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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.

Harald Hanche-Olsen
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Julian
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  • I suspect that there exists some $F$ such that $\phi(n, k)=1/(n+k)$ but $\lVert F\rVert=+\infty$, which means that you have already obtained a good bound. – Giuseppe Negro Jun 16 '12 at 10:55
  • More specifically I think that the operator $F\colon \ell^2\to \ell^2$,

    $$F(x_1, x_2, \ldots)=\begin{bmatrix} 1/2 & 1/3 & 1/4 & \ldots \ 1/3 & 1/4 & 1/5 &\ldots \ 1/4 & 1/5& 1/6 & \ldots \ \ldots & \ldots & \ldots & \ldots \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \ x_3 \ \vdots \end{bmatrix}$$

    is well-defined, unbounded, and satisfies $\phi(n, k)=1/(n+k)$. Well-definitess looks like the hardest part, and I suspect that Hardy's discrete inequality http://en.wikipedia.org/wiki/Hardy%27s_inequality plays a role in proving it.

     – Giuseppe Negro Jun 16 '12 at 11:06
  • Please don't use <…> for inner products – use \langle…\rangle instead. (I fixed it for you.) – Harald Hanche-Olsen Jun 16 '12 at 11:10
  • @GiuseppeNegro: On the contrary, I think it's a theorem that you cannot explicitly define an everywhere defined, unbounded operator on a Hilbert space: There are variants of ZF set theory without the axiom of choice in which any everywhere defined linear operator is bounded. – Harald Hanche-Olsen Jun 16 '12 at 11:15
  • @HaraldHanche-Olsen: Oh, thank you. This means that the operator $F$ mentioned above for sure is not well defined. – Giuseppe Negro Jun 16 '12 at 11:23
  • @GiuseppeNegro Can you explain why this $F$ is not bounded? I have $\lVert F e_n\lVert^2=\sum_{k=n}^{\infty} \frac{1}{k^2}\leq \sum_{k=1}^{\infty} \frac{1}{k^2}=\frac{\pi^2}{6}$. A fellow student pointet out that it's possbile to prove that $\pi$ is an upper bound for $\lVert F\lVert$. Is he correct? – Julian Jun 16 '12 at 11:27
  • @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. – Giuseppe Negro Jun 16 '12 at 11:35

1 Answers1

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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.

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Norbert
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