Open Access

Some criteria for concave conformal mappings

Journal of Inequalities and Applications20152015:119

https://doi.org/10.1186/s13660-015-0640-5

Received: 25 February 2015

Accepted: 24 March 2015

Published: 4 April 2015

Abstract

The main purpose of this paper is to derive some criteria for concave conformal mappings.

Keywords

concave mapping meromorphic function Jack’s lemma

MSC

30C55

1 Introduction

A conformal, meromorphic function f on the punctured unit disk
$${\mathbb{U}}^{*}:= \bigl\{ z\in{\mathbb{C}}: 0< \vert z \vert <1 \bigr\} =: \mathbb{U}\backslash\{0\} $$
is said to be a concave mapping if \(f(\mathbb{U}^{*})\) is the complement of a compact, convex set.
Let Σ denote the class of analytic functions of the form
$$ f(z)=\frac{1}{z}+\sum_{k=0}^{\infty }b_{k}z^{k} \quad \bigl(z\in\mathbb{U}^{*}\bigr), $$
(1.1)
then the necessary and sufficient condition for f to be a concave mapping is
$$ 1+\Re \biggl(\frac{zf''(z)}{f'(z)} \biggr)< 0\quad (z\in\mathbb{U}), $$
(1.2)
where
$$\frac{zf''(z)}{f'(z)}=-2-2b_{1}z^{2}-6b_{2}z^{3}- \bigl(12b_{3}+2b_{1}^{2} \bigr)z^{4}- \cdots. $$

Recently, Bhowmik et al. [1], Chuaqui et al. [2], Ibrahim and Sokół [3] derived some interesting properties of concave conformal mappings. In this paper, we aim at proving several criteria for the function \(f\in\Sigma\) to be a concave mapping.

To prove our main results, we need the following two lemmas.

Lemma 1.1

(Jack’s lemma [4])

Let \(h(z)=a_{n}z^{n}+a_{n+1}z^{n+1}+\cdots\) be a non-constant analytic function in \(\mathbb{U}\). If \(\vert h(z)\vert \) attains its maximum value on the circle \(\vert z\vert =r<1\), then
$$z_{0}h'(z_{0})=kh(z_{0}), $$
where k is a real number with \(k\geq n\).

Lemma 1.2

(See [5])

Let Ω be a set in the complex plane and suppose that Φ is a mapping from \(\mathbb{C}^{2}\times\mathbb{U}\) to which satisfies \(\Phi(ix,y;z)\notin\Omega\) for \(z\in\mathbb{U}\) and for all real x, y such that \(y\leq-\frac{1+x^{2}}{2}\). If the function \(p(z)=1+c_{1}z+c_{2}z^{2}+\cdots\) is analytic in \(\mathbb{U}\) and \(\Phi (p(z),zp'(z);z )\in\Omega\) for all \(z\in\mathbb {U}\), then \(\Re(p (z))>0\).

2 Main results

We first give the following result.

Theorem 2.1

Suppose that \(f\in\Sigma\) with \((zf'(z))'\neq 0\). If f satisfies the condition
$$ \biggl\vert \frac{zf''(z)}{f'(z)}-\frac {z(2f''(z)+zf'''(z))}{f'(z)+zf''(z)}\biggr\vert < \lambda\quad \biggl(0<\lambda\leq\frac{1}{2} \biggr), $$
(2.1)
then f is concave in \(\mathbb{U}^{*}\).

