In the paper "Hilbert's inequality", Montgomery and Vaughan proved that a generalization of the discrete Hilbert transform is bounded in $\ell^2$. The inequality reads as follows
$$ \Big| \sum\_{k\neq n}\frac{a\_k \overline{b\_n}}{\lambda\_k-\lambda\_n} \Big| \leq \frac{\pi}{\delta} \Big(\sum\_{k=1}^{\infty} |a\_k |^2 \Big)^{1/2}\Big( \sum\_{n=1}^{\infty} |b\_n |^2 \Big)^{1/2}, $$ 
where $\{a\_k\}, \{ b\_n \}\in \ell^2 $, $ \lambda\_n $ is an increasing sequence of real numbers such that
$$ \delta:= \inf\_{k n}| \lambda\_k-\lambda\_{k+1}|. $$
Of course $\delta$ is assumed to be strictly positive. Also the constant appearing in the inequality $\pi/\delta$ is optimal. Quite surprisingly all proofs I managed to find use strongly the Hilbert space structure of $\ell^2$. 


Therefore I would like to ask if anything is known for the this inequality when considered on $\ell^p, p\neq 2$. Namely, is it true
$$\Big| \sum\_{k\neq n}\frac{a\_k \overline{b\_n}}{\lambda\_k-\lambda\_n} \Big| \leq C(p,\delta) \Big(\sum\_{k=1}^{\infty} |a\_k |^p \Big)^{1/p}\Big( \sum\_{n=1}^{\infty} |b\_n |^q \Big)^{1/q}, $$
where $1<p<\infty$, $q$ is the conjugate exponent of $p$ and $C(p,\delta)>0 ?$


(I wouldn't venture so far as to ask for an optimal constant in this case, given the difficulty of the problem for the classical discrete Hilbert transform.)



A generalization of discrete Hilbert's transform (Montgomery's inequality)
One can transfer the continuous $L^p$ theory to this discrete setting without difficulty.


Let's normalise $\sum\_k |a\_k|^p = \sum\_n |b\_n|^q = 1$. Consider the two quantities


$$ X\_1 := \sum\_{k \neq n} \frac{a\_k \overline{b\_n}}{\lambda\_k - \lambda\_n}$$


$$ X\_2 := \sum\_{k, n} p.v. \int\_{{\bf R}^2} \varphi(s) \varphi(t) \frac{a\_k \overline{b\_n}}{(\lambda\_k+s) - (\lambda\_n+t)}\ ds dt$$


where $\varphi$ is a bump function of total mass 1. It is not difficult to show that
$$ p.v. \int\_{\bf R} p.v. \int\_{\bf R} \varphi(s) \varphi(t) \frac{1}{(\lambda\_k+s) - (\lambda\_n+t)}\ dt$$
is equal to $\frac{1}{\lambda\_k - \lambda\_n} + O\_\delta( |k-n|^{-2} )$ when $k \neq n$ and $O\_\delta(1)$ when $k=n$, so we have $X\_1-X\_2 = O\_{p,\delta}(1)$ by Schur's test. One can also write $X\_2$ as
$$ p.v. \int\_{\bf R} \int\_{\bf R} \frac{f(x) g(y)}{x-y}\ dx dy$$
where
$$ f(x) := \sum\_k a\_k \varphi(x-\lambda\_k)$$
and
$$ g(y) := \sum\_n b\_n \varphi(x-\lambda\_n)$$
so from the $L^p$ boundedness of the continuous Hilbert transform we have $X\_2 = O\_{p,\delta}(1)$, and the claim follows.


