Convolution with measures on hypersurfaces
Convolution with measures on hypersurfaces
复制标题
与超曲面上的测量进行卷积
DOI：
10.1017/s0305004100004552
复制
发表时间：
2000
影响因子：
0.8
通讯作者：
D. Oberlin
中科院分区：
文献类型：
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作者：
D. Oberlin
设S是n中光滑超曲面，具有表面积测度ds和高斯曲率κ（s）.定义卷积算子T的公式在这里为适当的函数f在n上。我们感兴趣的是T的Lp − Lq映射性质。写[Sscr ]为T的类型集，这里的集合公式众所周知（参见[O 1]），[Sscr ]包含在顶点为（0，0），（1，1）和（n/（n+1），1/（n+1））的闭三角形[Tscr ]中。本文主要研究以下公式的估计：估计（1）很有趣，因为它是T的L（n+1）/n − Ln+1有界性的弱替代。例如，如果S是紧的并且（1）成立，那么众所周知的论证表明[Sscr ]与全三角形[Tscr ]的差别最多为点（n/（n + 1），1/（n + 1））。我们的主要结果是一个条件足以暗示（1）。它的陈述需要以下定义。
Let S be a smooth hypersurface in ℝn with surface area measure ds and Gaussian curvature κ(s). Define the convolution operator T by formula here for suitable functions f on ℝn. We are interested in the Lp − Lq mapping properties of T. Write [Sscr ] for the type set of T, the set formula here It is well known (see, e.g. [O1]) that [Sscr ] is contained in the closed triangle [Tscr ] with vertices (0, 0), (1, 1) and (n/(n+1), 1/(n+1)). This paper is concerned with estimates of the form formula here The estimate (1) is interesting because it serves as a weak substitute for the L(n+1)/n − Ln+1 boundedness of T. For example, if S is compact and (1) holds, then well-known arguments show that [Sscr ] differs from the full triangle [Tscr ] by at most the point (n/(n + 1), 1/(n + 1)). Our main result is a condition sufficient to imply (1). Its statement requires the following definition.