On classes of meromorphic locally univalent functions defined by differential inequalities

TL;DR

This paper investigates meromorphic functions in the unit disk, providing an elementary proof for a univalence condition and discussing related known results and open problems.

Contribution

It offers a new elementary proof for a univalence criterion for meromorphic functions in the unit disk.

Findings

Provided a sufficient univalence condition for meromorphic functions

Included known results related to univalence

Presented open problems for future research

Abstract

n this article we consider functions meromorphic in the unit disk. We give an elementary proof for a condition that is sufficient for the univalence of such functions which also contains some known results. We include few open problems for further research.

Equations

$${\mathcal{A}}=\{f:\,\mbox{ $f$ is analytic in ${\mathbb{D}}$, $f(0)=f^{\prime}(0)-1=0$}\}.$$

$$\left|\frac{z}{f(z)}-z{\left(\frac{z}{f(z)}\right)}^{\prime }-1\right|<\lambda ,z\in \mathbb{D},$$

$$|U_{f}(z)-\mu|<\lambda~{}\mbox{ in ${\mathbb{D}}$},$$

$$U_f(z):={\left(\frac{z}{f(z)}\right)}^2{f}^{\prime }(z)=\frac{z}{f(z)}-z{\left(\frac{z}{f(z)}\right)}^{\prime },z\in \mathbb{D}.$$

$$f(z)=c_m(z-z_0{)}^m+\cdots (c_m\ne 0,m\ge 1)$$

$$U_f(z)=\frac{m{z}_0}{c_m(z-z_0{)}^{m+1}}+\cdots$$

$$\lambda \Omega (z)=\frac{z}{f(z)}-z{\left(\frac{z}{f(z)}\right)}^{\prime }-\mu =1-\mu +(a_3-a_2^2)z^2+\cdots =1-\mu +O(z^2)$$

$$\varphi (z)=\frac{\Omega (z)-a}{1-\overline{a}\Omega (z)}$$

$$\Omega (z)=\frac{\varphi (z)+a}{1+\overline{a}\varphi (z)}.$$

$$\frac{z}{f(z)}-z{\left(\frac{z}{f(z)}\right)}^{\prime }-\mu =\lambda \left(\frac{a+z^2\omega (z)}{1+\overline{a}{z}^2\omega (z)}\right),z\in \mathbb{D},$$

$$\frac{z}{f(z)}$$

$$f(z)=\frac{z}{1+cz-\lambda z{\int }_0^z\frac{(1-|a{|}^2)\omega (t)}{1+\overline{a}{t}^2\omega (t)}dt},$$

$$f(z)=\frac{z}{1+cz-\lambda z{\int }_0^z\frac{(1-|a{|}^2)}{1+\overline{a}{t}^2}dt}.$$

$$\frac{z_1^2}{1+\overline{a}{z}_1^2}=\frac{-1}{\lambda (1-|a{|}^2)}.$$

$$|\lambda (1-|a{|}^2)+\overline{a}|>1,$$

$$g_{\omega }(z)=\frac{1}{z}\left[1+cz-\lambda z{\int }_0^z\frac{(1-|a{|}^2)\omega (t)}{1+\overline{a}{t}^2\omega (t)}dt\right].$$

$$0\ne \frac{g_{\omega }(z_1)-g_{\omega }(z_2)}{z_1-z_2}=\frac{-1}{z_1{z}_2}-\frac{\lambda }{z_1-z_2}{\int }_{z_2}^{z_1}\frac{(1-|a{|}^2)\omega (t)}{1+\overline{a}{t}^2\omega (t)}dt,$$

$$\left|\frac{(1-|a{|}^2)\omega (z)}{1+\overline{a}{z}^2\omega (z)}\right|\le \frac{1-|a{|}^2}{1-|a|}=1+|a|,z\in \mathbb{D},$$

$$\left|\lambda \frac{z_1{z}_2}{z_1-z_2}{\int }_{z_2}^{z_1}\frac{(1-|a{|}^2)\omega (t)}{1+\overline{a}{t}^2\omega (t)}dt\right|<\lambda (1+|a|)=\lambda +|1-\mu |$$

$$\frac{z}{f_0(z)}=1-z+\frac{\lambda (1-a^2)z}{2\sqrt{a}}\log \left(\frac{1-(\sqrt{a}/(1+\sqrt{a}))(1-z)}{1+(\sqrt{a}/(1-\sqrt{a}))(1-z)}\right),a=\frac{1-\mu }{\lambda }.$$

