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Let $X$ be a reflexive and separable Banach space. Let $(h_n)$ be a sequence dense in $\overline{B^*_1}$ (the closed ball in $X^*$ of radius $1$). Set $$ d(x,y) := \sum_{n=1}^\infty 2^{-n} |(x-y, h_n)|, $$ where $(x-y, h_n)=(x-y, h_n)_{X,X^*}$ is the dual pairing on $X$. Then $d$ is symmetric, satisfies the triangle inequality and $d(x,y) = 0 \implies x = y$, hence $d$ is a metric.

On closed bounded (by the norm) balls of $X$, the metric $d$ induces the weak topology.

Can the topology induced by $d$ be described in a functional way in relation to the weak topology or the bounded weak-topology? Would any answer change if $d$ is defined using the ratio $\frac{|(x-y,h_n)|}{1 + |(x-y,h_n)|}$ ?

This seems to be something that should be in Megginson's book on Banach spaces, I can only see something related in exercise 2.86-88 of section 2.7.

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    $\begingroup$ For $E$ a Banach space, the bounded weak-$\ast$ topology on $E^*$ is not metrizable unless $E$ is finite-dimensional. The reason is that a base of neighbourhoods of $0$ in $E^*$ is defined by the polars of the norm-compact absolutely convex subsets of $E$. If the bounded weak-$\ast$ topology is first-countable, then this implies that there are countably many norm-compact absolutely convex sets covering $E$. By the Baire category theorem, this means one of them has nonempty interior, so $E$ is locally compact and therefore finite-dimensional. $\endgroup$
    – Robert Furber
    Commented Jul 23, 2020 at 15:29
  • $\begingroup$ I've not read the book, but I bet Megginson is using "complete" in the sense of uniform spaces, not metric spaces (because that makes it a true statement). If $E$ is a separable infinite-dimensional Banach space, then both the weak-$\ast$ and the bounded weak-$\ast$ topology restrict to the same, metrizable, topology on any bounded subset of $E^*$, but are not metrizable themselves. $\endgroup$
    – Robert Furber
    Commented Jul 23, 2020 at 15:38
  • $\begingroup$ @RobertFurber Megginson has in Definition 2.3.1 the following: "A topological space is topologically complete if some complete metric induces its topology." $\endgroup$
    – Jorge E. Cardona
    Commented Jul 23, 2020 at 17:17
  • $\begingroup$ @RobertFurber what exactly do you mean by "the polars of the norm-compact absolutely convex subsets of $E^\star$"? $\endgroup$
    – Jorge E. Cardona
    Commented Jul 23, 2020 at 17:22
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    $\begingroup$ Proving that the bounded weak-$\ast$ topology is not completely metrizable is easier, actually. By Banach-Alaoğlu the closed ball of radius $n$ in $E^*$ is compact, $E^*$ is the union of these closed balls, so by the Baire category theorem if $E^*$ is completely metrizable in the bounded weak-$\ast$ topology then one of these balls has nonempty interior, so $E^*$ is locally compact and therefore finite-dimensional. $\endgroup$
    – Robert Furber
    Commented Jul 23, 2020 at 20:32

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