TITLE: A generalization of discrete Hilbert's transform (Montgomery's inequality)
QUESTION [7 upvotes]: 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.)

REPLY [8 votes]: 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.