Relation between the negative Laplacian operator and its square root

83 Views Asked by At

Let $\Omega \subset \mathbb{R}^n (n\geq 1)$ be an open bounded set with regular boundary.

Denote by $-\Delta$ the negative Laplacian operator with domain $$D(-\Delta):=H^2(\Omega) \cap H^1_0(\Omega),$$ and by $-\Delta^{\frac{1}{2}}$ its square root with domain $$ D(-\Delta^{\frac{1}{2}}):=H^1_0(\Omega). $$ Is there a relation (an inequality) between $\Vert -\Delta u \Vert_{L^2(\Omega)}$ and $\Vert -\Delta^{\frac{1}{2}} u \Vert_{L^2(\Omega)},$ where $u\in D(-\Delta)$?