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.

  • $\begingroup$ The last paragraph is only true if $\Omega$ is nice enough. $\endgroup$ – gerw Mar 25 at 7:17

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