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 $\Leftarrow$ ?

  • 2
    $\begingroup$ hint: $\langle x_n - x, x_n - x \rangle = \dotsb \to 0$. $\endgroup$
    – user251257
    Commented Nov 23, 2016 at 20:57

1 Answer 1


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 $x_n\to x$.

  • $\begingroup$ 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? $\endgroup$
    – user372904
    Commented Nov 23, 2016 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. $\endgroup$ Commented Nov 23, 2016 at 21:07
  • $\begingroup$ Okay, great! Thanks for your help! $\endgroup$
    – user372904
    Commented Nov 23, 2016 at 21:36

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