Timeline for Domains of type (A) are Lipschitz?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 22 at 19:11 | comment | added | Pietro Majer | Btw the question in the title (does (A) imply Lip?) is different from the question in body (does Lip imply (A)? ). | |
| Aug 22 at 16:44 | comment | added | Bogdan | Sorry for the ambiguous formulation. I meant a bounded domain with a uniform Lipschitz boundary (no cusps). I have edited my post. | |
| Aug 22 at 12:28 | comment | added | mlk | I mean the second quote in the question, which was talking about domains of "class $C^1$ or Lipschitz". And you are right that without the comma the sentence can only be read in one way, but that could have been a typo, so maybe we have to wait for clarification. | |
| Aug 22 at 12:15 | comment | added | Iosif Pinelis | @mlk : Sorry, I did not read your previous comment carefully enough. On the other hand, I can hardly read "Lipschitz connected, open and bounded" (with a comma already there) as "Lipschitz, connected, open and bounded". Also, what do you mean by "the second example"? | |
| Aug 22 at 12:10 | comment | added | mlk | It is the image of a Lipschitz function, but not the graph; in general one uses Lipschitz-domains precisely to avoid the type of cusp you have in your example, as they tend to cause all kind of problems in PDE. But you are right, it is not clear if the question meant (Lipschitz connected) or Lipschitz, connected domains. However I have never seen the former anywhere, while the latter is standard and also the one used in the second example. | |
| Aug 22 at 12:03 | comment | added | Iosif Pinelis | @mlk : The domain was said to be, not Lipschitz, but Lipschitz connected, which seems to mean that any two points of it can be connected by a Lipschitz path $\gamma\colon[0,1]\to\Omega$ with the same Lipschitz constant for all such paths. Also, I think the boundary is the graph of a Lipschitz function: parametrize the boundary by the arclength. | |
| Aug 22 at 11:48 | comment | added | mlk | That is not a Lipschitz-domain though, as no matter how you rotate it, the boundary is not the graph of a Lipschitz-function in any neighborhood of the origin. | |
| Aug 22 at 11:42 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 168 characters in body
|
| Aug 22 at 11:35 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |