# The Koch snow flake, Holder exponents of conformal mappings

The Koch snow flake $$K$$ is a domain of $$\mathbb{C}$$, complex plane. Though, I do not state the precise definition, you can see the picture in wikipedia Koch snow flake.

The Koch snow flake $$K$$ is a quasidisk. Let $$\mathbb{D}$$ be an open unit disk and let $$\phi:\mathbb{D} \to K$$ be a conformal mapping. It is known that $$\phi$$ and the inverse map $$\phi^{-1}$$ are Holder continuous: there exist $$\alpha \in (0,1]$$, $$\beta \in (0,1]$$, and $$L_1,L_2 \in (0,\infty)$$ such that \begin{align*} |\phi(z_1)-\phi(z_2)| &\le L_1 |z_1-z_2|^{\alpha},\quad z_1,z_2 \in \mathbb{D}, \\ |\phi^{-1}(w_1)-\phi^{-1}(w_2)| &\le L_2 |w_1-w_2|^{\beta},\quad w_1,w_2 \in K. \end{align*}

My question

Is there a study for quantitative estimates on the Holder exponents $$\alpha$$ and $$\beta$$?

I think that there is such study because the Koch snow flake is a famous fractal set.

By an argument by Benoît Kloeckner, $$\alpha$$ must be less than or equal to $$\log3 / \log 4$$. Is there a reasonable lower bound for $$\alpha$$ and $$\beta$$?

Let $$C$$ be a closed Jordan curve.

Lasley considers the following condition on $$C$$.

Definition. Let $$w_1$$ and $$w_3$$ be points on $$C$$ and let $$w_2$$ be on the arc of small diameter between $$w_1$$ and $$w_3$$. Then, $$C$$ is said to be a $$c$$-quasiconformal curve if there exist positive constants $$c$$ and $$\delta$$ such that \begin{align*} \frac{|w_1-w_2|+|w_2-w_3|}{|w_1-w_3|} \le c \end{align*} for any such $$w_1,w_2,w_3$$ with $$|w_1-w_3| \le \delta$$.

Lasley prove

Theorem. Suppose that $$f$$ maps $$\mathbb{D}$$ conformally onto the interior $$\Omega$$ of a $$c$$-quasiconformal curve $$C$$. Then, $$f$$ is $$\alpha$$-Holder continuous. Here,
\begin{align*} \alpha=\frac{2 (\text{arcsin}(1/c))^2}{\pi^2 -\pi \text{arcsin}(1/c)}. \end{align*}

If $$\Omega=K$$, the Koch snowflake, $$c=\cdots$$.

• 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 '19 at 14:47
• @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 '19 at 16:16
• @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 '19 at 16:37
• @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 '19 at 16:59
• @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) Jul 6 '19 at 9:59

The Koch snow flake is the union of three Koch curves of Hausdorff dimension $$D=\ln 4/\ln 3$$ and Hölder exponent $$\beta=\log 2/\log 3$$.
(For reference, Pietro Majer's argument for $$\alpha=\log 3/\log 4$$ is in this 2016 MO answer.)
• Thank you for your comment. U.R Freiberg and M.R. Lancia show in Prop 2.2 that the domain of the energy from on Koch curve (not on Koch snow flake) is continuously embedded into the space of Holder continuous functions with exponent $\log2/\log3$. However, does this result lead us to an estimate of the Holder exponents of conformal mappings $\phi$ and $\phi^{-1}$? Jul 6 '19 at 12:16
• @DanieleTampieri Thank you for showing me the paper by Vivaldi and Lancia. But, I do not know how to use it. By using their results, we can just see $\beta \le \log2/\log3$. Dec 24 '19 at 4:17
• Why can you deduce $\beta \ge \log2/ \log3$? Dec 24 '19 at 12:09