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How can we prove this inequality?

For $q=\frac{np}{n-p}$ and $1\leq p<n$, there is a constant $c=c(n,p)$ such that if $u\in W^{1,p}(B_r)$, then $$\Bigg(\frac{1}{|B_r|}\int_{B_r}|u-\overline{u}_{B_r}|^q\Bigg)^{\frac{1}{q}}\leq cr\Bigg(\frac{1}{|B_r|}\int_{B_r}|Du|^p\Bigg)^{\frac{1}{p}}$$

I try to prove this inequality by using poincare inequality, G-N-S inequality and Jensen but I cannot get the conclusion. Can you give me a solution or hint?

Thank you!

kokodail
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1 Answers1

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Hint: First prove that $$\|u-\overline{u}_{B_r}\|_p\le c\|\nabla u\|_p,\ \forall u\in W^{1,p}(B_r)\tag{1}.$$

You can prove $(1)$ by contradiction and you will need Rellich-Kondrachov theorem. To conclude, you can use Sobolev inequality: $$\|u\|_{p^\star}\le c\|u\|_{1,p},\ \forall\ u \in W^{1,p}(B_r),$$

which will imply that $$\|u-\overline{u}_{B_r}\|_{p^\star}\le c(\|u-\overline{u}_{B_r}\|_p+\|\nabla u\|_p)\le c\|\nabla u\|_p,\ \forall\ u\in W^{1,p}(B_r).$$

Now try to match the constants and you will reach the desired inequality.

Tomás
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  • Isn't the last line meant to read: $$|u-\overline{u}{B_r}|{p^\star}\le c(|u-\overline{u}_{B_r}|_p+|\nabla u|_p)\le \color{red}{(c^2+c)}|\nabla u|_p,\ \forall\ u\in W^{1,p}(B_r).$$ – F.White Jul 17 '17 at 07:19
  • Also why isn't it $|u-\bar{u}{B_r}|{p^\star}\leq c|u-\bar{u}{B_r}|_p + |\nabla(u\color{red}{-\bar{u}{B_r}})|_p$ – F.White Jul 17 '17 at 07:20
  • The Sobolev inequality is wrong on Balls, just take $u=1$ to find the contradiction. – Guy Fsone Apr 11 '21 at 22:18