Weak topology on an infinite-dimensional normed vector space is not metrizable
Let $X$ be a normed space. You can show that if the weak topology of $X$ admits a countable base of open sets at $0$, then $X$ is finite dimensional:
- Prove the existence of a countable set $\{\zeta_n\}$ in $X^*$ such that every $\zeta \in X^*$ is a finite linear combination of the $\zeta_n$.
- Derive from this that $X^*$ is finite dimensional.
- Deduce that $X$ is finite dimensional.
The original answer is correct, and here I only want to elaborate every steps in details (to earn some contribution points if possible :)) Hopefully they are correct.
We claim:
(1) $\text{dim} X^* \geq \text{dim} X$, and equality holds iff $\text{dim} X < \infty$:
This is a well known theorem, e.g. one possible proof is on pages 244-248 of Jacobson's {Lectures in Abstract Algebra: II. Linear Algebra.}
(2) Dual space of $X^*$ of normed vector space $X$ is Banach with norm $||\cdot ||_{X^*}$:
Let $\{T_n\}\subset X^*$ a Cauchy sequence. Then for each fixed $x$, the sequence $\{T_nx\}\subset \Phi$ is a Cauchy sequence, which converges by completeness to some element of $\Phi$ denoted $Tx$. The map $x\mapsto Tx$ is linear; we have to check that it is continuous and that $\lVert T_n-T\lVert_{X^*}\to 0$.
We get $n_0$ such that if $n,m\geq n_0$ then for each $x$ $\lVert T_nx-T_mx\rVert_\Phi\leq\lVert x\rVert $ and letting $m\to+\infty$ we obtain $\lVert T_nx-Tx\rVert_\Phi \leq\lVert x\rVert $ so $\lVert Tx\rVert\leq \lVert x\rVert+ \lVert T_{n_0}\rVert\lVert x\rVert$ and $T$ is continuous.
Fix $\varepsilon>0$. We can find $N$ such that if $n,m\geq N$ and $x\in E$ then $\lVert T_nx-T_mx\rVert_\Phi\leq \varepsilon\lVert x\rVert$. Letting $m\to \infty$, we get for $n\geq N$ and $x\in X$ that $\lVert T_nx-Tx\rVert_\Phi\leq \varepsilon\lVert x\rVert$, and taking the supremum over the $x\neq 0$ we get for $n\geq N$ that $\lVert T-T_n\rVert_{X^*}\leq \varepsilon$.
(3) Every proper subspace of a normed vector space has empty interior.
We only need to show that the only subspace of a normed vector space $X$ that has a non-empty interior, is $X$ itself. Suppose subspace $S$ has a nonempty interior. Then it contains some ball $B(x,r) = \{y : \|y-x\| < r\}$. Now the idea is that every point of $V$ can be translated and rescaled to put it inside the ball $B(x,r)$. Namely, if $z \in V$, then set $y = x + \frac{r}{2 \|z\|} z$, so that $y \in B(x,r) \subset S$. Since $S$ is a subspace, we have $z = \frac{2 \|z\|}{r} (y-x) \in S$. So $S=V$.
(4) Finite-dimensional subspace of normed vector space is closed.
Suppose we have a convergent sequence $\{x_n\}$ such that $||x_n-x||\rightarrow 0$ and thus it is a Cauchy sequence as $||x_n-x_m||\leq ||x_n-x||+||x_m-x||$ . Since $\forall n, x_n=\sum_{i=1}^{K} \alpha_{n,i} x'_i, \alpha_{n,i} \in \Phi$, we know $\forall i, \{\alpha_{n,i} \}_n$ forms a Cauchy sequence in $\Phi$ as well and thus converge to some $\alpha_i \in \Phi$. Clearly, $x= \sum_{i=1}^{K} \alpha_{n,i} x' _i$ is inside the subspace.
