1. Introduction
In 1983, Mané, Sad and Sullivan proved the following surprising result. Suppose that
Then each fλ can be extended to a quasiconformal homeomorphism of into . See [6] and see [10] and [1] for further results.
We shall consider a different but related set of assumptions. Let Λ be a domain in C. We write :=. Let the function be defined for. We assume that
We will often write instead of . The assumption (3) corresponds to (b) in (1) whereas assumption (2) is quite different from (a). The initial condition (c) has no counterpart.
We need the Hartogs theorem of the theory of several complex variables, see e.g. [2, p.140]. We write it in a form adapted to our present context.
Proposition 1.1. Let (2) and (3) be satisfied. Then the function
is holomorphic except in z = ∞ and therefore continuous in every compact subset of.
2. Univalence
A complex-valued function is called univalent if it is injective and meromorphic in a domain in We define
Since by assumption (2) we can therefore write
Since the coefficients ak in (2) are holomorphic in A it follows that the coefficients are defined and holomorphic for see [7, p.58]. The Grunsky inequality states that
see [4] [7, p.60] [3, p.122]. It easily follows from (5) and (6) that
see e.g. [7, p.59].
Theorem 2.1. Let (2) and (3) be satisfied. Then U is relatively closed. For every component C of the open set Λ \ U we have
Proof. (i) Let Then there are such that as It follows from Proposition 1.1 that
Now fλn is univalent by (4). Since fλo is non-constant by assumption (2) it follows that fλ0 is univalent in so that.
(ii) If C is unbounded then obviously. Now let C be a bounded component and suppose that so that. Since C is a component of Λ \ U we have.It follows, by (4) and (7), that 1 for. Since the are holomorphic in C we conclude by the maximum principle that the inequality 1 also holds for. Hence we obtain from (7) that f λ is univalent in D * so that by (4). This contradicts.
Remark 2.2. . If A is simply connected then Theorem 2.1 implies that every component of the open kernel U° is simply connected. If in addition then Λ \ U is a domain.
Example 2.3. Let p(λ) be a non-constant entire function and let. Then (2) is satisfied with . We have
If 1 then has the zero in so that fλ is not univalent. If 1 then 0 so that fλ is univalent. Hence we have. Since p is arbitrary this provides us with a huge variety of closed sets where fλ is univalent. If p is a non-constant polynomial then U is compact.
If we choose p(λ) = λ2- 1 then U is the classical lemniscate . If we choose p(λ) = exp(λ2) then U is the unbounded closed set which consists of two quarter-planes that meet at 0.
3. Quasiconformality
We assume that our conditions (2) and (3) are satisfied. Let U be defined by (4) and b k,l (λ) by (5).
Proposition 3.1. (Schiffer and Springer). Let. The function fλ has a quasiconformal extension to if and only if there exists K < 1 such that
See [8] [9] [5] [7,Th.9.12,Th.9.13].
Theorem 3.2. Let Λo be a subdomain of Λ and let fλ be univalent infor. If there exists such that has a quasiconformal extension to then fλ has a quasiconformal extension tofor every.
Let V be any component of U°. An obvious consequence of this theorem is that f λ has a quasiconformal extension either for all or for no. This raises an interesting question: If f λ has a quasiconformal extension for some component V, does it follow that f λ has quasiconformal extension for all
Proof. Let. Then there is a simply connected domain G with G and . Let g map conformally onto G and let.
Since fλ is univalent in we obtain from (6) that for. Furthermore, since and φ is holomorphic in Λo it follows that for. Finally we have
by our assumption and by Proposition 3.1.
The hyperbolic metric in is defined by