Some Properties for Subordinations of Analytic Functions

Abstract

Let the class of functions of $f\left(z\right)$ of the form $f\left(z\right)=z+{\sum }_{k=2}^{\infty }{a}_k{z}^k$, which are denoted by $\mathcal{A}$ and called analytic functions in the open-unit disk. There are many interesting properties of the functions $f\left(z\right)$ in the class $\mathcal{A}$ concerning the subordinations. Applying the three lemmas for $f\left(z\right)\in \mathcal{A}$ provided by Miller and Mocanu and by Nunokawa, we consider many interesting properties of $f\left(z\right)\in \mathcal{A}$ with subordinations. Furthermore, we provide simple examples for our results. We think it is very important to consider examples of the results.

1. Introduction

Let $\mathcal{A}$ be the class of functions $f\left(z\right)$ of the form:

$$f\left(z\right)=z+\sum _{k=2}^{\infty }{a}_k{z}^k$$

(1)

which are analytic in the open-unit disk $\mathbb{U}=\{z\in \mathbb{C}:|z|<1\}.$ Given the two analytic functions $f\left(z\right)$ and $g\left(z\right)$, the function $f\left(z\right)$ is said to be subordinate to $g\left(z\right)$ in $\mathbb{U}$ and written as $f\left(z\right)\prec g\left(z\right)$ if there exists a Schwarz function $w\left(z\right)$ analytic such that $f\left(z\right)=g\left(w\right(z\left)\right)$ with $w\left(0\right)=0$ and $\left|w\right(z\left)\right|\le 1,$$z\in \mathbb{U}.$ In particular, if $g\left(z\right)$ is univalent in $\mathbb{U},$ then $f\left(z\right)\prec g\left(z\right)$ if and only if $f\left(0\right)=g\left(0\right)$ and $f\left(\mathbb{U}\right)\subseteq g\left(\mathbb{U}\right)$ (cf. [1,2]). We note that if $f\left(z\right)\in \mathcal{A}$ satisfies

$$\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\prec \frac{1+(1-2\alpha )z}{1-z}(z\in \mathbb{U})$$

(2)

for some real $\alpha (0\le \alpha <1),$ then $f\left(z\right)$ is said to be the starlike function of order $\alpha$ in $\mathbb{U}$ and, if $f\left(z\right)\in \mathcal{A}$ satisfies

$$1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\prec \frac{1+(1-2\alpha )z}{1-z}(z\in \mathbb{U})$$

(3)

for some real $\alpha (0\le \alpha <1),$ then $f\left(z\right)$ is said to be the convex of order $\alpha$ in $\mathbb{U}.$

Furthermore, let $p\left(z\right)$ be analytic in $\mathbb{U}$ and $p\left(0\right)=1.$ Then, if $p\left(z\right)$ satisfies

$$p\left(z\right)\prec {\left(\frac{1+z}{1-z}\right)}^{\alpha }(z\in \mathbb{U})$$

(4)

for some real $\alpha (\alpha >0),$ then $p\left(z\right)$ satisfies

$$\left|argp\left(z\right)\right|<\frac{\pi }{2}\alpha (z\in \mathbb{U}).$$

(5)

If $f\left(z\right)\in \mathcal{A}$ satisfies

$$\left|arg{f}^{\prime }\left(z\right)\right|<\frac{\pi }{2}\alpha (z\in \mathbb{U})$$

(6)

for some real $\alpha (0<\alpha \le 1),$ then we say that $f\left(z\right)$ is the strongly univalent function of order $\alpha$ in $\mathbb{U}.$ If $f\left(z\right)\in \mathcal{A}$ satisfies

$$\left|arg\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right|<\frac{\pi }{2}\alpha (z\in \mathbb{U})$$

(7)

for some real $\alpha (0<\alpha \le 1),$ then we say that $f\left(z\right)$ is the strongly starlike function of order $\alpha$ in $\mathbb{U}.$ Further, if $f\left(z\right)\in \mathcal{A}$ satisfies

$$\left|arg\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)\right|<\frac{\pi }{2}\alpha (z\in \mathbb{U})$$

(8)

for some real $\alpha (0<\alpha \le 1),$ then we say that $f\left(z\right)$ is the strongly convex function of order $\alpha$ in $\mathbb{U}$ (cf. [2]).

