# Seminorm makes $W^{2,p}(\Omega)\cap W^{1,p}_0(\Omega)$ a Banach space

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 $$u$$.

• You could mimic the proof that the $W^{1,p}$-seminorm is a norm on $W^{1,p}_0$. – daw Oct 16 '19 at 15:07
• That elliptic regularity result is only true for smooth $\Omega$. – daw Oct 16 '19 at 15:07
• @daw Sorry. Yes. I assume $\Omega$ to be sufficiently nice, such as convex and $C^2$. – John Oct 16 '19 at 15:13
• @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 $\partial \Omega$. – John Oct 16 '19 at 15:16

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: $$1=\|u_n\|_{W^{1,p}}\to\|u\|_{W^{1,p}}=0$$.