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I am trying to undestand the Corollary 2.2 from Osgood, Phillips and Sarnak (see http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.486.558&rep=rep1&type=pdf), that is, if $u \in W^{1}(S^{2})'$ with mean value zero and antipodal symmetry then $$\int_{S^{2}}|\nabla_0 u_s|^{2}\frac{dA_0}{4\pi} \le \frac{1}{2}\int_{S^{2}}|\nabla_0 u|^{2}\frac{dA_0}{4\pi},$$ where $u_s$ is the symmetric decreasing rearrangement of $u$.

They suggest to use the usual symmetrization procedure with the isoperimetric inequality for $S^{2}$ ( that is, $L^{2} \ge A(4\pi-A)$), replaced by a sharper form holding for domains with antipodal symmetry: $$L^{2} \ge 2A(4\pi-A).$$

From Pólya-Szego there is a following result: If $D$ is a open domain in $\mathbb{R}^{n}$ then $$\int_{D^{*}} |\nabla u^{*}|dx \le \int_{D}|\nabla u|dx,$$ where $D^{*}$ is the symmetric rearrangement of $D$, that is, $D^{*}$ is the open centered ball whose volume agrees with $D$.

How can I match the Polya-Szego inequality with the sharp isoperimetric inequality to prove the first inequality?

Thanks in Advance.

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