By claim 2 we know $(X^*,||\cdot||_{X^*})$ is a Banach space. Suppose the Banach space $(X^*,||\cdot||_{X^*})$ has a countable basis $\{v_n; n\in\mathbb N\}$. Let us denote $X^*_n=[v_1,\dots,v_n]$, the linear span by the first n basis, which is a subspace as $\forall x,y\in X^*_n, \alpha,\beta \in \Phi, $ we know $\alpha x + \beta y = \sum_{i=1}^{n } (\alpha \lambda_{x,i}+ \lambda_{y,i}) v_i \in X^*_n$ where $x=\sum_{i=1}^{n }\lambda_{x,i}v_i, y=\sum_{i=1}^{n }\lambda_{y,i}v_i$. Then we have:
- $X^*=\bigcup\limits_{n=1}^\infty X^*_n$
- $X^*_n$ is a finite-dimensional subspace of $X^*$, hence it is closed by claim 4.
- $X_n^*$ is a proper subspace of $X^*$, so it has empty interior by claim 3.
So we see that $ (\overline{X}^*_n)^\circ = X_n^\circ=\emptyset$, which means that $X^*_n$ is nowhere dense. So $X^*$ is a countable union of nowhere dense subsets and thus is of first category, which contradicts the Baire's category theorem. Therefore, we only need to show that $(X^*,||\cdot||_{X^*})$ has at most countable basis, which would then imply $\text{dim} X^* < \infty$ since otherwise we would have countable basis and it derives contradictions as shown above. Then by claim 1 we know $\text{dim} X < \infty$.
Now we show there exists a countable set $F\subset X^*$ such that $\forall f\in X^*$ is a (finite) linear combination of elements in $F$. Notice a collection of neighborhoods of the form $$B=\left\{x\in X: |f_i(x)|<r_i, f_i\in X^*, r_i>0, 1\leq i\leq K \right\} $$ forms local base of the weak topology, which means arbitrary neighborhood of zero contains some neighborhood of this form. Then suppose $\{A_\alpha \}_{\alpha\in\mathbb{N}}$ forms a countable local base of weak topology, we know $\forall A_\alpha, \exists B_\alpha = \left\{x\in X: |f^\alpha_i(x)|<r^\alpha_i, f^\alpha_i\in X^*, r^\alpha_i>0, 1\leq i\leq K^\alpha\right\} \subset A_\alpha$. We claim $F= \cup_{\alpha\in\mathbb{N}} \left\{f^\alpha_i, 1\leq i\leq K^\alpha \right\} $, which is countable and we need to show $\forall f\in X^*$, $f$ is a (finite) linear combination of elements in $F$. Notice $f$ is continuous and thus $\{x\in X, |f(x)|<r \}$ forms a zero neighnorhood and contains some local base $A_\alpha$. Notice $$B_\alpha= \left\{x\in X: |f^\alpha_i(x)|<r^\alpha_i, f^\alpha_i\in X^*, r^\alpha_i>0, 1\leq i\leq K^\alpha\right\} \subset A_\alpha$$ which means $\forall x\in B_\alpha, |f(x)|<r$. Now we denote $$B_\alpha^n = \left\{x\in X: |f^\alpha_i(x)|<\frac{r^\alpha_i}{n}, f^\alpha_i\in X^*, r^\alpha_i>0, 1\leq i\leq K^\alpha\right\}$$ Then clearly $B_\alpha^n \subset \frac{1}{n} B_\alpha \subset\frac{1}{n} A_\alpha \subset \{x\in X: |f(x)|<\frac{r}{n} \}$, since $\forall x\in B_\alpha^n, |f^\alpha_i(nx)|=n|f^\alpha_i(x)| < {r^\alpha_i}$ and $\forall x\in A_\alpha, |f(\frac{1}{n}x)|=\frac{1}{n}|f(x)| < \frac{r}{n}$ by linearity. Therefore, $\forall x\in B_\alpha^n, |f(x)| < \frac{r}{n}$. Let $n\rightarrow \infty,$ we know $$\forall x\in B^\infty_\alpha = \left\{x\in X: |f^\alpha_i(x)|=0, f^\alpha_i\in X^*, 1\leq i\leq K^\alpha\right\}, |f(x)|=0$$ which is to say $f_1(x)=\cdots = f_{K^\alpha}(x)=0\Rightarrow f(x)=0$. Then by results of lemma 3.9 in Rudin Functional Analysis, we know $f$ is the linear combination of $f_1(x), \cdots, f_{K^\alpha}(x)$. Done