Proving $A: l_2 \to l_2$ is a bounded operator 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.
 A: *

*We show that $\left|\sum_{j=1}^nx_j\right|\leq \left(\sum_{j=1}^nx_j^2\sqrt j\right)^{1/2}\left(\sum_{j=1}^n\frac 1{\sqrt j}\right)^{1/2}$, applying Cauchy-Schwarz inequality $\sum_j a_jb_j\leq \sqrt{\sum_ja_j}\sqrt{\sum_jb_j}$ to $a_j=x_j\sqrt j$ and $b_j=\frac 1{\sqrt j}$.

*We have  $$\sum_{j=1}^n\frac 1{\sqrt j}\leq \sum_{j=1}^n\int_{j-1}^jx^{-1/2}dx=\sum_{j=1}^n2(\sqrt j-\sqrt{j-1})=2\sqrt n.$$

*Using the last inequality 
\begin{aligned}
\frac 1{n^2}\left(\sum_{j=1}^nx_j\right)^2&\leq\frac 1{n^2}\sum_{j=1}^nx_j^2\sqrt j2\sqrt n\\
&=2n^{-3/2}\sum_{j=1}^nx_j^2\sqrt j,
\end{aligned}
so 
\begin{aligned}
\sum_{n=1}^{+\infty}\frac 1{n^2}\left(\sum_{j=1}^nx_j\right)^2&\leq 2\sum_{1\leq j\leq n\leq +\infty}n^{-3/2}x_j^2\sqrt j\\
&=2\sum_{j=1}^{+\infty}\sum_{n=j}^{+\infty}n^{-3/2}x_j^2\sqrt j\\
&\leq 2\sum_{j=2}^{+\infty}\sum_{n=j}^{+\infty}\int_{n-1}^nt^{-3/2}dtx_j^2\sqrt j+2\sum_{n=1}^{+\infty}n^{-3/2}x_1^2\\
&=2\sum_{j=2}^{+\infty}\sum_{n=j}^{+\infty}[-2t^{-1/2}]_{n-1}^nx_j^2\sqrt j+2\sum_{n=1}^{+\infty}n^{-3/2}x_1^2\\
&=4\sum_{j=2}^{+\infty}\sum_{n=j}^{+\infty}((n-1)^{-1/2}-n^{-1/2})x_j^2\sqrt j+2\sum_{n=1}^{+\infty}n^{-3/2}x_1^2\\
&=4\sum_{j=2}^N(j-1)^{-1/2}x_j^2\sqrt j+2x_1^2\sum_{n=1}^{+\infty}
n^{-3/2}\\
&\leq 2\max(2\sqrt 2,\sum_{n=1}^{+\infty}
n^{-3/2})||x||_2^2.
\end{aligned}

A: How about using that 
$$ (x_1+\ldots +x_n)^2 = \sum_i x_i^2 + \sum_{i,j} x_ix_j $$
in conjunction with the well-known
$$ 2x_ix_j \leq x_i^2+x_j^2$$
coming from the selfevident inequality $0 \leq (x_i-x_j)^2$. This will provide the bound (EDIT: No, it won't, see the comments)
$$ \frac{(x_1+\ldots +x_n)^2}{n^2} \leq 2\sum_{i=1}^n \frac{x_i^2}{n^2}$$ 
hence 
$$ \sum_{n=1}^{N} \frac{(x_1+\ldots +x_n)^2}{n^2} \leq  \sum_{n=1}^N \sum_{i=1}^n 2 \frac{x_i^2}{n^2}  \leq \sum_{i=1}^N x_i^2 \big(2 \sum_{n=i}^N \frac{1}{n^2}\big) < 2 \zeta(2) \sum_{i=1}^N x_i^2 $$
