A generalization of discrete Hilbert's transform (Montgomery's inequality)

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 $$1, $$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.)

• 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. Mar 22, 2020 at 17:37
• @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. Mar 22, 2020 at 17:52

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.

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 $$0 we get the $$L^p$$ boundedness properties from those of the Hilbert transform.

• Thank you very much for the answer. I think a similar reasoning can be found also here link Mar 23, 2020 at 13:20

In the same idea a paper "SHARP NORM INEQUALITIES FOR THE TRUNCATED HILBERT TRANSFORM" by Enrico Laeng .

• Thank you for your contribution, but which part of the paper you think it is related to the Montgomery inequality ? Apr 2, 2020 at 16:24
• I read this paper a long time ago, but maybe you can generalize Apr 5, 2020 at 12:28