Starlikeness Associated With The Exponential Function

TL;DR

This paper explores the class of functions related to the exponential function within complex analysis, providing new conditions for functions to be considered starlike associated with the exponential, expanding understanding of differential subordinations.

Contribution

It investigates properties of admissible functions linked to the exponential function and derives new sufficient conditions for functions to be starlike related to the exponential.

Findings

Derived new sufficient conditions for exponential starlikeness

Extended differential subordination theory to exponential functions

Provided applications to normalized analytic functions

Abstract

Given a domain $\Omega$ in the complex plane $\mathbb{C}$ and a univalent function $q$ defined in an open unit disk $\mathbb{D}$ with nice boundary behaviour, Miller and Mocanu studied the class of admissible functions $\Psi(\Omega,q)$ so that the differential subordination $\psi(p(z),zp(z),z^2p''(z);z)\prec h(z)$ implies $p(z)\prec q(z)$ where $p$ is an analytic function in $\mathbb{D}$ with $p(0)=1$, $\psi:\mathbb{C}^3\times \mathbb{D}\to\mathbb{C}$ and $\Omega=h(\mathbb{D})$. This paper investigates the properties of this class for $q(z)=e^z$. As application, several sufficient conditions for normalized analytic functions $f$ to be in the subclass of starlike functions associated with the exponential function are obtained.

Equations

$${\mathcal{S}}^{\ast }(\phi )=\left\{f\in \mathcal{H}:\frac{z{f}^{\prime }(z)}{f(z)}\prec \phi (z),z\in \mathbb{D}\right\}.$$

$$\psi (p(z),z{p}^{\prime }(z),z^2{p}^{\prime \prime }(z);z)\prec h(z)$$

$$E(q)=\{\zeta \in \partial \mathbb{D}:\underset{z\to \zeta }{\lim }q(z)=\infty \}$$

$$\psi (r,s,t;z)\notin \Omega$$

$$r=q(\zeta )\text{ is finite, }s=m\zeta q^{\prime }(\zeta )\text{ and }\mathrm{Re}\left(1+\frac{t}{s}\right)\ge m\mathrm{Re}\left(1+\frac{\zeta q^{\prime \prime }(\zeta )}{q^{\prime }(\zeta )}\right)$$

$$\psi (p(z),z{p}^{\prime }(z),z^2{p}^{\prime \prime }(z);z)\in \Omega$$

$$\psi (p(z),zp(z),z^2{p}^{\prime \prime }(z);z)\prec h(z)\Rightarrow p(z)\prec e^z$$

$$\mathrm{Re}\left(1+\frac{\zeta q^{\prime \prime }(\zeta )}{q^{\prime }(\zeta )}\right)=\mathrm{Re}(1+e^{i\theta })=1+\cos \theta .$$

$$\{\begin{array}{l}\begin{array}{r}\psi (r,s,t;z)\notin \Omega \text{whenever}r=q(\zeta )=e^{e^{i\theta }}\\ s=m\zeta q^{\prime }(\zeta )=m{e}^{i\theta }r\\ \text{and}\mathrm{Re}(1+t/s)\ge m(1+\cos \theta )\end{array}\end{array}$$

$$\psi (e^{e^{i\theta }},m{e}^{i\theta }{e}^{e^{i\theta }};z)\notin \Omega$$

$$\psi (p(z),z{p}^{\prime }(z),z^2{p}^{\prime \prime }(z);z)\in \Omega \Rightarrow p(z)\prec e^z.$$

$$\psi (p(z),z{p}^{\prime }(z),z^2{p}^{\prime \prime }(z);z)\prec e^z\Rightarrow p(z)\prec e^z.$$

$$h(z)=2\left(\frac{2z+1}{2+z}\right).$$

$$|\psi (r,s,t;z)|$$

$$\ge e^{-1}((1+2e)m-1)\ge 2$$

$$|p(z)+(1+2e)z{p}^{\prime }(z)|<2\Rightarrow p(z)\prec e^z.$$

$$|(\psi (r,s,t;z){)}^2-1|$$

$$\ge (1+\sqrt{2})e|s|((1+\sqrt{2})e|s|-2)$$

$$=(1+\sqrt{2})m{e}^{1+\cos \theta }((1+\sqrt{2})m{e}^{1+\cos \theta }-2)$$

$$\ge (1+\sqrt{2})(\sqrt{2}-1)=1$$

$$|(1+(1+\sqrt{2})ez{p}^{\prime }(z){)}^2-1|<1\Rightarrow p(z)\prec e^z.$$

$$|\psi (r,s,t;z)-1|=|s|=|m{e}^{i\theta }{e}^{e^{i\theta }}|=m{e}^{\cos \theta }\ge m{e}^{-1}\ge e^{-1}$$

$$\left|z{p}^{\prime }(z)\right|<e^{-1}\Rightarrow p(z)\prec e^z.$$

$$|\psi (r,s,t;z)-1|$$

$$=e^{\cos \theta }|e^{e^{i\theta }}-1+(2+e)m{e}^{i\theta }|$$

$$\ge e^{-1}((2+e)m-|e^{e^{i\theta }}-1|)$$

$$\ge e^{-1}((2+e)m-1-e^{\cos \theta })$$

$$\ge e^{-1}((2+e)m-1-e)\ge e^{-1}$$

$$|p^2(z)-p(z)+(1+e)z{p}^{\prime }(z)|<e^{-1}\Rightarrow p(z)\prec e^z.$$

$$|\psi (r,s,t;z)-1|=|s/r^2|=|m{e}^{i\theta }{e}^{-e^{i\theta }}|=m{e}^{-\cos \theta }\ge m{e}^{-1}\ge e^{-1}$$

$$\left|\frac{z{p}^{\prime }(z)}{p^2(z)}\right|<\frac{1}{e}\Rightarrow |\log p(z)|<1.$$

