# eqiuvalent norms in $H_0^2$

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;

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There's also a similar result for functions without zero boundary data, but it's harder to prove: see this question – user53153 Feb 12 '13 at 18:21

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.
thanks a lot, the only thing i can't see is the poincare inequality for $H_0^2$. I thought that it was $∥u∥_{H^2_0}(D)^2≤C\sum_{|a|=2}∥D^au∥^2L_2(D)$..i must be missing something! – nikosp Feb 12 '13 at 18:48
In my notation, $\|D^2 u\|_{L^2(D)}$ means $$\|D^2 u\|_{L^2(D)} = \left( \sum_{|\alpha| = 2} \|D^\alpha u\|_{L^2(D)}^2 \right)^{1/2}.$$ This is the notation used throughout Evans's Partial Differential Equations. – Henry T. Horton Feb 12 '13 at 18:54