2
$\begingroup$

$\Omega$ 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 $\Delta v\in L^2(\Omega)$?

  • 0
    I guess you meant $\nabla u$ instead of $du$!2012-09-21

1 Answers 1

3

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 $\Delta v =-\lambda v=-f(v)v \in L^2(\Omega)$.