# Bounding double sum off diagonal

Let $$\{a_n\}_{n\in\mathbb{N}}\subset\mathbb{C}$$ and $$N\in\mathbb{N}$$. I have to prove the following bound $$\sum_{n\leq N}\sum_{\substack{m\leq N\\ m\neq n}}|a_ma_n|\left(\log\frac{m}{n}\right)^{-2}\leq\sum_{n\leq N}|a_n|^2\sum_{1\leq m I think I have to use the fact that $$1-\frac{1}{x}\leq \log x$$ to get $$\left(\log \frac{m}{n}\right)^{-2}\leq\left(\frac{m}{n-m}\right)^2$$ but then I don't know how to reduce the term off the diagonal to a sum only of the diagonal terms. Any advice is welcome, thank you!

Edit. The above inequality is not true for $$N=2$$ as pointed out by Doyun Nam. What I eventually wanted to show is the following bound $$\sum_{n\leq N}\sum_{\substack{m\leq N\\ m\neq n}}|a_ma_n|\left(\frac{m}{n}\right)^{-2}\ll \sum_{n\leq N}n^2|a_n|^2$$ Maybe is easier to directly prove such bound without pasing first through the first inequality that I have written (which turned out to be wrong).

• I'm not sure your inequality is true. I checked in the case $N=2$. And I think at that time, this inequality doesn't hold. Commented Feb 1, 2019 at 8:43
• which $\{a_n\}\subset\mathbb{C}$ have you considered?
– asd
Commented Feb 1, 2019 at 8:45
• When $N=2$, I get the inequality $$|a_2 a_1|(\log2)^{-2} + |a_1 a_2| (\log\frac{1}{2})^{-2} \leq |a_2|^2 .$$ And If $|a_1|$ goes to infinity, then the left-hand side goes to infinity. Commented Feb 1, 2019 at 8:49
• Yeah, that's strange. Would $\sum_{m\neq n}$ on the left hand side mean the same as the double sum I have written?
– asd
Commented Feb 1, 2019 at 9:19
• The last inequality I have added is an asymptotic bound: it doesn't have to hold pointwise but only as $N\to\infty$. In other words it says that the left hand side is a "big oh" of the right hand side as $N$ tends to infinity.
– asd
Commented Feb 1, 2019 at 11:40

For each $$n$$ put $$b_n=|a_n|$$ . Then $$\{b_n\}$$ is a sequence of non-negative real numbers and we have to show that
$$\sum_{n\leq N}\sum_{\substack{m\leq N\\ m\neq n}} b_mb_n\left(\frac{m}{n}\right)^{-2}\ll \sum_{n\leq N}n^2b_n^2$$
$$\sum_{n\leq N}\sum_{\substack{m\leq N\\ m\neq n}} b_mb_n\left(\frac{m}{n}\right)^{-2}\ll \sum_{n\leq N}n^2b_n^2$$ $$\sum_{n\leq N}\sum_{\substack{m\leq N }} b_mb_n \left(\frac{m}{n}\right)^{-2}\ll \sum_{n\leq N}(n^2+1)b_n^2$$ $$\left( \sum_{n\leq N} b_nn^2\right)\left(\sum_{\substack{m\leq N }} b_mm^{-2}\right)\ll \sum_{n\leq N}(n^2+1)b_n^2$$ Clearly, this inequality holds asymptotically iff $$\sum_{m=1}^\infty b_mm^{-2}<\infty$$.