$$\frac{z}{f(z)}-z{\left(\frac{z}{f(z)}\right)}^{\prime }-\mu =\lambda \Omega (z),$$

$$\Omega (z)=\frac{1-\mu }{\lambda }+\sum _{k=2}^{\infty }{c}_k{z}^k,$$

$$\frac{z}{f(z)}=1+\sum _{k=1}^{\infty }{b}_k{z}^k=1-a_{2}z+(a_2^2-a_3)z^2+\cdots .$$

$$|b_k|\le \frac{\lambda }{k-1}\left(1-\frac{|1-\mu {|}^2}{{\lambda }^2}\right)$$

$$b_k(1-k)=\lambda c_k,k\ge 2.$$

$$\frac{z}{f_k(z)}=1+cz+\lambda z{\int }_0^z\frac{(1-|a{|}^2)t^{k-2}}{1-\overline{a}{t}^k}dt=1+cz+\lambda (1-|a{|}^2)\sum _{j=2}^{\infty }(\overline{a}{)}^j\frac{z^{(j+1)k}}{(j+1)k-1}.$$

$$b_k=\frac{\lambda }{k-1}\left(1-\frac{|1-\mu {|}^2}{{\lambda }^2}\right).$$

$$|b_2|=\left|a_2^2-a_3\right|\le \lambda \left(1-\frac{|1-\mu {|}^2}{{\lambda }^2}\right)$$

$$\sum _{k=2}^{\infty }|b_k{|}^2(k-1{)}^2\le {\lambda }^2\left(1-\frac{|1-\mu {|}^2}{{\lambda }^2}\right).$$

$$\left|1-\mu -\sum _{k=2}^{\infty }(k-1)b_k{z}^k\right|\le \lambda ,z\in \mathbb{D},$$

Full text

File: 1905.01690.tex, printed: 19-3-2024, 4.16

1. Preliminaries and Main Results

We denote the unit disk by ${\mathbb{D}}=\{z\in{\mathbb{C}}:\,|z|<1\}$ and let

[TABLE]

The family ${\mathcal{S}}$ of univalent functions in $\mathcal{A}$ together with its many subfamilies, for which the image domains have special geometric properties, have been investigated in details. See [3, 4]. Throughout, $\mathcal{B}$ denotes the class of functions $\omega$, analytic in ${\mathbb{D}}$ such that $|\omega(z)|\leq 1$ for $z\in{\mathbb{D}}$. The well known inequality $\sum_{k=0}^{\infty}|c_{k}|^{2}\leq 1$ for $\Omega(z)=\sum_{k=0}^{\infty}c_{k}z^{k}\in\mathcal{B}$ will be used to get the proof of Theorem 4. Recently, in [8, Theorem 2(b)], the second and the third authors proved among other things the following result which extends the earlier known result for analytic functions.

Theorem A. * Let $f$ be meromorphic in ${\mathbb{D}}$ such that $f(0)=f^{\prime}(0)-1=0$. If for all $z\in{\mathbb{D}}$ the inequality*

[TABLE]

*is valid for some $\lambda\in(0,1]$, then $f$ is univalent in ${\mathbb{D}}$. *

The proof of Theorem in [8] was elegant and was also different from the other known methods. In this article, we shall consider slightly more general situation. For $0<\lambda$ and $\mu\in{\mathbb{C}}$ such that $|1-\mu|<\lambda$, we consider the family ${\mathcal{U}}(\lambda,\mu)$ of meromorphic functions $f$ satisfying the inequality

[TABLE]

where

[TABLE]

Note that the center $1$ has been replaced by $\mu$. First we consider the problem of determining conditions on $\lambda$ and $\mu$ so that functions in ${\mathcal{U}}(\lambda,\mu)$ are univalent in ${\mathbb{D}}$.

As with the case of analytic functions, for notational simplicity, we let ${\mathcal{U}}(\lambda):={\mathcal{U}}(\lambda,1)$, ${\mathcal{U}}:={\mathcal{U}}(1)$. In the analytic case, it was well-known that ${\mathcal{U}}\subsetneq{\mathcal{S}}$ (see [1, 2]).

