• Research
• Open Access

# Some properties concerning close-to-convexity of certain analytic functions

Journal of Inequalities and Applications20122012:245

https://doi.org/10.1186/1029-242X-2012-245

• Received: 1 August 2012
• Accepted: 8 October 2012
• Published:

## Abstract

Let $f\left(z\right)$ be an analytic function in the open unit disk $\mathbb{D}$ normalized with $f\left(0\right)=0$ and ${f}^{\prime }\left(0\right)=1$. With the help of subordinations, for convex functions $f\left(z\right)$ in $\mathbb{D}$, the order of close-to-convexity for $f\left(z\right)$ is discussed with some example.

MSC:30C45.

## Keywords

• analytic
• starlike
• convex
• close-to-convex
• subordination

## 1 Introduction

Let $\mathcal{A}$ be the class of functions $f\left(z\right)$ of the form
$f\left(z\right)=z+\sum _{n=2}^{\mathrm{\infty }}{a}_{n}{z}^{n}$
which are analytic in the open unit disk $\mathbb{D}=\left\{z\in \mathbb{C}||z|<1\right\}$. A function $f\left(z\right)\in \mathcal{A}$ is said to be convex of order α if it satisfies

for some real α ($0\leqq \alpha <1$). This family of functions was introduced by Robertson  and we denote it by $\mathcal{K}\left(\alpha \right)$.

A function $f\left(z\right)\in \mathcal{A}$ is called starlike of order α in $\mathbb{D}$ if it satisfies

for some real α ($0\leqq \alpha <1$).

This class was also introduced by Robertson  and we denote it by ${\mathcal{S}}^{\ast }\left(\alpha \right)$. By the definitions for the classes $\mathcal{K}\left(\alpha \right)$ and ${\mathcal{S}}^{\ast }\left(\alpha \right)$, we know that $f\left(z\right)\in \mathcal{K}\left(\alpha \right)$ if and only if $z{f}^{\prime }\left(z\right)\in {\mathcal{S}}^{\ast }\left(\alpha \right)$.

Marx  and Strohhäcker  showed that $f\left(z\right)\in \mathcal{K}\left(0\right)$ implies $f\left(z\right)\in {\mathcal{S}}^{\ast }\left(\frac{1}{2}\right)$.

This estimate is sharp for an extremal function
$f\left(z\right)=\frac{z}{1-z}.$
Jack  posed a more general problem: What is the largest number $\beta =\beta \left(\alpha \right)$ so that
$\mathcal{K}\left(\alpha \right)\subset {\mathcal{S}}^{\ast }\left(\beta \left(\alpha \right)\right).$

MacGregor  determined the exact value of $\beta \left(\alpha \right)$ for each α ($0\leqq \alpha <1$) as the infimum over the disc $\mathbb{D}$ of the real part of a specific analytic function. It has been conjectured that this infimum is attained on the boundary of $\mathbb{D}$ at $z=-1$.

Wilken and Feng  asserted MacGregor’s conjecture: If $0\leqq \alpha <1$ and $f\left(z\right)\in \mathcal{K}\left(\alpha \right)$, then $f\left(z\right)\in {\mathcal{S}}^{\ast }\left(\beta \left(\alpha \right)\right)$, where
(1)
Ozaki  and Kaplan  investigated the following functions: If $f\left(z\right)\in \mathcal{A}$ satisfies

for some convex function $g\left(z\right)$, then $f\left(z\right)$ is univalent in $\mathbb{D}$. In view of Kaplan , we say that $f\left(z\right)$ satisfying the above inequality is close-to-convex in $\mathbb{D}$.

It is well known that the above definition concerning close-to-convex functions is equivalent to the following condition:

for some starlike function $g\left(z\right)\in \mathcal{A}$.

Let us define a function $f\left(z\right)\in \mathcal{A}$ which satisfies

for some real α ($0\leqq \alpha <1$) and for some starlike function $g\left(z\right)$ in $\mathbb{D}$.

Then we call $f\left(z\right)$ close-to-convex of order α in $\mathbb{D}$ with respect to $g\left(z\right)$.

It is the purpose of the present paper to investigate the order of close-to-convexity of the functions which satisfy $f\left(z\right)\in \mathcal{K}\left(\alpha \right)$ and $0\leqq \alpha <1$.

## 2 Preliminary

To discuss our problems, we have to give here the following lemmas.

