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I have a complicated proof involving real interpolation + restricted bounds on operators, but I can only imagine there is a simpler proof of this statement involving some combinations of basic inequalities, such as Cauchy-Schwartz.

Jacob Denson
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  • Interesting ineq, may I know where it comes from? – Paresseux Nguyen Jul 10 '21 at 07:42
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    Let's present why this question is NON-trivial. Firstly, if we replace $\frac{1}{k+n}$ by $1$ for all terms. The resulted ineq $$ \sum_k \sum_n a_k b_n \lesssim ( \sum a_k^2)^{1/2}( \sum b_n^2)^{1/2}$$ fails delibrately. Now if we look closely, we the see sum $k+n$ is associated to the product $a_kb_n$. So we expect this ineq should be trivial if we use power series. However, the "trivial" approach by power series yields something not really promising, that is:$$| \sum_{k,n} \frac{a_kb_n}{k+n+1}x^{k+n+1}| \le | a|_2 | b|_2 \int_0^{|x|} \frac{1}{1-t}dt $$ – Paresseux Nguyen Jul 10 '21 at 09:15
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    It is not promising because the ineq's constant goes to infinity as $|x|$ converges to $1$, and we want $|x|=1$ – Paresseux Nguyen Jul 10 '21 at 09:17
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    This is one of the inequalities known as Hilbert's Inequality. – JimmyK4542 Jul 10 '21 at 09:24
  • @JimmyK4542: Thank you, I'm kinda embarrassed of my lack of knowledge – Paresseux Nguyen Jul 10 '21 at 09:27
  • Don't feel bad. I didn't know what it was called until I needed a bound on the operator norm of the Hilbert matrix for my research. If I remember correctly, It took me a few hours of Googling to find that link. – JimmyK4542 Jul 10 '21 at 09:45
  • Thanks for the reference! This was a question from a University of Madison analysis past qualifying exam I'm preparing to lead a set of preparation sessions for, and I wasn't sure how to solve this question. – Jacob Denson Jul 10 '21 at 15:36

1 Answers1

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To avoid the ambiguity about the convergence, assume $a_n,b_n \ge 0$ and $\sum a_n^2 =1$.
We have: $$ \begin{align} \left(\sum_{k\ge 1}\sum_{n \ge 1} \frac{a_kb_n}{n+k}\right)^2 &\le \left( \sum_{k \ge 1} a_k^2 \right)\left( \sum_{k \ge 1} \left( \sum_{n \ge 1} \frac{b_n}{n+k} \right)^2 \right) \\ & \le \sum_{k \ge 1} \left( \sum_{n \ge 1} \frac{b_n^2\sqrt{n}}{n+k} \right) \underbrace{ \left( \sum_{n \ge 1} \frac{1}{\sqrt{n}(n+k)}\right)}_{=\sum_{n=1}^{k} \frac{1}{\sqrt{n}(n+k)}+ \sum_{n \ge k+1} \frac{1}{\sqrt{n}(n+k)}} \\ & \le \sum_{k \ge 1} \left( \sum_{n \ge 1} \frac{b_n^2\sqrt{n}}{n+k} \right)\left( 2\frac{1}{\sqrt{k}}+2\frac{1}{\sqrt{k}} \right) \\ &= 4\sum_{n \ge 1} b_n^2 \sqrt{n}\left( \sum_{k \ge 1} \frac{1}{\sqrt{k}(k+n)} \right) \\ & \le 4\sum_{n \ge 1} b_n^2 \sqrt{n} \frac{4}{\sqrt{n}} (\text{ like above}) \\ & \le 16 \sum_{n \ge 1} b_n^2 \end{align} $$ Hence the conclusion.

Paresseux Nguyen
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