I have the following question, that a friend of mine asked me yesterday:

Let $H$ be a Hilbert space, with norm $\|\cdot\|$. Let $(x_k)_{k \ge 0}$ be some sequence in $H$. Assume that $x_k$ is bounded (so that $\|x_k\| \le C$ for all $k$), and that $(x_k)$ has bounded L2 variance, i.e., $$\sum_{k \ge 0} \|x_{k+1}-x_k\|^2 < +\infty.$$

Does it then follow that the sequence $(x_k)$ converges?

If not for all Hilbert spaces, then

Does it at least follow that $(x_k)$ converges for $H=\mathbb{R}^n$?


If $\sum_k \|x_{k+1}-x_k\| <\infty$ then we have that $$ x_n = x_1 + \sum_{k=1}^{n-1} (x_{k+1}-x_k) $$ where the sum converges absolutely, hence converges (in any Banach space).

This suggests a counter-example: pick your favourite sequence in $\mathbb R$ which is square summable, but not summable-- as suggested over at MathOverflow, the harmonic sequence $(1/n)$ works. Let $$ x_1 = 1, x_2=1-1/2, x_3=1-1/2-1/3, $$ and so forth, until say $-2<x_n<-1$. Then add terms $1/k$ until we get $1<x_m<2$ (for some $m>n$). Now subtract terms until again we are between -2 and -1, then add, and so forth. Then $(x_n)$ is bounded, and for all $k$ we have $x_{k+1}-x_k=\pm 1/(k+1)$ so $\sum_k |x_{k+1}-x_k|^2<\infty$. But as $(x_n)$ slowly oscillates between 1 and -1, it does not converge.

  • $\begingroup$ Thanks for your nice example Matthew. I have deleted the MO question, and am waiting to gain some rep here, so that I can also upvote your answer! $\endgroup$
    – suvrit
    Nov 5 '11 at 11:53

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