Theorem 1**.**

Let $\lambda\in(0,1]$ and $|1-\mu|<\lambda$. All members of the family $\mathcal{U}(\lambda,\mu)$ are functions meromorphic and locally univalent in ${\mathbb{D}}$ if and only if $|\mu|\geq\lambda.$

Proof. It is a simple exercise to see that if $f\in{\mathcal{U}}(\lambda,\mu)$, then $f(z)\neq 0$ in ${\mathbb{D}}\backslash\{0\}$, because otherwise $f(z_{0})=0$ for some $z_{0}\in{\mathbb{D}}\backslash\{0\}$ which would then imply that

[TABLE]

in a neighborhood of $z_{0}$ so that

[TABLE]

implying that $U_{f}(z)$ has a pole of order $m+1$ which would clearly be a contradiction to the fact that $U_{f}(z)$ is bounded. Moreover, for $|\mu|\geq\lambda$, the inequality $|U_{f}(z)-\mu|<\lambda$ for $z\in{\mathbb{D}}$ implies $|U_{f}(z)|>0$ for $z\in{\mathbb{D}}$ and hence, in either way $f^{\prime}(z)\neq 0$ for $z\in{\mathbb{D}}$ and $f(z)\neq 0$ in ${\mathbb{D}}\backslash\{0\}$.

To prove the other direction of our assertion, we let $f(z)=z+\sum_{n=2}^{\infty}a_{n}z^{n}$ in ${\mathcal{U}}(\lambda,\mu)$ and consider

[TABLE]

as $z\rightarrow 0$, where $\Omega$ is analytic in ${\mathbb{D}}$ such that $\Omega(0)=(1-\mu)/\lambda$, $\Omega^{\prime}(0)=0$ and $|\Omega(z)|<1$ in ${\mathbb{D}}$. For simplicity, we let $\Omega(0)=a$. Then

[TABLE]

is a Schwarz function, i. e., $\phi(0)=0$ and $|\phi(z)|<1$ for $z\in{\mathbb{D}}$. Hence,

[TABLE]

Since $\Omega^{\prime}(0)=(1-|a|^{2})\phi^{\prime}(0)=0$, we have $\phi^{\prime}(0)=0$. Therefore, we can write $\phi(z)=z^{2}\omega(z)$, where $\omega\in\mathcal{B}.$ Consequently,

[TABLE]

where $a=(1-\mu)/\lambda$ and $\omega\in\mathcal{B}$. Note that, in the neighborhood of $z=0$, we have the representation

[TABLE]

As with the standard procedure, the integration of the differential equation (3) delivers that each $f\in{\mathcal{U}}(\lambda,\mu)$ has the representation

[TABLE]

where $c=-a_{2}$, $a=\frac{1-\mu}{\lambda}$ and $\omega\in\mathcal{B}$. Now, we let $\omega(z)=1$, $z\in{\mathbb{D}}$, in this representation and obtain that

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It follows that there exists a $z_{1}\in{\mathbb{D}}$ such that $f^{\prime}(z_{1})=0$ if and only if

[TABLE]

This is equivalent to

[TABLE]

or equivalently $|\mu|<\lambda$, which is a contradiction to the local univalency of $f$. Hence, the rest of the assertion is proved. ∎

Theorem 2**.**

*Let $f$ be meromorphic in ${\mathbb{D}}$ such that $f(0)=f^{\prime}(0)-1=0$. Then $f\in U(\lambda,\mu)$ is univalent in ${\mathbb{D}}$ if either (a) $|1-\mu|<\lambda\leq 1/2$ or (b) $\lambda\in(1/2,1]$ and $|1-\mu|\leq 1-\lambda$. *

Proof. By using the representation (4), we can write $f=1/g_{\omega}$ for $f\in\mathcal{U}(\lambda,\mu)$, where

[TABLE]

We see that we have to prove

[TABLE]

where $z_{1},z_{2}\in{\mathbb{D}}\setminus\{0\}$, $z_{1}\neq z_{2}$, and $\omega\in\mathcal{B}$. Since $a=(1-\mu)/\lambda$ and

[TABLE]

we get

[TABLE]

and thus, (5) holds whenever $|1-\mu|\leq 1-\lambda$. Hence, $f$ is univalent if $|1-\mu|\leq 1-\lambda$. Therefore, every $f\in\mathcal{U}(\lambda,\mu)$ is univalent in ${\mathbb{D}}$ whenever $|1-\mu|<\lambda\leq\frac{1}{2}$ or $|1-\mu|\leq 1-\lambda$ with $\lambda\in(\frac{1}{2},1]$. This completes the proof. ∎

