Are there $L^2$ functions whose Laplacian is in $L^2$ yet its gradient is not in $\mathbf{L}^2$?

Question

Let $\Omega$ be a bounded smooth domain. Is it true that $H_\Delta^1(\Omega):=\{ u\in L^2(\Omega) : \Delta u\in L^2(\Omega)\}\subset H^1(\Omega):=\{ u\in L^2(\Omega) : \nabla u\in \mathbf{L}^2(\Omega )\}$?

Answer 1

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 $H^1$.