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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 $\Leftarrow$ ?

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    $\begingroup$ hint: $\langle x_n - x, x_n - x \rangle = \dotsb \to 0$. $\endgroup$ – user251257 Nov 23 '16 at 20:57
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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 $x_n\to x$.

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  • $\begingroup$ Wow, thanks for the quick anwer. Is it most of the times the same way of "proof method" to show such statements? Basically, playing around with the inner product properties and the limit? $\endgroup$ – TigerLa Nov 23 '16 at 21:06
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    $\begingroup$ Generally looking at the square of the norm is useful whenever the norm comes from an inner product. $\endgroup$ – carmichael561 Nov 23 '16 at 21:07
  • $\begingroup$ Okay, great! Thanks for your help! $\endgroup$ – TigerLa Nov 23 '16 at 21:36

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