$$|\psi (r,s,t;z)-1|=|s/r|=|m{e}^{i\theta }|=m\ge 1$$

$$1+\frac{z{p}^{\prime }(z)}{p(z)}\prec 1+z\Rightarrow p(z)\prec e^z$$

$$|\psi (r,s,t;z)|$$

Full text

1. Introduction and Preliminaries

Let $\mathcal{H}[a,n]$ denote the class of analytic functions defined in the open unit disk $\mathbb{D}:=\{z\in\mathbb{C}\colon|z|<1\}$ of the form $f(z)=a+a_{n}z^{n}+a_{n+1}z^{n+1}+\cdots$, where $n$ is a positive integer and $a\in\mathbb{C}$. Set $\mathcal{H}_{1}:=\mathcal{H}[1,1]$. Let $\mathcal{H}$ be the subclass of $\mathcal{H}[0,1]$ consisting of functions $f$ normalized by the condition $f(0)=f^{\prime}(0)-1=0$. Let $\mathcal{S}$ be a subclass of $\mathcal{H}$ containing univalent functions. Given any two analytic functions in $\mathbb{D}$, we say that $f$ is subordinate to $g$, written as $f\prec g$, if there exists a Schwarz function $w$ that is analytic in $\mathbb{D}$ with $w(0)=0$ and $|w(z)|<1$ satisfying $f(z)=g(w(z))$ for all $z\in\mathbb{D}$. In particular, if $g$ is univalent, then $f\prec g$ if and only if $f(0)=g(0)$ and $f(\mathbb{D})\subset g(\mathbb{D})$. Some special classes of univalent functions are of great significance in geometric function theory due to their geometric properties. By considering the analytic function $\varphi\in\mathcal{H}_{1}$ with positive real part in $\mathbb{D}$ that maps $\mathbb{D}$ onto regions which are starlike with respect to a point $\varphi(0)=1$ and symmetric with respect to the real axis, in 1994, Ma and Minda [8] gave a unified treatment of various subclasses of starlike functions in terms of subordination by studying the class

[TABLE]

For special choices of $\varphi$, the class $\mathcal{S}^{*}(\varphi)$ reduces to widely-known subclasses of starlike functions. For example, when $-1\leq B<A\leq 1$, $\mathcal{S}^{*}[A,B]:=\mathcal{S}^{*}((1+Az)/(1+Bz))$ is the class of Janowski [6] starlike functions, $\mathcal{S}^{*}_{P}:=\mathcal{S}^{*}(1+2(\log((1+\sqrt{z})/(1-\sqrt{z})))^{2}/\pi^{2})$ is the class consisting of parabolic starlike functions [15], $\mathcal{S}^{*}_{L}:=\mathcal{S}^{*}(\sqrt{1+z})$ is the class of lemniscate starlike functions [18] and $\mathcal{S}^{*}_{q}:=\mathcal{S}^{*}(z+\sqrt{1+z^{2}})$ is the class of starlike functions associated with lune [14]. In 2015, Mendiratta et al. [9] also introduced the class $\mathcal{S}^{*}_{e}=\mathcal{S}^{*}(e^{z})$ of starlike functions associated with the exponential function satisfying the condition $|\log(zf^{\prime}(z)/f(z))|<1$ for $z\in\mathbb{D}$.

The study of differential subordination which is a generalized form of differential inequalites began with a prodigious article “Differential subordination and univalent functions” by S. Miller and P. Mocanu [10] in 1981. After that the theory of differential subordination brought a revolutionary change and attracted many researchers to use this technique for the study of univalent functions. Given a complex function $\psi(r,s,t;z)\colon\mathbb{C}^{3}\times\mathbb{D}\to\mathbb{C}$ and a univalent function $h$ in $\mathbb{D}$, if $p$ is an analytic function in $\mathbb{D}$ that satisfies the second-order differential subordination

[TABLE]

then $p$ is called a solution of the differential subordination. The univalent function $q$ is said to be a dominant of the solutions of the differential subordination if $p\prec q$ for all $p$ satisfying (1.1). A dominant $\tilde{q}$ that satisfies $\tilde{q}\prec q$ for all dominants $q$ of (1.1) is said to be the best dominant of (1.1). The best dominant is unique upto a rotation of $\mathbb{D}$. Moreover let $\mathcal{Q}$ denote the set of analytic and univalent functions $q$ in $\overline{\mathbb{D}}\setminus E(q)$, where

[TABLE]

and are such that $q^{\prime}(\zeta)\not=0$ for $\zeta\in\partial\mathbb{D}\setminus E(q)$. The following definition of admissible functions and the fundamental theorem laid the foundation stone in the theory of differential subordination.

Definition 1.1*.*

[11, p. 27] Let $\Omega$ be a domain in $\mathbb{C}$, $q\in\mathcal{Q}$ and $n$ be a positive integer. Define $\Psi_{n}(\Omega,q)$ to be the class of admissible functions $\psi\colon\mathbb{C}^{3}\times\mathbb{D}\to\mathbb{C}$ that satisfies the admissibility condition:

[TABLE]

whenever

[TABLE]

where $z\in\mathbb{D}$, $\zeta\in\partial\mathbb{D}\setminus E(q)$ and $m\geq n$ is a positive integer. We write $\Psi_{1}(\Omega,q)$ as $\Psi(\Omega,q)$.

Theorem 1.2**.**

[11, p. 28]** Let $\psi\in\Psi_{n}(\Omega,q)$ with $q(0)=a$. If $p\in\mathcal{H}[a,n]$ satisfies

[TABLE]

then $p(z)\prec q(z)$.

Miller and Mocanu [11] in their monograph discussed the class of admissible functions $\Psi(\Omega,q)$ when the function $q$ maps $\mathbb{D}$ onto a disk or a half-plane. These two special classes together with Theorem 1.2 lead to several important and interesting results in the theory of differential subordinations. However the aim of this paper is to consider differential implications with the superordinate function $q(z)=e^{z}$. In Section 2, the admissibility class $\Psi(\Omega,e^{z})$ is obtained, by deriving its admissibility condition. Examples are provided to illustrate the obtained results.

In 2015, Mendiratta et al. [9] estimated bounds on $\beta$ for which $p(z)\prec e^{z}$ whenever $1+\beta zp^{\prime}(z)/p(z)$ is subordinate to $e^{z}$, $(1+Az)/(1+Bz)$ and $\sqrt{1+z}$. In 2018, Kumar and Ravichandran [7] extended the result of Mendiratta et al. and obtained bounds on $\beta$ for $1+\beta zp^{\prime}(z)/p^{j}(z)$ $(j=0,2)$. They also estimated the bounds on $\beta$ such that $p(z)\prec e^{z}$ whenever $1+\beta zp^{\prime}(z)/p^{j}(z)$ $(j=0,1,2)$ is subordinate to $1+\sin z$ and $1+(z(k+z))/(k(k-z))$, where $k=\sqrt{2}+1$. Also, Gandhi et al.[4] obtained bounds on $\beta$ for which $p(z)\prec e^{z}$ whenever $1+\beta zp^{\prime}(z)/p^{j}(z)$ $(j=0,1,2)$ is subordinate to $z+\sqrt{1+z^{2}}$. Motivated by their works and that of [12, 3, 13, 16, 2, 19, 17, 1, 5], in Section 3, the problem

[TABLE]

is established for special cases of Janowski starlike functions $h$. In Section 4, the above said problem is solved for various expressions when $h$ in particular, is also an exponential function. The results of [9] are not only generalized but new differential implications are also obtained in last two sections. Additionally, the applications of the results obtained yield sufficient conditions for functions $f\in\mathcal{H}$ to belong to the class $\mathcal{S}^{*}_{e}$.

