Dual of weak topologies

Question

The following question is a follow up to an earlier question I asked (Isometric isomorphism and annihilators of annihilators). So, here is the question:

Let $(X,{\cal T}_{||\cdot||})$ be a normed space. And let ${\cal S} \subseteq (X,{\cal T}_{||\cdot||})^{*}$ be a subspace of its dual. For $\psi \in {\cal S}$, let $\rho_{\psi}(x \in X) = |\psi(x)|$, a subnorm.

Let ${\cal T}_{\cal S} \subseteq {\cal T}_{||\cdot||}$ be the minimal topology for which all $\{\rho_{\psi}, \psi \in {\cal S}\}$ are continuous.

${\bf Claim}$: $(X,{\cal T}_{\cal S})^{*} = {\cal S}$

For instance:

a) If $S = (X,{\cal T}_{||\cdot||})^{*}$, ${\cal T}_{\cal S} = \sigma(X;X^*)$ is the weak topology, and trivially $(X,\sigma(X,X^*))^* = (X,{\cal T}_{||\cdot||})^* = {\cal S}$

b) If $X = V^*$ with its standard topology, and $S = j(X) \subseteq V^{**}$, then ${\cal T}_{\cal S} = \sigma(X^*;X)$ is the weak-$^*$ topology, and $(V^*,\sigma(X^*;X))^* = j(X) = S$.

The latest claim in b) has been proved in this thread (What is the topological dual of a dual space with the weak* topology?), and it think it can be copied step-by-step to prove the more general ${\bf Claim}$ I made above.

So, is the ${\bf Claim}$ correct?

Answer 1

The claim is correct. Suppose that $f \colon X \to \mathbb R$ is a linear functional continuous with respect to $\mathcal T_S$. Then the set $f^{-1}((-1,1))$ is open in $\mathcal T_S$ and contains the origin. Hence, there exist $g_1,\dots,g_N \in S$ and $ε>0$ such that $$ \bigcap_{i=1}^N g_i^{-1}((-ε,ε))\subset f^{-1}((-1,1)). \tag{1} $$ In particular, $$ \bigcap_{i=1}^N \ker g_i \subset \ker f \tag{2}$$ and so by a linear algebra lemma $f \in \operatorname{span} \{g_1,\dots, g_N\} \subset S$.

To see why $(2)$ holds, suppose that $x \in \bigcap_i \ker g_i = Y$. Since $X$ is infinite dimensional, $Y\neq \{0\}$ (for if not, then the map $x \mapsto (g_1(x),\dots,g_N(x)) $ is injective from $X$ to $\mathbb R^N$) so that for any $n \in \mathbb N$ we have that $nx \in Y $ and thus, by $(1)$, $|f(nx)|<1$. This implies that $f(x)=0$.