$ denotes the completion of $E$, their dual spaces are the same: the correspondence $f \mapsto f|_{E}$ is indeed a surjective linear isometry from $\overline{E}^*$ to $E^*$. It seems well-known that the corresponding weak$^*$ topologies $\sigma(E^*,E)$ and $\sigma(E^*,\overline{E})$ are the same as well. This is, for instance, also pointed out in a corollary of this answer and suggested by this one. However, I have failed while trying to prove this equality by myself. In all my attempts, it is not clear why the topology $\sigma(E^*,E)$ is the smaller one to guarantees the continuity of all evaluation maps at points of $\overline{E}$, once $\overline{E}$ has strictly more vectors than $E$ (although it contains $E$ densely). Am I missing some key property of general topology or functional analysis?

lucasreal
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    The topologies $\sigma(E^\ast, E)$ and $\sigma(E^\ast, \overline{E})$ are *not* the same, since the former has its space of continuous linear functionals being $E$ while the latter has its space of continuous linear functionals being $\overline{E}$. The answers you linked to are simply false. (In fact, if $E$ is countably infinite-dimensional, then the topology $\sigma(E^\ast, E)$ on $E^\ast$ is first-countable.) Though they are “close” to be being the same, in that the two topologies coincide if you restrict to norm bounded subsets of $E^\ast$, which is probably what caused the confusion. – David Gao Aug 14 '24 at 15:08
  • @DavidGao Thanks for your answer, I will think about that. However, I am now wondering how I can conclude that $\sigma(E^*,E)$ is not a metric topology even if $E$ is countably infinite-dimensional. Most arguments that I am aware of (especially those I have just linked) claim that it is enough to show this for Banach spaces. – lucasreal Aug 14 '24 at 17:19
  • You cannot conclude that because it is false. $\sigma(E^\ast, E)$ is metrizable if $E$ is countably infinite dimensional. – David Gao Aug 14 '24 at 18:48
  • @DavidGao, I am only aware that, for separable spaces, $\sigma(E^,E)$ is metrizable over the unit ball of $E^$. However, Brézis's Functional Analysis book remarks that the weak and weak$^*$ topologies are never metrizable in infinite-dimensional normed spaces, as quoted by this answer for example. Unfortunately, I only know how to prove this for Banach spaces. – lucasreal Aug 14 '24 at 20:05
  • I’m not sure the exact context here since I don’t have access to the book, but the claim is true for weak topology, since the dual space is always Banach and an infinite-dimensional Banach space has Hamel dimension at least continuum. Then you can apply the argument in the first answer you linked to in your post. (It’s only the Corollary part of that answer that’s false. Everything else is correct.) For weak$^\ast$ topology, it’s usually only applied to Banach spaces. If $E$ is infinite-dimensional Banach, then $\sigma(E^\ast, E)$ is not metrizable, per the same reason. If this is all the… – David Gao Aug 15 '24 at 00:44
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    … book claims, then it is certainly all correct. The question you linked to in your comment seems to be only talking about Banach spaces as well. If the book claims $\sigma(E^\ast, E)$ is never metrizable for all infinite-dimensional normed $E$, then it is definitely false. Again, the entirety of $E^\ast$ under $\sigma(E^\ast, E)$ is metrizable if $E$ is countably infinite dimensional. The proof is just noting that the topology is generated by countably many seminorms, coming from a countable basis of $E$. (I can write this down in more details as an answer if you want.) – David Gao Aug 15 '24 at 00:48
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    For a concrete example you can look at $E=c_{00}$ with the $\ell_2$ norm. Then $E^*=\ell_2$ and the weak star topology is the topology of pointwise convergence on $\ell_2$ which is metrizable. – Evangelopoulos Foivos Aug 15 '24 at 05:18

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