Differential Inequalities and Univalent Functions

TL;DR

This paper studies a class of univalent functions defined by a differential inequality, proving that the harmonic mean of two such functions remains in the class, and explores their geometric properties like starlikeness.

Contribution

It introduces a new class of univalent functions defined via a differential inequality and proves the harmonic mean of two functions in this class also belongs to it.

Findings

Harmonic mean of two functions in the class remains in the class.

Several functions in the class are shown to be starlike.

Conjecture that not all functions in the class are starlike.

Abstract

Let ${\mathcal M}$ be the class of analytic functions in the unit disk $\ID$ with the normalization $f(0)=f'(0)-1=0$, and satisfying the condition $$\left |z^2\left (\frac{z}{f(z)}\right )''+ f'(z)\left(\frac{z}{f(z)} \right)^{2}-1\right |\leq 1, \quad z\in \ID. $$ Functions in $\mathcal{M}$ are known to be univalent in $\ID$. In this paper, it is shown that the harmonic mean of two functions in ${\mathcal M}$ are closed, that is, it belongs again to ${\mathcal M}$. This result also holds for other related classes of normalized univalent functions. A number of new examples of functions in $\mathcal{M}$ are shown to be starlike in $\ID$. However we conjecture that functions in $\mathcal{M}$ are not necessarily starlike, as apparently supported by other examples.

Tables

Table 1. Table 1 . Values of A ​ ( θ ) 𝐴 𝜃 A(\theta) for certain choices of θ 𝜃 \theta

$n$	value of $A\left(\theta \right)$	$n$	value of $A\left(\theta \right)$
1	$-0.0258011$	8	$-0.000243709$
2	$-0.0103986$	9	$-0.000154718$
3	$-0.00437311$	10	$-0.0000989276$
4	$-0.00211511$	11	$-0.0000628326$
5	$-0.00113174$	12	$-0.0000388937$
6	$-0.00064961$	13	$-0.000022708$
7	$-0.00039145$	14	$-0.0000116051$

Equations

$$\left|z^2{\left(\frac{z}{f(z)}\right)}^{\prime \prime }+f^{\prime }(z){\left(\frac{z}{f(z)}\right)}^2-1\right|\le 1,z\in \mathbb{D}.$$

$$\left|f^{\prime }(z){\left(\frac{z}{f(z)}\right)}^2-1\right|\le 1,z\in \mathbb{D},$$

$$\left|{\left(\frac{z}{f(z)}\right)}^{\prime \prime }\right|\le 2,z\in \mathbb{D}.$$

$$\left|{\mathcal{M}}_f(z)\right|\le 1,\text{mbox}and\left|{\mathcal{N}}_f(z)\right|\le 1,z\in \mathbb{D},$$

$${\mathcal{M}}_f(z)=z^2{\left(\frac{z}{f(z)}\right)}^{\prime \prime }+f^{\prime }(z){\left(\frac{z}{f(z)}\right)}^2-1$$

$${\mathcal{N}}_f(z)=-z^3{\left(\frac{z}{f(z)}\right)}^{\prime \prime \prime }+f^{\prime }(z){\left(\frac{z}{f(z)}\right)}^2-1.$$

$$\mathcal{N}⊊\mathcal{M}⊊\mathcal{P}⊊\mathcal{U}.$$

$$F(z)=\frac{2f(z)g(z)}{f(z)+g(z)},$$

$$\frac{1}{F(z)}-\frac{1}{z}=\frac{1}{2}\left[\left(\frac{1}{f(z)}-\frac{1}{z}\right)+\left(\frac{1}{g(z)}-\frac{1}{z}\right)\right],$$

$$\left|F^{\prime }(z){\left(\frac{z}{F(z)}\right)}^2-1\right|$$

$$f^{\prime }(z){\left(\frac{z}{f(z)}\right)}^2-1=-z^2{\left(\frac{1}{f(z)}-\frac{1}{z}\right)}^{\prime }.$$

$${\mathcal{M}}_f(z)$$

$${\mathcal{M}}_F(z)=\frac{1}{2}\left({\mathcal{M}}_f(z)+{\mathcal{M}}_g(z)\right),$$

