# Dual space $E'$ is metrizable iff $E$ has a countable basis

I saw that it was already asked, but the book where I'm studying is slightly different. Recall some definition, if $E$ it's $\mathbb{K}$-vector space and let $\mathcal{E}$ be a vector subspace of the algebraic dual of $E$, wich is the vector space of all linear forms on $E$. We say that $(E,\mathcal{E})$ is a dual couple if $\mathcal{E}$ separates points of $E$, that is, $u(x)=u(y)$ $\forall u \in \mathcal{E}$ implies $x=y$. We have the following result:

1. "Suppose $(E,\mathcal{E})$ is a dual couple , $u_1,...,u_n , u \in \mathcal{E}$. Then $u=\sum_{k=1}^n \lambda_k u_k$ iff $u_1(x)=...=u_n(x)=0$ implies $u(x)=0$"

Exercise (a): Let $E$ be an infinite-dimensional locally convex space. Prove that the weak* topology $\sigma(E',E)$ on $E'$ is metrizable if and only if $E$ has a countable algebraic basis.

Hint (a): Let $U_n=\lbrace \max\lbrace p_{x_1},...,p_{x_N} \rbrace < 1/M \rbrace$. If $x \in E$, every ball $\lbrace p_x < \epsilon \rbrace$ contains some $U_n$; that is, $|u(x_j| < 1/n$ $\forall j$ implies $|u(x)| <1$. Hence $u(x_1)= \cdot \cdot \cdot u(x_N)=0$ implies $u(x)=0$ and $x \in \mathrm{span}(x_1,...,x_n)$.

Proof (Exercise (a)). We assume that the weak* topology $\sigma(E',E)$ on $E'$ it's metrizable. Let $x_1,...,x_N \in E \setminus \lbrace 0 \rbrace$. Note that the weak* topology is defined by seminorm family $\mathcal{F}= \lbrace p_x(u)=|u(x)| : x \in E \rbrace$, where \begin{align*} \displaystyle U_n := \lbrace u \in E' : \max \lbrace p_{x_1}(u),...,p_{x_N}(u) \rbrace < 1/n \rbrace \end{align*} It is an open neighborhood in weak* topology. By hypothesis $\forall x \in E$, $\exists n \in \mathbb{N}$ and exists open ball $B_{E'}:=\lbrace u \in E' : p_x(u):=|u(x)| < \epsilon \rbrace$ such that $U_n \subset B_{E'}$. Now, if $u \in U_n$ we have $|u(x_j)| < 1/n$ $\forall j$ and then $|u(x)|< 1$ with $\epsilon=1$, and by (1) we have: \begin{align*} u(x_1)=...=u(x_N)=0 \Longrightarrow 0=u(x)=\sum_{j=1}^n \lambda_j u(x_j) \end{align*} by linearity $x \in \mathrm{span}(x_1,...,x_n)$.
Reciprocally, just reverse the above implications, using results in (1)?

Someone can help me? can you check if proof of exercise (a) is correct? Thank you.

• There are few things I do not understand: Why do you get $|u(x)|<1$ and not $|u(x)| \leq \max_{1\leq j\leq N} |\lambda_j$? I also do not see why the hypothesis implies $U_n \subset B_{E'}$. I also do not really understand the formulation of the hint, where in the definition of $U_n$, $n$ does not appear on the right-hand-side. – Christian Aug 26 '16 at 14:38
• Hi @Christian thanks for the reply, the exercise suggestion says exactly so as. I have tried to carry out the exercise based on this, but I find it difficult to understand. However, the book is "Linear Functional Analysis" by J. Cerdà. – user288972 Aug 26 '16 at 14:45
• Perhaps the hypothesis that $U_n \subset B_{E'}$ it's true since weak* topology is waekest topology on $E'$ that make continuous each linear functional? – user288972 Aug 26 '16 at 14:47
• Okay, so $x_1,\ldots, x_N$ are not chosen in advance but using that the family $\{\{\max\{p_{x_1},\ldots, p_{x_N}\}<1/M \colon x_1,\ldots, x_N\in E, M\in\mathbb{N}\}$ is a fundamental system of neighbourhoods of zero of the weak$^*$ topology? – Christian Aug 26 '16 at 14:54
• @Christian yes it's correct – user288972 Aug 26 '16 at 14:57

Assume that the weak$^*$-topology on $E'$ is metrizable, then there is a countable basis $\{V_n\}$ of neighbournoods of zero. Each of these contains a set of the form $$U_n:=\{u\in E'\colon \max\{p_{x_{n,1}}, \ldots, p_{x_{n,\varepsilon_n}}\} < 1/M_n\},$$ where $N_n,M_n\in\mathbb{N}$, since the latter is a basis of neighbourhoods of zero for the weak$^*$-topology. So for every $x\in E$ and every $\varepsilon>0$, the set $$\{u\in E'\colon |u(x)|<\varepsilon\}$$ contains a set $U_n$. Hence $u(x_{n,1})=\ldots u(x_{n,N_n})=0$ implies $u(x)=0$ and therefore $x\in\operatorname{span}\{x_{n,1},\ldots,x_{n,N_n}\}$. Therefore $$\bigcup_{n\in\mathbb{N}}\{x_{n,1},\ldots,x_{n,N_n}\}$$ is a countable basis of $E$.
To prove the other implication, let $\{x_n\}$ be a countable basis of $E$ and set $$U_{m,n} := \left\{u\in E'\colon \max_{1\leq i\leq n} |u(x_i)| < \frac{1}{m}\right\}.$$ We have to show that $\{U_{m,n}\colon m,n\in\mathbb{N}\}$ is a basis of neighbourhoods of zero. Let $x\in E$, then there is an $N\in\mathbb{N}$ such that $x=\sum_{i=1}^{N} \lambda_i x_i$ for some $\lambda_i$. We get $$|u(x)| \leq \sum_{i=1}^{N} |\lambda_i| |u(x_i)| \leq C \max_{1\leq i \leq N} |u(x_i)|,$$ where $C:=\sum_{i=1}^{N} |\lambda_i|$. Hence every set of the form $$\{u\in E'\colon |u(x)| < \varepsilon\}$$ and therefore also finite intersections of these, contains a set of the form $U_{n,m}$.
• Where did you use the local convexity of $E$? – Damian Sobota Feb 10 at 15:09
• @DamianSobota: I think I did not. The topology on $E$ is only used for the definition of $E'$ but this should work for every dual pair $(E,F)$. – Christian Feb 10 at 15:39
• Exactly. And that's what makes me confused. If we consider $E=L_p([0,1])$ for $0<p<1$, then $E'=\{0\}$, so $E'$ is weakly* metrizable, but the dimension of $E$ is uncountable. – Damian Sobota Feb 10 at 15:46
• @DamianSobota: We know that the linear functional $\langle x,\cdot\rangle$ is in the span of $\langle x_{n,1},\cdot\rangle, \ldots,\langle x_{n,N_n},\cdot\rangle$ (for $E=L_p[0,1]$ this also true since all of them are the zero functional), i.e. $\langle x, \cdot\rangle=\sum\lambda_i \langle x_{n,i}, \cdot\rangle$ and $\langle x-\sum\lambda_ix_{n,i},u\rangle=0$ for all $u\in E'$. If $(E,E')$ separates points, we may conclude that $x=\sum\lambda_ix_{n,i}$. – Christian Feb 10 at 16:28