# Topological dual of $(X,\sigma(X,\Phi))$

Let $X$ be a normed vector space and $\Phi$ a vector subspace of $X^*$. I would like to prove that if a linear functional $\phi$ on $X$ is continuous for the topology $\sigma(X,\Phi)$ (that is the initial topology for $\Phi$) then $\phi \in \Phi$.

If $\phi$ is such a linear functional then for all $\epsilon>0$ there exists $\zeta_i$ and $\epsilon_i$ such that $\phi^{-1}(]-\epsilon,\epsilon[) \supset \bigcap\limits_{i=1}^n \zeta_i^{-1}(]-\epsilon_i,\epsilon_i[)$. So I hope that the inclusion, for some $\epsilon>0$, implies that $\phi$ is a linear combination of the $\zeta_i$'s.

Is it true? If so how to prove it?

## 1 Answer

Let $\varepsilon=1$: we can find an integer $N$ and $f_i\in X^*$, $\varepsilon_i>0$ such that $$\bigcap_{i=1}^Nf_i^{-1}((-\varepsilon_i,\varepsilon_i))\subset \phi^{-1}((-1,1)).$$ Let $x\in\bigcap_{j=1}^N\ker f_i$. Then for all $\alpha\in\Bbb R$, $\alpha x\in\bigcap_{j=1}^N\ker f_i$. This gives $|\phi(\alpha x)|<1$ and $|\phi(x)|\leq |\alpha|^{-1}$ for all $\alpha$. So $\phi(x)=0$.

We have proved that $\bigcap_{j=1}^N\ker f_i\subset\ker \phi$. We conclude by this result.