Let $(X,\|\cdot\|)$ be a normed space and the weak topology $\tau_{w}$ has a countable base at 0, hence metrisable. I want to prove that this implies $X$ is finite.

I have already shown the existence of a countable set $U\subset X^{\ast}$ s.t. each $\Lambda\in X^{\ast}$ is a finite linear combination of elements of $U$. That is, $\Lambda(x)=\sum_{i=1}^{k}\lambda_{i}\Lambda_{i}(x)$, with $\lambda_{i}\in\mathbb{F}$, the underlying field.

Now, one idea that has been hinted is to use Baire's theorem to show that $X^{\ast}$ is of finite dimension and then to conclude from there. But I don't see how one would go about using Baire's theorem since we would first have to show that $(X^{\ast},\tau_{w})$ is complete and I don't immediately see that.

Certainly, we can say that since $\tau_{w}$ is metrisable $\exists$ and invariant $d$ on $X^{\ast}$. Then we can let $(\Lambda_{n})_{n\in\mathbb{N}}$ be Cauchy. However, I'm not sure that there is enough information to prove that for all $\epsilon>0$ there exists $N\in\mathbb{N}$ such that $d(\Lambda_{n},\Lambda)<\epsilon$ whenever $n>N$, i.e. that $X^{\ast}$ is complete.

On the other hand, we can also prove that $X^{\ast}$ is finite if the closed unit ball in $X^{\ast}$ is weak-compact. Certainly, we know that the closed unit ball $B^{\ast}_{1}:=\{\Lambda\in X^{\ast}:\|\Lambda\|_{X^{\ast}}\le 1\}$ is weak$^{\ast}$-compact, courtesy of the Banach-Alaoglu theorem. But obviously this does not equate to weak-compactness.

My other idea was to use the set $U=\{\Lambda(x),\Lambda_{1}(x),...,\Lambda_{k}(x)\}$ in another way. Since $\Lambda_{1}(x)=...=\Lambda_{k}(x)=0\implies\Lambda(x)=0$ we can write

$$\begin{aligned}\mathscr{N}&=\{x\in X:\Lambda_{1}(x)=...=\Lambda_{k}(x)=0\} \\ &=\{x\in X:\Lambda(x)=\Lambda_{1}(x)=...=\Lambda_{k}(x)=0\} \end{aligned}$$

And if

$$ V:=\{x\in X:|f_{i}(x)|<r_{i}, \text{ for }1\le i\le n\text{ and }r_{i}>0\}$$

is a weak neighbourhood of $0$, then, since $x\mapsto(\Lambda(x),\Lambda_{1}(x),...,\Lambda_{k}(x))$ maps $X$ into $\mathbb{F}$ with null space $\mathscr{N}$, it follows that $\dim X\le n+\dim\mathscr{N}$. Moreover, $\mathscr{N}\subset V$, but I still don't know how to proceed from here.


Here is an idea: $X$ is a normed space so $X^{*}$ is complete in the operator norm. Since there is a metric $d$ on $X$, such that the induced topology coincides with the weak topology, we may define balls $B_k=\left \{ x:d(x,0)<1/k \right \}$, and then of course, there is a finite collection of functionals $F_k=\left \{ \Lambda_1,\cdots ,\Lambda _{n_{k}} \right \}$ such that $U_k=\left \{ x:\vert \Lambda_i x\vert <\epsilon_k;1\le i\le {n_{k}} \right \}\subseteq B_k$.

Let $F=\bigcup_{k\ge1}F_k$. Then $F$ is countable. I don't think it's too hard to show that $F$ spans $X^{\star}$, and then since $X^{* }$ is complete in the operator norm, Baire says that it is finite dimensional. But then, so is $X^{** }$ and since $X$ injects into $X^{** }$, you'll have the result.

To show that $F$ spans $X^{*}$, let $f$ be any functional and now, using the fact that $N=\left \{ x:\vert \Lambda x\vert <1 \right \}$ is a weak neighbrhood of $0$, find a $k$ large enough so that $B_k\subset N$, and finally letting $x\in \bigcap_{\Lambda \in F_k}\ker \Lambda $,show that $x\in \ker f$.

  • $\begingroup$ To show that $F$ spans $X^{\ast}$, doesn't that just follow from the fact that each $f\in X^{\ast}$ is a finite linear combination of elements of $F$? And also, how exactly does it follow fom Baire's theorem that $X^{\ast}$ is finite dimensional? $\endgroup$ – Jason Born Oct 15 '16 at 11:19
  • $\begingroup$ Yes it does but you have to prove it, which amounts to showing that $f$ shares its kernel with all members of $F$. To use Baire, you can argue as follows: consider the subspaces $X_k^{*}=[F_1,\cdots, F_k]$. Then $X^{*}=\bigcup_kX_k^{*}$ and each of the sets in the union is closed,with empty interior. $\endgroup$ – Matematleta Oct 15 '16 at 15:14

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