Timeline for Refinement of mean value conjecture for complex polynomials?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 2, 2021 at 17:11 | vote | accept | Stefan Steinerberger | ||
| S May 27, 2021 at 16:36 | history | bounty ended | Stefan Steinerberger | ||
| S May 27, 2021 at 16:36 | history | notice removed | Stefan Steinerberger | ||
| May 27, 2021 at 11:45 | answer | added | Peter Mueller | timeline score: 10 | |
| May 24, 2021 at 14:36 | comment | added | Stefan Steinerberger | Somehow "proving that there exists" seems a lot harder than "there is exactly the same number of roots and critical points". The second statement seems to match so well with the classical complex analysis toolbox... | |
| May 22, 2021 at 15:29 | comment | added | Luka Thaler | Using your terminology, the "weak" Smale's mean value conjecure can be viewd in the following way: You are looking for the solution of the equation $zg'(z)/g(z)=-1$ that satisfies $|g(z)|\leq 1$. Now from the discusion above we can see that the conjecure holds if there is a component of the set $\{|g(z)|=1\}$ on which $zg'(z)/g(z)\notin (-\infty,-1)$. It seems to me that this does not simplify the problem or do think otherwise? | |
| S May 21, 2021 at 0:33 | history | bounty started | Stefan Steinerberger | ||
| S May 21, 2021 at 0:33 | history | notice added | Stefan Steinerberger | Draw attention | |
| May 19, 2021 at 16:11 | comment | added | Luka Thaler | I was a bit sceptical about your previous comment since it was hard to belive that the proof would be so simple :D Anyhow I agree that it suffices to exclude that $zg'(z)/g(z)$ is real and $<-1$ for all z on the boudnary ;) | |
| May 19, 2021 at 16:04 | comment | added | Stefan Steinerberger | Interesting! So I guess the desired inequality becomes $|z g'(z)| < | g(z) + z g'(z)| + |g(z)|$ on the boundary. Dividing by $|g(z)|$ (which never vanishes), we get $|z g'(z)/g(z)| < |1 + z g'(z)/g(z)| + 1$. This is true unless $z g'(z)/g(z)$ is on a real number $\leq -1$. If $z g'(z)/g(z) = -1$, we would have found a critical point of $z g(z)$ on the boundary, so it suffices to exclude $z g'(z)/g(z)$ being real and $<-1$... | |
| May 19, 2021 at 14:50 | comment | added | Luka Thaler | The following stronger version on Rouche thm, given by Estermann, could be more useful for this problem: THM: Suppose that $f$ and $g$ are holomorphic on a domain $D$, that $C$ is a simple closed contour in $D$ and that the following "strict" inequality holds $|f(z)-g(z)|< |f(z)|+|g(z)|$ for all $z\in C$. Then $f$ and $g$ have the same number of zeros inside $C$. Now in your case take $f(z)=g(z)+zg'(z)$. | |
| May 19, 2021 at 14:36 | comment | added | Stefan Steinerberger | Right, it's not just Rouche but it looks somewhat `friendlier' and similar to other problems I have seen solved via a combination of the Schwarz Lemma, Riemann mapping, Rouche, the Argument Principle, Koebe, .... | |
| May 19, 2021 at 14:32 | comment | added | Luka Thaler | Nice approach! The following is not a counterexample but just an example that shows a potential problem if you whish to use Rouche theorem directly. Take $g(z)=1-\frac{1}{2}z$. The set $A=B$ is the disc of radius $2$ centered at $z=2$. Now $g'(z)=-1/2$ and therefore $|zg'(z)|\leq |g(z)|$ fails on the boudnary $\partial B$, for example at $z=4$ . | |
| May 19, 2021 at 0:32 | history | asked | Stefan Steinerberger | CC BY-SA 4.0 |