Timeline for Is there a differentiable map surjective from low to high dimension？
Current License: CC BY-SA 4.0
35 events
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| Oct 4, 2021 at 13:45 | vote | accept | weak solution | ||
| Oct 4, 2021 at 13:42 | history | edited | weak solution | CC BY-SA 4.0 |
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| Oct 4, 2021 at 13:41 | comment | added | weak solution | @YCor I'm wrong. Thanks for your solution! | |
| Oct 4, 2021 at 12:55 | comment | added | YCor | @weaksolution yes it does: if $f$ is differentiable, then it satisfies the limsup condition: the limsup at $x$ is actually equal to the operator norm of the differential at $x$. | |
| Oct 4, 2021 at 12:54 | comment | added | weak solution | @YCor I want to explain that if $f$ satisfies the limsup condition ,we can use measure theory to proove it must not be a surjective,but it may not be met, so the problem can't be solve by this. | |
| Oct 4, 2021 at 12:13 | comment | added | YCor | I don't see any rude behavior. But you're asking about the existence of a surjective differentiable map $f$, then say that the case $f$ satisfies the limsup condition follows from measure theory, and then say that "the above condition may be not meet". I have no idea what you mean by this, since precisely the differentiable case is covered by this limsup condition. It will be useful if you edit so as to clarify. For the moment, your question is completely answered in comments and answers. | |
| Oct 4, 2021 at 12:08 | comment | added | weak solution | So sorry for my poor Englilsh level. And i just learned how to @ someone, so I never responded to the comments before. I've been a little busy lately, so I'll take time to review the answers and comments when i am not so busy. Again, I'm sorry for my rude behavior. | |
| Oct 4, 2021 at 12:01 | comment | added | weak solution | @YCor I'm sorry that my English is so poor.It is just a assumption,not a assertion. | |
| Oct 4, 2021 at 11:01 | comment | added | YCor | I'm confused by your edit, which maintains a confusing assertion: if $f$ is differentiable, then the given limsup is finite at every point. | |
| Oct 4, 2021 at 10:56 | history | edited | weak solution | CC BY-SA 4.0 |
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| Oct 4, 2021 at 5:07 | history | edited | Rodrigo de Azevedo | CC BY-SA 4.0 |
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| Oct 3, 2021 at 8:36 | history | edited | Ben McKay | CC BY-SA 4.0 |
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| Oct 3, 2021 at 8:15 | history | edited | Ben McKay | CC BY-SA 4.0 |
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| Oct 3, 2021 at 5:45 | history | reopened |
abx weak solution Francois Ziegler bof Pietro Majer |
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| Oct 3, 2021 at 4:41 | history | edited | weak solution | CC BY-SA 4.0 |
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| Oct 3, 2021 at 4:22 | history | edited | weak solution | CC BY-SA 4.0 |
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| Oct 3, 2021 at 4:01 | review | Reopen votes | |||
| Oct 3, 2021 at 5:47 | |||||
| S Oct 3, 2021 at 2:58 | history | suggested | mathworker21 | CC BY-SA 4.0 |
The previous version was confusing, since it seemed like it was asking two questions.
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| Oct 3, 2021 at 2:41 | comment | added | Moishe Kohan | This is Lemma 7.25 from Rudin's "Real and complex analysis" that the measure of the image is zero. | |
| Oct 3, 2021 at 2:23 | history | closed |
Alex M. Daniele Tampieri paul garrett Ryan Budney Kevin Walker |
Not suitable for this site | |
| Oct 2, 2021 at 23:57 | answer | added | Nicolast | timeline score: 10 | |
| Oct 2, 2021 at 23:19 | review | Suggested edits | |||
| S Oct 3, 2021 at 2:58 | |||||
| Oct 2, 2021 at 23:14 | comment | added | Nicolast | If the question was really about differentiable functions with discontinuous derivatives, maybe it could state it more clearly ? And mention that the answer is well-known in $\mathcal C^1$ regularity (cf &YCor's answer)? | |
| Oct 2, 2021 at 12:25 | history | became hot network question | |||
| Oct 2, 2021 at 9:09 | answer | added | YCor | timeline score: 25 | |
| Oct 2, 2021 at 8:44 | comment | added | Ben McKay | @WlodAA: as I say in my previous comment, I don't know a reference to prove such a result, and I don't see how to prove it. Would you mind giving an outline proof in an answer? | |
| Oct 2, 2021 at 8:13 | comment | added | Ben McKay | Sard actually called his paper "The measure of the critical values of differentiable maps", but he really does require continuous differentiability, so I can't find a reference for differentiable but not continuously differentiable functions. | |
| Oct 2, 2021 at 8:13 | answer | added | abx | timeline score: 34 | |
| Oct 2, 2021 at 8:07 | comment | added | weak solution | sard's theorem requires the derived function of f is continuous.peano cuves and hilbert cuves only need f is continuous, before giving a strict solution,please do not underestimate the difficulty of the problem and review the questions carefully. | |
| Oct 2, 2021 at 7:31 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor Math Jaxing and formatting
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| Oct 2, 2021 at 7:18 | review | Close votes | |||
| Oct 2, 2021 at 17:15 | |||||
| Oct 2, 2021 at 4:41 | comment | added | Bma | The argument at the following link appears to generalize to any $C^1$ $f: \mathbb{R}^n \to \mathbb{R}^m$ where $n < m$, showing that this is impossible when the derivative is continuous. This is more or less the proof of Sard's theorem I know. See here: math.stackexchange.com/questions/1905299/… | |
| Oct 2, 2021 at 4:31 | comment | added | Bma | This is impossible when $f$ is smooth, because the set of critical values of $f$ in $\mathbb{R}^m$ will be the entire range of $f$, which has measure zero by Sard's theorem. There is a form of Sard's theorem which applies to $C^r$ functions, $r < \infty$, but the minimal $r$ to which it applies depends on $n$ and $m$. | |
| S Oct 2, 2021 at 4:25 | review | First questions | |||
| Oct 2, 2021 at 6:57 | |||||
| S Oct 2, 2021 at 4:25 | history | asked | weak solution | CC BY-SA 4.0 |