We know the norm $\|\Delta u\|_{L^2(\Omega)}$ is an equivalent norm of $H^2$ norm in space $H_0^1(\Omega)\cap H^2(\Omega)$ where $\Omega$ open bounded with smooth boundary.
Now let's generalize this problem. Given $\Omega\subset \mathbb R^N$ open bounded with smooth boundary. Define the elliptic operator $$ Lu:=-\partial_j(a_{ij}\partial_i u)+cu $$ where $A:=(a_{ij})$ satisfies uniform elliptic condition, i.e., $A\xi\xi\geq \theta |\xi|^2$ for $\theta>0$. We also assume $A$, $c\in C^\infty(\bar \Omega)$ with $c\geq 0$ so that $L$ always have a unique solution.
Now we define $$ H_L^m(\Omega):=\{u\in H^m(\Omega),\, L^{(l)}u\in H_0^1(\Omega),\,0\leq l<m/2\} $$ where $L^{(0)}u=u$, $L^{(1)}u=Lu$, $L^{(2)}u=L(Lu)$.
I want to prove that the norm $H_0^m$ and $H^m$ is an equivalent norm on $H_L^m$. That is, I want to prove that there exists $C_1>0$ $C_2>0$ such that for all $u\in H_L^m$, we have $$ C_1\|u\|_{H_0^m(\Omega)}\leq \|u\|_{H^m(\Omega)}\leq C_2\|u\|_{H_0^m(\Omega)}\tag 1$$
What I tried so far:
It is easy to have $$C_1\|u\|_{H_0^m(\Omega)}\leq \|u\|_{H^m(\Omega)}$$ Now for the other inequality, we only need to prove that $H_L^m$ is a Banach space under norm $H_0^m$ and by open mapping theorem we done.
To start, we notice that $H_L^0=L^2$ and $H_L^1=H_0^1$, $H_L^2=H_0^1\cap H^2$, hence the case that $m=0,1,2$ is obvious.
Now, we start to use induction to assume this result $(1)$ hold for all $m-1$ and we work on $m$ case.
We take $(u_n)\subset H_L^m$ as a Cauchy sequence in $H_0^m$ norm. i.e., $$ \|u_n-u_m\|_{H_0^m(\Omega)}\to 0 \tag 2$$ and I want to show there exists $u\in H_L^m$ such that $$ \|u_n-u\|_{H_0^m(\Omega)}\to 0 $$
In order to obtain an candidate for the limit of sequence $(u_n)$, we formulate the following trivial PDE \begin{cases} L(u_n-u_)m=L(u_n-u_m)\\ u_n-u_m=0 \end{cases} and by regularity we have $$ \|u_n-u_m\|_{H^m(\Omega)}\leq C\|Lu_n-Lu_m\|_{H^{m-2}}\leq C\|Lu_n-Lu_m\|_{H_0^{m-2}}$$ and the last inequality was obtained by induction assumption.
Now if we could prove that $$ \|Lu_n-Lu_m\|_{H_0^{m-2}}\to 0 \tag 3$$ we would be done. I think we could proving $(3)$ by using fact $(2)$ above but I can not prove it. The difficulties are $\|Lu\|_{H_0^{m-2}}$ have more terms then $\|u\|_{H_0^m}$ and I can not get rid of them.
Any help is really welcome. Thank you!
How do you define $\|\cdot\|_{H^m_0}$? For even $m$ (in order to fit with the original problem) you could say $\|\cdot\|_{H^m_0}=\|L^{m/2}\cdot\|_{L^2}$. In that case you directly get $$\|Lu\|_{H^{m-2}_0}=\|L^{m/2-1}Lu\|_{L^2}=\|L^{m/2}u\|_{L^2}=\|u\|_{H^m_0}.$$ You could also use regularity estimates for elliptic operators to get the same result: Although you will not find the result verbatim, take a look at Chapter 13 of Wloka's PDE or Chapter 2 of Gazzola, Grunau and Sweers' Polyharmonic boundary value problems.