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.
Weak convergence, together with convergence of norms, implies strong convergence in a Hilbert space.
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Ricardo Gomes
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5So every sequence of unit vectors converges to every unit vector? – Jonas Meyer Oct 22 '13 at 19:51
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5Clearly false. Easy to build a counterexample: let $v\in H$, $v\ne 0$, then take $X_n=(-1)^n v$. – TZakrevskiy Oct 22 '13 at 19:51
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Thank you TZakrevskiy, thats correct ! That was question in my exam paper.. and was incorrect. – Ricardo Gomes Oct 31 '13 at 19:59
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1I edited the question into what it should have been, so that it can serve as a reference for this fact. – Jan 05 '15 at 03:17
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Does strong convergence implies convergence in norm? – na1201 Mar 14 '17 at 06:19
1 Answers
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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 $\langle x_n,x\rangle\to \lVert x\rVert^2$.
Davide Giraudo
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Thank you Davide Giraudo. Yes ! that is the correct version. but in my exam paper was wrong. – Ricardo Gomes Oct 31 '13 at 20:01
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@manhattan You mean convergence of the norms? Sure, by the triangle inequality. – 0xbadf00d Mar 19 '19 at 21:05
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$x_n \rightharpoonup x$ is equivalent to [$\left\langle x_n, y \right\rangle \to \left\langle x, y\right\rangle$ for each $y \in H$] due to the Riesz Representation Theorem, (https://proofwiki.org/wiki/Weak_Convergence_in_Hilbert_Space) which doesn't apply to general inner product spaces and requires completeness. (I guess it's possible you might define weak convergence on Hilbert spaces this way too) For counterexamples of Riesz for non-complete $H$, see https://math.stackexchange.com/questions/678656/necessity-of-completeness-of-the-inner-product-space-in-riesz-representation-the. – George Coote Jan 31 '22 at 14:34
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@Davide Giraudo, this result still works for a reflexive space? – Rodolfo Ferreira de Oliveira Dec 10 '23 at 18:39
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For extra information, we have $\langle x, x_n \rangle \to \langle x,x \rangle$ because of weak convergence and because $y \mapsto \langle x, y \rangle$ or $\langle y , x \rangle$ is a linear functional (depending on the convention) ! – Neckverse Herdman Jan 29 '24 at 19:18