Timeline for Elliptic estimates in unweighted Sobolev spaces
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 30 at 13:49 | comment | added | Willie Wong | @TarekAcila: you should ask a different question. Your problem is completely different; this question is essentially about scaling of interior regularity. Your case requires also considering boundary problems. | |
| Aug 25 at 20:32 | comment | added | Tarek Acila | @WillieWong Grisvard's Elliptic Problems in Nonsmooth Domains has a version of this inequality but only for $W^{2,p}$. | |
| Aug 25 at 20:25 | comment | added | Tarek Acila | @WillieWong What if one has a bounded smooth domain and not the whole space? Any references that state this result (if it is true)? I assume that boundary conditions have to be considered (Dirichlet, Neumann) and will affect the right-hand side of the inequality. CONTEXT: I intend to use this with homogenous boundary conditions to define “equivalent norms” of the $H^k$ spaces given that the homogenous boundary conditions are satisfied. | |
| Jul 10, 2023 at 17:02 | comment | added | Willie Wong | Actually, in the case where you have the whole space, if $L$ has sufficiently regular coefficients you have: given any unit ball $B$, let $B'$ denote the ball of the same center but twice the radius, then $\|u\|_{H^k(B)} \leq C ( \| Lu \|_{H^{k-m}(B')} + \|u\|_{L^2(B')})$ by standard bounded elliptic estimate, with $C$ a uniform constant depending only on the properties of $L$. Next you just sum over a collection of balls with finite overlap and you are done. | |
| Jul 10, 2023 at 16:51 | comment | added | Willie Wong | This holds provided that $L$ has suitably regular coefficients. If you look at Gilbarg and Trudinger Theorem 9.11, it contains most of what you need for the proof; all you need is some additional control on the regularity of $L$ as $|x|\to\infty$ to ensure that you get a uniform bound as $\Omega'$ increases toward the whole space. | |
| Jul 6, 2023 at 19:37 | history | edited | Chris | CC BY-SA 4.0 |
added 191 characters in body
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| Jul 6, 2023 at 19:36 | comment | added | Chris | That makes sense - but I don't see how it's a counterexample to this inequality. If $u \sim |x|^{2-n}$ near infinity, then the right-hand side will be infinite due to the presence of the $H^0$ norm, so there is no contradiction. Also I've edited the question regarding the geometry at infinity of the manifold. | |
| Jul 6, 2023 at 19:03 | comment | added | user378654 | Generally not: do a basic computation where you solve $\Delta u = f$ on $\mathbb{R}^n$ for some smooth compactly supported nonnegative $f$. You can check explicitly that it behaves like the fundamental solution at infinity in arbitrarily strong topology, i.e. like $|x|^{2-n}$. But this function is not square integrable at infinity in some dimensions ($n \leq 4$), and its derivative is not square integrable in some dimensions $n \leq 2$. If you have a higher-order operator on a manifold, the same is possible in higher dimensions too (and depends on volume growth at infinity, etc.) | |
| Jul 6, 2023 at 16:59 | history | asked | Chris | CC BY-SA 4.0 |