Two propositions about weak* convergence and (weak) convergence Let $E$ be a normed space. We have the usual definitions:
1) $f, f_n \in E^*$, $n \in \mathbb{N}$, then $$f_n \xrightarrow{w^*} f :<=> \forall x \in E: f_n(x) \rightarrow f(x)$$ and in this case we say that $(f_n)$ is $weak^*$-$convergent$ to $f$.
2)$x, x_n \in E$, $n \in \mathbb{N}$, then $$x_n \xrightarrow{w} x :<=> \forall f \in E^*: f(x_n) \rightarrow f(x)$$ and in this case we say that $(x_n)$ is $weakly\ convergent$ to $x$.
Now for the two propositions I want to prove or disprove the following statements.
Let $f, f_n \in E^*$, $n \in \mathbb{N}$, such that $f_n \xrightarrow{w^*} f$ and let $x, x_n \in E$. Consider:
[edit: Thanks for pointing out my mistake!]
a) $x_n \rightarrow x$ => $f_n(x_n) \rightarrow f(x)$,
b) $x_n \xrightarrow{w} x$ => $f_n(x_n) \rightarrow f(x)$.
So far, I think that even b) is true which would imply that a) is also true. My reasoning is that, by assumption, we have $f_m(x_n) \rightarrow f_m(x)$ for every fixed $m \in \mathbb{N}$ as well as $f_m(x) \rightarrow f(x)$ for every $x \in E$. Hence we have $$\lim_{m\to \infty} \lim_{n\to \infty} f_m(x_n) = \lim_{m\to \infty} f_m(x) = f(x)$$ which should be the same as $$\lim_{n\to \infty} f_n(x_n) = f(x).$$
However, I'm a little bit suspicious because the setting seems to imply that a) ist true, but b) is not. Is my argument too sloppy?

a) 
$$
\|f_n(x_n) - f(x) \| \leq \|f_n(x_n) - f_n(x)\| + \| f_n(x) - f(x) \| \leq \|f_n\| \|x_n - x\| +   \| f_n(x) - f(x) \|.
$$
By the principle of uniform boundedness, $\sup_n \|f_n\|$ is finite, so you're done.
b) Not true in general. Let $\{x_n\}$ be an orthonormal basis in some Hilbert space. Then $x_n \rightarrow 0$ weakly and, by reflexivity, $\langle x_n, \cdot \rangle \rightarrow 0$ in the weak-* topology. Evidently, $\langle x_n, x_n \rangle$ does not go to $0$.
