# The discrete Hardy-Littlewood-Sobolev inequality

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, doi: 10.3934/dcds.2014.34.1951).

## 1 Answer

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 $$C>0$$.

• More generally, in the regime $\frac{1}{p} + \frac{1}{q} + \alpha > 2$ one can proceed using Young's convolution inequality and Holder's inequality. – Terry Tao Oct 12 '18 at 16:51