Convergence in weak topology implies convergence in norm topology

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

Solution 1:

$\def\p#1{\left\langle#1\right\rangle}$That is not true. Consider $\ell^2$ with the standard inner product $$ \p{x,y} = \sum_{i=0}^\infty x_i\bar y_i $$ and $e_n = (0,\ldots, 0, 1, 0,\ldots)\in \ell^2$ ($1$ at position $n$). Then for $x \in \ell^2$ we have $$ \p{e_n, x} = \bar x_n \to 0 = \p{0, x}$$ so $e_n \to 0$ weakly. But $\|e_n\| = 1 \not \to 0$, so $e_n \not\to 0$ in norm.

What is true in every Hilbert space is that $x_n \to x$ weakly and $\|x_n\| \to \|x\|$ implies $\|x_n - x\|\to 0$, that is $x_n \to x$ in norm. To see that, write \begin{align*} \|x_n - x\|^2 &= \|x_n\|^2 - 2\Re\p{x_n, x} + \|x\|^2 \end{align*} Now the first term converges to $\|x\|^2$ by assumption and the second one to $2\Re\p{x,x} = 2\|x\|^2$ by weak convergence, so alltogether $$ \|x_n -x\|^2 = \|x_n\|^2 - 2\Re\p{x_n, x}+ \|x\|^2 \to \|x\|^2 - 2\|x\|^2 + \|x\|^2 = 0 $$

Solution 2:

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 $$ \|x-x_n\|^2=\langle x-x_n,x-x_n\rangle = \|x\|^2+\|x_n\|^2-2\text{Re}\,\langle x_n,x\rangle\to\|x\|^2+\|x\|^2-2\|x\|^2=0. $$