Why is the laplacian a closed operator in $W^{2,p}(\mathbb{R}^n)$?

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.

• Why do you want to show equivalence of these two norms? Isn't it enough to say that this is a bounded operator? – Michał Miśkiewicz Nov 18 '18 at 23:41
• Well, I know that if an operator is bounded then it is closed. I don’t know a proof of this fact, though. I would like the equivalence to conclude easily, but if the proof bounded $\implies$ closed is easier I would use that, instead. – tommy1996q Nov 19 '18 at 7:01
• As far as I know, this is more of a tautology than a proof. What do you mean by closed here? – Michał Miśkiewicz Nov 19 '18 at 9:28
• Let $B_1$ and $B_2$ be Banach spaces. A linear operator $T: B_1 \to B_2$ is closed if given a sequence $\{x_n\}$ converging in $B_1$ to $x$ such that $\{Tx_n \}$ converges in $B_2$ to $y$, then $x$ belongs to the domain of $T$ and $Tx=y$ – tommy1996q Nov 19 '18 at 13:04
• Continuous implies closed should follow from the closed graph theorem, but I should prove boundedness then – tommy1996q Nov 19 '18 at 13:04