2. Some Applications of Differential Subordinations

To consider some applications for subordinations, we introduce the following lemma from Miller and Mocanu [3].

Lemma 1. 

Let ${\beta }_0=1.21872\dots$ be the solution of $\beta \pi =\frac{3}{2}\pi -Ta{n}^{-1}\beta$ and let $\alpha =\alpha \left(\beta \right)=\beta +2Ta{n}^{-1}\left(\frac{\beta }{\pi }\right)$ for $0<\beta <{\beta }_0.$ If $p\left(z\right)$ is analytic in $\mathbb{U}$ with $p\left(0\right)=1,$ then

$$p\left(z\right)+z{p}^{\prime }\left(z\right)\prec {\left(\frac{1+z}{1-z}\right)}^{\alpha };(z\in \mathbb{U})$$

(9)

implies that

$$p\left(z\right)\prec {\left(\frac{1+z}{1-z}\right)}^{\beta };(z\in \mathbb{U}).$$

(10)

Remark 1. 

If $\beta =1$ in Lemma 1, then $\alpha \left(1\right)=\frac{3}{2}.$ Thus, Lemma 1 says that if the function $p\left(z\right)$ satisfies the following subordination:

$$p\left(z\right)+z{p}^{\prime }\left(z\right)\prec {\left(\frac{1+z}{1-z}\right)}^{\frac{3}{2}};(z\in \mathbb{U})$$

(11)

then

$$p\left(z\right)\prec \frac{1+z}{1-z};(z\in \mathbb{U}).$$

(12)

Now, we prove the following theorem.

Theorem 1. 

Let ${\beta }_0=1.21872\dots$ be the solution of $\beta \pi =\frac{3}{2}\pi -Ta{n}^{-1}\beta$ and let $\alpha =\alpha \left(\beta \right)=\beta +2Ta{n}^{-1}\left(\frac{\beta }{\pi }\right)$ for $0<\beta <{\beta }_0.$ If $p\left(z\right)$ is analytic in $\mathbb{U}$ with $p\left(0\right)=1,$ then

$$p\left(z\right)+z{p}^{\prime }\left(z\right)\prec \gamma +(1-\gamma ){\left(\frac{1+z}{1-z}\right)}^{\alpha };(z\in \mathbb{U})$$

(13)

implies that

$$p\left(z\right)\prec \gamma +(1-\gamma ){\left(\frac{1+z}{1-z}\right)}^{\beta };(z\in \mathbb{U})$$

(14)

where $0\le \gamma <1.$

Proof. 

Let us define a function $F\left(z\right)$ using

$$F\left(z\right)=\frac{p\left(z\right)-\gamma }{1-\gamma },(z\in \mathbb{U}).$$

(15)

Then, $F\left(z\right)$ is analytic in $\mathbb{U}$ with $F\left(0\right)=1$ and

$$z{F}^{\prime }\left(z\right)=\frac{z{p}^{\prime }\left(z\right)}{1-\gamma }.$$

(16)

Therefore, Lemma 1 implies that if

$$F\left(z\right)+z{F}^{\prime }\left(z\right)=\frac{p\left(z\right)+z{p}^{\prime }\left(z\right)-\gamma }{1-\gamma }\prec {\left(\frac{1+z}{1-z}\right)}^{\alpha },(z\in \mathbb{U})$$

(17)

then

$$F\left(z\right)=\frac{p\left(z\right)-\gamma }{1-\gamma }\prec {\left(\frac{1+z}{1-z}\right)}^{\beta },(z\in \mathbb{U}).$$

(18)

The subordination (17) implies (13) and the subordination (18) is the same as (14). □

Letting $\beta =1$ in Theorem 1, we obtain the following corollary.

Corollary 1. 

If $p\left(z\right)$ is analytic in $\mathbb{U}$ with $p\left(0\right)=1$ satisfies

$$p\left(z\right)+z{p}^{\prime }\left(z\right)\prec \gamma +(1-\gamma ){\left(\frac{1+z}{1-z}\right)}^{\frac{3}{2}};(z\in \mathbb{U})$$

(19)

for some real γ$(0\le \gamma <1)$ then

$$p\left(z\right)\prec \frac{1+(1-2\gamma )z}{1-z};(z\in \mathbb{U})$$

(20)

and $Rep\left(z\right)>\gamma (z\in \mathbb{U}).$

In Corollary 1, considering $p\left(z\right)=\frac{f\left(z\right)}{z}$ for the function $f\left(z\right)$ in the class $\mathcal{A},$ we have the following.