Lemma 1 Let $p\left(z\right)=1+{\sum }_{n=1}^{\mathrm{\infty }}{c}_{n}{z}^{n}$ be analytic in $\mathbb{D}$ and suppose that
$p\left(z\right)\prec \frac{1-\alpha z}{1+\beta z}\phantom{\rule{1em}{0ex}}\mathit{\text{in}}\phantom{\rule{0.5em}{0ex}}\mathbb{D},$

where means the subordination, $0<\alpha <1$ and $0<\beta <1$.

Then we have
$\frac{1-\alpha }{1+\beta }
This shows that
$Rep\left(z\right)>0\phantom{\rule{1em}{0ex}}\mathit{\text{in}}\phantom{\rule{0.5em}{0ex}}\mathbb{D}.$

A proof is very easily obtained.

Lemma 2 Let $p\left(z\right)=1+{\sum }_{n=1}^{\mathrm{\infty }}{c}_{n}{z}^{n}$ be analytic in $\mathbb{D}$, and suppose that there exists a point ${z}_{0}\in \mathbb{D}$ such that
$Rep\left(z\right)>c\phantom{\rule{1em}{0ex}}\mathit{\text{for}}\phantom{\rule{0.5em}{0ex}}|z|<|{z}_{0}|$
and
$Rep\left({z}_{0}\right)=c,\phantom{\rule{2em}{0ex}}p\left({z}_{0}\right)\ne c$
for some real c ($0). Then we have
$Re\frac{{z}_{0}{p}^{\prime }\left({z}_{0}\right)}{p\left({z}_{0}\right)}\leqq \left\{\begin{array}{cc}-\frac{1-c}{2c}\hfill & \mathit{\text{when}}\phantom{\rule{0.5em}{0ex}}\frac{1}{2}\leqq c<1,\hfill \\ -\frac{c}{2\left(1-c\right)}\hfill & \mathit{\text{when}}\phantom{\rule{0.5em}{0ex}}0
Proof Let us put
$q\left(z\right)=\frac{p\left(z\right)-c}{1-c},\phantom{\rule{2em}{0ex}}q\left(0\right)=1.$
Then $q\left(z\right)$ is analytic in $\mathbb{D}$ and
and
$Req\left({z}_{0}\right)=0,\phantom{\rule{2em}{0ex}}q\left({z}_{0}\right)\ne 0.$
Then, from [, Theorem 1], we have
$\frac{{z}_{0}{q}^{\prime }\left({z}_{0}\right)}{q\left({z}_{0}\right)}=i\ell ,$
where
and
$\ell \leqq -\frac{1}{2}\left(a+\frac{1}{a}\right)\leqq -1\phantom{\rule{1em}{0ex}}\text{when}argq\left({z}_{0}\right)=-\frac{\pi }{2},$

where $q\left({z}_{0}\right)=±ia$ and $a>0$.

For the case $argq\left({z}_{0}\right)=\frac{\pi }{2}$, we have
$\frac{{z}_{0}{q}^{\prime }\left({z}_{0}\right)}{q\left({z}_{0}\right)}=\frac{{z}_{0}{p}^{\prime }\left({z}_{0}\right)}{p\left({z}_{0}\right)-c}=i\ell$
If we put
$h\left(x\right)=\frac{1+{x}^{2}}{{c}^{2}+{\left(1-c\right)}^{2}{x}^{2}}\phantom{\rule{1em}{0ex}}\left(x>0\right),$
then it is easy to see that
and
This shows that
For the case $argq\left({z}_{0}\right)=-\frac{\pi }{2}$, applying the same method as above, we have the same conclusion

This completes the proof of the lemma. □

Our next lemma is

Lemma 3 Let $p\left(z\right)=1+{\sum }_{n=1}^{\mathrm{\infty }}{c}_{n}{z}^{n}$ be analytic in $\mathbb{D}$ and suppose that there exists a point ${z}_{0}\in \mathbb{D}$ such that
$Rep\left(z\right)>c\phantom{\rule{1em}{0ex}}\mathit{\text{for}}\phantom{\rule{0.5em}{0ex}}|z|<|{z}_{0}|$
and
$Rep\left({z}_{0}\right)=c,\phantom{\rule{2em}{0ex}}p\left({z}_{0}\right)\ne c$

for some real c ($c<0$).

