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!
Best Answer
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.