Absolutely Continuous Conjugacies of Blaschke Products
โ Scribed by D.H. Hamilton
- Publisher
- Elsevier Science
- Year
- 1996
- Tongue
- English
- Weight
- 539 KB
- Volume
- 121
- Category
- Article
- ISSN
- 0001-8708
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โฆ Synopsis
Introduction
Let S be the unit circle. We call a (measureable) bijection . : S ร S absolutely continuous if for any A/S, *(A)=0 if and only if *(.(A))=0, where * denotes normalised Lebesgue (arc) measure on S. We consider (finite) nontrivial (i.e. not 1:1 or constant) Blaschke products
which of course map S onto S. Mappings f j : S ร S, j=1 and 2, are conjugate (by .) if there is a bijection . : S ร S, so that f 2 =. b f 1 b . &1 . If f has no absolutely continuous conjugations to another Blaschke product, except by Mobius transformations (i.e. bilinear transformations or the complex conjugate of one) then we say that f is strongly rigid. Schub and Sullivan [17] prove Theorem SS. Nontrivial Blaschke products f : S ร S, with fixed point w, |w| <1, are strongly rigid.
This was the hyperbolic case. In the case that the attractive point is on the boundary then we need not have rigidity. We find that rigidity is characterized by f being ergodic i.e. there does not exist A/S, 0<*(A)<1, such that f &1 (A)=A. It is not hard to show that for finite Blaschke products f ergodicity is equivalent to the Julia set J( f )=S. We prove:
Theorem 1. Let f: S ร S be a nontrivial finite Blaschke product. Then f is strongly rigid if and only if the Julia set J( f )=S.
๐ SIMILAR VOLUMES
Let A be a contraction on Hilbert space H and ฯ a finite Blaschke product. In this paper, we consider the problem when the norm of ฯ(A) is equal to 1. We show that (1) ฯ(A) = 1 if and only if A k = 1, where k is the number of zeros of ฯ counting multiplicity, and (2) if H is finite-dimensional and A