2. The Admissibility Condition

In this section, we describe the admissible class $\Psi(\Omega,q)$ with examples, where $\Omega$ is a domain in $\mathbb{C}$ and $q(z)=e^{z}$. Note that $q$ is a univalent function in $\overline{\mathbb{D}}$ with $q(\mathbb{D})=\Delta$ and $q(0)=1$, where $\Delta:=\{w\in\mathbb{C}\colon|\log w|<1\}$. Thus $q\in\mathcal{Q}$ with $E(q)=\emptyset$ and hence the class $\Psi(\Omega,q)$ is well-defined.

For $|\zeta|=1$, $q(\zeta)\in q(\partial\mathbb{D})=\partial q(\mathbb{D})=\{w\in\mathbb{C}:|\log w|=1\}$. This gives $|\log q(\zeta)|=1$ so that $\log q(\zeta)=e^{i\theta}$, where $\theta\in[0,2\pi)$ and hence $q(\zeta)=e^{e^{i\theta}}$. But $q(\zeta)=e^{\zeta}$ which implies that $\zeta=e^{i\theta}$. Also $\zeta q^{\prime}(\zeta)=e^{i\theta}e^{e^{i\theta}}$ and

[TABLE]

Thus the admissibility condition reduces to

[TABLE]

where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$. Therefore the class $\Psi(\Omega,e^{z})$ consists of those functions $\psi\colon\mathbb{C}^{3}\times\mathbb{D}\to\mathbb{C}$ that satisfy the admissibility condition given by (2.1). If $\psi\colon\mathbb{C}^{2}\times\mathbb{D}\to\mathbb{C}$, then the admissibility condition (2.1) reduces to

[TABLE]

where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$. As a particular case of Theorem 1.2, we have the following

Theorem 2.1**.**

Let $p\in\mathcal{H}_{1}$.

• (i)

If $\psi\in\Psi(\Omega,e^{z})$, then

[TABLE]

• (ii)

If $\psi\in\Psi(\Delta,e^{z})$, then

[TABLE]

We close this section with some examples illustrating Theorem 2.1.

Example 2.2*.*

Let $\psi(r,s,t;z)=r+(1+2e)s$ and $h:\mathbb{D}\to\mathbb{C}$ be defined by

[TABLE]

Then $\Omega=h(\mathbb{D})=\{w\in\mathbb{C}:|w|<2\}$. To prove $\psi\in\Psi(\Omega,e^{z})$, we need to show that the admissibility condition (2.1) is satisfied. Consider

[TABLE]

whenever $r=e^{e^{i\theta}}$, $s=me^{i\theta}r$ and $\operatorname{\mathrm{Re}}\left(1+t/s\right)$ $\geq$ $m(1+\cos\theta)$, where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$. Therefore $\psi(r,s,t;z)$ $\notin$ $\Omega$ and hence $\psi\in\Psi(\Omega,e^{z})$. By Theorem 2.1, it follows that if $p\in\mathcal{H}_{1}$, then

[TABLE]

Example 2.3*.*

If $\psi(r,s,t;z)=1+(1+\sqrt{2})es$ and $h(z)=\sqrt{1+z}$, then $\Omega=h(\mathbb{D})=\{w\in\mathbb{C}:|w^{2}-1|<1\}$. Consider

[TABLE]

whenever $r=e^{e^{i\theta}}$, $s=me^{i\theta}r$ and $\operatorname{\mathrm{Re}}\left(1+t/s\right)$ $\geq$ $m(1+\cos\theta)$, where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$. Thus $\psi(r,s,t;z)$ $\notin$ $\Omega$ and therefore $\psi\in\Psi(\Omega,e^{z})$. By Theorem 2.1, it is easily seen that if $p\in\mathcal{H}_{1}$, then

[TABLE]

Example 2.4*.*

Let $\psi(r,s,t;z)=1+s$ and suppose that $\Omega=\{w\in\mathbb{C}\colon|w-1|<e^{-1}\}$. In order to prove $\psi\in\Psi(\Omega,e^{z})$, note that

[TABLE]

whenever $r=e^{e^{i\theta}}$, $s=me^{i\theta}r$ and $\operatorname{\mathrm{Re}}\left(1+t/s\right)$ $\geq$ $m(1+\cos\theta)$, where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$. Therefore $\psi(r,s,t;z)$ $\notin$ $\Omega$ which implies $\psi\in\Psi(\Omega,e^{z})$. For any $p\in\mathcal{H}_{1}$, we obtain

[TABLE]

Similarly, if we take $\psi(r,s,t;z)=r^{2}-r+(1+e)s+1$ with the same $\Omega$ as defined earlier, then

[TABLE]

whenever $r=e^{e^{i\theta}}$, $s=me^{i\theta}r$ and $\operatorname{\mathrm{Re}}\left(1+t/s\right)$ $\geq$ $m(1+\cos\theta)$, where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$. By Theorem 2.1, it is easy to deduce that for any $p\in\mathcal{H}_{1}$, we have

[TABLE]

In the similar fashion, by taking $\psi(r,s,t;z)=1+s/r^{2}$ and $\Omega$ as above, it is easily seen that

[TABLE]

whenever $r=e^{e^{i\theta}}$, $s=me^{i\theta}r$ and $\operatorname{\mathrm{Re}}\left(1+t/s\right)$ $\geq$ $m(1+\cos\theta)$, where $\theta\in[0,2\pi)$ and $m\geq 1$. This implies that $\psi(r,s,t;z)$ $\notin$ $\Omega$ and hence $\psi\in\Psi(\Omega,e^{z})$. Thus for any $p\in\mathcal{H}_{1}$, we have

[TABLE]

Example 2.5*.*

Let $\psi(r,s,t;z)=1+s/r$ and $\Omega=h(\mathbb{D})=\{w\in\mathbb{C}\colon|w-1|<1\}$, where $h(z)=1+z$. Consider

[TABLE]

whenever $r=e^{e^{i\theta}}$, $s=me^{i\theta}r$ and $\operatorname{\mathrm{Re}}\left(1+t/s\right)$ $\geq$ $m(1+\cos\theta)$, where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$. Therefore $\psi(r,s,t;z)$ $\notin$ $\Omega$ and hence $\psi\in\Psi(\Omega,e^{z})$. Using Theorem 2.1, in terms of subordination, the result can be written as

[TABLE]

where $p\in\mathcal{H}_{1}$. Since $\psi(q(z),zq^{\prime}(z),z^{2}q^{\prime\prime}(z);z)=1+z=h(z)$ and $\psi\in\Psi(\Omega,q)$, where $q(z)=e^{z}$, it follows that $e^{z}$ is the best dominant by [11, Theorem 2.3e, p. 31].

