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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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    @steve I edited the problem - see if you think it's clearer now.2011-12-27

1 Answers 1

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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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    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.2011-12-28