I am trying to undestand the Corollary 2.2 from Osgood, Phillips and Sarnak (see http://citeseerx.ist.psu.edu/viewdoc/download?doi=, 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.