Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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 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
show 2 more comments

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
    
Thank you very much! –  Julian Jun 17 '12 at 9:50
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.