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