Corollary 2. 

If the function $f\left(z\right)$ in the class $\mathcal{A}$ satisfies

$$f^{\prime }\left(z\right)\prec \gamma +(1-\gamma ){\left(\frac{1+z}{1-z}\right)}^{\frac{3}{2}};(z\in \mathbb{U})$$

(21)

for some real γ$(0\le \gamma <1)$ then

$$\frac{f\left(z\right)}{z}\prec \frac{1+(1-2\gamma )z}{1-z};(z\in \mathbb{U})$$

(22)

and $Re\left(\frac{f\left(z\right)}{z}\right)>\gamma (z\in \mathbb{U}).$

In Corollary 1, ensuring $p\left(z\right)=f^{\prime }\left(z\right)$ for the function $f\left(z\right)$ in the class $\mathcal{A},$ we have the following.

Corollary 3. 

If the function $f\left(z\right)$ in the class $\mathcal{A}$ satisfies

$$f^{\prime }\left(z\right)+z{f}^{\prime \prime }\left(z\right)\prec \gamma +(1-\gamma ){\left(\frac{1+z}{1-z}\right)}^{\frac{3}{2}};(z\in \mathbb{U})$$

(23)

for some real γ$(0\le \gamma <1),$ then

$$f^{\prime }\left(z\right)\prec \frac{1+(1-2\gamma )z}{1-z};(z\in \mathbb{U})$$

(24)

and $Re\left(f^{\prime }\left(z\right)\right)>\gamma (z\in \mathbb{U}).$

Further, in Corollary 1, letting $p\left(z\right)=\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}$ for the function $f\left(z\right)$ in the class $\mathcal{A},$ we have the following corollary.

Corollary 4. 

If the function $f\left(z\right)$ in the class $\mathcal{A}$ satisfies

$$2\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}+\frac{z^2{f}^{″}\left(z\right)}{f\left(z\right)}-{\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)}^2\prec \gamma +(1-\gamma ){\left(\frac{1+z}{1-z}\right)}^{\frac{3}{2}};(z\in \mathbb{U})$$

(25)

for some real γ$(0\le \gamma <1),$ then

$$\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\prec \frac{1+(1-2\gamma )z}{1-z};(z\in \mathbb{U})$$

(26)

and $f\left(z\right)$ is the starlike function of order γ in $\mathbb{U}.$

To consider the next problem, let ${\mathcal{P}}_n$ be the class of functions $p\left(z\right)$ that are analytic in $\mathbb{U}$ with

$$p\left(z\right)=1+\sum _{k=n}^{\infty }{c}_n{z}^n(n\in \mathbb{N}=\{1,2,3,\cdots \}).$$

(27)

For $p\left(z\right)\in {\mathcal{P}}_n,$ Nunokawa [4,5] derives the following lemma.

Lemma 2. 

Let a function $p\left(z\right)$ be in the class ${\mathcal{P}}_n.$ If there exists a point $z_0\left(\right|z_0|<1)$ such that

$$\left|arg\left(p\right(z\left)\right)\right|<\frac{\pi }{2}\beta \left(\right|z|<|z_0\left|\right)$$

(28)

and

$$\left|arg\left(p\left(z_0\right)\right)\right|=\frac{\pi }{2}\beta$$

(29)

for some real $\beta >0,$ then

$$\frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)}=\frac{2im}{\pi }arg\left(p\left(z_0\right)\right)$$

(30)

for some $m\ge \frac{n}{2}\left(a+\frac{1}{a}\right)>n,$ where

$${\left(p\left(z_0\right)\right)}^{\frac{1}{\beta }}=\pm ia(a>0).$$

(31)

Applying the Lemma 2, we derive the following theorem.

Theorem 2. 

If the function $p\left(z\right)$ in the class ${\mathcal{P}}_n$ satisfies

$${\left\{\frac{p\left(z\right)-\alpha }{1-\alpha }+\frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)-\alpha }\right\}}^2-1\prec \frac{4{(n+1)}^{2}z}{{(1-z)}^2},(z\in \mathbb{U})$$

(32)

for some real $\alpha (0\le \alpha <1),$ then

$$Re\left(p\right(z\left)\right)>\alpha ,(z\in \mathbb{U}).$$

(33)

Proof. 

