Complete convergence of the Hilbert transform

TL;DR

This paper establishes a complete convergence theorem for the Hilbert transform, providing bounds on the measure of the set where the transform exceeds a threshold, under certain summability and invariance conditions.

Contribution

It introduces a new complete convergence result for the Hilbert transform in both sequence and measure space settings, extending classical pointwise convergence results.

Findings

Proves a bound on the sum of measures where the Hilbert transform exceeds a threshold.

Extends classical results to a broader class of sequences and measure-preserving transformations.

Provides a unified approach to convergence in sequence and measure space contexts.

Abstract

Suppose that $\{a_j\}\in \ell^1$, and suppose that for any sequence $(t_n)$ of integers there exits a constant $C_1>0$ such that $$\sharp\left\{k\in\mathbb{Z}:\sup_{n\geq 1}\left|\sum_{i\in \mathcal{B}_n-t_n} \!\!\!\raise{1.9ex}\hbox{$\scriptsize\prime$}\; \frac{a_{k+i}}{i}\right|>\lambda\right\}\\ \leq C_1\sharp\left\{k\in\mathbb{Z}:\sup_{n\geq 1}\left|\sum_{i\in \mathcal{B}_n} \!\!\raise{1.9ex}\hbox{$\scriptsize\prime$}\; \frac{a_{k+i}}{i}\right|>\lambda\right\},$$ for all $\lambda >0$, where $\mathcal{B}_n=\{-n, -(n-1), -(n-2),\dots , n-2, n-1, n\}$. Then there is a constant $C_2>0$ which does not depend on the sequence $\{a_j\}$ such that $$\sum_{n=1}^\infty\sharp\left\{k\in\mathbb{Z}:\left|\sum_{i=-n}^{n} \!\!\raise{1.9ex}\hbox{$\scriptsize\prime$}\; \frac{a_{k+i}}{i}\right|>\lambda\right\}\leq\frac{C_2}{\lambda}\sum_{i=-\infty}^{\infty}|a_i|$$ for all $\lambda>0$. Let $(X,\mathscr{B},\mu )$ be a measure space, $\tau :X\to X$ an invertible measure-preserving transformation, and suppose that $f\in L^1(X)$ such that for any sequence $(t_n)$ of integers there exists a constant $C_1>0$ such that $$\mu\left\{ x: \sup_{n\geq 1}\left|\sum_{i\in \mathcal{B}_n-t_n}\!\!\!\raise{1.9ex}\hbox{$\scriptsize\prime$}\; \frac{f(\tau^ix)}{i}\right| >\lambda \right\}\leq C_1\mu\left\{x: \sup_{n\geq 1}\left|\sum_{i\in \mathcal{B}_n}\!\!\raise{1.9ex}\hbox{$\scriptsize\prime$}\; \frac{f(\tau^i x)}{i}\right|>\lambda \right\} $$ for all $\lambda >0$, where $\mathcal{B}_n=\{-n, -(n-1), -(n-2),\dots , n-2, n-1, n\}$. Then there exists a constant $C_2>0$ which does not depend on $f$ such that $$\sum_{n=1}^\infty\mu\left\{x:\left|\sum_{i=-n}^{n}\!\!\raise{1.9ex}\hbox{$\scriptsize\prime$} \;\frac{f(\tau^ix)}{i}\right|>\lambda\right\}\leq\frac{C_2}{\lambda}\|f\|_1$$ for all $\lambda >0$.