Then we have
$Re\frac{{z}_{0}{p}^{\prime }\left({z}_{0}\right)}{p\left({z}_{0}\right)}>-\frac{c}{2\left(1-c\right)}>0.$
(2)
Proof Let us put
$q\left(z\right)=\frac{p\left(z\right)-c}{1-c},\phantom{\rule{2em}{0ex}}q\left(0\right)=1.$
Then $q\left(z\right)$ is analytic in $\mathbb{D}$. If $p\left(z\right)$ satisfies the hypothesis of the lemma, then there exists a point ${z}_{0}\in \mathbb{D}$ such that
and
$Req\left({z}_{0}\right)=0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}q\left({z}_{0}\right)\ne 0,$

then $p\left(z\right)$ satisfies the conditions of the lemma.

For the case $argq\left({z}_{0}\right)=\frac{\pi }{2}$, applying the same method as in the proof of Lemma 2, we have
$Re\frac{{z}_{0}{p}^{\prime }\left({z}_{0}\right)}{p\left({z}_{0}\right)}=-\frac{\left(1-c\right)ca\ell }{{c}^{2}+{\left(1-c\right)}^{2}{a}^{2}}\geqq -\frac{c\left(1-c\right)}{2}\left(\frac{1+{a}^{2}}{{c}^{2}+{\left(1-c\right)}^{2}{a}^{2}}\right).$
Putting
$h\left(x\right)=\frac{1+{x}^{2}}{{c}^{2}+{\left(1-c\right)}^{2}{x}^{2}}\phantom{\rule{1em}{0ex}}\left(x>0\right),$
it follows that
${h}^{\prime }\left(x\right)=\frac{\left(2c-1\right)x}{{\left({c}^{2}+{\left(1-c\right)}^{2}{x}^{2}\right)}^{2}}<0\phantom{\rule{1em}{0ex}}\left(x>0\right).$
(3)

Therefore, from (3) we obtain (2) .

For the case $argq\left({z}_{0}\right)=-\frac{\pi }{2}$, applying the same method as above, we have the same conclusion as in the case $argq\left({z}_{0}\right)=\frac{\pi }{2}$. □

## 3 The order of close-to-convexity

Now, we discuss the close-to-convexity of $f\left(z\right)$ with the help of lemmas.

Theorem 1 Let $f\left(z\right)\in \mathcal{A}$, and suppose that there exists a starlike function $g\left(z\right)$ such that
1. (i)

and
1. (ii)

Then we have
$Re\frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}>c\phantom{\rule{1em}{0ex}}\mathit{\text{in}}\phantom{\rule{0.5em}{0ex}}\mathbb{D}.$

This means that $f\left(z\right)$ is close-to-convex of order c in $\mathbb{D}$.

Proof Let us put
$p\left(z\right)=\frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)},\phantom{\rule{2em}{0ex}}p\left(0\right)=1.$
Then it follows that
$1+\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}=\frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}+\frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)}.$
1. (i)
For the case $\frac{1}{2}\leqq c<1$, if there exists a point ${z}_{0}\in \mathbb{D}$ such that

and
$Rep\left({z}_{0}\right)=c,$
then, applying Lemma 2 and the hypothesis of Theorem 1, we have
$p\left({z}_{0}\right)\ne c$
and
$Re\frac{{z}_{0}{p}^{\prime }\left({z}_{0}\right)}{p\left({z}_{0}\right)}\leqq -\frac{1-c}{2c}.$
Thus, it follows that
$\begin{array}{rcl}1+Re\frac{{z}_{0}{f}^{″}\left({z}_{0}\right)}{{f}^{\prime }\left({z}_{0}\right)}& =& Re\frac{{z}_{0}{g}^{\prime }\left({z}_{0}\right)}{g\left({z}_{0}\right)}+Re\frac{{z}_{0}{p}^{\prime }\left({z}_{0}\right)}{p\left({z}_{0}\right)}\\ \leqq & Re\frac{{z}_{0}{g}^{\prime }\left({z}_{0}\right)}{g\left({z}_{0}\right)}-\frac{1-c}{2c},\end{array}$
which contradicts the hypothesis of Theorem 1. (ii) For the case $0, applying the same method as above, we also have that

This completes the proof of the theorem. □

Applying Theorem 1, we have the following corollary.