We suppose that there exists a point $z_0\left(\right|z_0|<1)$ such that

$$\mathrm{Re}\left\{\frac{p\left(z\right)-\alpha }{1-\alpha }\right\}>0\left(\right|z|<|z_0|<1)$$

(34)

and

$$\mathrm{Re}\left\{\frac{p\left(z_0\right)-\alpha }{1-\alpha }\right\}=0.$$

(35)

If

$$\frac{p\left(z_0\right)-\alpha }{1-\alpha }\ne 0,$$

(36)

then Lemma 2 provides

$$\begin{array}{cc}\hfill \frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)-\alpha }& =\frac{2im}{\pi }arg\left(\frac{p\left(z_0\right)-\alpha }{1-\alpha }\right)\hfill \\ \hfill & =\frac{2im}{\pi }arg\left(p\left(z_0\right)-\alpha \right)\hfill \end{array}$$

(37)

for some real $m\ge \frac{n}{2}\left(a+\frac{1}{a}\right)>n$ with

$${\left(\frac{p\left(z_0\right)-\alpha }{1-\alpha }\right)}^{\frac{1}{\beta }}=\pm ia(a>0).$$

(38)

It follows from the above that

$$\begin{array}{cc}\hfill {\left\{\frac{p\left(z_0\right)-\alpha }{1-\alpha }+\frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)-\alpha }\right\}}^2-1& ={(\pm ia\pm im)}^2-1\hfill \\ \hfill & \le -{\left(a+\frac{n(a^2+1)}{2a}\right)}^2-1.\hfill \end{array}$$

(39)

We consider a function $h\left(a\right)$ provided by

$$\begin{array}{c}\hfill h\left(a\right)=a+\frac{n(a^2+1)}{2a}(a>0).\end{array}$$

(40)

Then, $h\left(a\right)$ satisfies

$$\begin{array}{c}\hfill h\left(a\right)\ge h\left(\sqrt{\frac{n}{n+2}}\right)=\sqrt{n(n+2)}.\end{array}$$

(41)

This implies that

$$\begin{array}{cc}\hfill {\left\{\frac{p\left(z_0\right)-\alpha }{1-\alpha }+\frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)-\alpha }\right\}}^2-1& \le -{\left(a+\frac{n(a^2+1)}{2a}\right)}^2-1\hfill \\ \hfill & =-{(n+1)}^2.\hfill \end{array}$$

(42)

On the other hand, we consider a function $g\left(z\right)$ provided by

$$\begin{array}{c}\hfill g\left(z\right)=\frac{4{(n+1)}^{2}z}{{(1-z)}^2}(z\in \mathbb{U}).\end{array}$$

(43)

The function $g\left(z\right)$ maps $\mathbb{U}$ onto the domain with the slit $(-\infty ,-{(n+1)}^2).$ This contradicts our condition (31). Therefore, we have that

$$\mathrm{Re}\left(\frac{p\left(z\right)-\alpha }{1-\alpha }\right)>0$$

(44)

for all $z\in \mathbb{U}.$ This shows us that

$$\mathrm{Re}\left(p\right(z\left)\right)>\alpha ,(z\in \mathbb{U}).$$

(45)

□

Considering $p\left(z\right)=\frac{f\left(z\right)}{z}$ for the function $f\left(z\right)$ in the class $\mathcal{A},$ we have the following corollary.

Corollary 5. 

If the function $f\left(z\right)$ in the class $\mathcal{A}$ satisfies

$${\left\{\frac{f\left(z\right)-\alpha }{z(1-\alpha )}+\frac{z{f}^{\prime }\left(z\right)-f\left(z\right)}{f\left(z\right)-z}\right\}}^2-1\prec \frac{16z}{{(1-z)}^2};(z\in \mathbb{U})$$

(46)

for some real α$(0\le \alpha <1),$ then

$$Re\frac{f\left(z\right)}{z}>\alpha ;(z\in \mathbb{U}).$$

(47)

Causing $p\left(z\right)=f^{\prime }\left(z\right)$ for the function $f\left(z\right)$ in the class $\mathcal{A},$ thus we obtain the following corollary.

Corollary 6. 