Example 2.6*.*

Let $\psi(r,s,t;z)=2s+t$ and $\Omega=\{w\in\mathbb{C}\colon|w|<1/e\}$. Then

[TABLE]

whenever $r=e^{e^{i\theta}}$, $s=me^{i\theta}r$ and $\operatorname{\mathrm{Re}}\left(1+t/s\right)$ $\geq$ $m(1+\cos\theta)$, where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$. This shows that $\psi(r,s,t;z)$ $\notin$ $\Omega$ and $\psi\in\Psi(\Omega,e^{z})$. By Theorem 2.1, the required result is

[TABLE]

3. Subordination Associated with the Janowski Function

For $-1\leq B<A\leq 1$, we consider the subordination $\psi(p(z),zp^{\prime}(z),z^{2}p^{\prime\prime}(z);$ $z)\prec(1+Az)/(1+Bz)$ implying $p(z)\prec e^{z}$ for $z\in\mathbb{D}$. In particular, we first estimate the bound on $\beta$ such that the first order differential subordination $1+\beta zp^{\prime}(z)/p^{n}(z)\prec 1+(1-\alpha)z$ (where $n$ is any non-negative integer and $0\leq\alpha<1$) implies $p(z)\prec e^{z}$. Throughout this paper, we will assume that $\beta$ is a positive real number and $r$, $s$, $t$ are same as referred in the admissibility condition (2.1).

Theorem 3.1**.**

If $n$ is a non-negative integer, $0\leq\alpha<1$ and $p\in\mathcal{H}_{1}$ satisfies the subordination

[TABLE]

then $p(z)\prec e^{z}$.

Proof.

Let $h(z)=1+(1-\alpha)z$, where $z\in\mathbb{D}$, $0\leq\alpha<1$ and $\Omega=h(\mathbb{D})=\{w\in\mathbb{C}\colon|w-1|<1-\alpha\}$.

Case (i). If $n=0$, consider the function $\psi(r,s,t;z)=1+\beta s$. Then

[TABLE]

Theorem 2.1 is applicable if we show that $\psi\in\Psi(\Omega,e^{z})$, that is, $\psi(r,s,t;z)\not\in\Omega$ whenever $r=e^{e^{i\theta}}$, $s=me^{i\theta}r$ and $\operatorname{\mathrm{Re}}\left(1+t/s\right)$ $\geq$ $m(1+\cos\theta)$, where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$. A simple calculation yields

[TABLE]

Hence $\psi(r,s,t;z)$ $\notin$ $\Omega$ which gives $\psi\in\Psi(\Omega,e^{z})$. Using Theorem 2.1, we get $p(z)\prec e^{z}$.

Case (ii). When $n\not=0$, the function $\psi(r,s,t;z)=1+\beta s/r^{n}$ satisfies

[TABLE]

whenever $r=e^{e^{i\theta}}$, $s=me^{i\theta}r$ and $\operatorname{\mathrm{Re}}\left(1+t/s\right)$ $\geq$ $m(1+\cos\theta)$, where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$. Therefore as argued in Case (i), $\psi(r,s,t;z)$ $\notin$ $\Omega$ which implies $\psi\in\Psi(\Omega,e^{z})$. Hence by Theorem 2.1, we have the desired result. ∎

Remark 3.2*.*

For the case $n=1$, Theorem 3.1 reduces to [9, Theorem 2.8b, p. 376] when $A=1-\alpha$ and $B=0$.

Consequently, if a function $f\in\mathcal{H}$ satisfies the subordination

[TABLE]

*where $0\leq\alpha<1$ and the bound on $\beta$ is defined as in Theorem 3.1, then $f\in\mathcal{S}^{*}_{e}$. *

Next, the bound on $\beta$ is determined such that the first order differential subordination $1+\beta zp^{\prime}(z)/p^{n+1}(z)\prec(2+z)/(2-z)$ (where $n$ is any non-negative integer) implies $p(z)\prec e^{z}$.

Theorem 3.3**.**

If $n$ is any non-negative integer and $p\in\mathcal{H}_{1}$ satisfies the subordination

[TABLE]

then $p(z)\prec e^{z}$.

Proof.

By considering the function $\psi(r,s,t;z)=1+\beta s/r^{n+1}$ and $\Omega=\{w\in\mathbb{C}\colon|(2w-2)/(w+1)|<1\}$, it suffices to show $\psi\in\Psi(\Omega,e^{z})$. For this, note that

[TABLE]

Since the real-valued function $g(x)=2x/(2+x)$ is increasing for $x\geq 0$ and $\beta me^{-n\cos\theta}\geq 2$, it is easy to deduce that

[TABLE]

whenever $r=e^{e^{i\theta}}$, $s=me^{i\theta}r$ and $\operatorname{\mathrm{Re}}\left(1+t/s\right)$ $\geq$ $m(1+\cos\theta)$, where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$. Hence by making use of Theorem 2.1 we get the required result. ∎

Remark 3.4*.*

The case $n=0$ in Theorem 3.3 is similar to [9, Theorem 2.8b, p. 376] for $A=1/2$ and $B=-1/2$.

As a result, we have

If a function $f\in\mathcal{H}$ satisfies the subordination

[TABLE]

where $\beta\geq 2e^{n}$ and $n$ is any non-negative integer, then $f\in\mathcal{S}^{*}_{e}$.

The next theorem provides a bound on $\alpha$ and $\beta$ such that the first order differential subordination $(1-\alpha)p(z)+\alpha p^{2}(z)+\beta zp^{\prime}(z)\prec 1+z$ implies $p(z)\prec e^{z}$.

