# A counterexample on the existence of some sequence in Hilbert space

I want to find a uniformly bounded sequence $\{x_n\}$ in $l^2(\mathbb{C})$ such that

$x_n$ does not converge to zero in weak topology, i.e., $\exists ~y\in l^2(\mathbb{C}),$ such that $\langle y, x_n\rangle\not\to 0$,

but $\{x_n\}$ satisfies the following condition:

$$\lim_m\lim_n\langle x_{n+m},x_n\rangle=0$$ or the stronger condition:

$$\lim_n\langle x_{n+m},x_n\rangle=0, \forall m\geq 1.$$

Thanks in advance!

Remarks:

1, Jacob Schlather has solved it for the case $\{x_n\}$ is not uniformly bounded, I have added the assumption that $\{x_n\}$ is uniformly bounded, which I forgot to add before.

2, This is one ''remark'' in page 85 of the book-- H.Furstenberg, Recurrence in ergodic theory and combinatorial number theory, Princeton Univ. Press, unless I misunderstand the meaning in the book.

It says: "It should be noted that the analogous result for ordinary convergence does not hold".

Lemma 4.9. Let $\{x_n\}$ be a bounded sequence of vectors in Hilbert space and suppose that $$D-\lim_m(D-\lim_n\langle x_{n+m}, x_n\rangle)=0$$ Then with respect to the weak topology, $$D-\lim_nx_n=0$$

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I changed several occurrences of $<\bullet>$ to $\langle\bullet\rangle$. That is standard usage. –  Michael Hardy Dec 31 '12 at 20:12
I changed the spelling of the title so that people who are really seeking counteremamples would not get sent here. –  rschwieb Dec 31 '12 at 20:34
Thanks Michael! –  ougao Dec 31 '12 at 21:39
Why do you think such a sequence exists? –  GEdgar Dec 31 '12 at 22:32
@GEdgar, one reason is that I can not prove that it does not exist, the other reason is that this is one ''remark'' in page 85 of the book-- H.Furstenberg, Recurrence in ergodic theory and combinatorial number theory, Princeton Univ. Press, unless I misunderstand the meaning in the book. –  ougao Dec 31 '12 at 23:18

## 2 Answers

Let $x_n = e_n$ if $n$ is not a square, and $x_n = e_1$ otherwise. Then $\{x_n\}$ does not converge weakly to $0$ (choose $y=e_1$ as a test function). On the other hand, $\langle x_{n}, x_{n+m} \rangle = 0$ unless both $n$ and $n+m$ are squares. For any fixed $m>0$, this can't happen for say, $n > m^2$, and so $\displaystyle\lim_{n\to\infty} \langle x_{n}, x_{n+m} \rangle = 0$.

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thanks, this is exactly what I want, thanks to your observation that for any fixed positive interger $m$, $m=k^2-s^2=(k+s)(k-s)$ can have only finite many integer solutions $(k,s)$! In my first attempt, I consider the case $x_n=e_1+e_n$ or $e_n$ depend on the type $n=2^k$ or not, now I see this is also a counterexample, since for any fixed $m=2^k-2^s=2^s(2^{(k-s)}-1)$ also have finite integer solutions $(k,s)$. Shame to fail to see this fact till now.. –  ougao Jan 1 '13 at 0:05

Consider the sequence $x_n=ne_n$ where $e_n$ is the standard basis vector. Note that for $y=\sum_{n=1}^\infty e_n/n$ we have $\langle x_i,y\rangle=1$ and also $\langle x_i,e_1\rangle=0$ for $i\neq 1$, so the sequence does not converge weakly. But for any $i \neq j$ we have that $\langle x_i,x_j \rangle=0$.

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thank you for your answer, I have add a missing assumption, if no one answer it in this case, I would accept your answer. –  ougao Dec 31 '12 at 21:43