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Let $L^2=L^2(\mathbb{R}^n)$

$-\Delta:D(-\Delta)\subset L^2\to\subset L^2$ by $-\Delta u=\mathcal{F}^{-1}(|\xi|^2\widehat{u}(\xi))$ and $D(-\Delta)=H^{2,2}=\left\{u\in L^2: \mathcal{F}^{-1}((1+|\xi|)^2\widehat{u}(\xi))\in L^2\right\}$ ($L^2$-Sobolev space of order 2 [Wong Introduction pseudo differential operator])

I have read that the Laplacian is bounded in the Sobolev space (for example, I read it here Laplace operator defined on a Sobolev space) but I cannot see this in a simple way. I don't know if I'm misunderstanding something about the norms. My question is:

Question 1. Is $-\Delta$ a bounded operator? i.e. $\left\|-\Delta u\right\|_{L^2}\leq C\left\|u\right\|_{L^2}$?

eraldcoil
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  • This question has already been asked multiple times, see e.g. https://math.stackexchange.com/questions/1358255/is-the-laplacian-an-unbounded-operator. – stange Jul 02 '23 at 06:59
  • It is that from what I understood, only $\left|-\Delta u\right|{L^2}\leq C \left|u\right|{H^{2,2}}$ is valid but not $\left|-\Delta u\right|{L^2}\leq C \left|u\right|{L^2}$. Does this mean that it would not be bounded with the $L^2$ norm but with the Sobolev norm? but not – eraldcoil Jul 02 '23 at 18:13
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    Yes, since if you're considering $D(-\Delta)\subset L^2$, $-\Delta u$ is not defined for every $u\in L^2$, but only for those which satisfy additionally $u\in H^2$, hence it can't be bounded in that case. – stange Jul 02 '23 at 18:20
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    but $D(-\Delta)$ is dense in $L^2$ so $-\Delta$ would be bounded over $L^2$? – eraldcoil Jul 02 '23 at 18:27
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    Yes, $H^2$ is dense in $L^2$, but $-\Delta u$ is not defined for every $u\in L^2$. While you may approximate $u\in L^2$ with $(u_j)\subset H^2$, this does not mean that $-\Delta u_j\rightarrow -\Delta u$, as again, the right hand-side might not even exist, $-\Delta u_j$ might possibly even explode in the $L^2$ norm as $j\rightarrow\infty$. – stange Jul 02 '23 at 18:32

1 Answers1

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No, it is not bounded with respect to $L^2$ norms. For example, fix a non-trivial, say Schwartz or a smooth compactly supported function $f$, and consider for each $r>0$, the rescaling $f_r(x)=f(rx)$. Then, by a change of variables, you see that \begin{align} \|f_r\|_{L^2(\Bbb{R}^n)}=\frac{1}{r^{n/2}}\|f\|_{L^2(\Bbb{R}^n)}. \end{align} Also, by the chain rule, $\Delta(f_r)(x)=r^2\cdot(\Delta f)(rx)=r^2\cdot (\Delta f)_r(x)$, so \begin{align} \|\Delta(f_r)\|_{L^2(\Bbb{R}^n)}=r^2\cdot\frac{1}{r^{n/2}}\|\Delta f\|_{L^2(\Bbb{R}^n)}. \end{align} So, if $-\Delta$ was bounded with respect to $L^2$ norms, then the inequality would have to hold for each $f_r$, and hence for each $r>0$, \begin{align} r^2\|\Delta f\|_{L^2(\Bbb{R}^n)}&\leq C\|f\|_{L^2(\Bbb{R}^n)}. \end{align} By taking $r\to\infty$, you see that this is a contradiction.

On the other hand, it is bounded $H^2(\Omega)\to L^2(\Omega)$ when you use the Sobolev norm rather than merely the $L^2$ norm on $H^2(\Omega)$.

peek-a-boo
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  • I think I just got it. summarizing $-\Delta: H^{2.2}\subset L^2\to\subset L^2 $ $\left|-\Delta u\right|{L^2}=\leq \left|u\right|{H^{2,2}}$ Therefore, $-\Delta$ is a bounded operator between the spaces $H^{2,2}$ (with norm $\left|\cdot\right|{H^{2,2}}$) and $L^2$ (with norm $\left|\cdot \right|{L^2}$. – eraldcoil Jul 02 '23 at 18:50
  • @eraldcoil If you're happy with the answer, then please accept it. – stange Jul 03 '23 at 09:16