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Ahmed Sulejmani
Ahmed Sulejmani
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In dimension $1$ the most common reason for the non-existence of a biholomorphic mapping of some unbounded domain to any bounded one is the Liouville theoremAhlfors and Beurling (any bounded mapping turns out to be constant)Ahlfors, howeverLars; Beurling, a classical construction by AhlforsArne Conformal invariants and Beurling shows that there are nonfunction-Liouville unbounded planar domains which still fail to be biholomorphic to a bounded onetheoretic null-sets. Acta Math. 83, (such domains allow nonconstant1950). 101–129.) provided an explicit example of a domain in $\mathbb C$ which allows non-constant bounded holomorphic function butfunctions but allows no injective bounded holomorphic ones)functions. Therefore this domain is not biholomorphic to any bounded domain but allows a "light" mapping to a bounded domain, where "light mapping" means that the preimages of every point are totally disconnected. My question is about the higher dimensional situationsuch examples in several variables. Do there exist unbounded domainsa domain in $\mathbb C^{n}$, $n>1$ which areis not biholomorphic to any bounded domain for reasons other than the Liouville theorembut allows a light mapping to a bounded domain? To exclude trivial cases, such as theClearly a Cartesian product of atwo copies of Ahlfors-Beurling domain with $\mathbb C$, let me state my question otherwise: If a unbounded domain allowsdomains will allow a "light" mapping to a bounded domain whichbut there is onto and no point in the image has, say uncountable preimagesome chance (or maybe contains a Liouville setI don't know) does it follow that the domain isit will eventually be biholomorphic to a bounded domain? If. A somehow simpler question is whether on some domain the algebra of bounded holomorphic functions dodoes not separate points but does locally separate them then the answer would be no, but is this possible in the. If such domain case? The only instancesexists that would provide an example I know when bounded holomorphic functions fail to separate points are exactly the cases when one can employrequest. (I actually doubt the Liouville theoremexistence of a domain as in some formthe second question).

In dimension $1$ the most common reason for the non-existence of a biholomorphic mapping of some unbounded domain to any bounded one is the Liouville theorem (any bounded mapping turns out to be constant), however, a classical construction by Ahlfors and Beurling shows that there are non-Liouville unbounded planar domains which still fail to be biholomorphic to a bounded one (such domains allow nonconstant bounded holomorphic function but no injective bounded holomorphic ones). My question is about the higher dimensional situation. Do there exist unbounded domains which are not biholomorphic to any bounded domain for reasons other than the Liouville theorem? To exclude trivial cases, such as the Cartesian product of a Ahlfors-Beurling domain with $\mathbb C$, let me state my question otherwise: If a unbounded domain allows a mapping to a bounded domain which is onto and no point in the image has, say uncountable preimage (or maybe contains a Liouville set) does it follow that the domain is biholomorphic to a bounded domain? If the bounded holomorphic functions do not separate points but locally separate them then the answer would be no, but is this possible in the domain case? The only instances I know when bounded holomorphic functions fail to separate points are exactly the cases when one can employ the Liouville theorem in some form.

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 domain in $\mathbb C$ which allows non-constant bounded holomorphic functions but allows no injective bounded holomorphic functions. Therefore this domain is not biholomorphic to any bounded domain but allows a "light" mapping to a bounded domain, where "light mapping" means that the preimages of every point are totally disconnected. My question is about such examples in several variables. Do there exist a domain in $\mathbb C^{n}$, $n>1$ which is not biholomorphic to any bounded domain but allows a light mapping to a bounded domain? Clearly a Cartesian product of two copies of Ahlfors-Beurling domains will allow a "light" mapping to a bounded domain but there is some chance (I don't know) that it will eventually be biholomorphic to a bounded domain. A somehow simpler question is whether on some domain the algebra of bounded holomorphic functions does not separate points but does locally separate them. If such domain exists that would provide an example I request. (I actually doubt the existence of a domain as in the second question).

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Ahmed Sulejmani
Ahmed Sulejmani
  • 141
  • 4

Bounded mappings on unbounded domains

In dimension $1$ the most common reason for the non-existence of a biholomorphic mapping of some unbounded domain to any bounded one is the Liouville theorem (any bounded mapping turns out to be constant), however, a classical construction by Ahlfors and Beurling shows that there are non-Liouville unbounded planar domains which still fail to be biholomorphic to a bounded one (such domains allow nonconstant bounded holomorphic function but no injective bounded holomorphic ones). My question is about the higher dimensional situation. Do there exist unbounded domains which are not biholomorphic to any bounded domain for reasons other than the Liouville theorem? To exclude trivial cases, such as the Cartesian product of a Ahlfors-Beurling domain with $\mathbb C$, let me state my question otherwise: If a unbounded domain allows a mapping to a bounded domain which is onto and no point in the image has, say uncountable preimage (or maybe contains a Liouville set) does it follow that the domain is biholomorphic to a bounded domain? If the bounded holomorphic functions do not separate points but locally separate them then the answer would be no, but is this possible in the domain case? The only instances I know when bounded holomorphic functions fail to separate points are exactly the cases when one can employ the Liouville theorem in some form.