if $v\in W_0^{1,p}(\Omega)$ then $\Delta v \in W^{-1,p'}(\Omega)$

Question

if $v\in W_0^{1,p}(\Omega)$ then $\Delta v \in W^{-1,p'}(\Omega)$

Here $C_0^\infty$ is dense in $ W_0^{1,p}(\Omega)$

and $ W^{-1,p'}(\Omega)$ is dual space of the $W_0^{1,p}(\Omega)$

what is mean $\Delta v \in W^{-1,p'}(\Omega)$

what we have to prove can some one explain me

Answer 1

Let $v \in W_0^{1,p}$. We have to show $\Delta v \in W^{-1,p'}$.

$$\|\Delta v\|_{W^{-1,p'}} =\sup_{\|\phi\|_{W_0^{1,p'}} \leq 1} |\langle \Delta v ,\phi \rangle| = \sup\left|\int \nabla v\nabla \phi \,dx\right| \leq \sup |\nabla v|_{L^p} |\nabla \phi|_{L^{p'}} \leq |v|_{W^{1,p}}. $$