MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

this turned out to be wrong, see the comment of fedja below for the counter example.

$\sum_{m>0,n>0,k>0,m+n>k}\frac{1}{(m+n-k)^{1/2}\dot (m+n)^{1/2}}a_m a_n b_k$$\le$ C$\\|A\|_{l^2}^2$*$\|B\|_{l^2}$.

where A is the sery $\{a_m\},m=1,2,3...$, B is the sery $b_k, k=1,2,3...$. C is a constant which is independent of A and B. $\| \|_{l^2}$ denotes the $l^2$ Hilbert norm of the sequence.

share|cite|improve this question
    
Isn't this some kind of Cauchy-Schwartz argument? – J. Fabian Meier Aug 8 '11 at 12:18
    
sorry, can you be more specific? because I don't quite see that, what I'm thinking now is some extension of Hilbert's double series theorem. – Shaoming Guo Aug 8 '11 at 12:23
    
This question could use a more specific title, and some motivation. – j.c. Aug 8 '11 at 13:02
6  
Anyway, there is no chance: if both sequences are $N^{-1/2}$ up to $N$, we have about $N^3$ terms each of which is at least $N^{-5/2}$, so the sum is $\ge N^{1/2}$. – fedja Aug 8 '11 at 14:23
1  
Harmonic sequence is much better than just $\ell^2$. It is $\ell^{1+\epsilon}$. And it is rather meaningless to say that it holds for one pair of sequences, when you have a free constant $C$. – Willie Wong Aug 8 '11 at 17:25

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.