Equivalent norms in $H^2(\Omega)\cap H^1_0(\Omega)$

Question

Let $\Omega \subset \mathbb{R}^3$ a smooth and bounded domain, and define in the space $H=H^2(\Omega)\cap H^1_0(\Omega)$ the norm $$\|u\|_H = \|\Delta u\|_2$$ for envery $u\in H.$

My question is, the norm defined above and the norm $$\|u\|_H = \left( \|\Delta u\|^2_2 + \|\nabla u\|^2_2 \right)^\frac{1}{2}$$

are equivalents? How can I show this?

Thanks for any help!

Answer 1

Yes, these norms are equivalent. One inequality is obvious, so let's prove $\|\nabla u\|_2^2+\|\Delta u\|_2^2\leq C\|\Delta u\|_2^2$. Since $u$ vanishes on the boundary, integration by parts gives $$ \|\nabla u\|_2^2=\int_\Omega u(-\Delta u)\,dx\leq \frac 1 2\|u\|_2^2+\frac 1 2\|\Delta u\|_2^2. $$ By the Poincaré inequality, $\|u\|_2\leq C\|\Delta u\|_2$. Thus $\|\nabla u\|_2^2\leq \tilde C\|\Delta u\|_2$.

Answer 2

WARNING - THIS IS NOT CORRECT. This answer is not correct, but it does contain some perspective. Following my usual policy, according to which we learn at least as much from mistakes as from correct solutions, I am not deleting it.

This answer contains the first obvious attempt one may perform to solve this problem, and the wall one hits. I am not sure it is really a mistake, but for sure it is a difficulty. MaoWao's answer corrects this approach and produces a right answer.

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NOTE. Here $C$ denotes a generic positive constant whose value may change from line to line.

Apply the inequality of Poincaré to the derivatives $\partial_j u$, you will see that (The following formula is not properly justified , but I am not really sure it is wrong - see comments) $$\lVert \nabla f \rVert_2\le C\lVert D^2 f\rVert_2, $$ where $D^2 f(x)=(\partial_{ij}f)_{i, j =1, 2, 3}$ is the Hessian matrix. And now it remains to show that $$ \lVert D^2 f\rVert_2\le C\lVert \Delta f\rVert_2, $$ which is done via integration by parts when $f\in C^\infty$, then extended by density. This last computation is in the book on PDEs of Evans, in the very beginning of the chapter on elliptic regularity.