Timeline for Intermediate value theorem for the Jacobian determinant restricted to a curve
Current License: CC BY-SA 3.0
11 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Oct 24, 2014 at 16:03 | answer | added | Alexander Kuleshov | timeline score: 4 | |
| Mar 29, 2014 at 22:24 | comment | added | Loïc Teyssier | @AlexDegtyarev: I think you were referring to a theorem by Darboux. en.wikipedia.org/wiki/Darboux%27s_theorem_%28analysis%29 | |
| Mar 29, 2014 at 11:11 | history | edited | Alexander Kuleshov | CC BY-SA 3.0 |
deleted a meaningless statement
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| Mar 28, 2014 at 8:55 | comment | added | Alex Degtyarev | @JoeSilverman The only hope that I see is that the restriction of the Jacobian to a curve is some kind of ordinary derivative. But standard proofs of such things usually assume class $C^1$, so one should go all the way over the proofs to see of they work in the more general setting. Boring :) | |
| Mar 28, 2014 at 8:52 | comment | added | Alex Degtyarev | @JoeSilverman This I don't know, as for ordinary derivatives $f'(x)$ there is an intermediate value theorem, even without the assumption that it's continuous. (Don't remember the exact name.) So, the question about the Jacobian may be not as simple as it seems. | |
| Mar 28, 2014 at 1:27 | comment | added | Joe Silverman | @AlexDegtyarev Ah, well, if $J_f$ is allowed to be (mildly) discontinuous, then can't one construct a counterexample by taking a standard example of a map $f$ that is differentiable, but not $C^1$. | |
| Mar 27, 2014 at 21:07 | comment | added | Alex Degtyarev | @JoeSilverman I guess some people by "differentiable" mean just differentiable, not necessarily $C^1$. Personally, I also disapprove such maps :) | |
| Mar 27, 2014 at 20:45 | review | First posts | |||
| Mar 27, 2014 at 20:48 | |||||
| Mar 27, 2014 at 20:43 | comment | added | Joe Silverman | What am I missing? $J_f$ is a continuous function on $\mathbb{R}^N$, so it is continuous when restricted to any continuous curve connecting $a$ to $b$. That curve is just the image of some continuous $\gamma:[0,1]\to\mathbb{R}^N$. So $F=J_f\circ\gamma:[0,1]\to\mathbb{R}$ is continuous and satisfies $F(0)<0$ and $F(1)>0$, so by the intermediate value theorem it must vanish for some value in $(0,1)$. | |
| Mar 27, 2014 at 20:28 | history | asked | Alexander Kuleshov | CC BY-SA 3.0 |