Let $p>1$, $q>1$, $0<\lambda<1$ be such that
$\frac{1}{p}+\frac{1}{q}+\lambda=2$. Suppose that 
$(a\_{k})\in \ell^{p}(\mathbb{Z})$ and $(b\_{k})\in \ell^{q}(\mathbb{Z})$.
It is known ([1,2,3]) that 
$$\sum\_{j\neq k}\frac{a\_{j}b\_{k}}{|j-k|^{\lambda}}\leq C\_{p,q} \parallel a \parallel\_{p}\parallel b \parallel\_{q}.$$
If $\frac{1}{p}+\frac{1}{q}=1$ and $\lambda=1$, then the estimate fails.
Namely we have 
$$\sum\_{j\neq k,\, j,k=1,...,N }\frac{a\_{j}b\_{k}}{|j-k|}\geq C\log{N} \parallel a \parallel\_{p}\parallel b \parallel\_{q}.$$


Question: What estimates do we still have when $\frac{1}{p}+\frac{1}{q}= 1$
and $\lambda>1$ ? I expect the inequality to hold. Does it?


Observe that when $\frac{1}{p}+\frac{1}{q}<1$ and $\lambda>1$ the inequality fails.
A counterexample is $a\_{k}=b\_{k}=1$ for which we have
$\parallel a \parallel\_{p}\parallel b \parallel\_{q}=N^{\frac{1}{p}+\frac{1}{q}}$, while $\sum\_{j\neq k,\, j,k=1,...,N }\frac{1}{|j-k|^{\lambda}}\geq C N$.


[1] G. H. Hardy, J. E. Littlewood, and G. Polya. Inequalities, volume 2. Cambridge at the University Press, 1952.


[2] Congming Li, John Villavert, An extension of the Hardy-Littlewood-Pólya inequality, Acta Mathematica Scientia, 31 (6), (2011), 2285-2288.


[3] Ze Cheng,Congming Li, An Extended Discrete Hardy-Littlewood-Sobolev Inequality, Discrete Contin. Dyn. Syst. 34 (5), (2014), 1951-1959 ([arXiv:1306.1649](https://arxiv.org/abs/1306.1649), [doi: 10.3934/dcds.2014.34.1951](http://dx.doi.org/10.3934/dcds.2014.34.1951)).



The discrete Hardy-Littlewood-Sobolev inequality
Write $$\sum\_{j\ne k,\,j,k=1,\dots,N} \frac{|a\_kb\_j|}{|j-k|^{\alpha}}=\sum\_{d=1}^{N}\frac{1}{d^{\alpha}} \Big(\sum\_{l=1}^{N-d} |a\_lb\_{l+d}| + \sum\_{l=1}^{N-d} |a\_{l+d}b\_{l}|\Big).$$


By Holder's inequality one has
$$\sum\_{l=1}^{N-d} |a\_lb\_{l+d}|\le \Big(\sum\_{l=1}^{N-d} |a\_l|^p \Big)^{1/p}\Big(\sum\_{l=1}^{N-d} |b\_{l+d}|^q\Big)^{1/q}\le \Big(\sum\_{k=1}^{N} |a\_k|^p \Big)^{1/p}\Big(\sum\_{k=1}^{N} |b\_{k}|^q\Big)^{1/q}$$ 
for every $d\in \{1,\dots,N\}$.


Similarly,
$$\sum\_{l=1}^{N-d} |a\_{l+d} b\_{l}|\le \Big(\sum\_{k=1}^{N} |a\_k|^p \Big)^{1/p}\Big(\sum\_{k=1}^{N} |b\_{k}|^q\Big)^{1/q}.$$ 


Therefore, if $\alpha>1$, then you get the estimate
$$\sum\_{j\ne k,\,j,k=1,\dots,N} \frac{|a\_kb\_j|}{|j-k|^{\alpha}} \le 2\zeta(\alpha)\Big(\sum\_{k=1}^{N} |a\_k|^p \Big)^{1/p}\Big(\sum\_{k=1}^{N} |b\_{k}|^q\Big)^{1/q}.$$


If $\alpha=1$, then you get
$$\sum\_{j\ne k,\,j,k=1,\dots,N} \frac{|a\_kb\_j|}{|j-k|^{\alpha}} \le C (\log N)\Big(\sum\_{k=1}^{N} |a\_k|^p \Big)^{1/p}\Big(\sum\_{k=1}^{N} |b\_{k}|^q\Big)^{1/q}$$
for some constant $C>0$.


