I am self-learning real analysis from Folland and got stuck on weak topology convergence. He defines weak topology as follows:
If $X$ is a normed vector space, the weak topology on $X$ is the weak topology generated by $\{f\,|\, f\in X^{*}\}$.
Then he asserts the following which I am trying to prove:
If $\langle x_\alpha \rangle$ is a net in $X$, then in the weak topology $\langle x_\alpha \rangle \to x \iff f(x_\alpha) \to f(x) \,\, \forall \, f \in X^*$.
Proof: $\implies$ Let $U$ be any ngh of $f(x)$.
Q1. Does this mean $V:=f^{-1}(U)$ is a ngh of $x$ (in the weak topology?).
Assuming Q1, we have $\exists \, \alpha_0$ such that $x_\alpha \in V \, \, \forall \, \alpha \geq \alpha_0$. Hence $f(x_\alpha) \in U \, \, \forall \, \alpha \geq \alpha_0$. Thus, $f(x_\alpha) \to f(x)$.
Q2. How can I go about the reverse implication?
You are. The reason you can choose $\alpha_0$ is that $V$ is a weak-neighbourhood of $x$.
The converse is not too different. Given a neighbourhood $V$ of $x$, by definition of the weak topology there exists $f\in X^*$ and $U\subset \mathbb C$ such that $x\in f^{-1}(U)\subset V$. Since $f(x_\alpha)\to f(x)$, there exists $\alpha_0$ such that $f(x_\alpha)\in U$ for all $\alpha\geq\alpha_0$. Which implies that $x_\alpha\in V$ for all $\alpha\geq\alpha_0$.