Equivalence of weak-* convergence in Banach spaces. Let $X$ be a Banach space and $f,(f_n)_{n \in \mathbb N} \in X^*$.  
$f_n \xrightarrow{w^*} f$ if and only if  
a) $\sup_{n \in \mathbb N} \|f_n\| < \infty$ and
 b) $\exists A \subset X: \operatorname{span}(A) \subset X$ is dense and $\lim_{n \rightarrow \infty}f_n(x)=f(x)$ for all $x \in A$.
Any clues are appreciated.

Hint. "only if": (1) Weak$^*$ly convergent sequences are bounded (which follows from the uniform boundedness theorem) and (2) $X$ is dense in $X$.
For "if": Let $x \in X$ and $\def\eps{\varepsilon}\eps > 0$. Choose $a \in A$ such that $\def\norm#1{\left\|#1\right\|}\norm{x-a} < \eps$. We have that
\begin{align*}
  \def\abs#1{\left|#1\right|}\abs{f_n(x) - f(x)} &\le \abs{f_n(x) - f_n(a)}
   + \abs{f_n(a) - f(a)} + \abs{f(a) -  f(x)}\\
   &\le \sup_n\norm{f_n} \cdot \norm{x-a} + \abs{f_n(a) - f(a)}
     + \norm f \cdot \norm{x-a}\\
   &\le \bigl(\sup_n \norm{f_n} + \norm f\bigr)\cdot \eps + \abs{f_n(a)-f(a)}
\end{align*}
Now use $f_n(a) \to f(a)$.
