$u\in H^1(\Omega)$ implies $\Delta u\in H^{-1}(\Omega)$?

Question

Let $\Omega\subset\mathbb{R}^n$ be a bounded domain with smooth boundary.

Is it true that if $u \in H^1(\Omega)$, then $\Delta u \in H^{-1}(\Omega)$?

Thanks

Answer 1

Hint: First check that the Laplacian operator on $u$ corresponds to multiplication of the Fourier transform $\hat u$ by ${\xi_1}^2+\dots+{\xi_n}^2$ (up to a sign that depends on the convention in the definition of $\Delta$). Then compare that with the definition of the Sobolev spaces in terms of Fourier transforms.

Answer 2

If $u\in H^1(\Omega)$ then, by definition, $\nabla u\in L^2(\Omega)^n$. On the othe hand, by definition, we have that $$\tag{1}\langle \Delta u,v\rangle=-\langle\nabla u,\nabla v\rangle, \ \forall \ v\in C_0^\infty(\Omega)$$

To conclude, you have to use $(1)$, the fact that $C_0^\infty(\Omega)$ is dense in $H_0^1(\Omega)$ and the fact that $\nabla u\in L^2(\Omega)^n$. Can you do this?

Remark: Also, I let for you to give the correct meaning of the expression in $(1)$.