If $u\in H_0^1$ with $\Delta u\in L^2$, then $\left\|\Delta u\right\|_{L^2}\le C\left\|u\right\|_{H^1}$

Question

Let

• $d\in\mathbb N$
• $\Lambda\subseteq\mathbb R^d$ be bounded and open
• $u\in H_0^1(\Lambda)$ admit a weak Laplacian $\Delta u\in L^2(\Lambda)$

Using the definition of the weak Laplacian and the Poincaré inequality, I was able to show that $$C_1\left\|u\right\|_{H^1(\Lambda)}\le\left\|\Delta u\right\|_{L^2(\Lambda)}\tag1$$ for some $C_1>0$. Now, I want to show that $$\left\|\Delta u\right\|_{L^2(\Lambda)}\le C_2\left\|u\right\|_{H^1(\Lambda)}\tag2$$ for some $C_2>0$. How can we do that?

Answer 1

No, we can't.

Consider the case $d=1$ and $\Lambda = [0,\pi]$. The Dirichlet eigenfunctions on this domain are given by (up to normalization) $u_n(x) = \sin(nx)$ for $n \ge 1$. You can compute $\| \Delta u_n \|_{L^2} = n^2 \|u_n \|_{L^2}$ whereas $\|u_n \|_{H^1}$, regardless of the specific definition you use for this last norm, is bounded from above by some constant times $\|u_n \|_{L^2}+\| u'_n \|_{L^2} = (1+n)\|u_n\|_{L^2}$. As $n+1$ is $o(n^2)$, no such constant $C_2$ exists.

This example generalizes in any dimension.