Timeline for A generalization of Cauchy's mean value theorem.
Current License: CC BY-SA 2.5
22 events
| when toggle format | what | by | license | comment | |
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| Feb 26, 2010 at 17:24 | vote | accept | Petya | ||
| Feb 26, 2010 at 16:16 | comment | added | Petya | I still do not understand how to see easily the condition (1)... I construct another example based on your ideas (it is posted as another answer). Please, check it! I think, that there should be another solution. There should be such a function $g$ on a circle such that for any A,B the function $g+Asin+Bcos$ extends inside the disk without critical points... | |
| Feb 26, 2010 at 3:20 | comment | added | Anton Petrunin | After smoothing, the set ($\nabla h_1=\nabla h_2$) sits in arbitrary small neighborhood of zero so it is still has no common points with $\nabla h_3$. | |
| Feb 25, 2010 at 23:38 | comment | added | Petya | Anyway, after smoothing $\nabla h_1 (x)=\nabla h_2 (y)$ in a set with non-empty interior. So I do not see why (1) holds. | |
| Feb 25, 2010 at 23:15 | comment | added | Anton Petrunin | $C^1$ again. This is very particular curve. It is graph over round $S^1$... | |
| Feb 25, 2010 at 23:13 | comment | added | Anton Petrunin | Shure it is $C^1$. About (1), first check it for $f(t)=2\cdot\sin (2\cdot t)$. When you add the bump, you may still think that $f(t)=-f(-t)$. In this case $h(z)=-h(\bar z)$ thus the set of $\nabla h_1$ is the reflection in $y=0$ of the set of $\nabla h_2$. The graph of $h_{12}$ has a flat triangle each side exteded by a convex ruled surface two of them are nearly the same as the segment for the sine and the one which corresponds to the bump. The gradients of this piece has $y$-coordinate of fixed sign. Once you see this picture it is clear that they intersect only at $0$. | |
| Feb 25, 2010 at 22:44 | comment | added | Petya | Accordingly to Arnold's "Catastrophe theory" the boundary of a convex hull of a space curve is not C^1 in general, there could be aedges of singular points (z \ge |x| in suitable local coordinates). So, I do not understand why (1) or other condition should survive after smoothing. | |
| Feb 25, 2010 at 22:21 | comment | added | Petya | Is it C^1 at least? I want to understand your example and now I do not see why (1) holds. Sure I'll ask Yasha when I meet him. | |
| Feb 25, 2010 at 22:08 | comment | added | Anton Petrunin | Yes, "boundary of a convex hull it is not smooth", but it is not a problem. Everything survives after smoothing. It is a very good problem. I still think that one can do similar construction for 2 surfaces, but it is better to ask some specialist in h-principle (say Elyashberg). | |
| Feb 25, 2010 at 21:48 | comment | added | Petya | Just realize you change the comment. I'll think about it. First remark is: it seems, that a boundary of a convex hull could be non-smooth.. | |
| Feb 25, 2010 at 19:30 | comment | added | Anton Petrunin | Now it should be correct --- sorry for all this mess. It should be really simple for those who do h-principle... | |
| Feb 25, 2010 at 19:25 | history | edited | Anton Petrunin | CC BY-SA 2.5 |
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| Feb 25, 2010 at 15:53 | comment | added | Anton Petrunin | "The set of normal unit vectors is not, in general, a curve for a ruled surface." you are right, I will think a bit to correct the answer --- it does not change much. | |
| Feb 25, 2010 at 15:23 | comment | added | Petya | I want to remark the following: Consider two surfaces in 3-space and suppose they do not have parallel tangent planes. Let its intersection contains a compact component. If one surface is a graph of a function on the plane then this component is non homologous to zero in the second surface. I conclude from that observation that both surfaces from a counterexample should be very curved. | |
| Feb 25, 2010 at 12:22 | comment | added | Petya | I do not understand your construction. The set of normal unit vectors is not, in general, a curve for a ruled surface. | |
| Feb 25, 2010 at 6:18 | history | undeleted | Anton Petrunin | ||
| Feb 25, 2010 at 6:17 | history | edited | Anton Petrunin | CC BY-SA 2.5 |
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| Feb 25, 2010 at 5:29 | history | deleted | Anton Petrunin | ||
| Feb 25, 2010 at 5:28 | comment | added | Petya | You can delete it. Your Idea of counterexample was very good! Even if the conjecture is false, the real question is to find a positive statement generalizing Cauchy's m.v. theorem. | |
| Feb 25, 2010 at 5:09 | history | edited | Anton Petrunin | CC BY-SA 2.5 |
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| Feb 25, 2010 at 5:04 | vote | accept | Petya | ||
| Feb 25, 2010 at 5:10 | |||||
| Feb 25, 2010 at 4:34 | history | answered | Anton Petrunin | CC BY-SA 2.5 |