A "holomorphic" Peano curve? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T10:43:44Z http://mathoverflow.net/feeds/question/114893 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114893/a-holomorphic-peano-curve A "holomorphic" Peano curve? aglearner 2012-11-29T15:56:39Z 2012-11-29T18:57:58Z <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 $|z|&lt;1$ 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#114906 Answer by Qfwfq for A "holomorphic" Peano curve? Qfwfq 2012-11-29T17:40:13Z 2012-11-29T18:28:22Z <p>Define $$\phi(z):=\frac{1}{2\pi i}\int_{S^1}(\zeta-z)^{-1}\cdot \varphi(\zeta)d\zeta$$</p> <p>for $|z|&lt;1$, 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#114914 Answer by Alexandre Eremenko for A "holomorphic" Peano curve? Alexandre Eremenko 2012-11-29T18:57:58Z 2012-11-29T18:57:58Z <p>Here it is: MR0015154 Salem, R.; Zygmund, A. Lacunary power series and Peano curves. Duke Math. J. 12, (1945). 569–578. </p>