Brezis's proof of Theorem 3.28: the existence of bounded neighborhood in weak$^\star$ topology

Question

I'm reading the proof of Theorem 3.28. in Brezis's Functional Analysis.

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Theorem 3.28. Let $E$ be a Banach space. If $B_{E^{\star}}$ is metrizable in $\sigma\left(E^{\star}, E\right)$, then $E$ is separable.

Proof: Suppose $B_{E^{\star}}$ is metrizable in $\sigma\left(E^{\star}, E\right)$ and let us prove that $E$ is separable. Set $$ U_n=\left\{f \in B_{E^{\star}} ; d(f, 0)<1 / n\right\} $$

and let $V_n$ be a neighborhood of $0$ in $\sigma\left(E^{\star}, E\right)$ such that $V_n \subset U_n$. We may assume that $V_n$ has the form $$ V_n=\left\{f \in B_{E^{\star}} ; |\langle f, x\rangle|<\varepsilon_n \quad \forall x \in \Phi_n\right\} $$

with $\varepsilon_n>0$ and $\Phi_n$ is a finite subset of $E$. Set $$ D=\bigcup_{n=1}^{\infty} \Phi_n, $$

so that $D$ is countable. We claim that the vector space generated by $D$ is dense in $E$ (which implies that $E$ is separable). Indeed, suppose $f \in E^{\star}$ is such that $\langle f, x\rangle=0 \quad \forall x \in D$. It follows that $f \in V_n \quad \forall n$ and therefore $f \in U_n \quad \forall n$, so that $f=0$.

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My question: Previously, I asked a question about the role of $B_{E^\star}$ in above proof. Thanks to the answer there, I'm able to pin down which element of the proof that confuses me.

The gist of the proof is the existence of $V_n$ in the statement [let $V_n$ be a neighborhood of $0$ in $\sigma\left(E^{\star}, E\right)$ such that $V_n \subset U_n$]. So $V_n$ is a neighborhood of $0$ in the weak$^\star$ topology of $E^\star$, and $V_n$ is bounded in norm. However, this seems to contradict the following lemma, i.e.

Lemma: If $X$ is an infinite-dimensional t.v.s., then every weak$^*$ neighborhood of $E^*$ contains an infinite-dimensional affine subspace.

Could you elaborate on my confusion?

Answer 1

I think the confusion may arise because $V_n$ is not an open set in a tvs but rather an open set in a subset of a tvs with the subset topology. Let me elaborate further.

In the proof, one defines $U_n=\{f \in B_{E^\star}: d(f, 0)<\frac{1}{n}\}$ and this gives us open (bounded) sets in (the bounded subset, but not subspace) $B_{E^\star}$ with the subset topology relative to $E^\star$ with $\omega^\star$ topology.

This means that when we choose $V_n$ such that $V_n$ are (basic) open neighbourhoods of $0$ and $V_n\subseteq U_n$ what we are actually doing is the following:

1. Since $U_n$ is open in the subspace topology, $U_n=W_n \cap B_{E^\star}$ for $W_n$ an open set in $E^\star$ with $\omega^\star$ topology.
2. Since $W_n$ is an open set in $E^\star$ with $\omega^\star$ topology we can find $G_n$ a basic open set i.e. $G_n=\{f\in E^\star : |\langle f, x\rangle|<\epsilon \text{ for all } x\in \phi_n\}$ and $\phi_n$ a finite set of points such that $0\in G_n \subseteq W_n$.
3. Note that now if $V_n = G_n \cap B_{E^\star}$ then $V_n$ is open in $B_{E^\star}$ with the subspace topology and $0\in V_n \subseteq U_n$, with $V_n = \{f\in B_{E^\star} : |\langle f, x\rangle|<\epsilon \text{ for all } x\in \phi_n\}$.

Note then that the set that should contain in infinite dimensional affine space is $G_n$ (which indeed happens, as $G_n$ is open in $E^{\star}$ with the $\omega^\star$ topology) but not $V_n$, as this is just a subset of $G_n$.