TITLE: In Hilbert space: $x_n → x$ if and only if $x_n \to x$ weakly and $\Vert x_n \Vert → \Vert x \Vert$.
QUESTION [6 upvotes]: Assume that $H$ is a $\mathbb K$-Hilbert space, $(x_n)_{n \ge 1}$ a sequence in $H$ and $x ∈ H$.
  Show that $x_n → x$ if and only if $x_n \to x$ weakly and $\Vert x_n \Vert → \Vert x \Vert$.

I'm trying to prove this statement. 
The $\Rightarrow$ is basically clear, since a  strongly convergent sequence is also weakly convergent. But how can I show the other direction $\Leftarrow$ ?

REPLY [9 votes]: To show that $x_n\to x$, you must show that $||x_n-x||\to 0$ as $n\to\infty$. Since $x_n\to x$ weakly, we know that
$$\langle x_n,y\rangle \to \langle x,y\rangle$$
for all $y\in H$, so in particular $\langle x_n,x\rangle\to\langle x,x\rangle$. Hence
$$\langle x,x_n\rangle=\overline{\langle x_n,x\rangle}\to\overline{\langle x,x\rangle}=\langle x,x\rangle$$
as well.
Therefore
$$ ||x_n-x||^2=\langle x_n-x,x_n-x\rangle=||x_n||^2-\langle x_n,x\rangle-\langle x,x_n\rangle +||x||^2\to 2||x||^2-2\langle x,x\rangle=0$$
so $x_n\to x$.