7
$\begingroup$

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.

--

Thank you.

$\endgroup$
5
  • 6
    $\begingroup$ This should help: en.wikipedia.org/wiki/Hardy%27s_inequality $\endgroup$
    – user940
    Feb 19, 2012 at 16:02
  • $\begingroup$ @ByronSchmuland: Interesting! There was a question here(math.stackexchange.com/questions/98483/…), 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 $p$? $\endgroup$
    – Aryabhata
    Feb 20, 2012 at 18:56
  • 1
    $\begingroup$ @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. $\endgroup$
    – user940
    Feb 20, 2012 at 19:51
  • $\begingroup$ @Aryabhata: if you're interested, in math.stackexchange.com/questions/242123/…, I give a simplified proof of the Knopp's theorem $$\sum_{n=1}^{N}\left(\frac{n}{a_1^p+\ldots+a_n^p}\right)^{1/p}<(p+1)^{\frac{1}{p}}\sum_{n=1}^{N}\frac{1}{a_n}.$$ $\endgroup$ Nov 24, 2012 at 17:00
  • $\begingroup$ Why is no answer accepted? $\endgroup$ Oct 31, 2019 at 20:58

2 Answers 2

1
$\begingroup$
  1. 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 $b_j=\frac 1{\sqrt j}$.
  2. We have $$\sum_{j=1}^n\frac 1{\sqrt j}\leq \sum_{j=1}^n\int_{j-1}^jx^{-1/2}dx=\sum_{j=1}^n2(\sqrt j-\sqrt{j-1})=2\sqrt n.$$
  3. 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}
$\endgroup$
1
  • $\begingroup$ The constant is almost surely not the best; I guess we can determine it with Hardy inequality. $\endgroup$ Feb 26, 2012 at 14:35
0
$\begingroup$

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 $$ \sum_{n=1}^{N} \frac{(x_1+\ldots +x_n)^2}{n^2} \leq \sum_{n=1}^N \sum_{i=1}^n 2 \frac{x_i^2}{n^2} \leq \sum_{i=1}^N x_i^2 \big(2 \sum_{n=i}^N \frac{1}{n^2}\big) < 2 \zeta(2) \sum_{i=1}^N x_i^2 $$

$\endgroup$
5
  • $\begingroup$ How do you get the bound after "this will provide the bound"? $\endgroup$ Feb 19, 2012 at 16:33
  • $\begingroup$ you substitute $x_ix_j$ by $(x_i^2+x_j^2)/2$ in the first line and rearrange terms. Am I being sleepy and missing something ? $\endgroup$
    – testcase
    Feb 19, 2012 at 16:37
  • 1
    $\begingroup$ You sum for example for the term $x_i^2$ with respect to $j$, so the bound we will get won't be enough. $\endgroup$ Feb 19, 2012 at 16:40
  • $\begingroup$ testcase: $\sum_{i<j} 2 x_i x_j \leq \sum_{i<j} x_i^2 + x_j^2$, this cannot be bounded by $2\sum_{i} x_i^2$. $\endgroup$
    – sdcvvc
    Feb 19, 2012 at 16:42
  • $\begingroup$ $\|x\| =\sum_{i\geq 1}{{2i+1}\{i+1}}|x_i|\leq \sum_{i\geq 1}2|x_i|=2\|x\|$ $\endgroup$
    – user296169
    Jan 20, 2018 at 22:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .