# On sufficient conditions for Carathéodory functions with applications

## Abstract

In the present paper, we derive some interesting relations associated with the Carathéodory functions which yield sufficient conditions for the Carathéodory functions in the open unit disk $\mathbb{U}=\left\{z:|z|<1\right\}$. Some interesting applications of the main results are also obtained.

MSC:30C45, 30C80.

## 1 Introduction

Let P denote the class of functions of the form

$p\left(z\right)=\sum _{n=0}^{\mathrm{\infty }}{p}_{n}{z}^{n},$

which are analytic in the unit disc $\mathbb{U}=\left\{z:|z|<1\right\}$. The function $p\left(z\right)$ is called a Carathéodory function if it satisfies the condition

$Re\left(p\left(z\right)\right)>0.$

Moreover, let A denote the class of functions of the form

$f\left(z\right)=z+\sum _{n=2}^{\mathrm{\infty }}{a}_{n}{z}^{n},$
(1.1)

which are analytic in the unit disc $\mathbb{U}$.

A function $f\left(z\right)\in A$ is in K, the class of convex functions, if it satisfies

$Re\left(1+\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}\left(z\right)}{f\left(z\right)}\right)>0\phantom{\rule{1em}{0ex}}\left(z\in \mathbb{U}\right).$
(1.2)

Also, a function $f\left(z\right)\in A$ is in ${S}^{\lambda }$ ($|\lambda |<\frac{\pi }{2}$), the class of λ-spirallike functions, if it satisfies

$Re\left({e}^{i\lambda }\frac{z{f}^{\mathrm{\prime }}\left(z\right)}{f\left(z\right)}\right)>0\phantom{\rule{1em}{0ex}}\left(z\in \mathbb{U}\right).$
(1.3)

Moreover, we denote by ${S}^{\ast }={S}^{0}$ the class of starlike functions in $\mathbb{U}$.

Definition 1.1 Let $f\left(z\right)$ and $F\left(z\right)$ be analytic functions. The function $f\left(z\right)$ is said to be subordinate to $F\left(z\right)$, written $f\left(z\right)\prec F\left(z\right)$, if there exists a function $w\left(z\right)$ analytic in $\mathbb{U}$, with $w\left(0\right)=0$ and $|w\left(z\right)|\le 1$, and such that $f\left(z\right)=F\left(w\left(z\right)\right)$. If $F\left(z\right)$ is univalent, then $f\left(z\right)\prec F\left(z\right)$ if and only if $f\left(0\right)=F\left(0\right)$ and $f\left(\mathbb{U}\right)\subset F\left(\mathbb{U}\right)$.

Definition 1.2 Let $\mathbb{D}$ be the set of analytic functions $q\left(z\right)$ and injective on $\overline{\mathbb{U}}\mathrm{\setminus }E\left(q\right)$, where

$E\left(q\right)=\left\{\zeta \in \partial \mathbb{U}:\underset{z\to \zeta }{lim}q\left(z\right)=\mathrm{\infty }\right\}$

and ${q}^{\mathrm{\prime }}\left(\zeta \right)\ne 0$ for $\zeta \in \partial \mathbb{U}\mathrm{\setminus }E\left(q\right)$. Further, let ${\mathbb{D}}_{a}=\left\{q\left(z\right)\in \mathbb{D}:q\left(0\right)=a\right\}$.

Many authors have obtained several relations of Carathéodory functions, e.g., see ([113]).

In the present paper, we derive some relations associated with the Carathéodory functions which yield the sufficient conditions for Carathéodory functions in $\mathbb{U}$. Some applications of the main results are also obtained.

## 2 Main results

To prove our results, we need the following lemma due to Miller and Mocanu [[14], p.24]

Lemma 2.1 Let $q\left(z\right)\in {\mathbb{D}}_{a}$ and let

$p\left(z\right)=b+{b}_{n}{z}^{n}+\cdots$
(2.1)

be analytic in $\mathbb{U}$ with $p\left(z\right)\ne b$. If $p\left(z\right)\nprec q\left(z\right)$, then there exist points ${z}_{0}\in \mathbb{U}$ and ${\zeta }_{0}\in \partial \mathbb{U}\mathrm{\setminus }E\left(q\right)$ and on $m\ge n\ge 1$ for which

1. (i)

$p\left({z}_{0}\right)=q\left({\zeta }_{0}\right)$,

2. (ii)

${z}_{0}{p}^{\mathrm{\prime }}\left({z}_{0}\right)=m{\zeta }_{0}{q}^{\mathrm{\prime }}\left({\zeta }_{0}\right)$.

Theorem 2.1 Let

$P:\mathbb{U}\to \mathbb{C}$

with

$Re\left(\overline{a}P\left(z\right)\right)>0\phantom{\rule{1em}{0ex}}\left(a\in \mathbb{C}\right).$

If $p\left(z\right)$ is an analytic function in $\mathbb{U}$ with $p\left(0\right)=1$ and

$Re\left(p\left(z\right)+P\left(z\right)z{p}^{\mathrm{\prime }}\left(z\right)\right)>\frac{E}{2|a{|}^{2}Re\left(\overline{a}P\left(z\right)\right)},$
(2.2)

then

$Re\left(ap\left(z\right)\right)>0,$

where

$E=-\left(Re\left(a\right)\right){\left(Re\left(\overline{a}P\left(z\right)\right)\right)}^{2}+2Re\left(\overline{a}P\left(z\right)\right){\left(Im\left(a\right)\right)}^{2}+\left(Re\left(a\right)\right){\left(Im\left(a\right)\right)}^{2}$
(2.3)

with $Re\left(a\right)>0$.

Proof Let us define both $q\left(z\right)$ and $h\left(z\right)$ as follows:

$q\left(z\right)=ap\left(z\right)$

and

$h\left(z\right)=\frac{a+\overline{a}z}{1-z}\phantom{\rule{1em}{0ex}}\left(Re\left(a\right)>0\right),$

where $p\left(z\right)$ is defined by (2.1) since $q\left(z\right)$ and $h\left(z\right)$ are analytic functions in $\mathbb{U}$ with $q\left(0\right)=h\left(0\right)=a\in \mathbb{C}$ with

$h\left(\mathbb{U}\right)=\left\{w:Re\left(w\right)>0\right\}.$

Now, we suppose that $q\left(z\right)\nprec h\left(z\right)$. Therefore, by using Lemma 2.1, there exist points

${z}_{0}\in \mathbb{U}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\zeta }_{0}\in \partial \mathbb{U}\mathrm{\setminus }\left\{1\right\}$

such that $q\left({z}_{0}\right)=h\left({\zeta }_{0}\right)$ and ${z}_{0}{q}^{\mathrm{\prime }}\left({z}_{0}\right)=m{\zeta }_{0}{h}^{\mathrm{\prime }}\left({\zeta }_{0}\right)$, $m\ge n\ge 1$.

We note that

${\zeta }_{0}={h}^{-1}\left(q\left({z}_{0}\right)\right)=\frac{q\left({z}_{0}\right)-a}{q\left({z}_{0}\right)+\overline{a}}$
(2.4)

and

${\varsigma }_{0}{h}^{\mathrm{\prime }}\left({\varsigma }_{0}\right)=-\frac{|q\left({z}_{0}\right)-a{|}^{2}}{2Re\left(a-q\left({z}_{0}\right)\right)}.$
(2.5)

We have $h\left({\zeta }_{0}\right)=\rho i$ ($\rho \in \mathbb{R}$); therefore,

(2.6)

where

and

$C=-\frac{Re\left(\overline{a}p\left({z}_{0}\right)\right)}{2Re\left(a\right)}.$

We can see that the function $g\left(\rho \right)$ in (2.6) takes the maximum value at ${\rho }_{1}$ given by

${\rho }_{1}=Im\left(a\right)\left(1+\frac{Re\left(a\right)}{Re\left(\overline{a}p\left({z}_{0}\right)\right)}\right).$

Hence, we have

$\begin{array}{rcl}Re\left(p\left({z}_{0}\right)+P\left({z}_{0}\right)z{p}^{\mathrm{\prime }}\left({z}_{0}\right)\right)& \le & g\left({\rho }_{1}\right)\\ =& \frac{E}{2|a{|}^{2}Re\left(\overline{a}P\left(z\right)\right)},\end{array}$

where E is defined by (2.3). This is a contradiction to (2.2). Then we obtain $Re\left(ap\left(z\right)\right)>0$. □

Theorem 2.2 Let $p\left(z\right)$ be a nonzero analytic function in $\mathbb{U}$ and $p\left(0\right)=1$. If

${\gamma }_{1}
(2.7)

where

${\gamma }_{1}=-\frac{\sqrt{|a{|}^{2}+2{\left(Re\left(a\right)\right)}^{2}}-Im\left(a\right)}{Rea}$

and

${\gamma }_{2}=\frac{\sqrt{|a{|}^{2}+2{\left(Re\left(a\right)\right)}^{2}}+Im\left(a\right)}{Re\left(a\right)},$

then

$Re\left(ap\left(z\right)\right)>0,$

where $Re\left(a\right)>0$.

Proof Let us define both $q\left(z\right)$ and $h\left(z\right)$ as follows:

$q\left(z\right)=ap\left(z\right)$

and

$h\left(z\right)=\frac{a+\overline{a}z}{1-z}\phantom{\rule{1em}{0ex}}\left(Re\left(a\right)>0\right),$

where $p\left(z\right)$ is defined by (2.1) since $q\left(z\right)$ and $h\left(z\right)$ are analytic functions in $\mathbb{U}$ with $q\left(0\right)=h\left(0\right)=a\in \mathbb{C}$ with

$h\left(\mathbb{U}\right)=\left\{w:Re\left(w\right)>0\right\}.$

Now, we suppose that $q\left(z\right)\nprec h\left(z\right)$. Therefore, by using Lemma 2.1, there exist points

${z}_{0}\in \mathbb{U}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\zeta }_{0}\in \partial \mathbb{U}\mathrm{\setminus }\left\{1\right\}$

such that $q\left({z}_{0}\right)=h\left({\zeta }_{0}\right)$ and ${z}_{0}{q}^{\mathrm{\prime }}\left({z}_{0}\right)=m{\zeta }_{0}{h}^{\mathrm{\prime }}\left({\zeta }_{0}\right)$, $m\ge n\ge 1$.

We note that

${\zeta }_{0}{h}^{\mathrm{\prime }}\left({\zeta }_{0}\right)=-\frac{|q\left({z}_{0}\right)-a{|}^{2}}{2Re\left(a-q\left({z}_{0}\right)\right)}.$
(2.8)

We have $h\left({\zeta }_{0}\right)=\rho i$ ($\rho \in \mathbb{R}$); therefore,

$\begin{array}{rcl}Im\left(p\left({z}_{0}\right)+\frac{{z}_{0}{p}^{\mathrm{\prime }}\left({z}_{0}\right)}{p\left({z}_{0}\right)}\right)& =& Im\left(q\left({z}_{0}\right)+\frac{{z}_{0}{q}^{\mathrm{\prime }}\left({z}_{0}\right)}{q\left({z}_{0}\right)}\right)\\ =& Im\left(\frac{h\left({\zeta }_{0}\right)}{a}+\frac{m{\zeta }_{0}{h}^{\mathrm{\prime }}\left({\zeta }_{0}\right)}{h\left({\zeta }_{0}\right)}\right)\\ =& Im\left(\frac{\rho i}{a}-\frac{m|\rho i-a{|}^{2}}{2Re\left(a\right)\rho i}\right)\\ =& \frac{\rho }{|a{|}^{2}}Re\left(a\right)+\frac{m|\rho i-a{|}^{2}}{2\rho Re\left(a\right)}.\end{array}$

For the case $\rho >0$, we obtain

$\begin{array}{rcl}Im\left(p\left({z}_{0}\right)+\frac{{z}_{0}{p}^{\mathrm{\prime }}\left({z}_{0}\right)}{p\left({z}_{0}\right)}\right)& \ge & \frac{\rho }{|a{|}^{2}}Re\left(a\right)+\frac{|\rho i-a{|}^{2}}{2\rho Re\left(a\right)}\\ =& \frac{1}{2\rho Re\left(a\right)}\left[\left(1+2{\left(\frac{Re\left(a\right)}{|a|}\right)}^{2}\right){\rho }^{2}+2Im\left(a\right)\rho +|a|\right]\\ =& g\left(\rho \right).\end{array}$
(2.9)

We can see that the function $g\left(\rho \right)$ in (2.9) takes the minimum value at ${\rho }_{1}$ given by

${\rho }_{1}=\frac{|a{|}^{2}}{\sqrt{|a{|}^{2}+2{\left(Re\left(a\right)\right)}^{2}}}.$

Hence, we have

$\begin{array}{rcl}Im\left(p\left({z}_{0}\right)+\frac{{z}_{0}{p}^{\mathrm{\prime }}\left({z}_{0}\right)}{p\left({z}_{0}\right)}\right)& \ge & g\left({\rho }_{1}\right)\\ =& {\gamma }_{2}.\end{array}$

This is a contradiction to (2.7). Then we obtain $Re\left(ap\left(z\right)\right)>0$.

For the case $\rho <0$, we obtain

$\begin{array}{rcl}Im\left(p\left({z}_{0}\right)+\frac{{z}_{0}{p}^{\mathrm{\prime }}\left({z}_{0}\right)}{p\left({z}_{0}\right)}\right)& \le & \frac{\rho }{|a{|}^{2}}Re\left(a\right)+\frac{|\rho i-a{|}^{2}}{2\rho Re\left(a\right)}\\ =& \frac{1}{2\rho Re\left(a\right)}\left[\left(1+2{\left(\frac{Re\left(a\right)}{|a|}\right)}^{2}\right){\rho }^{2}+2Im\left(a\right)\rho +|a{|}^{2}\right]\\ =& g\left(\rho \right).\end{array}$
(2.10)

We can see that the function $g\left(\rho \right)$ in (2.10) takes the maximum value at ${\rho }_{2}$ given by

${\rho }_{2}=-\frac{|a{|}^{2}}{\sqrt{|a{|}^{2}+2{\left(Re\left(a\right)\right)}^{2}}}.$

Hence, we have

$\begin{array}{rcl}Im\left(p\left({z}_{0}\right)+\frac{{z}_{0}{p}^{\mathrm{\prime }}\left({z}_{0}\right)}{p\left({z}_{0}\right)}\right)& \le & g\left({\rho }_{2}\right)\\ =& {\gamma }_{1}.\end{array}$

This is a contradiction to (2.7). Then we obtain $Re\left(ap\left(z\right)\right)>0$. □

Theorem 2.3 Let $p\left(z\right)$ be a nonzero analytic function in $\mathbb{U}$ with $p\left(0\right)=1$. If

$|p\left(z\right)+\frac{z{p}^{\mathrm{\prime }}\left(z\right)}{p\left(z\right)}-1|<\frac{3Re\left(a\right)}{2|a|},$

then

$Re\left(\frac{a}{p\left(z\right)}\right)>0,$

where $Re\left(a\right)>0$.

Proof Let us define both $q\left(z\right)$ and $h\left(z\right)$ as follows:

$q\left(z\right)=ap\left(z\right)$

and

$h\left(z\right)=\frac{a+\overline{a}z}{1-z}\phantom{\rule{1em}{0ex}}\left(Re\left(a\right)>0\right),$

where $p\left(z\right)$ is defined by (2.1) since $q\left(z\right)$ and $h\left(z\right)$ are analytic functions in $\mathbb{U}$ with $q\left(0\right)=h\left(0\right)=a\in \mathbb{C}$ with

$h\left(\mathbb{U}\right)=\left\{w:Rew>0\right\}.$

Now, we suppose that $q\left(z\right)\nprec h\left(z\right)$. Therefore, by using Lemma 2.1, there exist points

${z}_{0}\in \mathbb{U}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\zeta }_{0}\in \partial \mathbb{U}\mathrm{\setminus }\left\{1\right\}$

such that $q\left({z}_{0}\right)=h\left({\zeta }_{0}\right)$ and ${z}_{0}{q}^{\mathrm{\prime }}\left({z}_{0}\right)=m{\zeta }_{0}{h}^{\mathrm{\prime }}\left({\zeta }_{0}\right)$, $m\ge n\ge 1$.

