# Is this a typo in Brezis's Ex 3.24?

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?

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