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Let us consider the following linear operator acting on $l_2$: $$ A(x_1,x_2,x_3,\ldots) ~\colon=~ \left(x_1,\frac{x_1+x_2}{2},\frac{x_1+x_2+x_3}{3},\ldots\right) $$

I need to show that $A$ is a bounded operator, that is $||Ax|| \leq C~||x||$ for some constant $C$ and all $x \in l_2$. In other words, I need to prove the inequality $$ \sum_{n=1}^{\infty}{\left( \frac{x_1+\ldots+x_n}{n} \right)^2} \leq C \sum_{n=1}^{\infty}{x_n^2} $$

I tried to use the fact that $$ \frac{x_1+\ldots+x_n}{n} \leq \sqrt{\frac{x_1^2+\ldots+x_n^2}{n}} $$ but it doesn't work because in that case we get $$ \sum_{n=1}^{\infty}{\left( \frac{x_1+\ldots+x_n}{n} \right)^2} \leq \sum_{n=1}^{\infty}~{\frac{x_1^2+\ldots+x_n^2}{n}} = \sum_{n=1}^{\infty}{\left( \frac{1}{n} + \frac{1}{n+1} + \cdots \right)x_n^2} $$ and coefficients of $x_n^2$ diverge.

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Thank you.

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    This should help: http://en.wikipedia.org/wiki/Hardy%27s_inequality2012-02-19
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    @ByronSchmuland: Interesting! There was a question here(http://math.stackexchange.com/questions/98483/series-inequality-involving-reciprocals-and-reciprocals-of-sums), for the case $p=-1$ (see wiki for what $p$ is). Apparently (according to the wiki) Hardy's inequality holds only for $p \gt 1$. Do you know if that has been improved to include other $p$?2012-02-20
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    @Aryabhata The case $p=-1$ was problem 11145 in the American Mathematical Monthly. The problem was published in April 2005 and the solution in October 2006. The published solution refers to the following result: for $p>0$ and positive $a_1,\dots, a_n$ we have $$\sum_{n=1}^\infty \left({n\over \sum_{j=1}^n 1/a_j}\right)^p\leq \left({p+1\over p}\right) \sum_{n=1}^\infty a_n^p.$$ The reference is K. Knopp *Uber Reihen mit positiven Gliedern* J. London Math. Soc. 3 (1928) 205-211.2012-02-20
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    @ByronSchmuland: Thanks!2012-02-20
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    @Aryabhata: if you're interested, in http://math.stackexchange.com/questions/242123/about-a-possible-hardy-type-inequality-for-negative-exponents/243817#243817, I give a simplified proof of the Knopp's theorem $$\sum_{n=1}^{N}\left(\frac{n}{a_1^p+\ldots+a_n^p}\right)^{1/p}<(p+1)^{\frac{1}{p}}\sum_{n=1}^{N}\frac{1}{a_n}.$$2012-11-24
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    @JackD'Aurizio: Thanks for the link!2012-11-24

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