We note that

${\zeta }_{0}{h}^{\mathrm{\prime }}\left({\zeta }_{0}\right)=-\frac{|q\left({z}_{0}\right)-a{|}^{2}}{2Re\left(a-q\left({z}_{0}\right)\right)}.$
(2.11)

We have $h\left({\zeta }_{0}\right)=\rho i$ ($\rho \in \mathbb{R}$).

Therefore,

$\begin{array}{rcl}\frac{|p\left({z}_{0}\right)+\frac{z{p}^{\mathrm{\prime }}\left({z}_{0}\right)}{p\left({z}_{0}\right)}-1|}{|p\left({z}_{0}\right)|}& =& |\frac{\rho i}{a}-\frac{m}{a}\frac{|a-i\rho {|}^{2}}{2Re\left(a\right)}-1|\\ \ge & \frac{1}{|a|}|\frac{m|a-i\rho {|}^{2}}{2Re\left(a\right)}+Re\left(a\right)|\\ \ge & \frac{1}{|a|}\left(\frac{|a-i\rho {|}^{2}}{2Re\left(a\right)}+Re\left(a\right)\right)\\ \ge & \frac{1}{2|a|Re\left(a\right)}\left(3{\left(Re\left(a\right)\right)}^{2}+{\left(Im\left(a\right)-\rho \right)}^{2}\right)\\ \ge & \frac{3Re\left(a\right)}{2|a|}.\end{array}$

This is a contradiction to (2.7). Then we obtain $Re\left(\frac{a}{p\left(z\right)}\right)>0$. □

## 3 Applications and examples

Putting $P\left(z\right)=\beta$ ($\beta >0$; real) in Theorem 2.1, we have the following corollary.

Corollary 3.1 If $p\left(z\right)$ is an analytic function in $\mathbb{U}$ with $p\left(0\right)=1$ and

$Re\left(p\left(z\right)+\beta z{p}^{\mathrm{\prime }}\left(z\right)\right)>\frac{E}{2\beta |a{|}^{2}Re\left(a\right)},$

then

$Re\left(ap\left(z\right)\right)>0,$

where

$E=-\left(Re\left(a\right)\right)\left[{\beta }^{2}{\left(Re\left(a\right)\right)}^{2}+\left(1+2\beta \right){\left(Im\left(a\right)\right)}^{2}\right]$

with $Re\left(a\right)>0$.

Putting $\beta =1$ in Corollary 3.1, we obtain the following corollary.

Corollary 3.2 If $p\left(z\right)$ is an analytic function in $\mathbb{U}$ with $p\left(0\right)=1$ and

$Re\left(p\left(z\right)+z{p}^{\mathrm{\prime }}\left(z\right)\right)>\frac{3}{2}-2{\left(\frac{Re\left(a\right)}{|a|}\right)}^{2},$

then

$Re\left(ap\left(z\right)\right)>0,$

where $Re\left(a\right)>0$.

Putting $p\left(z\right)=\frac{f\left(z\right)}{g\left(z\right)}$ and $P\left(z\right)=\frac{g\left(z\right)}{z{g}^{\mathrm{\prime }}\left(z\right)}$ in Theorem 2.1, we have the following corollary.

Corollary 3.3 Let $f\left(z\right)\in A$, $g\left(z\right)\in {S}^{\ast }$ and

$Re\left(\frac{{f}^{\mathrm{\prime }}\left(z\right)}{{g}^{\mathrm{\prime }}\left(z\right)}\right)>\frac{3}{2}-2{\left(\frac{Re\left(a\right)}{|a|}\right)}^{2}.$

Then

$Re\left(a\frac{f\left(z\right)}{g\left(z\right)}\right)>0,$

where $Re\left(a\right)>0$.