Remark 1**.**

By using Theorem 1, we see that at least for nonnegative real numbers $\mu$ the assertion of Theorem 2(b) is best possible. This follows from the fact that for $\mu\in(1-\lambda,\lambda)$, the family $\mathcal{U}(\lambda,\mu)$ contains a function that is not locally univalent in ${\mathbb{D}}$. In order to present a couple of precise functions, we consider the function $f_{0}$ defined by (see also Problem 2)

[TABLE]

Problem 1**.**

Do there exist families $\mathcal{U}(\lambda,\mu)$ consisting of univalent functions besides those mentioned in Theorem 2?

In the following we use the equation

[TABLE]

where

[TABLE]

with $\Omega\in\mathcal{B}$ to get sharp estimates for the coefficients of the representation

[TABLE]

Theorem 3**.**

For $f\in\mathcal{U}(\lambda,\mu)$ of the form (8) and $k\geq 2$, the inequalities

[TABLE]

are valid. These inequalities are best possible.

Proof. From (6) and (7) we derive the identities

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The well known inequalities $|c_{k}|\leq 1-|c_{0}|^{2}$ for $\Omega(z)=\sum_{k=0}^{\infty}c_{k}z^{k}$ with $\Omega\in\mathcal{B}$, $c_{0}=(1-\mu)/\lambda$ and $c_{1}=0$, imply the validity of our assertion.

For the proof of the sharpness, we set $\omega(z)=-z^{k-2}$ in (4) (with $c_{0}=a$) and consider the following functions $f_{k}$ for $k\geq 2$:

[TABLE]

Obviously, $f_{k}\in\mathcal{U}(\lambda,\mu)$, and further we get that the $k$-th coefficient $b_{k}$ of the function $z/f_{k}(z)$ satisfies

[TABLE]

The proof is complete. ∎

Remark 2**.**

In particular, in the case $k=2$ of Theorem 3, we get the sharp inequality

[TABLE]

for $f(z)=z+\sum_{n=2}^{\infty}a_{n}z^{n}\in\mathcal{U}(\lambda,\mu).$ **

The well known inequality $\sum_{k=0}^{\infty}|c_{k}|^{2}\leq 1$ for $\Omega(z)=\sum_{k=0}^{\infty}c_{k}z^{k}\in\mathcal{B}$ will be used to get the proof of Theorem 4.

Theorem 4**.**

If $f\in\mathcal{U}(\lambda,\mu)$ is of the form (8), then

[TABLE]

This inequality is best possible.

Proof. In view of the relations (6) and (7), the assumption gives

[TABLE]

from which the desired inequality follows because $\sum_{k=0}^{\infty}|c_{k}|^{2}\leq 1$ for $\Omega(z)=\sum_{k=0}^{\infty}c_{k}z^{k}\in\mathcal{B}$. Thus, it remains to prove the assertion of the sharpness. To that end, we consider the functions $f_{k},k\geq 2$, given by (9). Then their Taylor expansions are given by

[TABLE]

Hence, in these cases we get

[TABLE]

This completes the proof of the sharpness. ∎

Finding sharp estimates for the Taylor coefficients of the functions in $U(\lambda,\mu)$ turned out to be a challenge. As a first result in this direction we prove the next result which extends [8, Theorem 4].

Theorem 5**.**

If $f\in\mathcal{U}(\lambda,\mu)$ is analytic in the disk ${\mathbb{D}}_{p}=\{z:\,|z|<p\},\,p\in(0,1],$ and $a=(1-\mu)/\lambda$, then the inequality

[TABLE]

is valid. This estimate is best possible for $\mu\in(1-\lambda,1].$

Proof. We assume on the contrary that $|a_{2}|>A_{2}$. In other words, we can assume that there exists an $r\in(0,1)$ such that

[TABLE]

Using Brouwer’s fixed point theorem, we shall prove that then the function

[TABLE]

has a pole in the disk $\overline{{\mathbb{D}}}_{rp}:=\{z:\,|z|\,\leq\,pr\}$. To that end, we consider the function

[TABLE]

and we show that it has a fixed point in the disk ${\mathbb{D}}_{pr}$. For $|z|\leq pr$, we get

[TABLE]

Since $F$ is a continuous function that maps the convex compact set $\overline{{\mathbb{D}}_{pr}}$ into itself, Brouwer’s fixed point theorem implies that $F$ has a fixed point in $\overline{{\mathbb{D}}_{pr}}$ which is a contradiction to the initial assumptions of Theorem 5. Hence $|a_{2}|\leq A_{2}$ is valid.

