Assume you have a sequence $x_n$ of functions in a seperable Hilbert space, where $\|x_n\|$ converges to some real number $r$. On top assume $x_n$ is weakly convergent to some $x$. Is $r=\|x\|$?
My attempt: $\|x\|^2 = \sum\limits_{i=1}^\infty |\langle x, e_i\rangle|^2 = \lim \limits_{n\rightarrow \infty} \sum\limits_{i=1}^\infty |\langle x_n, e_i\rangle|^2 = \lim\limits_{n\rightarrow \infty}\| x_n\|^2 = r^2$
Yet I don't know why I can pull the limit in the series.
Of course, this is not true. In general, you only have $\|x\| \le \liminf \|x_n\|=r$. A counterexample to equality is the sequence $(e_i)$ itself, it converges weakly to zero.
You cannot exchange weak limit and summation. In order to do that, you would need that $\langle e_i, x_n -x \rangle\to 0$ uniformly in $i$.