I'm reading section 3.6 Separable spaces in Brezis's Functional Analysis.
Separability properties are closely related to the metrizability of the weak topologies. Let us recall that a topological space $X$ is said to be metrizable if there is a metric on $X$ that induces the topology of $X$.
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Theorem 3.28. Let $E$ be a separable Banach space. Then $\color{red}{B_{E^{\star}}}$ is metrizable in the weak$^{\star}$ topology $\sigma\left(E^{\star}, E\right)$. Conversely, if $B_{E^{\star}}$ is metrizable in $\sigma\left(E^{\star}, E\right)$, then $E$ is separable.
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Theorem 3.29. Let $E$ be a Banach space such that ${E^{\star}}$ is separable. Then $\color{blue}{B_E}$ is metrizable in the weak topology $\sigma\left(E, E^{\star}\right)$. Conversely, if $B_E$ is metrizable in $\sigma\left(E, E^{\star}\right)$, then $E^{\star}$ is separable.
My understanding:
- A Banach space is separable if and only if its closed unit ball is separable.
- It seems the proof of the first direction of Theorem 3.28. generalizes without problem to […. $E$ be a separable Banach space. Then $\color{red}{{E^{\star}}}$ is metrizable …].
- I don't know if the generalization [… ${{B_{E^{\star}}}}$ is separable. Then $\color{blue}{E}$ is metrizable …] is true for Theorem 3.29.
To sum up, could you confirm if below statements are correct?
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S1: Let $E$ be a Banach space such that $B_E$ is separable. Then $\color{red}{E^{\star}}$ is metrizable in the weak$^{\star}$ topology $\sigma\left(E^{\star}, E\right)$. Conversely, if $B_{E^{\star}}$ is metrizable in $\sigma\left(E^{\star}, E\right)$, then $E$ is separable.
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S2: Let $E$ be a Banach space such that ${{B_{E^{\star}}}}$ is separable. Then $\color{blue}{E}$ is metrizable in the weak topology $\sigma\left(E, E^{\star}\right)$. Conversely, if $B_E$ is metrizable in $\sigma\left(E, E^{\star}\right)$, then $E^{\star}$ is separable.
Best Answer
As @DavidMitra mentioned in a comment, my claim is wrong. Indeed, the author of the book has a remark just right after the proof of Theorem 3.28, i.e.,