Timeline for Hölder continuity of Green function for simply connected domains
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Sep 11 at 19:35 | comment | added | Malik Younsi | I have not looked at Rakhmanov's paper. I am simply pointing out that several of your comments and answers include incorrect mathematical statements or mathematical objects that are not well-defined. I do have higher standards for this forum, even though it is indeed not a paper as you pointed out. But let's agree to disagree and put an end to this discussion. | |
| Sep 11 at 19:00 | comment | added | user528012 | I must add: this is a forum for exchange, it is not a paper, is it? People write with good intentions but not the same scrutiny as they would do in a paper. I know you can fix the proof if you need it. Apparently you were not the referee of Rakhmanov's paper! | |
| Sep 11 at 18:51 | comment | added | user528012 | What I say is: ignore the "min". Yes the other post has the unnecessary assumption of continuity up to the boundary, but you seem well equipped to fix the missing pieces by yourself. Hint: the closest point on the boundary of the image, $w_\star$ , to the point $z= F(\zeta)$ is accessible.... Then look at the answer by Kostya. Fill the gaps! BTW, see a similar inequality here | |
| Sep 11 at 18:44 | comment | added | Malik Younsi | 2. In the inequality that you refer to in your answer, you wrote that $F$ is assumed to be continuous up to the boundary. This is not the case here, so you cannot directly apply this inequality to answer your question without adding more justification. More details are necessary. It is possible that the proof can be fixed, but if I were to referee a paper that includes your proof as it is, I would reject the paper. You are welcome to disagree of course, but that is my opinion. | |
| Sep 11 at 18:43 | comment | added | Malik Younsi | Yes, I do understand that the distance to a compact set is well-defined. That is not the issue here. 1. In your answer you write $\min_{|\rho|=1} |F(\zeta)-F(\rho)|$. This is not well-defined since $F(\rho)$ may not be even defined for all $\rho$ with $|\rho|=1$. | |
| Sep 11 at 14:59 | comment | added | user528012 | Hi Malik, you may be right, but the inequality that is being used is really only between the distance from $\mathcal K$ and $z$. To clarify: the complement, $\mathcal K$, of the image $F(\{|\zeta|>1\})$ is compact so the distance is certainly well defined. | |
| Sep 10 at 19:10 | comment | added | Malik Younsi | See en.wikipedia.org/wiki/… | |
| Sep 10 at 18:51 | comment | added | Malik Younsi | This is true if and only if $\mathcal{K}$ is locally connected! If $\mathcal{K}$ is not locally connected then the argument you provide does not work. | |
| Sep 10 at 7:10 | comment | added | Malik Younsi | $F(\rho)$ may not even be defined...! | |
| Sep 10 at 2:17 | comment | added | user528012 | It means the minimum of the set of numbers $\{ |F(\zeta)-F(\rho)|: \rho\in S^1\}$. This is the same as the distance of $z = F(\zeta)$ from the boundary of the image of $F$. $\mathcal K$ (the complement in the $z$--plane of the image of $F$) is compact, because the complement of an unbounded open domain that contains a full neighbourhood of $\infty$. | |
| Sep 9 at 18:37 | comment | added | Malik Younsi | I don't understand what you mean by $\min_{|\rho|=1} |F(\zeta)-F(\rho)|$. What are your assumptions on $\mathcal{K}$? Note that in general the uniformizing map of the complement of $\mathcal{K}$ may not extend to the boundary. | |
| Sep 7 at 23:56 | vote | accept | CommunityBot | ||
| S Sep 7 at 9:22 | review | First answers | |||
| Sep 7 at 11:01 | |||||
| S Sep 7 at 9:22 | history | answered | user528012 | CC BY-SA 4.0 |