A "holomorphic" Peano curve? - MathOverflow most recent 30 from http://mathoverflow.net2013-05-19T10:43:44Zhttp://mathoverflow.net/feeds/question/114893http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/114893/a-holomorphic-peano-curveA "holomorphic" Peano curve?aglearner2012-11-29T15:56:39Z2012-11-29T18:57:58Z
<p>A Peano curve is a continuous map $[0,1]\to [0,1]^2$ whose image is the whole square. </p>
<p>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 $\phi$.</p>
http://mathoverflow.net/questions/114893/a-holomorphic-peano-curve/114906#114906Answer by Qfwfq for A "holomorphic" Peano curve?Qfwfq2012-11-29T17:40:13Z2012-11-29T18:28:22Z<p>Define $$\phi(z):=\frac{1}{2\pi i}\int_{S^1}(\zeta-z)^{-1}\cdot \varphi(\zeta)d\zeta$$</p>
<p>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 $z\in S^1$. [<strong>Edit</strong>: this construction doesn't work because $\phi$, as I defined it, may not be continuous up to the boundary - see the comments]</p>
http://mathoverflow.net/questions/114893/a-holomorphic-peano-curve/114914#114914Answer by Alexandre Eremenko for A "holomorphic" Peano curve?Alexandre Eremenko2012-11-29T18:57:58Z2012-11-29T18:57:58Z<p>Here it is:
MR0015154
Salem, R.; Zygmund, A.
Lacunary power series and Peano curves.
Duke Math. J. 12, (1945). 569–578. </p>