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Let $H$ a hilbert space with an orthonormal basis $(e_n)_{n\in \mathbb{N}}$ and $F$ a linear operator, such that $\langle e_k,F e_n\rangle =:\phi(n,k)$. Find a good estimate for $\lVert F\lVert$ in terms of $\phi(n,k)$. Apply your estimate to the special case of $\phi(n,k)=\frac{1}{n+k}$.

I tried applying parsevals identity and hölder's inequality: \begin{align} \lVert F x\lVert^2&=\lVert \sum_{n=1}^{\infty}\langle x,e_n\rangle F e_n\lVert^2=\lVert\sum_{n=1}^{\infty}\langle x,e_n\rangle \sum_{k=1}^{\infty}\langle e_k,F e_n\rangle e_k \lVert^2 \\&=\sum_{k=1}^{\infty} \left(\sum_{n=1}^{\infty} \langle x,e_n\rangle \phi(n,k)\right)^2 \leq \sum_{k=1}^{\infty} \left [\left(\sum_{n=1}^{\infty} \langle x,e_n\rangle ^2\right)\left(\sum_{n=1}^{\infty}\phi(n,k)^2\right)\right]\\&= \lVert x\lVert^2 \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \phi(n,k)^2 \end{align}

Which implies $\lVert F\lVert \le \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \phi(n,k)^2$. But applied to $\phi(n,k)=\frac{1}{n+k}$ this sum doesn't converge at all.

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    @Julian: As of Harald's comment, operator $F$ of my previous comment is not well-defined (meaning that $Fx$ needs not be $\ell^2$ if $x$ is), so you should not consider it for your exercise. Sorry for the confusion I made.2012-06-16

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In fact the desired norm is the smallest constant $C$ such that for all $a,b\in\ell^2(\mathbb{N})$ we have $ \sum\limits_{m=1}^\infty\sum\limits_{n=1}^\infty \frac{a_n b_m}{m+n}\leq C\left(\sum\limits_{n=1}^\infty a_n^2\right)^{1/2}\left(\sum\limits_{m=1}^\infty b_m^2\right)^{1/2}. $ This is well known in narrow circles Hilbert's inequality. It is known that the smallest possible constant there is $\pi$, so $\Vert F\Vert=\pi$.

You can find several proofs of this fact in The Cauchy-Schwarz Master Class by J. Michael Steele (pages 155-165).

Also you should take a look at this discussion.