# Equivalence of norms in the space $H_\Delta(\Omega)$

Let $$\Omega \subset \mathbb{R}^n$$ is an open bounded domain with smooth boundary $$\Gamma$$. Consider the following space $$H_\Delta(\Omega)=\{u\in L^2(\Omega) : \Delta u \in L^2(\Omega)\},$$ with the norm $$\|u\|_{H_\Delta(\Omega)}^2 := \|u\|_{L^2(\Omega)}^2+\|\Delta u\|_{L^2(\Omega)}^2,$$

Some authors define the previous space as $$H^1_\Delta(\Omega)=\{u\in H^1(\Omega) : \Delta u \in L^2(\Omega)\},$$ with the norm $$\|u\|_{H^1_\Delta(\Omega)}^2 := \|u\|_{H^1(\Omega)}^2+\|\Delta u\|_{L^2(\Omega)}^2.$$

I'm wondering if the norms $$\|u\|_{H_\Delta(\Omega)}$$ and $$\|u\|_{H^1_\Delta(\Omega)}$$ are equivalent to $$\|u\|_{H^2(\Omega)}$$. What about the trace theorem in the spaces $$H_\Delta(\Omega)$$ and $$H^1_\Delta(\Omega)$$ ? Thank you.

• I'm asking if the norms $\|u\|_{H_\Delta(\Omega)} := \|u\|_{L^2(\Omega)} +\|\Delta u\|_{L^2(\Omega)}$ and $\|u\|_{H^2(\Omega)}$ are equivalent. – S. Maths Nov 10 '18 at 21:53