The Nevanlinna counting functions for Rudin's orthogonal functions

The Nevanlinna counting functions for Rudin's orthogonal functions
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Rudin 正交函数的 Nevanlinna 计数函数

DOI:
10.1090/s0002-9939-02-06671-6
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发表时间:
2001
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通讯作者:
T. Nakazi
T. Nakazi
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--
文献类型:
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作者:
T. Nakazi

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H∞和H2表示开单位圆盘D上的哈代空间.设Φ是H∞中的函数,||Φ|| ∞ = 1。若Φ是内函数且Φ(0)= 0,则{Φ n ; n = 0,1,2,.}在H2上是正交的鲁丁问,如果匡威是正确的和C。Sundberg和C.毕晓普指出,相反的情况是不正确的。因此存在一个函数Φ使得Φ不是内函数且{Φ n }在H2中正交。本文证明了:{Φ n }在H2中正交当且仅当在[0,1]上存在唯一的概率测度ν 0,其中1 ∈ supp ν 0,使得N Φ(z)= f1| z| log r/|z| dν 0(r),其中N Φ是Φ的Nevanlinna计数函数。如果Φ是内函数,则ν 0是r = 1处的狄拉克测度。
H∞ and H 2 denote the Hardy spaces on the open unit disc D. Let Φ be a function in H∞ and ||Φ||∞ = 1. If Φ is an inner function and Φ(0) = 0, then {Φ n ; n = 0,1,2,...} is orthogonal in H 2 . W.Rudin asked if the converse is true and C. Sundberg and C. Bishop showed that the converse is not true. Therefore there exists a function Φ such that Φ is not an inner function and {Φ n } is orthogonal in H 2 . In this paper, the following is shown: {Φ n } is orthogonal in H 2 if and only if there exists a unique probability measure ν 0 on [0,1] with 1 ∈ supp ν 0 such that N Φ (z) = f 1 |z| log r/ |z| dν 0 (r) for nearly all z in D where N Φ is the Nevanlinna counting function of Φ. If Φ is an inner function, then ν 0 a Dirac measure at r = 1.