Let $B_R$ be a ball of radius $R$ in $\mathbb{R}^d$. Under what conditions does the Sobolev embedding $\dot{W}^{1,p}_0(B_R) \hookrightarrow L^{p^*}(B_R)$ holds, where $p^* = \frac{np}{n-p}$?
Here $\dot{W}^{1,p}_0(B_R)$ is the homogeneous Sobolev space, and $1<p<\infty$. Thanks!
In the answer, the following version of the Poincare inequality is stated:
\begin{equation}
\int_\Omega |u|^p\,dx \leq C(\Omega,p,n)\int_\Omega |\nabla u|^p\,dx,
\end{equation}
if $\Omega$ is a bounded domain in $\mathbb{R}^n$. It manifests that $\dot{W}^{1,p}(\Omega) = W^{1,p}(\Omega)$, of which I am convinced.
Now, in the case $\Omega = B_R$, I am interested in getting explicit dependence of the constant with respect to $R$. For example, is it possible to get an estimate of the form
\begin{equation}
\bigg(\int_{B_R} |u|^{p^*}\,dx\bigg)^{1/p^*} \leq C(p,n) R^\beta\bigg(\int_{B_R} |\nabla u|^p\,dx\bigg)^{1/p},
\end{equation}
where $u \in W^{1,p}(B_R)$ with $p \in ]1,n[$ and $\beta \in \mathbb{R}$ is some scaling index?
(I know that, roughly speaking, for Poincare inequality over $B_R$ we will get an $R$ in the constant, while the Sobolev inequality corresponding to $W^{1,p}(B_R) \hookrightarrow L^{p^*}(B_R)$ should be independent of $R$. But, in the case without the average term, I wonder if we can still get something neat?)
Thank you!
Best Answer
If $\Omega$ is a bounded domain, then we have the following Poincaré inequality: There is a constant $C>0$ such that $$ \int_{\Omega}|u|^p\leq C\int_{\Omega}|\nabla u|^p $$ for all $u\in C_c^\infty(\Omega)$. In other words, $\|\nabla\cdot\|_p$ and $\|\cdot\|_p+\|\nabla\cdot\|_p$ are equivalent norms on $C_c^\infty(\Omega)$ and thus $W^{1,p}_0(\Omega)=\dot{W}^{1,p}_0(\Omega)$ with equivalent norms. Hence you can apply the classical embedding theorems for Sobolev spaces that Evan William Chandra mentioned in the comments to get the embedding you want.