Theorem 3.5**.**

Let $\alpha$, $\beta$ be positive real numbers satisfying $\alpha(e-1)+\beta e\geq e$ and $p\in\mathcal{H}_{1}$. If the following subordination

[TABLE]

holds, then $p(z)\prec e^{z}$.

Proof.

Let $\Omega=h(\mathbb{D})=\{w\in\mathbb{C}\colon|w-1|<1\}$, where $h(z)=1+z$. If $\psi(r,s,t;z)=(1-\alpha)r+\alpha r^{2}+\beta s$, the required subordination is proved if we show that $\psi\in\Psi(\Omega,e^{z})$ in view of Theorem 2.1. Observe that

[TABLE]

The second derivative test shows that the function $g$ attains its minimum value at $\theta=\pi$ for $\alpha>0$ and $\beta>0$. Therefore for all $\theta\in[0,2\pi)$

[TABLE]

by the given condition $\alpha(e-1)+\beta e\geq e$. Thus $\psi(r,s,t;z)\not\in\Omega$ whenever $r=e^{e^{i\theta}}$, $s=me^{i\theta}r$ and $\operatorname{\mathrm{Re}}\left(1+t/s\right)$ $\geq$ $m(1+\cos\theta)$, where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$. By Definition 1.1, $\psi\in\Psi(\Omega,e^{z})$ and the result is evident by Theorem 2.1. ∎

Thus if $\alpha(e-1)+\beta e\geq e$ and $f\in\mathcal{H}$ satisfies the following subordination

[TABLE]

*then $f\in\mathcal{S}^{*}_{e}$. *

Next, we determine the bounds on $\beta$ such that the first order differential subordinations $p(z)+\beta zp^{\prime}(z)/p^{n}(z)\prec(2+2z)/(2-z)$, where $n=0,1$ implies $p(z)\prec e^{z}$.

Theorem 3.6**.**

Let $p\in\mathcal{H}_{1}$. Then both the following conditions are sufficient for $p(z)\prec e^{z}$:

1. (a)

$p(z)+\beta zp^{\prime}(z)\prec(2+2z)/(2-z)$* for $\beta\geq(e+2-\sqrt{2}(e-1))/(e(\sqrt{2}-1))\approx 2.0323$.* 2. (b)

$p(z)+\beta zp^{\prime}(z)/p(z)\prec(2+2z)/(2-z)$* for $\beta\geq(e+2-\sqrt{2}(e-1))/(\sqrt{2}-1)\approx 5.52436$.*

Proof.

Define $h\colon\mathbb{D}\to\mathbb{C}$ by $h(z)=(2+2z)/(2-z)$ and suppose that $\Omega=h(\mathbb{D})=\{w\in\mathbb{C}\colon|(w-1)/(w+2)|<1/2\}$.

(a) As done earlier in the previous results, the function $\psi(r,s,t;z)=r+\beta s$ should satisfies $\psi(r,s,t;z)\not\in\Omega$ whenever $r=e^{e^{i\theta}}$, $s=me^{i\theta}r$ and $\operatorname{\mathrm{Re}}\left(1+t/s\right)$ $\geq$ $m(1+\cos\theta)$, where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$. Observe that

[TABLE]

It is easily verified that the minimum value of the function in the right hand side of the above equation occurs at $\theta=0$ and therefore we obtain

[TABLE]

since $\beta\geq(e+2-\sqrt{2}(e-1))/(e(\sqrt{2}-1))$.

(b) The required subordination is proved if we show that the function $\psi(r,s,t;z)=r+\beta s/r$ does not lie in $\Omega$. For $\beta\geq(e+2-\sqrt{2}(e-1))/(\sqrt{2}-1)$, using the same technique as in previous case, we have

[TABLE]

whenever $r=e^{e^{i\theta}}$, $s=me^{i\theta}r$ and $\operatorname{\mathrm{Re}}\left(1+t/s\right)$ $\geq$ $m(1+\cos\theta)$, where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$. Therefore using Theorem 2.1, we have $p(z)\prec e^{z}$. ∎

As a consequence, we obtain

If a function $f\in\mathcal{H}$ satisfies either of the following subordinations

1. (a)

$\dfrac{zf^{\prime}(z)}{f(z)}+\beta\dfrac{zf^{\prime}(z)}{f(z)}\left(1-\dfrac{zf^{\prime}(z)}{f(z)}+\dfrac{zf^{\prime\prime}(z)}{f^{\prime}(z)}\right)\prec\dfrac{2+2z}{2-z}$* for $\beta\geq\dfrac{e+2-\sqrt{2}(e-1)}{e(\sqrt{2}-1)}$* 2. (b)

$\dfrac{zf^{\prime}(z)}{f(z)}+\beta\left(1-\dfrac{zf^{\prime}(z)}{f(z)}+\dfrac{zf^{\prime\prime}(z)}{f^{\prime}(z)}\right)\prec\dfrac{2+2z}{2-z}$* for $\beta\geq\dfrac{e+2-\sqrt{2}(e-1)}{\sqrt{2}-1}$*

*then $f\in\mathcal{S}^{*}_{e}$. *

4. Subordination Associated With The Exponential Function

In this section, we consider the problem of determining the conditions under which the subordination $\psi(p(z),zp^{\prime}(z),$ $z^{2}p^{\prime\prime}(z);z)\prec e^{z}$ implies that $p(z)\prec e^{z}$ also holds. Alternatively, our aim is to show that $\psi\in\Psi\{e^{z}\}:=\Psi(\Delta,e^{z})$ for various choices of $\psi$, where $\Delta:=\{w\in\mathbb{C}\colon|\log w|<1\}$. The first theorem of this section estimates the bound on $\beta$ such that the first order differential subordination $1+\beta(zp^{\prime}(z))^{n}\prec e^{z}$ (where $n$ is any positive integer) implies $p(z)\prec e^{z}$. Recall that, for $z\not=0$

[TABLE]

Theorem 4.1**.**

If $n$ is any positive integer and $p\in\mathcal{H}_{1}$ satisfies the subordination

[TABLE]

then $p(z)\prec e^{z}$.

Proof.

