On the ranges of analytic functions

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 $\left\vert A \right\vert\, \geqslant \,2\rho \, > \,0$,

$$\mathop {\lim \,\inf }\limits_{i \to \infty } \,\left\vert {f({P_... ... \,1\,{\text{where}}\,{P_i}\, \to \,P\, \in \,A\,{\text{and}}\,{P_i}\, \in \,U.$$

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

$$(1\, - \,\left\vert z \right\vert)\left\vert {{f_n}' (z)} \right\... ... \leqslant \,1/n\,{\text{for}}\,z\, \in \,U,\,{\text{where}}\,n\, > \,N(\rho ).$$

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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