$${\mathcal{N}}_f(z)$$

$${\mathcal{N}}_F(z)=\frac{1}{2}\left({\mathcal{N}}_f(z)+{\mathcal{N}}_g(z)\right),$$

$$\frac{z}{f(z)}=1+(1-\alpha )z+\alpha z^m,$$

$$\sum _{k=2}^{\infty }(k-1{)}^2|b_k|=(m-1{)}^2\alpha =1,$$

$$\frac{z{f}^{\prime }(z)}{f(z)}=\frac{1-\alpha (m-1)z^m}{1+(1-\alpha )z+\alpha z^m}.$$

$$\frac{e^{i\theta }{f}^{\prime }(e^{i\theta })}{f(e^{i\theta })}=\frac{A(\theta )+iB(\theta )}{|1+(1-\alpha )e^{i\theta }+\alpha e^{im\theta }{|}^2},$$

$$A(\theta )$$

$$A(\theta )=1-\frac{1}{(m-1{)}^3}-\frac{m(m-2)}{(m-1{)}^2}D(\theta ),$$

$$D(\theta )=-\cos \theta +\frac{1}{m}\cos (m\theta )+\frac{\cos (m-1)\theta }{m-1}.$$

$$A(\theta )=\frac{7}{8}-\frac{3}{4}\left[-\cos \theta +\frac{1}{3}\cos 3\theta +\frac{1}{2}\cos 2\theta \right],$$

$$A(\theta )$$

$$f_3(z)=\frac{z}{1+\frac{3}{4}z+\frac{1}{4}{z}^3}=\frac{4z}{(1+z)(4-z+z^2)},$$

$$D^{\prime }(\theta )$$

$$=\sin \theta -2\sin \frac{(2m-1)\theta }{2}\cos \frac{\theta }{2}$$

$$\displaystyle=2\cos\frac{\theta}{2}\left[\sin\frac{\theta}{2}{\color[rgb]{1,0,0}{-}}\sin\frac{(2m-1)\theta}{2}\right]$$

$$=4\cos \frac{\theta }{2}\cos \frac{m\theta }{2}\sin \frac{(m-1)\theta }{2}.$$

$$D^{\prime }(\theta )=4\cos \frac{\theta }{2}\cos \frac{(2n+1)\theta }{2}\sin n\theta ,n\ge 2,$$

$$D(\theta )=-\cos \theta +\frac{1}{2n+1}\cos (2n+1)\theta +\frac{1}{2n}\cos (2n\theta ).$$

$$\left\{\begin{array}[]{lll}\theta_{j}=&\displaystyle\frac{(2j-1)\pi}{2n+1}&\mbox{for $j=1,2,\ldots,n$},\\[8.53581pt] \theta_{j}^{\prime}=&\displaystyle\frac{j\pi}{n}&\mbox{for $j=1,2,\ldots,n-1$,}\end{array}\right.$$

Full text

[TABLE]

Functions in $\mathcal{M}$ are known to be univalent in ${\mathbb{D}}$. In this paper, it is shown that the harmonic mean of two functions in ${\mathcal{M}}$ are closed, that is, it belongs again to ${\mathcal{M}}$. This result also holds for other related classes of normalized univalent functions. A number of new examples of functions in $\mathcal{M}$ are shown to be starlike in ${\mathbb{D}}$. However we conjecture that functions in $\mathcal{M}$ are not necessarily starlike, as apparently supported by other examples.

1. Introduction

Let ${\mathcal{H}}$ denote the family of analytic functions in the open unit disk ${\mathbb{D}}:=\{z\in{\mathbb{C}}:\,|z|<1\},$ and ${\mathcal{A}}$ its subclass of normalized functions $f(z)=z+a_{2}z^{2}+a_{3}z^{3}+\cdots$. Further, let ${\mathcal{S}}$ denote the subclass of ${\mathcal{A}}$ consisting of functions $f$ univalent in ${\mathbb{D}}$. Denote by $\mathcal{S}^{\ast}$ and $\mathcal{C}$ respectively the subclasses of $\mathcal{S}$ consisting of starlike and convex functions. Functions $f\in\mathcal{S}^{*}$ map ${\mathbb{D}}$ onto starlike domains with respect to the origin, while $f\in\mathcal{C}$ whenever $f({\mathbb{D}})$ is a convex domain. Analytically, $f\in\mathcal{S}^{\ast}$ if ${\rm Re}\left(zf^{\prime}(z)/f(z)\right)>0$, while $f\in\mathcal{C}$ if ${\rm Re}\left(1+zf^{\prime\prime}(z)/f^{\prime}(z)\right)>0$.

