Here $B_r$ presents the open ball of radius r in $R^3$. So I hope to know how to prove the following inequality. $$ \int_{B_r} |u|\le Cr^{\frac{9}{5}}\left(\int_{B_r} |u|^2\right)^{\frac{1}{5}}\left(\int_{B_r} |u|^3\right)^{\frac{1}{5}} $$ I did not figure out where these exponents come from. Could anyone show me?
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2$\begingroup$ Write $|u| = |u|^{2/5} |u|^{3/5} 1^{3/5}$ and apply Holder's inequality with exponents $(1/5,1/5,3/5)$. $\endgroup$– Peter HumphriesCommented Mar 15, 2022 at 14:11
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$\begingroup$ @PeterHumphries Hi thanks and just now I have come out about it... :) $\endgroup$– Xeh DengCommented Mar 15, 2022 at 14:14
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$\begingroup$ @PeterHumphries Excuse me and in fact I am now stuck at this question bound, and it would be much appreciative if any suggestions could be given. Thanks!~ $\endgroup$– Xeh DengCommented Mar 15, 2022 at 14:17
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I think that I have got the answer… it is quite a simple result from the Hölder inequality, $u^{\frac{2}{5}}u^{\frac{3}{5}}1$ with exponents $\frac{1}{5}$, $\frac{1}{5}$ and $\frac{3}{5}$….
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$\begingroup$ @LSpice Yeah.. I was just typing and did not notice that someone has already posted it.. :) $\endgroup$– Xeh DengCommented Mar 15, 2022 at 14:19