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Let $(x_n)$ be a weakly convergent sequence in a Hilbert space $H$. If $\| x_n \| \to \| x \|$, show that $x_n$ converges strongly to $x$.

Context

This problem comes from a question in my exam paper; the original problem was incorrect.

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    $\begingroup$ So every sequence of unit vectors converges to every unit vector? $\endgroup$ – Jonas Meyer Oct 22 '13 at 19:51
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    $\begingroup$ Clearly false. Easy to build a counterexample: let $v\in H$, $v\ne 0$, then take $X_n=(-1)^n v$. $\endgroup$ – TZakrevskiy Oct 22 '13 at 19:51
  • $\begingroup$ Thank you TZakrevskiy, thats correct ! That was question in my exam paper.. and was incorrect. $\endgroup$ – Ricardo Gomes Oct 31 '13 at 19:59
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    $\begingroup$ I edited the question into what it should have been, so that it can serve as a reference for this fact. $\endgroup$ – user147263 Jan 5 '15 at 3:17
  • $\begingroup$ Does strong convergence implies convergence in norm? $\endgroup$ – manhattan Mar 14 '17 at 6:19
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The result you want to show should be: if $x_n$ converges to $x$ weakly and $\lVert x_n\rVert\to \lVert x\rVert$, then there is convergence in norm. To see that, expand $\lVert x_n-x\rVert^2$ and use the fact that $\langle x_n,x\rangle\to \lVert x\rVert^2$.

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  • $\begingroup$ Thank you Davide Giraudo. Yes ! that is the correct version. but in my exam paper was wrong. $\endgroup$ – Ricardo Gomes Oct 31 '13 at 20:01
  • $\begingroup$ Does strong convergence implies convergence in norm? $\endgroup$ – manhattan Mar 14 '17 at 6:19
  • $\begingroup$ @manhattan You mean convergence of the norms? Sure, by the triangle inequality. $\endgroup$ – 0xbadf00d Mar 19 '19 at 21:05

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