Sequence of arithmetic means of a square summable sequence is square summable

Question

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.

Answer 1

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 $$\left(\frac{1}{n} \sum_{i=1}^n x_i\right)^2\sim (2x_n)^2=\frac{4}{n\ln^2(n+1)}.$$