The required subordination is proved if we show that the function defined as $\psi(r,s,t;z)=1+\beta s^{n}$ satisfies the condition $\psi(r,s,t;z)\not\in\Omega$ whenever $r=e^{e^{i\theta}}$, $s=me^{i\theta}r$ and $\operatorname{\mathrm{Re}}\left(1+t/s\right)$ $\geq$ $m(1+\cos\theta)$, where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$. Consider

[TABLE]

where

[TABLE]

and

[TABLE]

Case (i). When $n$ is odd and $\beta\geq e^{n+1}+e^{n}$, we have

[TABLE]

for all $m\geq 1$. Therefore second derivative verifies that minimum value of $g$ is attained at $\theta=\pi$. If $\beta\geq e^{n+1}+e^{n}$, we obtain

[TABLE]

for all $\theta\in[0,2\pi)$. Thus $|\log\psi(r,s,t;z)|\geq 1$ and Theorem 2.1 gives $\psi\in\Psi\{e^{z}\}$.

Case (ii). If $n$ is even and

[TABLE]

for $\beta>0$, the minimum value of the function $g$ is attained at $\theta=\pi$. Therefore for all $\theta\in[0,2\pi)$ and $\beta\geq e^{n+1}-e^{n}$, we get

[TABLE]

This implies that $\psi\in\Psi\{e^{z}\}$. Hence Theorem 2.1 gives the desired differential subordination. ∎

Now, we estimate the bound on $\beta$ such that the first order differential subordination $1+\beta zp^{\prime}(z)/p^{n+1}(z)\prec e^{z}$ (where $n$ is any non-negative integer) implies $p(z)\prec e^{z}$.

Theorem 4.2**.**

If $p\in\mathcal{H}_{1}$ satisfies the subordination

[TABLE]

and $n$ is any non-negative integer, then $p(z)\prec e^{z}$.

Proof.

We apply Theorem 2.1 to show that $\psi\in\Psi\{e^{z}\}$, where $\psi(r,s,t;z)=1+\beta s/r^{n}$. Whenever $r=e^{e^{i\theta}}$, $s=me^{i\theta}r$ and $\operatorname{\mathrm{Re}}\left(1+t/s\right)$ $\geq$ $m(1+\cos\theta)$, where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$, note that

[TABLE]

where

[TABLE]

and

[TABLE]

Let $u(x)=x(-1+n)^{2}-\left(-xn+e^{n}(1-3n+n^{2})\right)\ln\left(e^{-n}(e^{n}+x)\right)$, where $x>0$ and $n$ is a non-negative integer. Natural logarithm being an increasing function implies that $\ln(e^{n}+x)>\ln(e^{n})$ for $x>0$, that is, $\ln(e^{n}+x)>n$ for $x>0$. This gives

[TABLE]

for $x>0$ and $n\not=1$. In particular for $n=1$, $u(x)=(x+e)(\ln(x+e)-1)>0$ for $x>0$. Therefore

[TABLE]

for $\beta>0$, which implies, using second derivative test, $g$ attains its minimum value at $\theta=0$. Hence for all $\theta\in[0,2\pi)$ and $\beta\geq e^{n+1}-e^{n}$

[TABLE]

Thus $|\log\psi(r,s,t;z)|\geq 1$ which implies $\psi\in\Psi\{e^{z}\}$. ∎

Remark 4.3*.*

For $n=0$, Theorem 4.2 reduces to [9, Theorem 2.8a, p. 376].

In the next two theorems, the bound on $\beta$ is computed such that the first order differential subordination $p(z)+\beta zp^{\prime}(z)/p^{n+1}$ $(z)\prec e^{z}$ (where $n=-1,0,1,2,\ldots$) implies $p(z)\prec e^{z}$.

Theorem 4.4**.**

Let $p\in\mathcal{H}_{1}$, then each of the following subordinations are sufficient for $p(z)\prec e^{z}$:

1. (a)

$p(z)+\beta zp^{\prime}(z)\prec e^{z}$* for $\beta\geq e^{2}+1\approx 8.38906$.* 2. (b)

$p(z)+\beta zp^{\prime}(z)/p(z)\prec e^{z}$* for $\beta\geq e+e^{-1}\approx 3.08616$.*

Proof.

(a) In order to prove the admissibility condition (2.1) for the function $\psi(r,s,t;z)=r+\beta s$, we need to show that $|\log\psi(r,s,t;z)|^{2}\geq 1$ whenever $r=e^{e^{i\theta}}$, $s=me^{i\theta}r$ and $\operatorname{\mathrm{Re}}\left(1+t/s\right)$ $\geq$ $m(1+\cos\theta)$, where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$. A simple computation gives

[TABLE]

Note that

[TABLE]

for $\beta>\beta^{*}\approx 3.4446$, where $\beta^{*}$ is the root of the equation $x(1-x)+(1-x+x^{2})\ln(-1+x)=0$. Therefore the minimum value of the function $g$ is clearly attained at $\theta=\pi$ for $\beta\geq e^{2}+1\approx 8.38906$. In that case, we have

[TABLE]

Hence $\psi\in\Psi\{e^{z}\}$.

(b) Using the same technique as above, for the function $\psi(r,s,t;z)=r+\beta s/r$, consider

[TABLE]

We observe that the second derivative of $g$ is positive on both of its critical points, therefore the absolute minimum of $g$ is attained at $\theta=\pi$ for $\beta\geq e+e^{-1}$. Hence we get

[TABLE]

Thus $\psi\in\Psi\{e^{z}\}$ and Theorem 2.1 completes the proof. ∎

Theorem 4.5**.**

Let $n$ be any positive integer and $\beta_{n}$ be a positive root of the equation

[TABLE]

If $p\in\mathcal{H}_{1}$ satisfies the subordination

[TABLE]

then $p(z)\prec e^{z}$.

Proof.

As argued in other cases, to prove the required subordinaton, it suffices to show that the function $\psi(r,s,t;z)=r+\beta s/r^{n+1}$ satisfies $\psi(r,s,t;z)\not\in\Delta$ whenever $r=e^{e^{i\theta}}$, $s=me^{i\theta}r$ and $\operatorname{\mathrm{Re}}\left(1+t/s\right)$ $\geq$ $m(1+\cos\theta)$, where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$. Note that

[TABLE]

where

[TABLE]

If $\beta>\beta_{n}$, where $\beta_{n}$ is a positive root of the equation (4.1), then

[TABLE]

Therefore the minimum value of $g$ is attained at $\theta=0$ by the second derivative test which implies for all $\theta\in[0,2\pi)$ and $\beta>\beta_{n}>0$

[TABLE]

for all positive integers $m$ and $n$. Hence $\psi\in\Psi\{e^{z}\}$ and Theorem 2.1 gives the desired result. ∎

Next the bound on $\beta$ is determined such that each of the first order differential subordination $p(z)+\beta(zp^{\prime}(z))^{2}/p^{n}(z)\prec e^{z}$ ($n=0,1,2$) implies $p(z)\prec e^{z}$.

