I have read that the laplacian is a closed operator in $W^{2,p}(\Omega)$,(that is, $\Delta : W^{2,p} \to L^p$) where $\Omega$ satisfies some conditions (I need the case $\Omega = \mathbb{R}^n$ so there shouldn't be any problems, since I think you don't need bounded domains). I need, in particular, to prove that the laplacian is closed in $W^{2,p}(\mathbb{R}^n)$ with $p \in (1,2)$. To do so, I would like t prove that the usual Sobolev norm and the norm defined by $|||u||| = ||u||_{L^p} +||\Delta u||_{L^p}$ are equivalent, and so I would conclude by completeness of $W^{2,p}$. The fact is, when I have $p=2$ and $n=1$ I can see that, because I have:
$$\int_{\mathbb{R}} (u')^2 = \int_{\mathbb{R}} (u')(u') = - \int_{\mathbb{R}} u u'' \leq ||u||_{L^2}^2 ||u''||_{L^2}^2 \leq \frac{1}{2} (||u||_{L^2}^2 + ||u''||_{L^2}^2)$$
and taking the square root I have:
$$||u'||_{L^2} \leq \frac{1}{\sqrt{2}}(||u||_{L^2} + ||u''||_{L^2})$$
I think there shouldn't be any problem to generalise this to dimension $n$, if $p=2$. However, if $p \neq 2$ I can't use the trick of writing $(u')^2$ as $(u')(u')$ and thus I don't know what to do.