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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?

1 Answers 1

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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.

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    Will the situation change, if we let $u\in H^2(\Lambda)$? Can we than find a $C_2>0$ with $$\left\|\Delta u\right\|_{L^2(\Lambda)}\le C_2\left\|u\right\|_{H^2(\Lambda)}\;?$$ – 2017-02-05
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    Indeed : by the usual definition of the Sobolev norm, you would have $\|u\|_{H^2} = (\sum_{\alpha : |\alpha| \le 2} \|D^{\alpha}u\|_{L^2}^2)^{1/2} \ge (\sum_{\alpha : |\alpha| = 2} \|D^{\alpha}u\|_{L^2}^2)^{1/2} \ge (\sum_{i=1}^d \|D_{ii}u\|_{L^2}^2)^{1/2}$. Interchanging sum and integral, then applying Cauchy-Schwartz (with the vectors $(D_{11}u, \dots, D_{dd}u)$ and $(1, \dots, 1)$) and then the triangle inequality, this is greater than some constant times $\| \Delta u \|_{L^2}$. – 2017-02-05
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    If, for example, $\partial\Lambda\in C^2$ or $\Lambda$ is convex, we're able to conclude $u\in H^2(\Lambda)$ as soon as we know that $u\in H_0^1(\Lambda)$ with $\Delta u\in L^2(\Lambda)$. Do you know whether the same conclusion can be made, if $\partial\Lambda$ is only Lipschitz (and, for example, $\Lambda$ is connected)? Or is there a counterexample? – 2017-02-05
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    I really don't know. If I remember correctly, the proof I know of the Poincaré inequality uses the hypothesis that the domain is a sufficiently smooth manifold with boundary to bring the considerations to the upper-half euclidean space, but this identification might possibly be taken less regular, I can't tell... This is not my field of expertise (if I have any) and I rarely think about minimal regularity conditions... Sorry. – 2017-02-06
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    You say above "this is greater than some constant times $\left\|\Delta u\right\|_{L^2}$". The constant is $1$, isn't it? – 2017-03-06
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    It is not 1, but $1/d$. Proof: $\sum_{i=1}^d \| D_{ii}u\|^2 = \sum_{i=1}^d \int |D_{ii}u(x)|^2dx = \int (\sum_{i=1} |D_{ii}u(x)|^2) dx = (1/d) \int (\sum_{i=1}^d 1^2)(\sum_{i=1} |D_{ii}u(x)|^2) dx \ge (1/d) \int (\sum_{i=1}^d 1.|D_{ii}u(x)|)^2 dx \ge (1/d) \int | \sum_{i=1}^d D_{ii}u(x)|^2 dx = (1/d) \int |\Delta u(x)|^2 dx = (1/d) \| \Delta u \|^2$. – 2017-03-06