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Let $(s_n)_{n \in \mathbb{N}}\in\ell^2(\mathbb{N})$ (i.e. $\displaystyle \sum_{n=0}^{\infty}\vert s_n\vert^2<\infty$). Define vectors $A=[A_1,\ldots,A_M]$ and $B=[B_1,\ldots,B_M]$ with coordinates $A_k=\displaystyle\sum_{i=k}^{M}|s_i|$ and $B_k=\displaystyle\sum_{i=M}^{2M-k}|s_i|,\ k=1,\ldots,M\ $ (note that the length of the vectors depends on $M$).

Is it true that

$$\frac{1}{\sqrt{M}}\Vert A-B\Vert_2\to0\ \text{as}\ M\to \infty ?$$

Sorry, I'm an economist and not a mathematician so my symbols might be off. Experimental data and model simulations suggest the above to be true with changing $M$ (i.e., error is less than $1/\sqrt{M}$ for all $M$, hence $1/\sqrt{\infty}=0$ at $M=\infty$), but can it be shown analytically?

I reason:

$$\frac{1}{\sqrt{M}}\Vert A_k-B_k\Vert_2=\frac{1}{\sqrt{M}}\left(\sum_{(i=something)}\Vert s_i\Vert_2-\sum_{(i=something else)}\Vert s_i\Vert_2\right)\to0$$

hence $\frac{1}{\sqrt{M}}\Vert A-B\Vert_2\to0$.

But I think this is wrong because $\frac{1}{\sqrt{M}}\Vert A-B\Vert_2=M \frac{1}{\sqrt{M}}\Vert A_k-B_k\Vert_2\nrightarrow 0$. What's wrong with my reasoning?

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To clarify the notation: each $s_n$ is an element of $\ell^2$? –  Nate Eldredge Dec 24 '11 at 22:48
@NateEldredge Yes, I should've written sequence $s_n$. So what I mean is $\sum_{n=-\infty}^\infty|s_n|^2 < \infty$. I'll edit to clarify this. –  steve Dec 25 '11 at 17:00
Ok, so $\{s_n\}$ is a single element of $\ell^2$. Then each $A_k$ is a number, not a vector (though $(A_1, \dots, A_M)$ is a vector). Now it makes sense. –  Nate Eldredge Dec 26 '11 at 3:15
@NateEldredge That is correct, sir. –  steve Dec 26 '11 at 15:22
@steve I edited the problem - see if you think it's clearer now. –  Pedro Milet Dec 27 '11 at 17:15
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1 Answer

I don't think that is true.

Let $s_n = 1/n$ (and let $s_0$ be your favorite number). Now $A_k$ and $B_k$ have opposing signs, so clearly $\frac{1}{\sqrt{M}}||A-B||_2 \geq \frac{1}{\sqrt{M}}||A||_2$.

Now for $k = 1,2, \ldots, M/2$ we have $$-A_k \geq \sum_{n=M/2}^M \frac{1}{n} > \int_{x=M/2}^M \frac{1}{x} dx = \ln(2),$$

so $$||A||_2^2 = \sum_{k=1}^M |A_k|^2 \geq \sum_{k=1}^{M/2} (\ln(2))^2 = \frac{M}{2}(\ln(2))^2$$

and $$\frac{1}{\sqrt{M}}||A||_2 \geq \frac{\ln(2)}{\sqrt{2}}$$

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$A_k$ and $B_k$ have opposing signs only in this example... they needn't generally, say if $s_n=1/|n|$. In that case, you can't say $\Vert A-B\Vert_2\geq \Vert A\Vert_2$... –  steve Dec 26 '11 at 22:52
I know, but this is a counter-example, not a proof. I showed that the statement (as it was phrased) is false by exhibiting one sequence in $l^2$ where the conclusion fails. Did you want the sequence to be non-negative? –  Pedro Milet Dec 27 '11 at 11:26
Looks like I've been trying to generalize when I shouldn't. You're right, my sequence is non-negative and symmetric. Damn, I should've written that in the question. Sorry about that... –  steve Dec 27 '11 at 15:06
Even with this supposition, I believe the claim is false. Take $s_n$ to be $1/n$ when $2^{2k} \leq n < 2^{2k+1}$ and $0$ when $2^{2k-1} \leq n < 2^{2k}$. Then for arbitrarily large $M$ we have $B = 0$; I'll work out the details later. –  Pedro Milet Dec 28 '11 at 0:30
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