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.)