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arXiv 1404.2196math.CA
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与平方相关的最大Beurling变换

The maximal Beurling transform associated with squares

Anna Bosch-Camós, Joan Mateu, Joan Orobitg

摘要:

已知改进的Cotlar不等式$B^{*}f(z) \le C M(Bf)(z)$、$z\in\mathbb C$对于Beurling变换$B$、最大Beurling变换$B^{*}f(z)=$、$\displaystyle\sup_{\varepsilon >0}\left|\int_{|w|>\varepsilon}f(z-w) \frac{1}{w^2} \,dw\right|$、$z\in\mathbb C$和Hardy-Littlewood最大算子$M$成立。本文考虑与正方形有关的最大Beurling变换,即$B^{*}_Sf(z)=\displaystyle\sup_{\varepsilon >0}\left|\int_{w\notin Q(0,\varepsilon)}f(z-w) \frac{1}{w^2} \,dw \right|$,$z\in\mathbb C$,$Q(0,\varepsilon)$是边长为$\varepsilon$的坐标轴平行的正方形.证明了$B_{S}^{*}f(z) \le C M^2(Bf)(z)$,$z\in\mathbb C$,其中$M^2=M \circ M$是Hardy-Littlewood极大算子的迭代,且$M^2$不能被$M$所取代.

英文摘要:

It is known that the improved Cotlar's inequality $B^{*}f(z) \le C M(Bf)(z)$, $z\in\mathbb C$, holds for the Beurling transform $B$, the maximal Beurling transform $B^{*}f(z)=$ $\displaystyle\sup_{\varepsilon >0}\left|\int_{|w|>\varepsilon}f(z-w) \frac{1}{w^2} \,dw\right|$, $z\in\mathbb C$, and the Hardy--Littlewood maximal operator $M$. In this note we consider the maximal Beurling transform associated with squares, namely, $B^{*}_Sf(z)=\displaystyle\sup_{\varepsilon >0}\left|\int_{w\notin Q(0,\varepsilon)}f(z-w) \frac{1}{w^2} \,dw \right|$, $z\in\mathbb C$, $Q(0,\varepsilon)$ being the square with sides parallel to the coordinate axis of side length $\varepsilon$. We prove that $B_{S}^{*}f(z) \le C M^2(Bf)(z)$, $z\in\mathbb C$, where $M^2=M \circ M$ is the iteration of the Hardy--Littlewood maximal operator, and $M^2$ cannot be replaced by $M$.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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