Theorem 4.6**.**

Let $p\in\mathcal{H}_{1}$. Then each of the following subordinations are sufficient for $p(z)\prec e^{z}$:

1. (a)

$p(z)+\beta(zp^{\prime}(z))^{2}\prec e^{z}$* for $\beta\geq e^{3}-e\approx 17.3673$.* 2. (b)

$p(z)+\beta(zp^{\prime}(z))^{2}/p(z)\prec e^{z}$* for $\beta\geq e^{2}-1\approx 6.38906$.* 3. (c)

$p(z)+\beta(zp^{\prime}(z))^{2}/p^{2}(z)\prec e^{z}$* for $\beta\geq e-e^{-1}\approx 2.3504$.*

Proof.

For different choices of $\psi$, we need to show that $\psi\in\Psi\{e^{z}\}$, that is, we must verify $|\log\psi(r,s,t;z)|\geq 1$ whenever $r=e^{e^{i\theta}}$, $s=me^{i\theta}r$ and $\operatorname{\mathrm{Re}}\left(1+t/s\right)$ $\geq$ $m(1+\cos\theta)$, where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$.

(a) Let $\psi(r,s,t;z)=r+\beta s^{2}$ and consider

[TABLE]

where

[TABLE]

and

[TABLE]

Since

[TABLE]

for $\beta>\beta^{*}\approx 3.7586$, where $\beta^{*}$ is a positive root of the equation $(e+2x)^{2}-(e^{2}+2xe+2x^{2})\ln(e+x)=0$, we can say that the minimum value of $g$ is obviously attained at $\theta=\pi$ for $\beta\geq e^{3}-e$. Therefore for $\beta\geq e^{3}-e$, we have

[TABLE]

Hence $\psi(r,s,t;z)\not\in\Delta$ and using Theorem 2.1 the result follows.

(b) Let the function be defined by $\psi(r,s,t;z)=r+\beta s^{2}/r$ and observe

[TABLE]

where

[TABLE]

It is easily verified that the minimum value of the function $g$ is attained at $\theta=\pi$ for $\beta\geq e^{2}-1$. In that case for all $\theta\in[0,2\pi)$ and $\beta\geq e^{2}-1$

[TABLE]

Therefore $|\log\psi(r,s,t;z)|\geq 1$ whenever $r=e^{e^{i\theta}}$, $s=me^{i\theta}r$ and $\operatorname{\mathrm{Re}}\left(1+t/s\right)$ $\geq$ $m(1+\cos\theta)$, where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$. Hence Theorem 2.1 yields the desired result.

(c) For the function $\psi(r,s,t;z)=r+\beta s^{2}/r^{2}$, it is easy to deduce that

[TABLE]

Since the second derivative of $g$ is positive on both of its critical points, $g$ attains absolute minimum at $\theta=\pi$ for $\beta\geq e-e^{-1}>0$. Therefore for all $\theta\in[0,2\pi)$ and for $\beta\geq e-e^{-1}$, we get

[TABLE]

Hence $\psi(r,s,t;z)\notin\Delta$ and thus $\psi\in\Psi\{e^{z}\}$. ∎

Next, we estimate the bound on $\beta$ such that each of the first order differential subordination $p^{2}(z)+\beta zp^{\prime}(z)/p^{n}(z)\prec e^{z}$ ($n=0,1,2$) implies $p(z)\prec e^{z}$.

Theorem 4.7**.**

Let $p\in\mathcal{H}_{1}$. Then each of the following subordinations are sufficient for $p(z)\prec e^{z}$:

1. (a)

$p^{2}(z)+\beta zp^{\prime}(z)\prec e^{z}$* for $\beta\geq e^{2}+e^{-1}\approx 7.75694$.* 2. (b)

$p^{2}(z)+\beta zp^{\prime}(z)/p(z)\prec e^{z}$* for $\beta\geq e+e^{-2}\approx 2.85362$.* 3. (c)

$p^{2}(z)+\beta zp^{\prime}(z)/p^{2}(z)\prec e^{z}$* for $\beta>\beta^{*}\approx 104.122$, where $\beta^{*}$ is a positive root of the equation*

[TABLE]

Proof.

(a) For the function $\psi(r,s,t;z)=r^{2}+\beta s$, Theorem 2.1 is applicable if we show that the function $\psi\in\Psi\{e^{z}\}$ whenever $r=e^{e^{i\theta}}$, $s=me^{i\theta}r$ and $\operatorname{\mathrm{Re}}\left(1+t/s\right)$ $\geq$ $m(1+\cos\theta)$, where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$. Consider

[TABLE]

where

[TABLE]

and

[TABLE]

To show $|\log\psi(r,s,t;z)|\geq 1$ note that $g^{\prime\prime}(\pi)>0$ for $\beta>\beta^{*}\approx 2.9432$, where $\beta^{*}$ is a root of the equation $2xe(1+xe)-(2+xe+x^{2}e^{2})\ln(-1+xe)=0$. Therefore minimum value of the function $g$ is obviously attained at $\theta=\pi$ for $\beta\geq e^{2}+e^{-1}$ and hence

[TABLE]

Consequently, Theorem 2.1 yields the result.

(b) The required subordination is proved if we show that $\psi\in\Psi\{e^{z}\}$, that is, if the admissibility condition (2.1) is satisfied. For the function $\psi(r,s,t;z)=r^{2}+\beta s/r$, observe

[TABLE]

For $\beta\geq e+e^{-2}$, it is easily verified using second derivative test that $g$ attains its minimum value at $\theta=\pi$ which implies

[TABLE]

Therefore $\psi(r,s,t;z)\notin\Delta$ whenever $r=e^{e^{i\theta}}$, $s=me^{i\theta}r$ and $\operatorname{\mathrm{Re}}\left(1+t/s\right)$ $\geq$ $m(1+\cos\theta)$, where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$. Using Theorem 2.1 we get $p(z)\prec e^{z}$.

