In a Hilbert space, let a sequence $(x_n)$ be weakly convergent to $x$ and be satisfied $\|x_{n+1}-x_n\|\to 0.$ I wonder if we can deduce that $(x_n)$ strongly converges to $x$, or at least it contains a subsequence strongly converges to $x$.

Thank you in advance for your help.


This is not true. In $\ell^2$, you can consider the following sequence: $$ e_1, \frac12 e_2, e_2, \frac12 e_2, \frac13 e_3, \frac23 e_3, e_3, \frac23 e_3, \frac13 e_3, \frac14 e_4, \frac24 e_4, \frac34 e_4, e_4, \frac34 e_4, \frac24 e_4, \frac14 e_4, \ldots$$ It is easy to check that this sequence has all desired properties.

Now, without a convergent subsequence: $$ e_1, e_1 + \frac12 e_2, e_1 + e_2, \frac12 e_1 + e_2, e_2, e_2 + \frac13 e_3, e_2 + \frac23 e_3, e_2 + e_3, \frac23 e_2 + e_3, \frac13 e_2 + e_3, e_3, \ldots $$

  • $\begingroup$ Thank you very much for your answer. So the sequence may not converge, however in your counterexample it still contains a convergent subsequence. $\endgroup$ – KTU Apr 23 '17 at 12:06
  • $\begingroup$ @KTU: Now I have written down a similar sequence without a convergent subsequence. $\endgroup$ – gerw Apr 23 '17 at 20:12
  • $\begingroup$ Very nice example +1 $\endgroup$ – JJR Apr 23 '17 at 21:28
  • $\begingroup$ @grew: very clear and interesting example. Many thanks! $\endgroup$ – KTU Apr 24 '17 at 1:53

Edited heavily using In Hilbert space: $x_n → x$ if and only if $x_n \to x$ weakly and $\Vert x_n \Vert → \Vert x \Vert$. and the fact that weak convergence is unique: If $$||x_{n+1}-x_n|| $$ converges to zero, then $(x_n)$ is Cauchy in H. So, $x_n$ converges to some $y$ in H. Using the other post, we have that $x_n$ converges weakly to $y$. But, weak convergence is unique. So, $x=y$ implying that $x_n$ converges to $x$ in H. Thus, $||x-x_n||$ tends to zero trivially giving us strong convergence.

  • $\begingroup$ earlier post: math.stackexchange.com/questions/2027861/… $\endgroup$ – Sean Nemetz Apr 23 '17 at 4:39
  • $\begingroup$ I don't see how you obtain the inequality $\|x-x_n\| \leq \|x-x_m\|$. $\endgroup$ – Jacky Chong Apr 23 '17 at 4:40
  • $\begingroup$ Oh, I see the problem, I dropped a limit. We had that $||x-x_n|| = ||x-x_{n+1} +x_{n+1} -x_n|| \leq ||x-x_{n+1}|| +||x_{n+1}-x_n||$ I took a limit and disregarded later. $\endgroup$ – Sean Nemetz Apr 23 '17 at 4:46
  • $\begingroup$ You still haven't said anything. I don't see how your inequality helps with your claim. $\endgroup$ – Jacky Chong Apr 23 '17 at 4:56
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    $\begingroup$ Thanks for your answer. However, $(x_n)$ is not a Cauchy sequence with just the property $\|x_{n+1}-x_n\|\to 0$. So the proof should be revised. $\endgroup$ – KTU Apr 23 '17 at 7:22

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