Let $X$ be a normed $\mathbb K$-linear space . Then the weak$*$ topology is metrizable iff $X$ is finite dimensional.
We know that if X is finite dimensional then the weak topology on X is metrizable and also X is finite dimensional iff $X^*$ is finite dimensional.
From here can we approach for proof of the above theorem?
Please someone give hints.
thank you..
I think you need $X$ to be a Banach space. Then first use the Banach-Steinhaus theorem to show that $(X^*,\sigma^*)$ is sequentially complete. If it is metrizable (hence a Frechet space) the open mapping theorem implies that the weak$^*$ topology coincides with the topology induced by the dual norm. For the unit ball $B$ of $X$ you thus get finitely many $x_1,\ldots,x_n\in X$ such that $\{x_1,\ldots,x_n\}^\circ \subset B^\circ$. Then apply the theorem of bipolars together with the fact that the absolutely convex hull $\Gamma(E)$ of a finite set is closed to get $B \subseteq \Gamma(\{x_1,\ldots,x_n\})$.
EDIT. As mentioned in the comment, for the space $X$ of scalar sequences with only finitely many non-zero terms endowed with the $\ell^2$-norm, the weak$^*$-topology on $X^*=\ell^2$ is the topology of coordinate-wise convergence which is metrizable.
Hint: in infinite dimensions weakly open neighborhood of origin always contain a (maximal) vector subspace.
It has already been mentioned in other answers that the statement in the question is false. However, there is still a characterisation of when the dual of a normed space is metrisable in the weak$^{*}$ topology. In particular, note that any normed space with an infinite countable Hamel basis, such as $c_{00}$, the normed space consisting of sequences with finitely many non-zero entries equipped with any norm, is an infinite-dimensional normed space whose dual is metrisable in the weak$^{*}$ topology.
Theorem. Let $X$ be a normed space. The following are equivalent.
$({\rm i})$ The space $X^{*}$ with the weak$^{*}$ topology is metrisable.
$({\rm ii})$ The space $X^{*}$ with the weak$^{*}$ topology is first countable.
$({\rm iii})$ The space $X$ has a countable Hamel basis.
Furthermore, if any (and hence every) one of the conditions $({\rm i})$-$({\rm iii})$ is satisfied and $\{e_{n} : n\in\mathbb{N} \}$ is a subset of $X$ which spans $X$, then a metric $d$ on $X^{*}$ inducing the weak$^{*}$ topology on $X^{*}$ is given by \begin{equation} d(f,g) = \sum_{n\in \mathbb{N}} \min \{|f(e_{n}) - g(e_{n})|, 2^{-n} \}. \end{equation}
Proof. $({\rm i}) \Rightarrow ({\rm ii})$. This follows from recalling that all metrisable topologies are first countable.
$({\rm ii}) \Rightarrow ({\rm iii})$. Suppose the weak$^{*}$ topology on $X^{*}$ is first countable. Then there exists a sequence $(U_{n})_{n\in\mathbb{N}}$ of weak$^{*}$ open neighbourhoods of $0$ such that $\bigcap_{n\in\mathbb{N}} U_{n} = \{0\}$. For each $n\in\mathbb{N}$ there exists a finite set $F_{n} \subseteq X$ and some $\varepsilon_{n} > 0$ such that \begin{equation} \{ f\in X^{*} : |f(x)| < \varepsilon_{n} \; \text{for all} \; x\in F_{n} \} \subseteq U_{n}. \end{equation} Let $F = \bigcup_{n\in\mathbb{N}} F_{n}$. Then $F$ is countable. We claim ${\rm span} \, F = X$. Let $x_{0}\in X$. Then \begin{equation} \{ f\in X^{*} : |f(x_{0})| < 1 \} \tag{1} \end{equation} is weak$^{*}$ open in $X^{*}$. By scaling $(1)$, there exists $n\in\mathbb{N}$ and $\alpha > 0$ such that \begin{equation} \{f\in X^{*} : |f(x)| < \varepsilon_{n} \; \text{for all} \; x\in F_{n} \} \subseteq U_{n} \subseteq \{f\in X^{*} : |f(x_{0})| < \alpha \}. \tag{2} \end{equation} Let $J:X \to X^{**}$ be the natural embedding of $X$ into $X^{**}$. Then ($2$) shows that $Jx_{0}$ is bounded (as a function) on the vector subspace $\bigcap_{x\in F_{n}} {\rm ker} \; Jx$. It follows that $\bigcap_{x\in F_{n}} {\rm ker} \; Jx \subseteq {\rm ker} \; Jx_{0}$. By this result we have $Jx_{0} \in {\rm span} \, J(F_{n}) = J({\rm span} \, F_{n})$. Then since $J$ is injective we see $x_{0} \in {\rm span} \, F_{n} \subseteq {\rm span} \, F$ as desired. Because $X$ is contained in the span of a countable set, we see that $X$ has a countable Hamel basis.
$({\rm iii}) \Rightarrow ({\rm i})$. Suppose $X$ has a countable Hamel basis. Then there exists a set $\{e_{n} : n \in \mathbb{N} \}$ which spans $X$. Define $\phi : X^{*} \to \prod_{n\in\mathbb{N}} \mathbb{K}$ by \begin{equation} \phi (f) := (f(e_{n}))_{n\in\mathbb{N}}. \end{equation} If $(f_{\alpha})_{\alpha \in A}$ is a net in $X^{*}$ and $f\in X^{*}$, then by linearity $(f_{\alpha})_{\alpha \in A}$ converges to $f$ in the weak$^{*}$ topology if and only if $(f_{\alpha}(e_{n}))_{\alpha \in A}$ converges to $f(e_{n})$ for all $n\in\mathbb{N}$, which occurs if and only if $(\phi (f_{\alpha}))_{\alpha \in A}$ converges to $\phi (f)$ in the product topology. Hence $\phi$ is a homeomorphism onto its image.
Recall that $\prod_{n\in\mathbb{N}} \mathbb{K}$ equipped with the product topology is metrisable with a metric $\rho$ that is compatible with the product topology on $\prod_{n\in\mathbb{N}} \mathbb{K}$ being given by \begin{equation} \rho ((\alpha_{n})_{n\in\mathbb{N}}, (\beta_{n})_{n\in\mathbb{N}} ) = \sum_{n\in\mathbb{N}} \min\{ |\alpha_{n} - \beta_{n}|, 2^{-n} \}. \end{equation} We then compose the metric $\rho$ with the homeomorphism $\phi$ to obtain a metric $d$ on $X^{*}$ given by \begin{equation} d(f,g) = \rho (\phi (f), \phi (g)) = \sum_{n\in \mathbb{N}} \min \{|f(e_{n}) - g(e_{n})|, 2^{-n} \} \end{equation} which is compatible with the weak$^{*}$ topology on $X^{*}$. In particular, we have that $X^{*}$ with the weak$^{*}$ topology is metrisable.
