Timeline for Bounded mappings on unbounded domains
Current License: CC BY-SA 3.0
9 events
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| Apr 1, 2013 at 22:48 | history | edited | Ahmed Sulejmani | CC BY-SA 3.0 |
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| Mar 29, 2013 at 2:58 | comment | added | Misha | Start with the AB example and ask if three are higher dim generalizations which admit light bounded holomorphic functions. | |
| Mar 29, 2013 at 2:55 | comment | added | Misha | Ahmed: I think you should edit your question along the lines of your comments. | |
| Mar 28, 2013 at 13:07 | comment | added | ahmed sulejmani | domain which allows non-constant bounded holomorphic functions (therefore the Liouville argument does not work) but allows no injective bounded holomorphic functions (and therefore this domain is again not biholomorphic to any bounded domain). My question was about such examples in several variables. | |
| Mar 28, 2013 at 13:05 | comment | added | ahmed sulejmani | Sorry for not expressing myself clearly enough. The typical situation I have in mind is a domain of the type $\mathbb C\setminus E$, where $E$ is a "small-set"- a set for which removable singularity for bounded holomorphic functions works. Then every bounded holomorphic function on $\mathbb C\setminus E$ extends to an entire bounded function which is necessarily constant by Liouville's theorem. Playing with small-sets Ahlfors and Beurling ( Ahlfors, Lars; Beurling, Arne Conformal invariants and function-theoretic null-sets. Acta Math. 83, (1950). 101–129.) provided an explicit example of a dom | |
| Mar 28, 2013 at 1:01 | comment | added | Alexandre Eremenko | On my opinion, the first sentence is completely meaningless. Most domains bounded or not can be mapped on bounded domains, and Liouville theorem has nothing to do with this. | |
| Mar 27, 2013 at 23:22 | comment | added | Misha | The first sentence is actually not completely meaningless since OP does not restrict to simply-connected domains. I think, he is asking for examples of higher-dimensional domains $D$ which: (1) admit, say, "light" holomorphic maps to bounded domains, but (2) at the same time $D$ is not biholomorphic to a bounded domain. (A map is called "light" if its fibers are totally disconnected.) | |
| Mar 27, 2013 at 3:10 | comment | added | Alexandre Eremenko | The very first sentence of the question is wrong: every simply connected region in the plane, other than the plane itself, can be mapped biholomorphically onto the unit disc. Liouville's theorem applies only to functions defined in the whole plane. The rest of the question makes little sense. | |
| Mar 27, 2013 at 0:25 | history | asked | Ahmed Sulejmani | CC BY-SA 3.0 |