Proof

Assume that
$$ \phi(z):=\frac{(1-\lambda)\frac {f'(z)}{f'(z)+zf''(z)}+1}{\lambda}-1 \quad \biggl(0< \lambda\leq \frac {1}{2}; z\in\mathbb{U} \biggr). $$
(2.2)
Then the function ϕ is analytic in \(\mathbb{U}\) with \(\phi(0)=0\). From (2.2), we know that
$$ \frac{f'(z)}{f'(z)+zf''(z)}=\frac{\lambda \phi(z)+\lambda-1}{1-\lambda}. $$
(2.3)
By differentiating both sides of (2.3) with respect to z logarithmically, we get
$$ \frac{zf''(z)}{f'(z)}-\frac {z(2f''(z)+zf'''(z))}{f'(z)+zf''(z)} =\frac{\lambda z\phi'(z)}{\lambda\phi(z)+\lambda-1}. $$
(2.4)
From (2.1) and (2.4), we find that
$$ \biggl\vert \frac{zf''(z)}{f'(z)}-\frac {z(2f''(z)+zf'''(z))}{f'(z)+zf''(z)}\biggr\vert = \lambda\biggl\vert \frac{\lambda z\phi'(z)}{\lambda\phi (z)+\lambda -1}\biggr\vert < \lambda. $$
(2.5)
Now, we can claim that \(\vert \phi(z)\vert <1\). If not, there exists a point \(z_{0}\in\mathbb{U}\) such that
$$\max_{\vert z\vert \leq \vert z_{0}\vert }\bigl\vert \phi(z)\bigr\vert =\bigl\vert \phi(z_{0})\bigr\vert =1. $$
By Lemma 1.1, we know that
$$ z_{0}\phi'(z_{0})=k \phi(z_{0})=ke^{i\theta } \quad (0\leq\theta< 2\pi; k\geq1). $$
(2.6)
For \(z=z_{0}\), we find from (2.4) and (2.6) that
$$ \biggl\vert \frac {z_{0}f''(z_{0})}{f'(z_{0})}-\frac {z_{0}(2f''(z_{0})+z_{0}f'''(z_{0}))}{f'(z_{0})+z_{0}f''(z_{0})}\biggr\vert = \lambda\biggl\vert \frac{k}{\lambda+(\lambda-1)e^{-i\theta }}\biggr\vert \geq \lambda. $$
(2.7)
But (2.7) contradicts (2.5). Thus, we deduce that \(\vert \phi(z)\vert <1\), which implies that
$$ \biggl\vert \frac{(1-\lambda)\frac {f'(z)}{f'(z)+zf''(z)}+1}{\lambda}-1\biggr\vert < 1, $$
(2.8)
or equivalently,
$$ \biggl\vert \frac{f'(z)}{f'(z)+zf''(z)}+1\biggr\vert < \frac {\lambda}{1-\lambda}. $$
(2.9)
From (2.9), we get
$$\Re \biggl(1+\frac{zf''(z)}{f'(z)} \biggr)< \lambda-1<0 \quad \biggl(0<\lambda\leq \frac{1}{2} \biggr), $$
which shows that the function f is concave in \(\mathbb{U}^{*}\). □

Theorem 2.2

Suppose that \(f\in\Sigma\) with \(f'(z)\neq 0\). If f satisfies the inequality
$$ \Re \biggl(\frac{z [ (f''(z)+zf'''(z) )f'(z)-z(f''(z))^{2} ]}{ f'(z) (zf''(z)+3f'(z) )} \biggr)< 1, $$
(2.10)
then f is concave in \(\mathbb{U}^{*}\).

