Can we show that $\left\|u\right\|_{L^2}^2+\left\|\Delta u\right\|_{L^2}^2\le C\left\|\Delta u\right\|_{L^2}^2$?

Question

Let

• $d\in\mathbb N$
• $\Lambda\subseteq\mathbb R^d$ be open
• $U:=\left\{u\in H_0^1(\Lambda):u\text{ admits a weak Laplacian }\Delta u\in L^2(\Lambda)\right\}$

Now, let $$\left\|u\right\|:=\left\|\Delta u\right\|_{L^2(\Lambda)}\;\;\;\text{for }u\in U$$ and $$|u|:=\sqrt{\left\|u\right\|_{L^2(\Lambda)}^2+\left\|u\right\|^2}\;\;\;\text{for }u\in U\;.$$

Can we show that there is a $C>0$ with $$|u|\le C\left\|u\right\|$$ for all $u\in U$?

Answer 1

For every $f \in L^2(\Omega)$ the (weak formulation of the) PDE $$ -\Delta u = f \text{ in } \Lambda; u = 0 \text{ on } \partial\Lambda$$ has a unique solution $u \in H_0^1(\Lambda)$ with $\|u\|_{H_0^1(\Lambda)} \le C \, \|f\|_{L^2(\Omega)}$.

Now, apply this result with $f = -\Delta u$.

Answer 2

Let $$\iota:\left(U,\left|\;\cdot\;\right|\right)\to\left(U,\left\|\;\cdot\;\right\|\right)$$ denote the identity. By definition, $\iota$ is a bounded linear operator. Since $\left(U,\left|\;\cdot\;\right|\right)$ and $\left(U,\left\|\;\cdot\;\right\|\right)$ are complete, we obtain that $\iota^{-1}$ is bounded, too, by the bounded inverse theorem.