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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.

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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
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
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

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