While reading this article on near $L^1$ estimates for the spherical lacunary maximal function, I came across the estimate $$ |\partial^{\gamma} (\widetilde{\sigma} \ast \sigma)(x)| \lesssim |x|^{-(1 + |\gamma|)} \chi_{|x| \leq 2}(x) \qquad \text{for $\gamma \in \mathbb{N}_{0}^{d}$ and $x \in \mathbb{R}^d$}.$$ Here, $\sigma$ is a measure on the unit sphere $\mathbb{S}^{d-1}$, given by $$\langle{\sigma, \phi}\rangle := \int \phi(y) \chi(y) d \theta(y),$$ with $\chi$ being a smooth function with a compact support contained inside a ball of radius $\leq 1/2$. Moreover, its refection $\widetilde{\sigma}$ is defined by the action $$\langle{\widetilde{\sigma}, \phi}\rangle := \langle{\sigma, \tilde{\phi}}\rangle, \qquad \text{where} \qquad \tilde{\phi}(x) = \phi(-x).$$ The authors mention that the formula is a standard one, but I have been unable to find a reference or a proof. Any help would be appreciated. Thanks!
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$\begingroup$ Could you please explain what do you mean by a reflection of a measure? $\endgroup$– Aleksei KulikovCommented Nov 19 at 5:50
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$\begingroup$ @AlekseiKulikov, I have edited the question to include the definitions. $\endgroup$– ZygmundCommented Nov 19 at 6:04
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1$\begingroup$ To be honest, now I am even more confused, as I do not seem to understand what is $\sigma$ (you say it is a measure on a sphere, but then talk about a function $\chi$ with a support in a ball of radius $\frac{1}{2}$). $\endgroup$– Aleksei KulikovCommented Nov 19 at 7:46
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$\begingroup$ What is the meaning of $\partial^{\gamma} (\widetilde{\sigma} \ast \sigma)(x)$, given that $\widetilde{\sigma} \ast \sigma$ is a measure? $\endgroup$– Iosif PinelisCommented Nov 19 at 13:02
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$\begingroup$ @AlekseiKulikov, sorry about the confusion. $d\theta$ is the standard spherical measure on $\mathbb{S}^{d-1}$. The purpose of $\chi$ is to localise the measure. If it is supported on a ball centered at the north pole with radius $\leq 1/2$, then we can see that $\sigma$ is supported on the half sphere. Does this clarify your doubt? $\endgroup$– ZygmundCommented Nov 21 at 2:33
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