L2-norm and H2-norm

Question

Let $\Omega \subset \mathbb{R}^n$ be a domain. Let $u\in H^2(\Omega)$. My question is that how do we bound $\|\nabla^2u\|_{L^2(\Omega)}$ with $\|u\|_{H^2(\Omega)}$. The former term is a Hessian matrix, may I know what does the $L^2-$norm tell us about? Correct me if I possibly stated something wrong in my question. Thank you.

Answer 1

$\def\n #1{{\left\lVert #1 \right\rVert}}$ The $H^2$-norm consists of the $L^2$-norms of all derivatives: $$ \n u_{H^2} = \sqrt{\n u_{L^2}^2 + \n{∇u}_{L^2}^2 + \n{∇^2u}_{L^2}^2} $$ If you drop the first two terms in the sum, it surely only gets smaller, so $$ \n{∇^2u}_{L^2} = \sqrt{\n{∇^2 u}_{L^2}^2} \leq \n u_{H^2}. $$