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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.

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  • $\begingroup$ Why do you want to show equivalence of these two norms? Isn't it enough to say that this is a bounded operator? $\endgroup$ – Michał Miśkiewicz Nov 18 '18 at 23:41
  • $\begingroup$ 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. $\endgroup$ – tommy1996q Nov 19 '18 at 7:01
  • $\begingroup$ As far as I know, this is more of a tautology than a proof. What do you mean by closed here? $\endgroup$ – Michał Miśkiewicz Nov 19 '18 at 9:28
  • $\begingroup$ 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 $Tx=y$ $\endgroup$ – tommy1996q Nov 19 '18 at 13:04
  • $\begingroup$ Continuous implies closed should follow from the closed graph theorem, but I should prove boundedness then $\endgroup$ – tommy1996q Nov 19 '18 at 13:04

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