How to prove this? (convergence in the weak*-topology) I want to prove the following theorem, but I'm not skilled in deal with weak and weak*-topologies:

Theorem. Let $X$ be a Banach space and $\sigma(X',X)$ the weak*-topology of $X'$ [$X'$ is the dual of $X$]. Then $f_n \overset{\star}{\rightharpoonup} f$ in $\sigma(X',X)$ iff $\langle f_n, x \rangle \overset{n \to \infty}{\longrightarrow} \langle f,x \rangle$ for all $x \in X$.

if I well understood, this is a relation between the convergence in the weak*-topology and the strong convergence in $X'$. For reference, this is for example Proposition 3.13 in Brezis's book on functional analysis (he doesn't prove it; only suggests that the reader should copy the proof of the "dual" theorem about the weak topology - that, anyway, is not carried out explicitly.).
May anyone post a detailed proof, in order to show me how to handle with such objects?
EDIT - I add the definition of weak*-topology I'm using. 

Definition. Let $X$ be a Banach space, $X'$ the dual of $X$, $\mathcal F \subset X''$ defined by $\mathcal F := \{f \to \langle f,x \rangle, x \in X\}$, $\mathcal F : X' \to \mathbb K^X$, $\mathbb K^X$ endowed with the product topology $\tau$. Then $\sigma( X', X) := \mathcal F^{-1}\tau$ is the weak*-topology of $X'$.

 A: The given definition of the weak$^\ast$ topology is a formal way to say that the weak$^\ast$ topology is the initial topology with respect to the family $\{f \mapsto f(x) : x \in X\}$. That means, a neighbourhood basis of $f \in X'$ is given by
$$V(f;\varepsilon; x_1,\,\dotsc,\,x_m) = \{g \in X' : \lvert g(x_i) - f(x_i)\rvert < \varepsilon \text{ for } 1 \leqslant i \leqslant m\},$$
where $\varepsilon > 0$, $m \in \mathbb{N}$ and $x_i \in X$ can be freely chosen.
An alternative, and often more useful way of describing the weak$^\ast$ topology is to say that it is the topology defined by the seminorms
$$p_F(f) = \sup \{\lvert f(x)\rvert : x \in F\}$$
where $F$ ranges over the finite (nonempty) subsets of $X$.
By definition of convergence, $f_n \overset{w^\ast}{\rightharpoonup} f$ if and only if every weak$^\ast$-neighbourhood of $f$ contains all but finitely many of the $f_n$.
By definition of the weak$^\ast$ topology, that means for all $x_1,\,\dotsc,\, x_m \in X$ and all $\varepsilon > 0$, there is an index $n_0\in \mathbb{N}$ such that for all $n \geqslant n_0$, we have
$$\lvert f_n(x_i) - f(x_i)\rvert < \varepsilon.$$
So, if $f_n \overset{w^\ast}{\rightharpoonup} f$, for every $x \in X$, for all $\varepsilon > 0$, we have $f_n \in V(f;\varepsilon; x)$ for all but finitely many $n$, and hence $f_n(x) \to f(x)$.
Conversely, if $f_n(x) \to f(x)$ for all $x \in X$, given the weak$^\ast$-neighbourhood $V(f;\varepsilon; x_1,\, \dotsc,\, x_m)$, we find $n_i \in \mathbb{N},\, 1 \leqslant i \leqslant m$, such that for all $n \geqslant n_i$, we have $\lvert f_n(x_i) - f(x_i)\rvert < \varepsilon$. For $n \geqslant n_0 := \max\limits_{1 \leqslant i \leqslant m} n_i$, we then have
$$f_n \in V(f;\varepsilon; x_1,\,\dotsc,\,x_m).$$
So for every weak$^\ast$-neighbourhood $U$ of $f$, we can find an $n_0 \in \mathbb{N}$ with $n \geqslant n_0 \Rightarrow f_n \in U$, and that means $f_n$ converges to $f$ in the weak$^\ast$ topology.