Proof

Define the function \(\varphi(z)\) by
$$ \varphi(z):=\frac{zf''(z)}{f'(z)}+2\quad (z\in\mathbb{U}). $$
(2.11)
It is easy to see that
$$\varphi(z)=-2b_{1}z^{2}-6b_{2}z^{3}- \bigl(12b_{3}+2b_{1}^{2} \bigr)z^{4}- \cdots $$
is analytic in \(\mathbb{U}\) with \(\varphi(0)=\varphi'(0)=0\). From (2.11), we obtain
$$ \frac{zf''(z)}{f'(z)}+3=1+\varphi(z)\quad (z\in\mathbb{U}). $$
(2.12)
Taking logarithmical derivatives of both sides of (2.12) with respect to z, we get
$$ \frac{z [ (f''(z)+zf'''(z) )f'(z)-z(f''(z))^{2} ]}{ f'(z) (zf''(z)+3f'(z) )}=\frac{z\varphi'(z)}{1+\varphi(z)}. $$
(2.13)
We now show that \(\vert \varphi(z)\vert <1\). If not, there exists a point \(z_{0}\in\mathbb{U}\) such that
$$\max_{\vert z\vert \leq \vert z_{0}\vert }\bigl\vert \varphi(z)\bigr\vert =\bigl\vert \varphi(z_{0}) \bigr\vert =1. $$
By Jack’s lemma, we know that
$$ z_{0}\varphi'(z_{0})=k \varphi(z_{0})=ke^{i\theta } \quad (0\leq\theta< 2\pi; k\geq2). $$
(2.14)
For \(z=z_{0}\), we have
$$\begin{aligned}& \Re \biggl(\frac{z_{0} [ (f''(z_{0})+z_{0}f'''(z_{0}) )f'(z_{0})-z_{0}(f''(z_{0}))^{2} ]}{ f'(z_{0}) (z_{0}f''(z_{0})+3f'(z_{0}) )} \biggr) \\& \quad =\Re \biggl( \frac{z_{0}\varphi'(z_{0})}{1+\varphi(z_{0})} \biggr) =\Re \biggl(\frac{ke^{i\theta}}{1+e^{i\theta}} \biggr)\geq \frac {k}{2}\geq1. \end{aligned}$$
(2.15)
But (2.15) is a contradiction to condition (2.10), which implies that \(\vert \varphi (z)\vert <1\). Consequently, we deduce from (2.11) that
$$\Re \biggl(1+\frac{zf''(z)}{f'(z)} \biggr)=\Re\bigl(\varphi(z)\bigr)-1\leq \bigl\vert \varphi(z)\bigr\vert -1< 0, $$
which implies that f is concave in \(\mathbb{U}^{*}\). □

Theorem 2.3

Suppose that \(f\in\Sigma\) with \(f'(z)\neq 0\). If f satisfies the condition
$$ \Re \biggl(\frac{zf'(z)}{(zf'(z))'} \biggl(\frac{(zf'(z))'}{f'(z)} \biggr)' \biggr)>\left \{ \begin{array}{l@{\quad}l} \frac{\delta}{2(\delta-1)} & (0\leqq\delta\leqq\frac{1}{2} ), \\ \frac{\delta-1}{2\delta} & (\frac{1}{2}\leqq\delta< 1 ), \end{array} \right . $$
(2.16)
then f is concave in \(\mathbb{U}^{*}\).

Proof

Suppose that
$$ \psi(z)=\frac{-\frac {zf''(z)}{f'(z)}-1-\delta}{1-\delta}\quad (0\leq\delta< 1; z\in \mathbb{U}). $$
(2.17)
Then ψ is analytic in \(\mathbb{U}\). From (2.17), we find that
$$ \frac{zf'(z)}{(zf'(z))'} \biggl(\frac {(zf'(z))'}{f'(z)} \biggr)'= \frac{(1-\delta)z\psi'(z)}{\delta+(1-\delta)\psi(z)}=\Phi \bigl(\psi(z),z\psi'(z);z \bigr), $$
(2.18)
where
$$\Phi(r,s;t)=\frac{(1-\delta)s}{\delta+(1-\delta)r}. $$
For the real numbers x and y satisfying the condition \(y\leq-\frac{1+x^{2}}{2}\), we know that
$$\begin{aligned} \Re \bigl(\Phi(ix,y;z) \bigr)&=\frac {(1-\delta)\delta y}{\delta^{2}+(1-\delta)^{2}x^{2}} \\ & \leq-\frac{(1-\delta)\delta}{2}\cdot\frac{1+x^{2}}{\delta ^{2}+(1-\delta)^{2}x^{2}} \\ &\leq \left \{ \begin{array}{l@{\quad}l} \frac{\delta}{2(\delta-1)} & (0\leqq\delta\leqq\frac{1}{2} ), \\ \frac{\delta-1}{2\delta} & (\frac{1}{2}\leqq\delta< 1 ). \end{array} \right . \end{aligned}$$
(2.19)
Now, we take
$$\Omega=\left \{\xi: \Re(\xi)> \left \{ \begin{array}{l@{\quad}l} \frac{\delta}{2(\delta-1)} & (0\leqq\delta\leqq\frac{1}{2} ) \\ \frac{\delta-1}{2\delta} & (\frac{1}{2}\leqq\delta< 1 ) \end{array} \right . \right \}, $$
then \(\Phi(ix,y;z)\notin \Omega\) for all real x, y such that \(y\leq-\frac{1+x^{2}}{2}\). Furthermore, by virtue of (2.16), we know that \(\Phi (\psi(z),z\psi'(z);z )\in\Omega\). Thus, by Lemma 1.2, we get \(\Re(\psi(z))>0\), which shows that f is concave in \(\mathbb{U}^{*}\). □
Finally, we correct an error of Theorem 2.1 in [3], the condition
$$\Re \biggl(\frac{zf'''(z)}{f''(z)} \biggr)< 0 \quad (z\in\mathbb{U}) $$
in it should be changed into
$$\Re \biggl(\frac{zf'''(z)}{f''(z)} \biggr)>-3\quad (z\in\mathbb{U}). $$

