Timeline for Is there a differentiable map surjective from low to high dimension？
Current License: CC BY-SA 4.0
19 events
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| Oct 4, 2021 at 18:49 | comment | added | weak solution | Thanks for your help. | |
| Oct 4, 2021 at 13:45 | vote | accept | weak solution | ||
| Oct 4, 2021 at 5:40 | history | edited | YCor | CC BY-SA 4.0 |
fixed username
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| Oct 3, 2021 at 18:25 | history | edited | YCor | CC BY-SA 4.0 |
added reference
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| Oct 3, 2021 at 15:45 | comment | added | Kosh | Well was a preamble to this one that is related to your answer. Because then the result follows immediately by Lemma 7.25 of Rudin in Real and Complex Analysis's. Just define the new function on all of R^m as you do, which will still be differentiable everywhere, and then invoke the result of Rudin as someone else also suggested. | |
| Oct 3, 2021 at 15:12 | comment | added | YCor | @Kosh this is a comment to the question, not to this specific answer. But indeed this limsup is finite at every point, for differentiable $f$, so this OP's claim is confusing. Maybe they mean this limsup is not necessarily (locally) bounded as function of $x$. | |
| Oct 3, 2021 at 15:00 | comment | added | Kosh | @MohanRamachandran thanks. The last thing I don't get, maybe you do, is why the OP says, when talking about limsup, "unfortunately the above conditions are not necessarily met" (when a function is differentiable everywhere, I understand). At every $x$ where the function is differentiable that limsup is finite. Indeed, the argument of that limsup is bounded by the norm of the differential at $x$ plus some $\epsilon$, in a neighborhood of $x$ that depends on $\epsilon$. Am I missing something? | |
| Oct 3, 2021 at 14:19 | comment | added | Mohan Ramachandran | Yes. For approximately differentiable points this is due to Federer Surface Area II TAMS 1944 vol 55 page 453 Theorem 5.2 . | |
| Oct 3, 2021 at 10:38 | history | edited | YCor | CC BY-SA 4.0 |
added more general case
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| Oct 3, 2021 at 8:18 | comment | added | Kosh | So in order to let your observation and Majer remark coexisting, the result is that if a function is differentiable everywhere then it is not necessarily the case that it is locally Lipschitz but still there exist countable many measurable subsets where it's restriction is Lipschitz. So a kind of local Lipschitzianity on certain measurable subsets rather than compact subsets. Am I right? | |
| Oct 2, 2021 at 23:55 | comment | added | Mohan Ramachandran | Yes. You can find the proof of a stronger result in Piotr Hajlasz lecture notes titled Geometric Analysis Lemma 2.8 page 18 | |
| Oct 2, 2021 at 22:08 | comment | added | Kosh | @MohanRamachandran can you give me a reference for this result? You are claiming that the function the OP is looking for cannot exists. Right? | |
| Oct 2, 2021 at 18:11 | comment | added | Mohan Ramachandran | However the set of points where F is differentiable is a countable union of sets on each of which F is Lipschitz , and by applying Lipschitz extension to each of the restrictions you are done. | |
| Oct 2, 2021 at 16:48 | comment | added | YCor | @PietroMajer You're right, I've added it to emphasize. | |
| Oct 2, 2021 at 16:48 | history | edited | YCor | CC BY-SA 4.0 |
added remark
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| Oct 2, 2021 at 16:42 | comment | added | Pietro Majer | But just differentiable does not imply Lipschitz, not even locally... | |
| Oct 2, 2021 at 13:21 | comment | added | Igor Belegradek | No, the proof I know is the same. The text develops Riemann integration on compact intervals, and eventually on $\mathbb R^n$, and does a bit of baby measure theory along the way. This problem illustrates a key idea in multivariable integration, which is why I liked assigning it. | |
| Oct 2, 2021 at 12:56 | comment | added | YCor | @IgorBelegradek wouldn't this rather fit a measure theory course? or else which analytic arguments do you have in mind? | |
| Oct 2, 2021 at 9:09 | history | answered | YCor | CC BY-SA 4.0 |