Univalence of the average of two analytic functions

TL;DR

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Contribution

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Abstract

Let $\mathcal{A}$ denote the set of all analytic functions $f$ in the unit disk $\ID=\{z:\,|z|<1\}$ of the form $f(z)=z+\sum_{n=2}^{\infty}a_nz^n.$ Let $\mathcal{U}$ denote the set of all $f\in \mathcal{A}$, $f(z)/z\neq 0$ and satisfying the condition $$ | f'(z) (\frac{z}{f(z)})^{2}-1 | < 1 {for $z\in \ID$}. $$ Functions in ${\mathcal U}$ are known to be univalent in $\ID$. For $\alpha \in [0,1]$, let $$ \mathcal{N}(\alpha)= \{f_\alpha :\, f_\alpha (z)=(1-\alpha)f(z)+\alpha \int_0^z\frac{f(t)}{t}\,dt, {$f\in\mathcal{A}$ with $|a_n|\leq n$ for $n\geq 2$}\}. $$ In this paper, we first show that the condition $\sum_{n=2}^{\infty}n|a_n|\leq 1$ is sufficient for $f$ to be in ${\mathcal U}$ and the same condition is necessary for $f\in {\mathcal U}$ in case all $a_n$'s are negative. Next, we obtain the radius of univalence of functions in the class $\mathcal{N}(\alpha)$. Also, for $f,g\in \mathcal{U}$ with $\frac{f(z)+g(z)}{z}\neq 0$ in $\ID$, $F(z)=(f(z)+g(z))/2$, and $G(z)=r^{-1}F(rz)$, we determine a range of $r$ such that $G\in {\mathcal U}$. As a consequence of these results, several special cases are presented.