Weak topology is induced by a dense subset

Question

Let $V$ be a normed space s.t. $V^*$ is separable, then the weak topology on $V$ is metrizable.

So the idea here is to pick a countable dense subset $D$ of $V^*$ and to say the initial topology of $D$ is the weak topology on V. Then we can take the standard metric for such problems: $d(v,w) = \sum_{f\in D}2^{-n} \frac{\lvert v-w\rvert _f}{1+\lvert v-w\rvert _f}$ for $\lvert v\rvert _f = \lvert f(v) \rvert$ the seminorm induced by $f$. And this works out nicely, since the family of seminorms in $D$ is sufficient as $V$ is normed.

Now how does one go about showing that $D$ induces the weak topology?

Lets recall the construction of the weak topology. We set $$\mathcal{O}_1 = \{f^{-1}(U) \mid f\in V^*, U\subseteq \mathbb{R} \text{ open}\}$$ $$\mathcal{O}_2 = \text{ all finite intersections of elements in } \mathcal{O}_1$$ then the weak topology is all unions of elements in $\mathcal{O}_2$. So it suffices to show that $$\mathcal{O}_1 = \tilde{\mathcal{O}_1}= \{f^{-1}(U) \mid f\in D, U\subseteq \mathbb{R} \text{ open}\}$$

Ok, lets try and do this, take $f\in V^*, U\subseteq \mathbb{R}$ open, arbitrary, and $(f_n)_n \subseteq D, f_n \rightarrow f$ (convergence in what topology?). Then $$f^{-1}(U) = \lim_{n\rightarrow \infty} f_n^{-1}(U)$$ and why is this in $\tilde{\mathcal{O}_1}$, and how do we interpret the limit of sets?

Answer 1

Let $V$ be a normed space such that $V^{*}$ is separable. Let $D = \{f_{n} : n\in\mathbb{N} \}$ be a countable dense subset of $V^{*}$, or even simply a countable subset which satisfies $\overline{{\rm span}} \, D = V^{*}$. Define $d\colon V\times V \to [0, \infty )$ by \begin{equation} d(x,y) = \sum_{n\in\mathbb{N}} 2^{-n} \frac{|f_{n}(x-y)| }{1 + |f_{n}(x - y)|}. \end{equation} Then $d$ is a metric on $V$. Furthermore, if $\tau_{d}$ denotes the topology on $V$ induced by the metric $d$ and if $\tau_{\text{weak}}$ denotes the weak topology on $V$, it can be shown that $\tau_{d} \subseteq \tau_{\text{weak}}$. To see this, fix $x_{0} \in V$ and $\varepsilon > 0$. For each $n\in\mathbb{N}$ define $\phi_{n} \colon V \to [0, \infty )$ by \begin{equation} \phi_{n} (x) := \sum_{k=1}^{n} 2^{-k} \frac{|f_{k}(x - x_{0})| }{1 + |f_{k}(x - x_{0})|}. \end{equation} Then $(\phi_{n})_{n\in\mathbb{N}}$ is a sequence of continuous functions on $V$ when equipped with the weak topology which converges uniformly to $\phi\colon V \to [0, \infty )$ defined by \begin{equation} \phi (x) := \sum_{n\in\mathbb{N}} 2^{-n} \frac{|f_{n}(x - x_{0})| }{1 + |f_{n}(x - x_{0})|}. \end{equation} As a result, $\phi$ is also a continuous function on $V$ with the weak topology. Consequently, the continuity of $\phi$ at $x_{0}$ gives some open neighbourhood $U$ of $x_{0}$ such that $x\in U$ implies $d(x, x_{0}) = \phi (x) < \varepsilon$. Hence the identity mapping of $(X, \text{weak})$ into $(X, \tau_{d})$ is continuous.

One may further hope that $\tau_{\text{weak}} \subseteq \tau_{d}$ and hence that $V$ with the weak topology is metrisable. However, the argument breaks down at this point. To see why, if we suppose that $\tau_{\text{weak}} \subseteq \tau_{d}$, then it follows from this result that $V$ is finite-dimensional. Consequently we have $\tau_{\text{weak}} \not\subseteq \tau_{d}$ whenever $V$ is infinite-dimensional.

However, the following result is true.

Let $V$ be a normed space such that $V^{*}$ is separable. Then the closed unit ball $B_{V}$ of $V$ with the induced weak topology is metrisable. Furthermore, if $D = \{f_{n} : n\in\mathbb{N} \}$ satisfies $\overline{{\rm span}} \, D = V^{*}$, then a metric $\rho$ on $B_{V}$ inducing the weak topology is given by \begin{equation} \rho (x,y) = \sum_{n\in\mathbb{N}} 2^{-n} \frac{|f_{n}(x-y)|}{1 + |f_{n}(x-y)|}. \end{equation}

One place to look for a proof of this result, presented in with a different metric, is Theorem 3.29 of "Functional Analysis, Sobolev Spaces and Partial Differential Equations" by Brezis. In that book it is described how to obtain a proof using methods which are more on the elementary side. Another place to look for a proof is Theorem V.5.2 of "Linear Operators, Part I" by Dunford and Schwartz. The proof there is essentially making use of the natural embedding into the second dual, which is a weak to weak$^{*}$ homeomorphism, combined with a corresponding metrisability result for bounded subsets of the weak$^{*}$ topology. Such an approach as used in the book by Dunford and Schwartz is also discussed in this post. It is also worth mentioning that all these sources are assuming the normed space is complete. However, this assumption is unnecessary and does not affect any of the proofs.