TITLE: Convergence in weak topology implies convergence in norm topology
QUESTION [9 upvotes]: In Hilbert space why does convergence in weak topology $x_n$ to $x$ imply that $x_n$ converges to $x$ in norm? 
Thank you very much for your answers. What if I put a condition on weak convergence i.e., suppose it also holds $\lim_{n\to \infty } \lVert x_n\rVert \to \lVert x\rVert$ then can I say that $x_n$ converges in norm? 
I doubt if it always holds! 
I think to understand this I need bit more explanation because I am struggling with the understanding of weak topologies. 
Thank you

REPLY [5 votes]: It is not true in general, as already mentioned, but it is true with the added condition. Suppose that $x_n\to x$ weakly, and that $\|x_n\|\to\|x\|$. Then
$$
\|x-x_n\|^2=\langle x-x_n,x-x_n\rangle = \|x\|^2+\|x_n\|^2-2\text{Re}\,\langle x_n,x\rangle\to\|x\|^2+\|x\|^2-2\|x\|^2=0.
$$