Concerning the sharpness, we see that for the numbers $\mu$ in question, the quantity $(1-\mu)/\lambda$ is nonnegative. We choose $\omega(z)=-1$ in the representation formula for $f\in\mathcal{U}(\lambda,\mu)$, and we get that the function $f=f_{0}$, where

[TABLE]

which is analytic in ${\mathbb{D}}_{p}$ and achieves equality in the estimate of our theorem. ∎

Note that the function $f_{0}$ given by (10) takes the form

[TABLE]

which may be simplified as

[TABLE]

In the case of $p=1$, we then ask in particular the following.

Problem 2**.**

Suppose that $f\in\mathcal{U}(\lambda,\mu)$ is analytic in the unit disk ${\mathbb{D}}$, and $a=(1-\mu)/\lambda>0$. Is

[TABLE]

Note that this problem has been solved in [5] for the case of $a=0$, i.e. for $\mu=1$ .

Remark 3**.**

If $\mu\notin(1-\lambda,1]$, it is possible to get an implicit sharp upper estimate for $|a_{2}|$ of functions $f\in\mathcal{U}(\lambda,\mu)$ analytic in the disk ${\mathbb{D}}_{p}$ in the following way: We consider again

[TABLE]

and assume that there exists a number $r<1$ such that

[TABLE]

Then we use the continuity of the function $F$ on the disk $\overline{{\mathbb{D}}_{pr}}$, the inclusion $F(\overline{{\mathbb{D}}_{pr}})\subset\overline{{\mathbb{D}}_{pr}}$, and Brouwer’s fixed point theorem to see that $F$ has a fixed point in $\overline{{\mathbb{D}}_{pr}}$. This contradicts the assumption that $f$ is analytic in the disk ${\mathbb{D}}_{p}$. Hence,

[TABLE]

To prove the sharpness of this inequality, we choose $z_{0}$, $|z_{0}|=p$, and $\omega_{0}\in\mathcal{B}$ such that

$\displaystyle\left|1-\lambda z_{0}\int_{0}^{z_{0}}\frac{(1-|a|^{2})\omega_{0}(t)}{1+\overline{a}t^{2}\omega_{0}(t)}\,dt\right|=$

[TABLE]

Now, let

[TABLE]

and consider

[TABLE]

The function $f_{0}$ shows that the above estimate is sharp. **

Problem 3**.**

*Calculate the above maximum. *

Remark 4**.**

Let $f(z)=z/g_{\omega}(z)\in\mathcal{U}(\lambda,\mu)$. See (4). Since for any $\omega\in\mathcal{B}$ there exists a positive constant $C$ such that

[TABLE]

the function $g_{\omega}$ is uniformly continuous in ${\mathbb{D}}$. Therefore, it has a continuous extension to $\overline{{\mathbb{D}}}$. This fact implies that $f$ has a continuous extension to $\overline{{\mathbb{D}}}\setminus\{z:\,g_{\omega}(z)=0\}$. Hence, it makes sense to ask for the univalence of this continuous extension. From the proof of Theorem 2, it is obvious that this extension is univalent if $|1-\mu|<\lambda\,\leq 1/2$ or if for $\lambda\in(1/2,1]$ the strict inequality $|1-\mu|<1-\lambda$ is valid.

On the other hand, Theorems 1 and 2 have the consequence that for $\lambda\in(1/2,1]$ the classes $\mathcal{U}(\lambda,\lambda)$ contain the interesting univalent slit mappings

[TABLE]

These functions have a pole at $z=p$, and their derivatives vanish at $z=1$ and $z=-1$. We conjecture that possibly these classes and those functions deserve further research. Much of the investigations carried out in [5, 6, 7, 8, 9, 10] on ${\mathcal{U}}(\lambda)$ and some other related classes could be considered for further research with an aim to obtain meromorphic analogue of these classes. **

Acknowledgments

The authors thank the referee for useful comments. The first author acknowledged the support from a USM research university grant 1001.PMATHS.8011038. The work of the second author is supported by Mathematical Research Impact Centric Support of Department of Science and Technology (DST), India (MTR/2017/000367).