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A Peano curve is a continuous map $[0,1]\to [0,1]^2$ whose image is the whole square.

I would like to know if on can obtain "holomorphic" Peano curves. Namely, is it possible to find a continuous map $\phi$ from the unit disk $|z|\le 1$ to $\mathbb C^1$ such that $\phi$ is holomorphic for $|z|<1$ and the image of the boundary $|z|=1$ has non-empty interior in $\mathbb C^1$ under the map $\phi$.

flag	asked Nov 29 at 15:56 aglearner 3,858●8●24

Alexandre Eremenko · Accepted Answer · 2012-11-29 18:57:58Z UTC

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Here it is: MR0015154 Salem, R.; Zygmund, A. Lacunary power series and Peano curves. Duke Math. J. 12, (1945). 569–578.

answered Nov 29 at 18:57

Alexandre Eremenko
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And here is the link: projecteuclid.org/… – Brian Rushton Nov 30 at 4:06

Qfwfq · Answer 2 · 2012-11-29 18:28:22Z UTC

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Define $$\phi(z):=\frac{1}{2\pi i}\int_{S^1}(\zeta-z)^{-1}\cdot \varphi(\zeta)d\zeta$$

for $|z|<1$, where $\varphi: S^1\to\mathbb{C}$ is a Peano curve (i.e. its image has nonempty interior), and $\phi(z):=\varphi(z)$ for $z\in S^1$. [Edit: this construction doesn't work because $\phi$, as I defined it, may not be continuous up to the boundary - see the comments]

edited Nov 29 at 18:28

answered Nov 29 at 17:40

Qfwfq
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In general there is no reason for this function $\phi$ to be contunuous on $\overline{\mathbb{D}}$! For example, if negative terms are present in the Fourier expansion of $g$, then certainly a continuous extension of the Cauchy Transform of $g$ to the circle will not be equal to $g$. – Malik Younsi Nov 29 at 18:16

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I see. You are right! I will not delete the answer because it might be an "instructive mistake" (I'm not an analyst, btw) - I will edit adding a disclaimer. – Qfwfq Nov 29 at 18:25

@Qfwfq : Yes, it is instructive. Thank you for the edit! – Malik Younsi Nov 30 at 12:48

A “holomorphic” Peano curve?

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