Sharp Weak Type Estimates for a Family of Soria Bases

被引:0
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作者
Dmitry Dmitrishin
Paul Hagelstein
Alex Stokolos
机构
[1] Odessa National Polytechnic University,Department of Applied Mathematics
[2] Baylor University,Department of Mathematics
[3] Georgia Southern University,Department of Mathematical Sciences
来源
The Journal of Geometric Analysis | 2022年 / 32卷
关键词
Maximal functions; Differentiation basis; Weak type inequalities; Covering lemmas; Primary 42B25;
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摘要
Let B\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathcal {B}}$$\end{document} be a collection of rectangular parallelepipeds in R3\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb {R}}^3$$\end{document} whose sides are parallel to the coordinate axes and such that B\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathcal {B}}$$\end{document} contains parallelepipeds with side lengths of the form s,2Ns,t\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$s, \frac{2^N}{s} , t $$\end{document}, where s,t>0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$s, t > 0$$\end{document} and N lies in a nonempty subset S of the natural numbers. We show that if S is an infinite set, then the associated geometric maximal operator MB\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$M_{\mathcal {B}}$$\end{document} satisfies the weak type estimate x∈R3:MBf(x)>α≤C∫R3|f|α1+log+|f|α2,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \left| \left\{ x \in {\mathbb {R}}^3 : M_{{\mathcal {B}}}f(x) > \alpha \right\} \right| \le C \int \nolimits _{{\mathbb {R}}^3} \frac{|f|}{\alpha } \left( 1 + \log ^+ \frac{|f|}{\alpha }\right) ^{2}, \end{aligned}$$\end{document}but does not satisfy an estimate of the form x∈R3:MBf(x)>α≤C∫R3ϕ|f|α\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \left| \left\{ x \in {\mathbb {R}}^3 : M_{{\mathcal {B}}}f(x) > \alpha \right\} \right| \le C \int \nolimits _{{\mathbb {R}}^3} \phi \left( \frac{|f|}{\alpha }\right) \end{aligned}$$\end{document}for any convex increasing function ϕ:[0,∞)→[0,∞)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\phi : \mathbb [0, \infty ) \rightarrow [0, \infty )$$\end{document} satisfying the condition limx→∞ϕ(x)x(log(1+x))2=0.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \lim _{x \rightarrow \infty }\frac{\phi (x)}{x (\log (1 + x))^2} = 0\;. \end{aligned}$$\end{document}
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