# $W_{0}^{2}(\Omega)=\{ f\in W_{0}^{1}(\Omega):\Delta f\in L^{2}(\Omega)\}?$

Let $$\Omega\subset\mathbb{R}^{n}$$ be an open bounded domain. Let $$W^{2}\left(\Omega\right)$$ be the usual Sobolev space $$W^{2}\left(\Omega\right)=\left\{ f\in L^{2}\left(\Omega\right):f,\partial_{i}f,\partial_{ij}^{2}f\in L^{2}\left(\Omega\right)\right\} .$$ Let $$W_{0}^{2}\left(\Omega\right)$$ be the closure of $$C_{0}^{\infty}\left(\Omega\right)$$ w.r.t. $$W^{2}\left(\Omega\right)$$-norm in $$W^{2}\left(\Omega\right)$$.

Question: is it true that $$W_{0}^{2}\left(\Omega\right)=\left\{ f\in W_{0}^{1}\left(\Omega\right):\Delta f\in L^{2}\left(\Omega\right)\right\} ?$$ That is, the Laplacian controls $$\partial_{ij}^{2}$$ in $$L^{2}\left(\Omega\right)$$.

No, this space is not the same as $$W_0^2$$. Note that $$W_0^2$$ and $$W_0^1 \cap W^2$$ are different spaces. In $$W_0^2$$ we additionally the information that $$\nabla f \in (H_0^1(\Omega))^n$$, i.e. that the first derivatives vanish on the boundary of $$\partial \Omega$$ in some sense (with the trace operator, so we have $$\text{Tr}(\nabla f) = \partial_\nu f=0$$ where $$\nu$$ is some outer normal to the sufficiently smooth domain $$\Omega$$).
You lose this important property in your space. But, your proposed space is indeed the same as $$W_0^1 \cap W^2$$, which was proven for example here.
• The last paragraph is only true if $\Omega$ is nice enough. – gerw Mar 25 at 7:17