Investigations into particular subclasses of ${\mathcal{A}}$ continued to be of recent interest. These include the class $\mathcal{U}$ consisting of functions $f\in\mathcal{A}$ satisfying

[TABLE]

as well as the class $\mathcal{P}$ of functions $f\in\mathcal{A}$ with

[TABLE]

The strict inclusion $\mathcal{P}\subsetneq\mathcal{U}\subsetneq\mathcal{S}$ holds within these classes (see [2, 5, 14] for a proof). There are several generalizations [7] of this result. For recent investigations on $\mathcal{U}$ and its generalization, we refer to [11, 12, 13] and the references therein.

In this paper, the phrase $f\in{\mathcal{U}}$ (respectively, $f\in{\mathcal{P}}$) in $|z|<r$ means that the defining inequality holds in $|z|<r$ instead of the full disk $|z|<1$. We also follow this standard convention for other classes. In [8] and [9]), the authors discussed the classes ${\mathcal{M}}$ and ${\mathcal{N}}$ of functions from ${\mathcal{A}}$ satisfying respectively the differential inequality

[TABLE]

where

[TABLE]

and

[TABLE]

These classes are also closely related to the class ${\mathcal{U}}$ in the sense of the strict inclusions

[TABLE]

A slightly general version of this result is given in [1].

In [10], Obradović, and Ponnusamy discussed “harmonic mean” of two univalent analytic functions. These are functions $F$ of the form

[TABLE]

or equivalently,

[TABLE]

where $f,g\in\mathcal{S}$. In particular, the authors in [10] determined the radius of univalency of $F$, and proposed the following two conjectures.

Conjecture 1**.**

1. (a)

The function $F$ defined by (1) is not necessarily univalent in ${\mathbb{D}}$ whenever $f,g\in\mathcal{S}$ such that $((f(z)+g(z))/z)\neq 0$ in ${\mathbb{D}}$. 2. (b)

The function $F$ defined by (1) is univalent in ${\mathbb{D}}$ whenever $f,g\in\mathcal{C}$ such that $((f(z)+g(z))/z)\neq 0$ in ${\mathbb{D}}$.

The authors in [10] showed that whenever $f,g\in\mathcal{U}$, then the function $F$ defined by (1) belongs to $\mathcal{U}$ in the disk $|z|<\sqrt{\frac{\sqrt{5}-1}{2}}\approx 0.78615$.

While Conjecture 1 remains open, the aim of this paper is to show that Conjecture 1  (a) does not hold when the class $\mathcal{S}$ is replaced by $\mathcal{U}$. Indeed, it does not hold true even for the classes ${\mathcal{M}},\,{\mathcal{N}},\,\mbox{and}~{}{\mathcal{P}}$. The second objective of the paper is to consider several examples in examining starlikeness of functions in the classes ${\mathcal{M}},\,{\mathcal{N}},\,\mbox{and}~{}{\mathcal{P}}$. We conclude with a conjecture that functions in the class ${\mathcal{M}}$ are not necessarily starlike in ${\mathbb{D}}$.

2. On the harmonic mean of univalent functions

Theorem 1**.**

Let $f,g\in\mathcal{U}$ satisfy $\frac{f(z)+g(z)}{z}\neq 0$ for $z\in{\mathbb{D}}$. Then the function $F$ given by (1) also belongs to the class $\mathcal{U}$.

Proof. From (2), it readily follows from the triangle inequality that the function $F$ satisfies

[TABLE]

Thus $F\in\mathcal{U}$. ∎

Moreover, we see that Theorem 1 holds true if the class $\mathcal{U}$ is replaced by the class ${\mathcal{M}}$.

