# In Hilbert space: $x_n → x$ if and only if $x_n \to x$ weakly and $\Vert x_n \Vert → \Vert x \Vert$.

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 $\Leftarrow$ ?

• hint: $\langle x_n - x, x_n - x \rangle = \dotsb \to 0$. Nov 23 '16 at 20:57

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 $x_n\to x$.