<s>[INST] Simple proof that $\sum_{k = 1}^\infty \sum_{n = 1}^\infty a_k b_n / (k + n) \lesssim (\sum_k a_k^2)^{1/2} (\sum_n b_n^2)^{1/2}$ 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.
 [/INST] 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.
</s>