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?

Your argument is wrong. Let $\delta_{ij}$ be Kronecker's $\delta$. Then you have
$$\lim_{i\to\infty}\lim_{j\to\infty} \delta_{ij} = \lim_{i\to\infty} 0 = 0,$$
but
$$\lim_{i\to\infty} \delta_{ij} = 1.$$
Note that this corresponds to the following situation: Let $E$ be a Hilbert space, $x_n$ be a orthonormal basis and $f_n(x) := (x_n, x)$, i.e., the functional generated by $x_n$. Then, $f_i(x_j) = \delta_{ij}$. This was already pointed out by Michael.