Example 3.1 Let $f\left(z\right)\in A$ satisfy the following relation:

$Re\left({f}^{\mathrm{\prime }}\left(z\right)\right)>\frac{3}{2}-2{\left(\frac{Re\left(a\right)}{|a|}\right)}^{2}.$

Then

$Re\left(a\frac{f\left(z\right)}{z}\right)>0,$

where $Re\left(a\right)>0$.

Example 3.2 Let $f\left(z\right)\in A$ satisfy the following relation:

$Re\left(\left(2+\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}\left(z\right)}{{f}^{\mathrm{\prime }}\left(z\right)}-\frac{z{f}^{\mathrm{\prime }}\left(z\right)}{f\left(z\right)}\right)\frac{z{f}^{\mathrm{\prime }}\left(z\right)}{f\left(z\right)}\right)>\frac{3}{2}-2{\left(\frac{Re\left(a\right)}{|a|}\right)}^{2}.$

Then

$Re\left(a\frac{z{f}^{\mathrm{\prime }}\left(z\right)}{f\left(z\right)}\right)>0,$

where $Re\left(a\right)>0$.

Remark 3.1

1. (i)

Putting $a={e}^{i\lambda }$ ($|\lambda |<\frac{\pi }{2}$) in Theorem 2.1, we have Theorem 1 due to Kim and Cho [3].

2. (ii)

Putting $a={e}^{i\lambda }$ ($|\lambda |<\frac{\pi }{2}$), $P\left(z\right)=\beta$ ($\beta >0$; real) in Theorem 2.1, we have Corollary 1 due to Kim and Cho [3].

3. (iii)

Putting $a=0$ and $P\left(z\right)=1$ in Theorem 2.1, we have the result due to Nunokawa et al. [15].

4. (iv)

Putting $a={e}^{i\lambda }$ ($|\lambda |<\frac{\pi }{2}$), $P\left(z\right)=1$ in Theorem 2.1, we have Corollary 2 due to Kim and Cho [3].

Putting $p\left(z\right)=\frac{z{f}^{\mathrm{\prime }}\left(z\right)}{f\left(z\right)}$ in Theorem 2.2, we have the following corollary.

Corollary 3.4 Let $f\left(z\right)\in A$. If

${\gamma }_{1}

where

${\gamma }_{1}=-\frac{\sqrt{|a{|}^{2}+2{\left(Re\left(a\right)\right)}^{2}}-Im\left(a\right)}{Re\left(a\right)}$

and

${\gamma }_{2}=\frac{\sqrt{|a{|}^{2}+2{\left(Re\left(a\right)\right)}^{2}}+Im\left(a\right)}{Re\left(a\right)},$

then

$Re\left(a\frac{z{f}^{\mathrm{\prime }}\left(z\right)}{f\left(z\right)}\right)>0,$

where $Re\left(a\right)>0$.

Putting $p\left(z\right)=\frac{z{f}^{\mathrm{\prime }}\left(z\right)}{f\left(z\right)}$ in Theorem 2.3, we have the following corollary.

Corollary 3.5 Let $p\left(z\right)$ be a nonzero analytic function in $\mathbb{U}$ with $p\left(0\right)=1$. If

$|\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}\left(z\right)}{{f}^{\mathrm{\prime }}\left(z\right)}|<\frac{3Re\left(a\right)}{2|a|},$

then

$Re\left(\frac{1}{a}\frac{z{f}^{\mathrm{\prime }}\left(z\right)}{f\left(z\right)}\right)>0,$

where $Re\left(a\right)>0$.

Remark 3.2 Putting $a={e}^{i\lambda }$ ($|\lambda |<\frac{\pi }{2}$) in Corollary 3.5, we have the result due to Kim and Cho [3].

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## Acknowledgements

Dedicated to Professor Hari M Srivastava.

The authors would like to express their gratitude to the referees for the valuable advices to improve this paper.

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### Authors’ contributions

All authors contributed equally to the paper. Also, all authors have read and approved the final version of the paper.

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Attiya, A.A., Nasr, M.A. On sufficient conditions for Carathéodory functions with applications. J Inequal Appl 2013, 191 (2013). https://doi.org/10.1186/1029-242X-2013-191