Compactness of the solution operator

Question

Let $\Omega$ be a smooth open bounded subset of $\mathbb{R}^n$. The bilinear form $$a(u,v)=\int_{\Omega}\frac{\partial u}{\partial x}\frac{\partial v}{\partial x}dx$$ is elliptic on $H^1(\Omega)/\mathbb{R}$. The correspoding problem $$a(u,v)=(f,v)$$ for all $v\in H^1(\Omega)/\mathbb{R}$ requires the compatibility condition $f\in\{g\in L^2(\Omega):\int_{\Omega}g=0\}$. Is the solution operator $$G:\{g\in L^2(\Omega):\int_{\Omega}g=0\}\to H^1(\Omega)/\mathbb{R}\to\{g\in L^2(\Omega):\int_{\Omega}g=0\}$$ compact?

Answer 1

The key question here is whether an embedding of the Sobolev space into the Lebesgue space $L^2(\Omega)$ be compact or not. Generally, for the embedding to be compact, it suffices to assume that domain $\Omega$ is bounded and satisfies a cone condition (for details, see "Sobolev spaces" by R. Adams).

Answer 2

Clearly, $$ (\nabla u,\nabla v)_{L^2(\Omega)}=(f,v)_{L^2(\Omega)}, \quad \text{for all $v$}, $$ and hence for $v=u$: $$ c|u|^2_{H^1(\Omega)}\le|\nabla u|^2_{L^2(\Omega)}=(\nabla u,\nabla u)_{L^2(\Omega)}=(f,u)_{L^2(\Omega)}\le |f|_{L^2 }|u|_{L^2}, $$ which implies that $f\mapsto u$ is compact from $L^2$ to $H^s$, for all $s<2$.