(c) As done in other cases, we need to show that $\psi(r,s,t;z)\notin\Delta$, where $\psi$ is defined as $\psi(r,s,t;z)=r^{2}+\beta s/r^{2}$. Consider

[TABLE]

where

[TABLE]

and

[TABLE]

We note that the minimum value of $g$ is attained at $\theta=0$ for $\beta>\beta^{*}\approx 104.122$, where $\beta^{*}$ is a positive root of the equation $6e^{6}+5xe^{3}-x^{2}+(-2e^{6}-5xe^{3}+x^{2})\ln(e^{3}+x)=0$. For $\theta\in[0,2\pi)$ and $\beta>\beta^{*}$, we have

[TABLE]

Therefore $\psi\in\Psi\{e^{z}\}$ and hence the result. ∎

Next, the bound on $\beta$ is ascertained such that each of the first order differential subordination $p^{n}(z)+\beta zp(z)p^{\prime}(z)\prec e^{z}$ ($n=1,2,3$) implies $p(z)\prec e^{z}$.

Theorem 4.8**.**

Let $p\in\mathcal{H}_{1}$, then each of the following subordinations are sufficient for $p(z)\prec e^{z}$:

1. (a)

$p(z)+\beta zp(z)p^{\prime}(z)\prec e^{z}$* for $\beta\geq e^{3}+e\approx 22.8038$.* 2. (b)

$p^{2}(z)+\beta zp(z)p^{\prime}(z)\prec e^{z}$* for $\beta\geq e^{3}+1\approx 21.0855$.* 3. (c)

$p^{3}(z)+\beta zp(z)p^{\prime}(z)\prec e^{z}$* for $\beta\geq e^{3}+e^{-1}\approx 20.4534$.*

Proof.

The subordination $p(z)\prec e^{z}$ is satisfied if we show that $\psi\in\Psi\{e^{z}\}$ for different choices of $\psi$. Equivalently, we need to verify the admissibility condition (2.1):

[TABLE]

whenever $r=e^{e^{i\theta}}$, $s=me^{i\theta}r$ and $\operatorname{\mathrm{Re}}\left(1+t/s\right)$ $\geq$ $m(1+\cos\theta)$, where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$.

(a) Let $\psi(r,s,t;z)=r+\beta rs$. A simple calculation yields

[TABLE]

where

[TABLE]

and

[TABLE]

It is easily verified that $g^{\prime\prime}(\pi)>0$ for $\beta>\beta^{*}\approx 7.7065$, where $\beta^{*}$ is the positive root of the equation $6x-2e+(e-2x)\ln((e-x)^{2})=0$. Since we are given that $\beta\geq e^{3}+e\approx 22.8038$, the minimum value of $g$ is attained at $\theta=\pi$ and for all $\theta\in[0,2\pi)$, we have

[TABLE]

Therefore $\psi\in\Psi\{e^{z}\}$.

(b) The function $\psi(r,s,t;z)=r^{2}+\beta rs$ satisfies

[TABLE]

where

[TABLE]

and

[TABLE]

Since $g^{\prime\prime}(\pi)>0$ for $\beta>\beta^{*}\approx 6.46722$, where $\beta^{*}$ is the root of the equation $x(2-3x)+(2-3x+2x^{2})\ln(-1+x)=0$, the minimum value of $g(\theta)$ is obviously attained at $\theta=\pi$ if $\beta\geq e^{3}+1\approx 20.0855$. Therefore for $\theta\in[0,2\pi)$, we get

[TABLE]

and hence $\psi\in\Psi\{e^{z}\}$.

(c) Let $\psi(r,s,t;z)=r^{3}+\beta rs$. With $r$, $s$ and $t$ stated above, $\psi$ takes the form $\psi(r,s,t;z)=e^{3e^{i\theta}}+\beta me^{i\theta}e^{2e^{i\theta}}$ which satisfies

[TABLE]

where

[TABLE]

and

[TABLE]

Clearly, $g$ attains its minimum value either at $\theta=0$ or $\theta=\pi$. Since $g^{\prime\prime}(\pi)>0$ for $\beta>\beta^{*}\approx 5.66489$, where $\beta^{*}$ is the root of the equation $xe(3+5xe)-(3-xe+2x^{2}e^{2})\ln(-1+xe)=0$, the minimum value of $g$ is attained at $\theta=\pi$ if $\beta\geq e^{3}+e^{-1}\approx 20.4534$. In that case, we have

[TABLE]

Therefore $\psi\in\Psi\{e^{z}\}$ by Theorem 2.1. ∎

Now, we estimate the bound on $\beta$ such that the first order differential subordination $p^{2}(z)+p(z)-1+\beta zp^{\prime}(z)\prec e^{z}$ implies $p(z)\prec e^{z}$.

Theorem 4.9**.**

Let $p\in\mathcal{H}_{1}$ and satisfies the subordination

[TABLE]

Then $p(z)\prec e^{z}$.

Proof.

Proceeding as in the previous theorems, we need to show that the function $\psi(r,s,t;z)=r^{2}+r-1+\beta s$ satisfies the admissibility condition (2.1). Whenever $r=e^{e^{i\theta}}$, $s=me^{i\theta}r$ and $\operatorname{\mathrm{Re}}\left(1+t/s\right)$ $\geq$ $m(1+\cos\theta)$, where $z\in\mathbb{D}$, $\theta\in[0,2\pi)$ and $m\geq 1$, $\psi(r,s,t;z)=e^{2e^{i\theta}}+e^{e^{i\theta}}-1+\beta me^{i\theta}e^{e^{i\theta}}$ satisfies

[TABLE]

where

[TABLE]

and

[TABLE]

Using second derivative test, it can be easily verified that $g$ attains its minimum value at $\theta=\pi$ for $\beta>0$. Therefore for $\theta\in[0,2\pi)$, we have

[TABLE]

Since logarithm is an increasing function and the condition $\beta\geq e^{2}+e^{-1}-e+1$ imply that

[TABLE]

Hence $\psi\in\Psi\{e^{z}\}$ and Theorem 2.1 gives the desired result. ∎

Remark 4.10*.*

As depicted in the previous section, results proved in this section also provide several sufficient conditions for a normalized analytic function to be in the class $\mathcal{S}^{*}_{e}$. These sufficient conditions can be obtained by simply putting $p(z)=zf^{\prime}(z)/f(z)$, where $f\in\mathcal{H}$.

Remark 4.11*.*

Since we were concerned with the starlikeness property in this paper, therefore we presented applications of our results only for the subclasses of $\mathcal{S}^{*}$. However, by setting $p(z)=f^{\prime}(z)$, $p(z)=f(z)/z$, $p(z)=2f(z)/z-1$, $p(z)=2\sqrt{f^{\prime}(z)}-1$, $p(z)=2zf^{\prime}(z)/f(z)-1$ and so forth in the theorems obtained, one can obtain many more differential implications.