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Transactions of the American Mathematical Society

ISSN 1088-6850(online) ISSN 0002-9947(print)

 
 

 

On the ranges of analytic functions


Author: J. S. Hwang
Journal: Trans. Amer. Math. Soc. 260 (1980), 623-629
MSC: Primary 30D40
DOI: https://doi.org/10.1090/S0002-9947-1980-0574804-0
MathSciNet review: 574804
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Abstract: Following Doob, we say that a function $f(z)$ analytic in the unit disk U has the property $K(\rho )$ if $f(0) = 0$ and for some $\operatorname {arc} A$ on the unit circle whose measure $\left | A \right | \geqslant 2\rho > 0$, \[ \lim \inf \limits _{i \to \infty } \left | {f({P_i})} \right | \geqslant 1 {\text {where}} {P_i} \to P \in A {\text {and}} {P_i} \in U.\] We recently have solved a problem of Doob by showing that there is an integer $N(\rho )$ such that no function with the property $K(\rho )$ can satisfy \[ (1 - \left | z \right |)\left | {{f_n}’ (z)} \right | \leqslant 1/n {\text {for}} z \in U, {\text {where}} n > N(\rho ).\] The function \[ {f_n}(z) = 1 + (1 - {z^n})/{n^2},\] shows that the condition ${f_n}(0) = 0$ is necessary and cannot be replaced by ${f_n}(0) = r{e^{i\alpha }}$, for $r > 1$. Naturally, we may ask whether this can be replaced by ${f_n}(0) = r{e^{i\alpha }}$, for $r < 1$? The answer turns out to be yes, when $n > N (r, \rho )$, where \[ N(r, \rho ) \doteqdot (1/(1 - r))\log (1/(1 - \cos \rho )).\] .


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Keywords: The range and analytic function
Article copyright: © Copyright 1980 American Mathematical Society