1
$\begingroup$

In a book I am currently reading, it states that: for $p\ge 2$, set $X$ for the Banach space $W^{2,p}(\Omega)\cap W^{1,p}_0(\Omega)$ equipped with the norm $$ \|u\|_X=\left( \int_\Omega\bigg(\sum_{i,j=1 }^n|D_{ij}u|^2\bigg)^{p/2}\,dx\right)^{1/p}. $$ Here $\Omega$ is a (sufficiently smooth) bounded open convex subset of $\mathbb{R}^n$, $W^{2,p}(\Omega)$ is the Sobolev $L^p$ space, $W^{1,p}_0(\Omega)$ is the closure of $C_0^\infty)(\Omega)$ function with respect to $W^{1,p}$-norm.

I was wondering, whether $(X,\|\cdot\|_X)$ is still a Banach space? In other words, is $\|\cdot\|_X$ an equivalent norm to the classical Sobolev norm?


I know for all $u\in X$, the elliptic regularity shows that for sufficiently large $\lambda$, $$ \|u\|_{W^{2,p}}\le C(\|-\Delta u+\lambda u\|_{L^p}), $$ which shows that $$ \|u\|_{W^{2,p}}\le C( \|u\|_{X}+\|u\|_{L^{p}}). $$ But I don't know how to eliminate the $L^p$ norm of $u$.

$\endgroup$
4
  • $\begingroup$ You could mimic the proof that the $W^{1,p}$-seminorm is a norm on $W^{1,p}_0$. $\endgroup$ – daw Oct 16 '19 at 15:07
  • $\begingroup$ That elliptic regularity result is only true for smooth $\Omega$. $\endgroup$ – daw Oct 16 '19 at 15:07
  • $\begingroup$ @daw Sorry. Yes. I assume $\Omega$ to be sufficiently nice, such as convex and $C^2$. $\endgroup$ – John Oct 16 '19 at 15:13
  • $\begingroup$ @daw I thought $W^{1,p}$-seminorm is a norm on $W^{1,p}_0$ due to the Poincare inequality. Here I cannot apply Poincare inequality to $Du$ since I only assume $u=0$ on $\partial \Omega$. $\endgroup$ – John Oct 16 '19 at 15:16
1
$\begingroup$

I claim that $\|\cdot\|_X$ is equivalent to $\|\cdot\|_{W^{2,p}}$ for $p\in (1,\infty)$. It remains to show that there exists $c>0$ such that $$ \|u\|_{W^{1,p}} \le c \|u\|_X \quad \forall u\in W^{2,p}\cap W^{1,p}_0. $$ Assume not. Then for every $n$ there is $u_n$ such that $$ \|u_n\|_{W^{1,p}} >n \|u_n\|_X. $$ Wlog $\|u_n\|_{W^{1,p}}=1$. Then $(u_n)$ is bounded in $W^{2,p}$, which is reflexive. After extracting subsequence, we have $u_n \rightharpoonup u$ in $W^{2,p}$ and by compact embeddings $u_n \to u$ in $W^{1,p}$.

By the construction of $u_n$, $\|u_n\|_X \to 0$, which implies that $D_{ij}u=0$ for all $i,j$. Hence $u$ is a polynomial of degree $1$. Due to the boundary conditions, we have $u=0$. This leads to a contradiction: $1=\|u_n\|_{W^{1,p}}\to\|u\|_{W^{1,p}}=0$.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.