Theorem 2**.**

Suppose $f,g\in\mathcal{M}$ satisfy $\frac{f(z)+g(z)}{z}\neq 0$ for $z\in{\mathbb{D}}$. Then the function $F$ given by (1) also belongs to the class $\mathcal{M}$.

Proof. Now

[TABLE]

Using this equality, it follows that

[TABLE]

In view of (2), this means that

[TABLE]

and use of the triangle inequality yields the desired result. ∎

Theorem 3**.**

Let $f,g\in\mathcal{N}$ satisfy $\frac{f(z)+g(z)}{z}\neq 0$ for $z\in{\mathbb{D}}$. Then the function $F$ given by (1) also belongs to the class $\mathcal{N}$.

Proof. As in the proof of Theorem 2, we see that

[TABLE]

Thus relation (2) gives

[TABLE]

and the proof of theorem readily follows. ∎

Finally, it is also readily shown that the above theorem holds true for the class $\mathcal{P}$.

3. Examples and a Conjecture

It is known that functions in the class ${\mathcal{U}}$ are not necessarily starlike. There are a number of examples displaying functions in ${\mathcal{U}}$ that are not starlike in ${\mathbb{D}}$, see for instance [6]. However, is ${\mathcal{M}}\subset{\mathcal{S}}^{*}$? This section discusses the latter problem.

Example 1**.**

To present a one-parameter family of functions in ${\mathcal{M}}$ that are also starlike, consider the function $f$ given by

[TABLE]

where $\alpha\in(0,1)$ and $m\in{\mathbb{N}}\backslash\{1\}=\{2,3,\ldots\}$ are such that $\alpha(m-1)^{2}=1.$ Then $z/f(z)\neq 0$ in $\mathbb{D}$ and

[TABLE]

and therefore, $f\in{\mathcal{M}}.$

Next, we show that $f$ is starlike whenever $m>1$ is an odd integer. Now, a simple calculation shows

[TABLE]

With $z=e^{i\theta},$ then

[TABLE]

where

[TABLE]

Note that $A(\theta)=A(-\theta)$. As $\alpha=1/(m-1)^{2},$ the expression for $A(\theta)$ reduces to

[TABLE]

where

[TABLE]

To show starlikeness, that is, $f\in\mathcal{S}^{*}$, it suffices to show that $A(\theta)\geq 0$ for $0\leq\theta\leq\pi.$ First we prove the assertion for the case $m=3$, while the general case is obtained separately. Setting $m=3$, $A(\theta)$ reduces to

[TABLE]

and from the identities $\cos 2\theta=2\cos^{2}\theta-1$ and $\cos 3\theta=4\cos^{3}\theta-3\cos\theta$,

[TABLE]

which shows that $A(\theta)\geq 0$. Thus, the function $f_{3}(z)$ given by

[TABLE]

is starlike in $\mathbb{D}$.

Next, we proceed to prove starlikeness for the general case. This requires more computations. First,

[TABLE]

We need to show that $A(\theta)\geq 0$ for $0\leq\theta\leq\pi$. It is convenient to set $m=2n+1$, $n\geq 2$ so that

[TABLE]

where $D(\theta)$ takes the form

[TABLE]

Clearly, $D^{\prime}(\theta)=0$ for $\theta=0,\pi$, and the critical points of $D(\theta)$ in the open interval $(0,\pi)$ are given by

[TABLE]

$n\geq 2$. Moreover, for each $n\geq 2$,

[TABLE]

In view of the above inequalities and after a careful scrutiny, it follows that

[TABLE]

where $0<\theta_{1}<\theta_{1}^{\prime}<\theta_{2}<\cdots<\theta_{j}<\theta_{j}^{\prime}<\theta_{j+1}<\cdots<\theta_{n}<\theta_{n}^{\prime}=\pi$. Therefore,

[TABLE]

Since

[TABLE]

then $D(0)\leq D(\pi)$. Moreover,

[TABLE]

and

[TABLE]

We deduce that $D\left(\theta_{j}\right)\leq D(\pi)$ and $D\left(\theta_{j}^{\prime}\right)\leq D(\pi)$ holds for each $j=1,2,\ldots,n$. Thus, $D(\theta)\leq D(\pi)$ for $\theta\in[0,\pi]$. This observation shows that

[TABLE]

Hence ${\rm Re}\,(e^{i\theta}f^{\prime}(e^{i\theta})/f(e^{i\theta}))\geq 0$, which implies that $f$ is starlike in ${\mathbb{D}}.$ Summarizing, for each $n\geq 1$, the function $f_{n}$ given by

[TABLE]

belongs ${\mathcal{M}}$, and $f_{n}$ is starlike in ${\mathbb{D}}$.

