Timeline for The Koch snow flake, Holder exponents of conformal mappings
Current License: CC BY-SA 4.0
30 events
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| S Dec 22, 2019 at 20:01 | history | bounty ended | CommunityBot | ||
| S Dec 22, 2019 at 20:01 | history | notice removed | CommunityBot | ||
| Dec 14, 2019 at 18:41 | history | edited | sharpe | CC BY-SA 4.0 |
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| S Dec 14, 2019 at 18:37 | history | bounty started | sharpe | ||
| S Dec 14, 2019 at 18:37 | history | notice added | sharpe | Canonical answer required | |
| S Jul 14, 2019 at 11:02 | history | bounty ended | CommunityBot | ||
| S Jul 14, 2019 at 11:02 | history | notice removed | CommunityBot | ||
| Jul 12, 2019 at 3:51 | history | edited | sharpe | CC BY-SA 4.0 |
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| Jul 8, 2019 at 15:51 | history | edited | sharpe | CC BY-SA 4.0 |
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| Jul 8, 2019 at 15:29 | comment | added | sharpe | @all Today, I found an interesting paper by F. D. Lesley. So, I update the contents of my question. | |
| Jul 8, 2019 at 10:50 | comment | added | Pietro Majer | Yes sorry, I had misread the question, and in fact I thought I had deleted my comment, but for some reason I failed again :/ | |
| Jul 7, 2019 at 17:37 | comment | added | Benoît Kloeckner | @PietroMajer: it is true that my affirmation is not universally true, but since here we consider a map from the domain delimited by the von Koch curve, it does apply. | |
| Jul 6, 2019 at 12:18 | history | edited | sharpe | CC BY-SA 4.0 |
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| Jul 6, 2019 at 10:11 | answer | added | Carlo Beenakker | timeline score: 5 | |
| Jul 6, 2019 at 9:59 | comment | added | Pietro Majer | @BenoîtKloeckner In fact "Hölder exponent greater than 1 implies that the function is constant" holds for functions on $\mathbb{R}^n$, or even on a length space, but I think it is not true here, due to the fact that any arc of $K$ has infinite length. In fact I think one can take $\beta=1/\alpha=\log4/\log3$ (so these exponents are also sharp) | |
| S Jul 6, 2019 at 9:14 | history | bounty started | sharpe | ||
| S Jul 6, 2019 at 9:14 | history | notice added | sharpe | Draw attention | |
| Jul 6, 2019 at 5:40 | comment | added | sharpe | @BenoîtKloeckner Thank you for your reply. I understood. | |
| Jul 5, 2019 at 7:24 | comment | added | Benoît Kloeckner | @sharpe the extension to an heomorphism of the boudary should be checked carfeully (for general domains, boundaries need not be topological circles, think of a slit disk for example), but yes that is the idea I had in mind. | |
| Jul 5, 2019 at 3:23 | history | edited | sharpe | CC BY-SA 4.0 |
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| Jul 4, 2019 at 17:26 | history | edited | sharpe | CC BY-SA 4.0 |
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| Jul 4, 2019 at 17:17 | comment | added | sharpe | @NikWeaver Thank you for your information. | |
| Jul 4, 2019 at 16:59 | comment | added | Nik Weaver | @BenoitKloeckner: yes, I meant $\ln 3/\ln 4$ --- in fact (embarrassingly) this correct value is given in my book Lipschitz Algebras (second edition). Sharpe, on p. 68 of this book there is a brief discussion of this example, in case that helps. | |
| Jul 4, 2019 at 16:45 | comment | added | sharpe | @BenoîtKloeckner I also thank you for teaching me the paper. I do not know much French, but I will read it. | |
| Jul 4, 2019 at 16:37 | comment | added | sharpe | @BenoîtKloeckner Maybe I understood. A conformal map $\phi : \mathbb{D} \to K$ is extended to a homeomorphism from $\bar{\mathbb{D}}$ to $\bar{K}$. Since $\partial K$ is regarded as the image of the unit circle under $\phi$, we have $\log4 /\log 3 \le 1/\alpha$. Hence, $\alpha \le \log3 /\log 4$. Is my understanding correct? | |
| Jul 4, 2019 at 16:16 | comment | added | Benoît Kloeckner | @sharpe: that $\log 3/\log 4$ is a upper bound can be seen by looking at the dimensions: the image of a metric space of Hausdorff dimension $d$ by a $\alpha$-Hölder map has dimension at most $d/\alpha$ (this follows from the definitions). That it can be achieved by a map from the circle to the Von Koch boundary is, I think, done in Assouad's Bulletin SMF paper archive.numdam.org/article/BSMF_1983__111__429_0.pdf (in French). | |
| Jul 4, 2019 at 16:11 | comment | added | Benoît Kloeckner | @NikWeaver: your guess for $\beta$ is too optimistic: having Hölder exponent greater than $1$ implies that the function is constant. For $\alpha$, you might be right, except that you should have written $\log 3/\log 4$ (there are $4$ images of scale $1/3$). $\alpha$ cannot be greater than this, that it can be achieved by a conformal map is not obvious. | |
| Jul 4, 2019 at 16:04 | comment | added | sharpe | @NikWeaver Thank you for your comment. Can you tell me a reference for the fact that the Holder exponent $\alpha$ for mapping the unit circle onto the boundary of the snowflake equals to $\log 2 /\log 3$. I didn't know that at all. | |
| Jul 4, 2019 at 14:47 | comment | added | Nik Weaver | I would guess it's the same as the Holder exponent for mapping the unit circle onto the boundary of the snowflake, so $\alpha = \ln 2/\ln 3$ and $\beta = \alpha^{-1}$. But it's a bit outside my area of expertise. | |
| Jul 4, 2019 at 8:24 | history | asked | sharpe | CC BY-SA 4.0 |