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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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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.

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