Example 2**.**

Consider

[TABLE]

We may rewrite $\phi$ as

[TABLE]

It is a simple exercise to see that $\phi(z)\neq 0$ in ${\mathbb{D}}$ and $f\in{\mathcal{M}}$. The Mathematica software is used to display the image of the unit disk under $f$ as shown in Figure 1. It apparently displays that $f({\mathbb{D}})$ is a starlike domain.

Example 3**.**

It is illustrative to present a general example showing that functions in ${\mathcal{U}}$ do not necessarily belong to ${\mathcal{S}}^{*}$. For $n\geq 3$, consider the function

[TABLE]

For $|b|\leq(n-2)/(n-1)$ and $\beta$ a real number, then

[TABLE]

and

[TABLE]

so that $f_{n}\in{\mathcal{U}}$ for each $n\geq 3$. On the other hand, $f_{n}$ is not in ${\mathcal{S}}^{*}$ when $0<b\leq(n-2)/(n-1)$ and $0<\beta<\arctan(b(n-1)/(n-2))$. This follows on account that

[TABLE]

Example 4**.**

Consider the function $f$ defined by

[TABLE]

where $\alpha\in(0,1)$ and $m\geq 3$ is an odd integer such that $\alpha m(m-1)=2.$ Then $z/f(z)\neq 0$ in $\mathbb{D}$ and

[TABLE]

and therefore, $f\in{\mathcal{P}}.$

As in Example 1,

[TABLE]

where

[TABLE]

Substituting $\alpha=2/(m(m-1))$ and $m=2n+1$ ($n\geq 1$), the last expression for $A(\theta)$ reduces to

[TABLE]

where

[TABLE]

To prove that $f$ is not starlike in ${\mathbb{D}}$, it suffices to show that $A(\theta)<0$ for some $\theta\in(-\pi,\pi)$. In the case of $m=3$ (i.e. $n=1$), it is a simple exercise to see that

[TABLE]

which is clearly negative for $\theta$ near $\pi$. Indeed, substituting $\cos\theta=-8/9$ or $\theta_{0}=6\pi/7$, it can be verified that $A(\theta)\approx-55/2187<0$, and $A(\theta_{0})\approx-0.25811<0$. Thus, the function

[TABLE]

belongs to ${\mathcal{P}}\backslash{\mathcal{S}}^{*}$.

To do away the problem for some other values of $n$, we proceed as follows. Set

[TABLE]

so that $\phi=(\pi-\theta)/2$. Then

[TABLE]

[TABLE]

and

[TABLE]

Thus, $A(\theta)$ given by (3) can be simplified leading to

[TABLE]

It is seen from the computer algebra system Mathematica that $A(\theta)<0$ for $n=1,2,\ldots,15$. For easy reference, Table 1 lists the values of $A(\theta)$ for $n=1,2,\ldots,14$.

Thus, we conclude that the above procedure helps us to show that for each $n\in\{1,2,\ldots,14\}$, the function $f_{n}$ given by

[TABLE]

is not starlike in ${\mathbb{D}}$. By a minor modification in the choice of $\theta$, one can show that $f_{n}$ is not starlike for some $n\geq 15$ although it is not clear whether $f_{n}$ is starlike for larger values of $n$.

The ideas and the motivations behind the above examples lead to the following conjecture:

Conjecture 2**.**

The class $\mathcal{M}$ is not contained in ${\mathcal{S}}^{*}$.

Acknowledgments

The first author gratefully acknowledged support from a Universiti Sains Malaysia research university grant 1001/PMATHS/8011101. The work of the third author is supported by Mathematical Research Impact Centric Support of DST, India (MTR/2017/000367).