Theorem 2.4

Suppose that \(f\in\Sigma\) with \(f'(z)\neq 0\). If f satisfies the inequality
$$ \Re \biggl(\frac{zf'''(z)}{f''(z)} \biggr)>-3\quad (z\in\mathbb{U}), $$
(2.20)
then f is concave in \(\mathbb{U}^{*}\).

Proof

Define the function \(\omega(z)\) by
$$ -1-\frac{zf''(z)}{f'(z)}=\frac{1+\omega (z)}{1-\omega(z)}. $$
(2.21)
Then ω is analytic in \(\mathbb{U}\) with \(\omega(0)=\omega'(0)=0\). From (2.21), we get
$$ \frac{zf''(z)}{f'(z)}=\frac{-2}{1-\omega(z)}. $$
(2.22)
Differentiating both sides of (2.22) logarithmically, we get
$$ \frac{zf'''(z)}{f''(z)}=\frac{z\omega '(z)}{1-\omega(z)}+\frac{zf''(z)}{f'(z)}-1. $$
(2.23)
Now, we show that \(\vert \omega(z)\vert <1\). If not, there exists a point \(z_{0}\in\mathbb{U}\) such that
$$\max_{\vert z\vert \leq \vert z_{0}\vert }\bigl\vert \omega(z)\bigr\vert =\bigl\vert \omega(z_{0}) \bigr\vert =1. $$
By Jack’s lemma, we know that
$$\begin{aligned} \Re \biggl(\frac {z_{0}f'''(z_{0})}{f''(z_{0})} \biggr)&= \Re \biggl(\frac{(k+1)\omega(z_{0})-3}{1-\omega(z_{0})} \biggr) \\ &=\Re \biggl(\frac{(k+1)(\cos\theta+i\sin\theta)-3}{1-\cos\theta -i\sin\theta} \biggr) \\ &=\frac{(k+4)\cos\theta-(k+4)}{(1-\cos\theta)^{2}+\sin^{2}\theta} \\ &=\frac{(k+4)(\cos\theta-1)}{(1-\cos\theta)^{2}+\sin^{2}\theta} \\ &=-\frac{k+4}{2} \\ &\leq-3, \end{aligned}$$
where \(k\geq2\), but this contradicts (2.20), which implies that \(\vert \omega(z)\vert <1\). Thus, f is concave in \(\mathbb{U}^{*}\). □

Declarations

Acknowledgements

The present investigation was supported by the National Natural Science Foundation under Grant Nos. 11301008 and 11226088, the Aid Program for Science and Technology Innovative Research Team in Higher Educational Institution of Hunan Province, the Foundation of Educational Committee of Henan Province under Grant No. 15A11006. The authors would like to thank the referees for their valuable comments and suggestions which essentially improved the quality of this paper.

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Authors’ Affiliations

(1)
School of Mathematics and Computing Science, Hunan First Normal University

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Copyright

© Wang and Li; licensee Springer. 2015