Timeline for Stokes theorem for Lipschitz forms
Current License: CC BY-SA 4.0
15 events
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| Oct 14 at 16:08 | comment | added | Piotr Hajlasz | @DeaneYang The proof is fairly easy. With a partition of unity argument it is reduced to the half-space. You do not need to multiply Lipschitz functions except for multiplication by smooth partition of unity. The proof extends also to the case of Sobolev forms in $W^{1,n}$, but then the restriction of $\omega$ to $\partial M$ is in the trace space $W^{1-\frac{1}{n},n}$. | |
| Oct 14 at 14:59 | comment | added | Deane Yang | @PiotrHajlasz, if I understand correctly, the crucial point is that the product of two Lipschitz functions is Lipschitz. I prefer to work with Sobolev spaces, but Holder spaces do have this advantage. | |
| Oct 13 at 17:35 | history | edited | Piotr Hajlasz | CC BY-SA 4.0 |
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| Oct 12 at 21:20 | history | edited | Piotr Hajlasz | CC BY-SA 4.0 |
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| Oct 12 at 19:12 | answer | added | Daniele Tampieri | timeline score: 5 | |
| Oct 11 at 14:02 | comment | added | M.G. | @PiotrHajlasz: The book is "Absolute Analysis" from the 70s, I think. IIRC, their version of Stokes' Thm is stated for once differentiable $\omega$ rather than $C^1$, which doesn't sound like what you are after, though. | |
| Oct 11 at 13:51 | comment | added | Piotr Hajlasz | @M.G. If you represent the form in the local coordinate sytem, the coeddicients are Lipschitz continuous on compact sets. What book ny Nevanlinna? | |
| Oct 11 at 13:43 | comment | added | M.G. | @PiotrHajlasz Sorry for my ignorance, but what is a Lipschitz differential form? The reason I'm asking is that IIRC in terms of regularity there is a stronger Stokes' Theorem (=weaker regularity assumption) stated and proved in a classical analysis book by F. and R. Nevanlinna, but the language might be different. | |
| Oct 11 at 12:41 | comment | added | Piotr Hajlasz | @NateRiver sets of finite perimeter/BV is precisely the geometric measure theory. However, if you can find a statement of the general Stokes theorem for Lipschitz forms (with any proof), I would appreciate it. The problem is that in the geometric measure theory they often use language of currents and for a non familiar reader it iwould not be clear that what they see contains the classical Sard for Lipschitz forms. | |
| Oct 11 at 12:31 | comment | added | Piotr Hajlasz | @mlk Taking a mollification works, but in the Lipschitz case it is not easy at all and not written anywhere. | |
| Oct 11 at 10:31 | comment | added | Nate River | I don’t think I have seen this anywhere except for geometric measure theory books too @Piotr Haijlasz. If you are okay with the theory of sets of finite perimeter/BV functions, I have some references in mind… | |
| Oct 11 at 8:49 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Fixed typo
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| Oct 11 at 8:14 | comment | added | coudy | Maybe Federer, geometric measure theory. | |
| Oct 11 at 7:33 | comment | added | mlk | Are you specifically looking for a "direct", self-contained proof? I guess you can show this by following the proof of Stokes with a bit more attention, but in practice taking the smooth Stokes for granted and applying a quick mollifier-argument would probably be far more efficient. Sadly I have no good reference for either. | |
| Oct 11 at 2:24 | history | asked | Piotr Hajlasz | CC BY-SA 4.0 |