Weak convergence and convergence of functionals in dual space Let $X$ be a normed space, $(x_n)_{n \in \mathbb{N}} \subset X$ and $(x_n')_{n \in \mathbb{N}} \subset X'$ such that $(x_n)_{n \in \mathbb{N}}$ weakly converges to $x \in X$ and $(x_n')_{n \in \mathbb{N}}$ converges pointwise to $x'$. 
What can you say about the convergence of $x'(x_n)$?
If I had that $X$ is a Banach space, I could use that then $x' \in X'$ (as a pointwise limit of linear and bounded functionals) and hence by definition of the weak convergence of $(x_n)_{n \in \mathbb{N}}$ we would receive that $\lim_{n \rightarrow \infty} x'(x_n)=x'(x)$. Is this right so far?
Now if I don't have that $X$ is Banach I'm wondering if I can show the same? 
I thought about something like this: $$\lim_{n \rightarrow \infty}x'(x_n) =  \lim_{n \rightarrow \infty} \lim_{m \rightarrow \infty} x_m'(x_n) = \lim_{m \rightarrow \infty} \lim_{n \rightarrow \infty} x_m'(x_n) = \lim_{m \rightarrow \infty} x_m'(x) = x'(x)$$ . But I'm not sure how I can argue that the limits are interchanged. Would it suffice if the limit of $x'_n$ would be uniformly instead of pointwise?
Thanks for your help!
 A: Let $X$ be the set of all 1-periodic trigonometric polynomials $$X=\{\sum_{n\in F} c_ne^{2\pi in\cdot}: F\subset\mathbb{Z}\hspace{2mm}\textrm{finite}, c_n\in\mathbb{C} \} $$ with the norm $$\displaystyle \|p\| = \left(\int_0^1 |p(t)|^2\ dt\right)^{1/2}.$$ Let $x_n(t) = e^{2\pi i nt}$ and $$x_m'(p) = \sum_{k=-m}^m \int_0^1 p(t)e^{-2\pi ikt}\ dt.$$ Then, $x_n\to 0$ weakly and $x_m'\to x'$ pointwise on $X$, where $x'(p) = p(0)$ for all $p\in X$. We have
$$\lim_{m\to\infty}\lim_{n\to\infty}x_m'(x_n) = 0, \hspace{5mm} 
  \lim_{n\to\infty} x'(x_n) =
  \lim_{n\to\infty}\lim_{m\to\infty}x_m'(x_n) = 1$$

Note that $\|x_m'\|^2 = 2m+1$, so $(\|x_m'\|)$ is not a bounded sequence. Also $x'$ is not a bounded linear functional on $X$.
Lastly, aside from an individual example, the theorem below might come in handy to get a more general result, which is nothing more than a conclusion of the uniform boundedness theorem.

Theorem: Let $Y,Z$ be a Banach spaces, $P_n:Y\to Z$ be bounded linear operators, $X$ be a dense subspace of $Y$. Suppose that the
sequence $(P_na)$ converges in $Z$ for each $a\in X$. Then, the
following are equivalent.
i. $(P_na)$ converges in $Z$ for all $a\in Y$.
ii. $\{P_n:n\in\mathbb{N}\}$ is a uniformly bounded set of operators.

For your problem, just take $Y$ be the completion of $X$, $Z=\mathbb{C}$,  and $P_n = x_n'$.
