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Is it true that if $x \in \ell^2$ then $\left(\frac{1}{n} \sum_{i=1}^n x_i\right)_{n} \in \ell^2$ ? I conjecture that this is false and the sequence $x_n = \frac{1}{\sqrt{n}\ln(n)}$ is a counterexample, but I cannot prove it.

Andrei Kh
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1 Answers1

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It is true by Hardy's inequality for $p=2$: $$\sum_{n=1}^{\infty}\left(\frac{1}{n} \sum_{i=1}^n x_i\right)^2 \leq \sum_{n=1}^{\infty}\left(\frac{1}{n}\sum_{i=1}^n |x_i|\right)^2 \stackrel{\text{Hardy}}{\leq} 4\sum_{n=1}^{\infty}|x_n|^2.$$

P.S. Your counterexample does not work: if $x_n = \frac{1}{\sqrt{n}\ln(n+1)}$ then, by Stolz-Cesaro Theorem, $$\begin{align}\lim_{n\to \infty}\frac{\frac{1}{n} \sum_{i=1}^n x_i}{x_n}&=\lim_{n\to \infty}\frac{\sum_{i=1}^n x_i}{nx_n}=\lim_{n\to \infty}\frac{x_{n+1}}{(n+1)x_{n+1}-nx_n}\\ &=\lim_{n\to \infty}\frac{1}{n+1 - \frac{n(n+1)^{1/2}\ln(n+2)}{n^{1/2}\ln(n+1)}}\\ &=\lim_{n\to \infty}\frac{1}{n+1 - n(1+\frac{1}{2n})}=2,\end{align}$$ anf it follows that $\left(\frac{1}{n} \sum_{i=1}^n x_i\right)_{n} \in \ell^2$ because $$\left(\frac{1}{n} \sum_{i=1}^n x_i\right)^2\sim (2x_n)^2=\frac{4}{n\ln^2(n+1)}.$$

Robert Z
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  • @Andrei Kh In my P.S. see the reason why your counterexample does not work. – Robert Z Aug 07 '18 at 17:41
  • Thank you for your answer. Why is the limit above equal to 2? I figured $\frac{x_{n+1}}{(n+1)y_{n+1} - ny_n} = \frac{1}{1-\frac{(n+1)^{1/2}\ln(n+1)}{n^{1/2}\ln(n+2)}}$ – Andrei Kh Aug 07 '18 at 17:53
  • (The idea behind $x_n$ was that the sum grows like $\sqrt{n}$ (actually a litte less that that) and that the square of the mean grows like $\frac{1}{n}$ and thus is not summable) – Andrei Kh Aug 07 '18 at 17:57
  • I think I don't see something obvious, but isn't that limit equal to $\infty$, since $\frac{(n+1)^{1/2}\ln(n+1)}{n^{1/2}\ln(n+2)} \sim 1$ ? – Andrei Kh Aug 07 '18 at 18:04
  • I edited my P.S. with more details. – Robert Z Aug 07 '18 at 18:08
  • Ah, I see it now. I got confused by the $y_n$ in yhe denominator. Thank you! – Andrei Kh Aug 07 '18 at 18:10
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    To the proposer: I noticed right away that your proposed counter-example fails: $\sum_{j=2}^n(1/(\sqrt j;\log j)\sim$ $\sim \int_2^n(1/(\sqrt x ;\log x)dx=$ $\int_2^n(2/\log x )d\sqrt x=$ $ 2\sqrt n/\log n-2\sqrt 2;/\log 2+\int_2^n1/(\sqrt x;\log^2 x)dx\sim$ $ 2\sqrt n /\log n.$.... So $(1/n)\sum_{j=2}^n x_j\sim 2x_n$ in this example. To Robert Z: +1 – DanielWainfleet Aug 07 '18 at 19:10