{"title":"双曲空间上的球面最大算子","authors":"Peng Chen, Minxing Shen, Yunxiang Wang, Lixin Yan","doi":"arxiv-2408.02180","DOIUrl":null,"url":null,"abstract":"In this article we investigate $L^p$ boundedness of the spherical maximal\noperator $\\mathfrak{m}^\\alpha$ of (complex) order $\\alpha$ on the\n$n$-dimensional hyperbolic space $\\mathbb{H}^n$, which was introduced and\nstudied by Kohen [13]. We prove that when $n\\geq 2$, for $\\alpha\\in\\mathbb{R}$\nand $1<p<\\infty$, if \\begin{eqnarray*}\n\\|\\mathfrak{m}^\\alpha(f)\\|_{L^p(\\mathbb{H}^n)}\\leq C\\|f\\|_{L^p(\\mathbb{H}^n)},\n\\end{eqnarray*} then we must have $\\alpha>1-n+n/p$ for $1<p\\leq 2$; or\n$\\alpha\\geq \\max\\{1/p-(n-1)/2,(1-n)/p\\}$ for $2<p<\\infty$. Furthermore, we\nimprove the result of Kohen [13, Theorem 3] by showing that\n$\\mathfrak{m}^\\alpha$ is bounded on $L^p(\\mathbb{H}^n)$ provided that\n$\\mathop{\\mathrm{Re}} \\alpha> \\max \\{{(2-n)/p}-{1/(p p_n)}, \\ {(2-n)/p} -\n(p-2)/ [p p_n(p_n-2) ] \\} $ for $2\\leq p\\leq \\infty$, with $p_n=2(n+1)/(n-1)$\nfor $n\\geq 3$ and $p_n=4$ for $n=2$.","PeriodicalId":501036,"journal":{"name":"arXiv - MATH - Functional Analysis","volume":"12 1","pages":""},"PeriodicalIF":0.0000,"publicationDate":"2024-08-05","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"0","resultStr":"{\"title\":\"The spherical maximal operators on hyperbolic spaces\",\"authors\":\"Peng Chen, Minxing Shen, Yunxiang Wang, Lixin Yan\",\"doi\":\"arxiv-2408.02180\",\"DOIUrl\":null,\"url\":null,\"abstract\":\"In this article we investigate $L^p$ boundedness of the spherical maximal\\noperator $\\\\mathfrak{m}^\\\\alpha$ of (complex) order $\\\\alpha$ on the\\n$n$-dimensional hyperbolic space $\\\\mathbb{H}^n$, which was introduced and\\nstudied by Kohen [13]. We prove that when $n\\\\geq 2$, for $\\\\alpha\\\\in\\\\mathbb{R}$\\nand $1<p<\\\\infty$, if \\\\begin{eqnarray*}\\n\\\\|\\\\mathfrak{m}^\\\\alpha(f)\\\\|_{L^p(\\\\mathbb{H}^n)}\\\\leq C\\\\|f\\\\|_{L^p(\\\\mathbb{H}^n)},\\n\\\\end{eqnarray*} then we must have $\\\\alpha>1-n+n/p$ for $1<p\\\\leq 2$; or\\n$\\\\alpha\\\\geq \\\\max\\\\{1/p-(n-1)/2,(1-n)/p\\\\}$ for $2<p<\\\\infty$. Furthermore, we\\nimprove the result of Kohen [13, Theorem 3] by showing that\\n$\\\\mathfrak{m}^\\\\alpha$ is bounded on $L^p(\\\\mathbb{H}^n)$ provided that\\n$\\\\mathop{\\\\mathrm{Re}} \\\\alpha> \\\\max \\\\{{(2-n)/p}-{1/(p p_n)}, \\\\ {(2-n)/p} -\\n(p-2)/ [p p_n(p_n-2) ] \\\\} $ for $2\\\\leq p\\\\leq \\\\infty$, with $p_n=2(n+1)/(n-1)$\\nfor $n\\\\geq 3$ and $p_n=4$ for $n=2$.\",\"PeriodicalId\":501036,\"journal\":{\"name\":\"arXiv - MATH - Functional Analysis\",\"volume\":\"12 1\",\"pages\":\"\"},\"PeriodicalIF\":0.0000,\"publicationDate\":\"2024-08-05\",\"publicationTypes\":\"Journal Article\",\"fieldsOfStudy\":null,\"isOpenAccess\":false,\"openAccessPdf\":\"\",\"citationCount\":\"0\",\"resultStr\":null,\"platform\":\"Semanticscholar\",\"paperid\":null,\"PeriodicalName\":\"arXiv - MATH - Functional Analysis\",\"FirstCategoryId\":\"1085\",\"ListUrlMain\":\"https://doi.org/arxiv-2408.02180\",\"RegionNum\":0,\"RegionCategory\":null,\"ArticlePicture\":[],\"TitleCN\":null,\"AbstractTextCN\":null,\"PMCID\":null,\"EPubDate\":\"\",\"PubModel\":\"\",\"JCR\":\"\",\"JCRName\":\"\",\"Score\":null,\"Total\":0}","platform":"Semanticscholar","paperid":null,"PeriodicalName":"arXiv - MATH - Functional Analysis","FirstCategoryId":"1085","ListUrlMain":"https://doi.org/arxiv-2408.02180","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"","JCRName":"","Score":null,"Total":0}
The spherical maximal operators on hyperbolic spaces
In this article we investigate $L^p$ boundedness of the spherical maximal
operator $\mathfrak{m}^\alpha$ of (complex) order $\alpha$ on the
$n$-dimensional hyperbolic space $\mathbb{H}^n$, which was introduced and
studied by Kohen [13]. We prove that when $n\geq 2$, for $\alpha\in\mathbb{R}$
and $1
1-n+n/p$ for $1
\max \{{(2-n)/p}-{1/(p p_n)}, \ {(2-n)/p} -
(p-2)/ [p p_n(p_n-2) ] \} $ for $2\leq p\leq \infty$, with $p_n=2(n+1)/(n-1)$
for $n\geq 3$ and $p_n=4$ for $n=2$.