Timeline for Estimation of the number of local extrema
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Feb 8, 2015 at 21:53 | comment | added | Pietro Majer | I think it would be good to include the precise definition of local extrema for this question, both for f on D and on S^1. | |
| Feb 6, 2015 at 12:49 | history | edited | Lev Balkanski | CC BY-SA 3.0 |
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| Feb 6, 2015 at 12:37 | comment | added | Lev Balkanski | Add: But the stronger form with $n/2+1$ extrema remains unclear to me. | |
| Feb 6, 2015 at 12:25 | comment | added | Lev Balkanski | By the way, I initially formulated the statement in a weaker form - "there exist at least $n/2+1$ critical points"; in this variant we may assume a finite number of critical points of $f$ and all level curves are "nice". Now I realize that my elementary proof works in fact without problems for this variant, so I'll may be go back to it. | |
| Feb 6, 2015 at 12:14 | comment | added | Alex Degtyarev | Right. These are certainly true for "decent" functions, at least, for those for which your pictures work :) Say, about an extremum, assuming a nice level curve nearby, $\nabla f$ is normal and pointing in the same direction (unless $0$), hence index is $1$. | |
| Feb 6, 2015 at 11:32 | comment | added | Lev Balkanski | The rotation number along the boundary is $n/2$, right. But the fact that "the index of ∇f is +1 at each extremum and negative at other critical points" needs a proof. On the other hand, I wish to avoid concepts like "index" and "rotation number". Anyway, if your arguments are correct, I will accept that the statement is "obvious" :) | |
| Feb 6, 2015 at 11:13 | comment | added | Alex Degtyarev | OK, I think I agree with your statement, but certainly not with the proof :) It appears that the index of $\nabla f$ is $+1$ at each extremum and negative at other critical points. It is also more or less clear that the rotation number along the boundary is $n/2$. (The lack of extrema inhibits "negative" rotation.) So, we're done. | |
| Feb 6, 2015 at 11:11 | comment | added | Lev Balkanski | But the boundary condition fails there. By the way a function $f$ satisfying the above conditions looks a little bit "strange". | |
| Feb 6, 2015 at 11:06 | comment | added | Pietro Majer | OK; actually I understood "inside" as "in the interior", in the open disk, $f$ has a unique critical point (the origin). At the boundary f has n/2 local maxima. | |
| Feb 6, 2015 at 11:05 | comment | added | Lev Balkanski | Add: At least at the local maxima of $\varphi$ the condition fails. | |
| Feb 6, 2015 at 10:55 | comment | added | Lev Balkanski | @ Pietro Majer: This is supposed to be a counter-example, right? But I don't think that it satisfies the boundary conditions, seems to me that even each local extremum of $\varphi$ is a local extremum of $f$ as well. | |
| Feb 6, 2015 at 10:52 | comment | added | Alex Degtyarev | Ah, sorry, I see that you inhibit extrema, not just critical points on the boundary. Then, indeed, there must be two. | |
| Feb 6, 2015 at 10:46 | comment | added | Alex Degtyarev | @PietroMajer maybe it's the same example :) $f=r^2+r^3\sin(n\theta)$ in the polar coordinates. If my math is correct, it has no critical points in $0<r<2/3$. Rescale it to the unit disk. | |
| Feb 6, 2015 at 10:39 | comment | added | Pietro Majer | If you allow degenerate minima, and (w.l.o.g.) $\phi>0$ we can take $f(x)=x^2\phi(x/|x|)$. | |
| Feb 6, 2015 at 10:29 | comment | added | Lev Balkanski | @ Alex Degtyarev: I don't think that "there may be just a single extremum unside" - if $P$ and $Q$ are an absolute minimum and an absolute maximum of the restriction, then the conditions guarantee at least 2 absolute extrema of $f$ inside $D$ - a minimum and a maximum. | |
| Feb 6, 2015 at 10:17 | history | edited | Lev Balkanski | CC BY-SA 3.0 |
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| Feb 6, 2015 at 10:15 | comment | added | Alex Degtyarev | Those pictures do not really look like a proof to me. Besides, to make the statement closer to the truth, one should count the extrema with some carefully defined multiplicities; otherwise, there may be just a single extremum unside. The multiplicity is, probably, the index of the gradient vector field, in which case the statement becomes just the Poincaré–Hopf theorem with boundary. | |
| Feb 6, 2015 at 10:07 | history | asked | Lev Balkanski | CC BY-SA 3.0 |