A problem involving Laplace operator $\Omega$ is a bounded open set in $\mathbb R^n$, consider the number
$ r = \inf \{ \left\| {du} \right\|_{{L^2}(\Omega )}^2:u \in H_0^1(\Omega ),{\left\| u \right\|_{{L^2}(\Omega )}} = 1\}$
If for some $v\in H_0^1(\Omega )$ the infimum is achieved, then is $\Delta v\in L^2(\Omega)$?

Let
$$
f, g: H_0^1(\Omega) \to \mathbb{R}, f(u)=\|\nabla u\|_{L^2(\Omega)}^2,\ g(u)=\|u\|_{L^2(\Omega)}^2.
$$
Then
$$
r=\inf\{f(u):\ u \in H_0^1(\Omega),\ g(u)=1\}.
$$
If
$$
r=f(v),
$$
where $v$ belongs to $H_0^1(\Omega)$ and satisfies $g(v)=1$, then, there is a $\lambda \in \mathbb{R}$ such that 
$$
Df(v)\cdot h=\lambda Dg(u)\cdot h \quad \forall h \in H_0^1(\Omega),
$$
i.e.
$$
\int_\Omega\nabla v\cdot\nabla h=\lambda\int_\Omega vh \quad \forall h \in H_0^1(\Omega).
$$
The latter shows that $v$ is a weak solution of the PDE
$$
-\Delta u=\lambda u, \ u \in H_0^1(\Omega).
$$
Hence $\Delta v =-\lambda v=-f(v)v \in L^2(\Omega)$.
