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Some criteria for concave conformal mappings

Abstract

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

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 \(\Re(p (z))>0\).

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 \(\mathbb{U}^{*}\). □

References

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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.

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Correspondence to Zhi-Gang Wang.

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Wang, Z., Li, M. Some criteria for concave conformal mappings. J Inequal Appl 2015, 119 (2015). https://doi.org/10.1186/s13660-015-0640-5

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MSC

  • 30C55

Keywords

  • concave mapping
  • meromorphic function
  • Jack’s lemma