eqiuvalent norms in $H_0^2$

Question

I have found that the $H^2(D)$ norm of a field with zero Cauchy data on $\partial D$ (i.e. in $H_0^2(D)$) is equivalent to the $L^2(D)$ norm of its Laplacian, where D is simply connected with smooth boundary in $\mathbb{R}^n$. How can i prove this using Poincare inequality;

Answer 1

Note that we have the following Poincaré inequality for $H^2_0(D)$: $$\|u\|_{H^2_0(D)} \leq C \|D^2 u\|_{L^2(D)}^2.$$ This is obtained by chaining the Poincaré inequality for $u$ with the Poincaré inequality for $Du$. Therefore we may take $$\|u\|_\ast^2 = \|D^2 u\|_{L^2(D)}^2$$ as our norm on $H^2_0(D)$, as it is equivalent to the standard $H^2_0(D)$ norm.

We claim that $$\|\Delta u\|_{L^2(D)} = \|D^2 u\|_{L^2(D)} = \|u\|_\ast$$ for any $u \in H_0^2(D)$. To see this, first consider $u \in C_0^\infty(D)$. Then integration by parts and commutativity of partial derivatives for smooth functions implies $$\int_D u_{x_i x_i} u_{x_j x_j} ~dx = -\int_D u_{x_i} u_{x_j x_j x_i} ~dx = - \int_D u_{x_i} u_{x_j x_i x_j} ~dx = \int_D u_{x_i x_j} u_{x_i x_j} ~dx$$ for all $1 \leq i, j \leq n$. Summing over all $i$ and $j$ then gives $$\|\Delta u\|_{L^2(D)} = \|D^2 u\|_{L^2(D)}$$ for all $u \in C_0^\infty(D)$. Since $C_0^\infty(D)$ is dense in $H_0^2(D)$, passing to limits we find that $$\|\Delta u\|_{L^2(D)} = \|D^2 u\|_{L^2(D)} \text{ for all } u \in H_0^2(D).$$ This gives the desired equality of norms.