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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.)

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  • $\begingroup$ What kind of inequality are you envisioning? If it involves $\ell^p$ and $\ell^q$ norms, you might as well use $\ell^2$. And I don't know how to imagine an inequality without the dual $\ell^q$ norm. $\endgroup$ – Lucia Mar 22 at 17:37
  • $\begingroup$ @Lucia The inequality involves $p$ and $q$ norms as you said (I edited the question so its more clear) but I don't really understand what you mean by saying you might as well use the $\ell^2$ norm. $\endgroup$ – an_ordinary_mathematician Mar 22 at 17:52
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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.

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Let me deal with a continuous situation. Let $\lambda:\mathbb R\rightarrow\mathbb R$ be an increasing $C^1$ diffeomorphism and let $u,v$ be in $L^2(\mathbb R)$. We have with $\phi=\lambda^{-1}$, $$ A=\iint \frac{u(y)\overline{u(x)}}{iπ(\lambda (x)-\lambda(y))} dx dy= \iint \frac{u(\phi(t))\overline{u(\phi(s))}}{iπ(s-t)}\phi'(t)\phi'(s) ds dt, $$ so that with $U(t)=u(\phi(t))\phi'(t)^{1/2}$, we find $$ A=\iint \frac{U(t)\phi'(t)^{1/2}\overline{U(s)\phi'(s)^{1/2}}}{iπ(s-t)}ds dt, $$ and thus assuming $ 0<m\le \phi'(t)\le M<+\infty $ we get the $L^p$ boundedness properties from those of the Hilbert transform.

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  • $\begingroup$ Thank you very much for the answer. I think a similar reasoning can be found also here link $\endgroup$ – an_ordinary_mathematician Mar 23 at 13:20
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In the same idea a paper "SHARP NORM INEQUALITIES FOR THE TRUNCATED HILBERT TRANSFORM" by Enrico Laeng .

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  • $\begingroup$ Thank you for your contribution, but which part of the paper you think it is related to the Montgomery inequality ? $\endgroup$ – an_ordinary_mathematician Apr 2 at 16:24
  • $\begingroup$ I read this paper a long time ago, but maybe you can generalize $\endgroup$ – Emmanuel Preissmann Apr 5 at 12:28

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