I'm doing Ex 3.24 in Brezis's book of Functional Analysis.

The purpose of this exercise is to sketch part of the proof of Theorem 3.29, i.e., if $E$ is a Banach space such that $B_{E}$ is metrizable with respect to $\sigma\left(E, E^{\star}\right)$, then $E^{\star}$ is separable. Let $d(x, y)$ be a metric on $B_{E}$ that induces on $B_{E}$ the same topology as $\sigma\left(E, E^{\star}\right)$. Set $$ U_{n}=\left\{x \in \color{blue}{B_{E}} ; d(x, 0)<\frac{1}{n}\right\} $$ Let $V_{n}$ be a neighborhood of $0$ for $\sigma\left(E, E^{\star}\right)$ such that $\color{blue}{V_{n} \subset U_{n}}$. We may assume that $V_{n}$ has the form $$ V_{n}=\left\{x \in \color{blue}{E} ;|\langle f, x\rangle|<\varepsilon_{n} \quad \forall f \in \Phi_{n}\right\} $$ with $\varepsilon_{n}>0$ and $\Phi_{n} \subset E^{\star}$ is some finite subset. Let $D=\cup_{n=1}^{\infty} \Phi_{n}$ and let $F$ denote the vector space generated by $D$. We claim that $F$ is dense in $E^{\star}$ with respect to the strong topology. Suppose, by contradiction, that $\overline{F} \neq E^{\star}$.

  1. Prove that there exist some $\xi \in E^{\star \star}$ and some $f_{0} \in E^{\star}$ such that $$ \left\langle\xi, f_{0}\right\rangle>1, \quad\langle\xi, f\rangle=0 \quad \forall f \in F, \quad \text{ and }\quad\|\xi\|=1 . $$
  2. Let $$ W=\left\{x \in B_{E} ;\left|\left\langle f_{0}, x\right\rangle\right|<\frac{1}{2}\right\} . $$ Prove that there is some integer $n_{0} \geq 1$ such that $V_{n_{0}} \subset W$.
  3. Prove that there exists $x_{1} \in B_{E}$ such that $$ \left\{\begin{array}{l} \left|\left\langle f, x_{1}\right\rangle-\langle\xi, f\rangle\right|<\varepsilon_{n_{0}} \quad \forall f \in \Phi_{n_{0}} \\ \left|\left\langle f_{0}, x_{1}\right\rangle-\left\langle\xi, f_{0}\right\rangle\right|<\frac{1}{2} \end{array}\right. $$
  4. Deduce that $x_{1} \in V_{n_{0}}$ and that $\left\langle f_{0}, x_{1}\right\rangle>\frac{1}{2}$.
  5. Conclude.

If $E$ is finite-dimensional the the weak topology coincides with the norm topology. Now consider the case $E$ is infinite dimensional. Then each weakly open set is unbounded, so the set $V_n$ defined by the author above is unbounded and thus $V_{n}$ can not be a subset of $U_{n}$. Hence I think it should be $$ V_{n}=\left\{x \in \color{blue}{B_E} ;|\langle f, x\rangle|<\varepsilon_{n} \quad \forall f \in \Phi_{n}\right\} $$

  • Could you confirm if my observation is correct?

  • I solve 3. as follows. It follows from $f_0 \notin \overline F$ that $f_0$ is linearly independent of any finite subset of $F$. This implies $\bigcap_{f\in \Phi_{n_0}} \ker f \not \subseteq \ker f_0$. This implies there is $0 \neq a \in \bigcap_{f\in \Phi_{n_0}} \ker f$ such that $a \notin \ker f_0$. The we can pick $t\in \mathbb R$ such that $x_1 := ta$ satisfies the requirement. Could you confirm if this argument is fine?

  • 2
    $\begingroup$ I think you are right, you must consider $x \in B_E$. $\endgroup$ Mar 6, 2022 at 9:08


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