Corollary 1 Let $f\left(z\right)\in \mathcal{A}$ be convex of order α ($0<\alpha <1$), and suppose that there exists a starlike function $g\left(z\right)$ such that
1. (i)

and
1. (ii)

Then we have
$Re\frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}>\beta \left(c\right)>\beta \left(\alpha \right)>\alpha \phantom{\rule{1em}{0ex}}\mathit{\text{in}}\phantom{\rule{0.5em}{0ex}}\mathbb{D}.$
Remark 1 For the case $0<\alpha , it is trivial that
$\alpha <\beta \left(\alpha \right)<\beta \left(c\right)<1.$
Example 1 Let $f\left(z\right)\in \mathcal{A}$ satisfy
(4)
where
$A=\frac{32\beta \left(\frac{1}{2}\right)-10}{8\beta \left(\frac{1}{2}\right)+10}\fallingdotseq 0.29605$
and $\beta \left(\frac{1}{2}\right)=\frac{1}{2log2}$. If we consider the starlike function $g\left(z\right)$ given by
$g\left(z\right)=\frac{z}{{\left(1+Az\right)}^{2}},$
then we have
$Re\frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}>\beta \left(\frac{1}{2}\right)\fallingdotseq 0.7213,$

which means that $f\left(z\right)$ is close-to-convex of order $\beta \left(\frac{1}{2}\right)$ in $\mathbb{D}$.

Next we show

Theorem 2 Let $f\left(z\right)\in \mathcal{A}$ and $g\left(z\right)\in \mathcal{A}$ be given by
$g\left(z\right)=\left\{\begin{array}{cc}\frac{z}{{\left(1+\beta z\right)}^{\frac{\alpha +\beta }{\beta }}}\hfill & \left(\beta \ne 0\right),\hfill \\ z{e}^{-\alpha z}\hfill & \left(\beta =0\right)\hfill \end{array}$
for some α ($0\leqq \alpha <1$) and some β ($0\leqq \beta <1$). Further suppose that for arbitrary r ($0),
$\underset{|z|=r}{min}\left(Re\frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}\right)={\left(Re\frac{{z}_{0}{f}^{\prime }\left({z}_{0}\right)}{g\left({z}_{0}\right)}\right)}_{|{z}_{0}|=r}\ne \frac{{z}_{0}{f}^{\prime }\left({z}_{0}\right)}{g\left({z}_{0}\right)}$
and
$1+Re\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\leqq -\frac{c}{2\left(1-c\right)}+\frac{1-\alpha }{1+\beta }$
for $c<0$. Then we have
$Re\frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}>c\phantom{\rule{1em}{0ex}}\mathit{\text{in}}\phantom{\rule{0.5em}{0ex}}\mathbb{D}.$
Proof Let us define the function $p\left(z\right)$ by
$p\left(z\right)=\frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)},\phantom{\rule{2em}{0ex}}p\left(0\right)=1$
for $c<0$. If there exists a point ${z}_{0}\in \mathbb{D}$ such that
and
$Rep\left({z}_{0}\right)=c$
for $c<0$, then from the hypothesis of Theorem 2, we have
$Rep\left({z}_{0}\right)\ne p\left({z}_{0}\right).$
Therefore, applying Lemma 1 and Lemma 3, we have
$\begin{array}{rcl}1+Re\frac{{z}_{0}{f}^{″}\left({z}_{0}\right)}{{f}^{\prime }\left({z}_{0}\right)}& =& Re\frac{{z}_{0}{p}^{\prime }\left({z}_{0}\right)}{p\left({z}_{0}\right)}+Re\frac{{z}_{0}{g}^{\prime }\left({z}_{0}\right)}{g\left({z}_{0}\right)}\\ >& -\frac{c}{2\left(1-c\right)}+\frac{1-\alpha }{1+\beta }.\end{array}$
This is a contradiction, and therefore we have

□

Remark 2 In view of the definition for close-to-convex functions, if $f\left(z\right)$ satisfies

then we can say that $f\left(z\right)$ is close-to-convex in $\mathbb{D}$. But c should be a negative real number in Theorem 2. Therefore, we cannot say that $f\left(z\right)$ is close-to-convex in $\mathbb{D}$ in Theorem 2.

## Declarations

### Acknowledgements

The authors thank the referees for their helpful comments and suggestions to improve our manuscript.

## Authors’ Affiliations

(1)
University of Gunma, Hoshikuki 798-8, Chuou-Ward, Chiba 260-0808, Japan
(2)
Department of Mathematics, Işık University, Meşrutiyet Koyu, Şile Kampusu, Istanbul, 34980, Turkey
(3)
Department of Mathematics, Kinki University, Higashi-Osaka Osaka, 577-8502, Japan
(4)
Department of Mathematics, Duzce University, Konuralp Yerleskesi, Duzce, 81620, Turkey

## References 