Let $\Omega$ be a bounded smooth domain. Is it true that $H_\Delta^1(\Omega):=\{ u\in L^2(\Omega) : \Delta u\in L^2(\Omega)\}\subset H^1(\Omega):=\{ u\in L^2(\Omega) : \nabla u\in \mathbf{L}^2(\Omega )\}$?
Asked
Active
Viewed 524 times
2
-
It is certainly not true for unbounded domains (e.g. $u(x)=\sum_i x_i$). – user31415926535 Jul 02 '15 at 15:40
-
@user31415926535 That function is not in $L^2(\Omega)$ for a general unbounded $\Omega$. – David C. Ullrich Jul 02 '15 at 16:22
-
Not that this quite answers your question, but it's certainly true for $\Omega=\mathbb R^n$; this is clear from looking at the Fourier transform. Hence it seems very likely true for sufficiently smooth domains. Not quite a proof... – David C. Ullrich Jul 02 '15 at 16:23
-
@DavidC.Ullrich silly me! – user31415926535 Jul 02 '15 at 17:18
1 Answers
3
No it is not: The mapping $u\mapsto (-\Delta u,\gamma u)$, where $\gamma$ is the trace operator, is an isomorphism from $H^1_\Delta(\Omega)$ onto $L^2(\Omega)\times H^{-1/2}(\partial\Omega)$ (see P. Grisvard, Elliptic Problems in Nonsmooth domains, Theorem 2.5.2.1). On the other hand, the trace operator $\gamma:H^1(\Omega)\rightarrow H^{1/2}(\partial\Omega)$ is surjective (see again for example P. Grisvard, Elliptic Problems in Nonsmooth domains, Theorem 1.5.1.2). Well, we are done, since $H^{1/2}(\partial\Omega)\subsetneq H^{-1/2}(\partial\Omega)$ and there is always a unique weak solution of the Dirichlet Laplace problem in $H^1$.
tks
- 400
- 1
- 12