If the function $f\left(z\right)$ in the class $\mathcal{A}$ satisfies

$${\left\{\frac{f^{\prime }\left(z\right)-\alpha }{1-\alpha }+\frac{z{f}^{\prime \prime }\left(z\right)}{f^{\prime }\left(z\right)-\alpha }\right\}}^2\prec \frac{16z}{{(1-z)}^2};(z\in \mathbb{U})$$

(48)

for some real α$(0\le \alpha <1),$ then

$${Ref}^{\prime }\left(z\right)>\alpha ;(z\in \mathbb{U}).$$

(49)

Using $p\left(z\right)=\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}$ for the function $f\left(z\right)$ in the class $\mathcal{A},$ we have the following corollary.

Corollary 7. 

If the function $f\left(z\right)$ in the class $\mathcal{A}$ satisfies

$${\left\{\frac{1}{f\left(z\right)}\left(\frac{z{f}^{\prime }\left(z\right)-\alpha f\left(z\right)}{1-\alpha }+\frac{z(f\left(z\right)f^{\prime }\left(z\right)+zf\left(z\right)f^{″}\left(z\right)-z{\left(f^{\prime }\left(z\right)\right)}^2)}{z{f}^{\prime }\left(z\right)-\alpha f\left(z\right)}\right)\right\}}^2-1\prec \frac{16z}{{(1-z)}^2}$$

(50)

for some real α$(0\le \alpha <1),$ then

$$Re\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)>\alpha ;(z\in \mathbb{U}).$$

(51)

Next, we derive the following theorem.

Theorem 3. 

If the function $p\left(z\right)$ in the class ${\mathcal{P}}_n$ satisfies

$$\frac{p\left(z\right)-\alpha }{1-\alpha }+\frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)-\alpha }\prec \frac{1+z}{1-z},(z\in \mathbb{U})$$

(52)

for some real $\alpha (0\le \alpha <1),$ then

$$Re\left(p\right(z\left)\right)>\alpha ,(z\in \mathbb{U}).$$

(53)

Proof. 

We consider that there exists a point $z_0\left(\right|z_0|<1)$ such that

$$\mathrm{Re}\left\{\frac{p\left(z\right)-\alpha }{1-\alpha }\right\}>0\left(\right|z|<|z_0|<1)$$

(54)

and

$$\mathrm{Re}\left\{\frac{p\left(z_0\right)-\alpha }{1-\alpha }\right\}=0.$$

(55)

If

$$\frac{p\left(z_0\right)-\alpha }{1-\alpha }\ne 0,$$

(56)

using Lemma 2 we have

$$\begin{array}{c}\hfill \frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)-\alpha }=\frac{2im}{\pi }arg\left(p\left(z_0\right)-\alpha \right)\end{array}$$

(57)

for some real $m\ge \frac{n}{2}\left(a+\frac{1}{a}\right)>n$ with

$$\frac{p\left(z_0\right)-\alpha }{1-\alpha }=\pm ia(a>0).$$

(58)

This provides

$$\begin{array}{cc}\hfill \frac{p\left(z_0\right)-\alpha }{1-\alpha }+\frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)-\alpha }& =\pm ia\pm im\hfill \\ \hfill & =\pm i(a+m).\hfill \end{array}$$

(59)

Noting that

$$\begin{array}{c}\hfill \mathrm{Re}\left(\frac{1+z}{1-z}\right)>0(z\in \mathbb{U}),\end{array}$$

(60)

we say that

$$\begin{array}{c}\hfill \frac{p\left(z\right)-\alpha }{1-\alpha }+\frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)-\alpha }isnotsubordinateto\frac{1+z}{1-z}(z\in \mathbb{U}).\end{array}$$

(61)

Therefore, there is no $z_0\left(\right|z_0|<1)$ as in (34) and (35). This implies that

$$\begin{array}{c}\hfill \left(\frac{p\left(z\right)-\alpha }{1-\alpha }\right)>0(z\in \mathbb{U})\end{array}$$

(62)

that is

$$\mathrm{Re}\left(p\right(z\left)\right)>\alpha ,(z\in \mathbb{U}).$$

(63)

□

Example 1. 

Let us consider a function $p\left(z\right)$ provided by

$$p\left(z\right)=\frac{1}{1-z}\in {\mathcal{P}}_1$$

(64)

and $\alpha =0.$ Then, $p\left(z\right)$ satisfies

$$p\left(z\right)+\frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)}=\frac{1+z}{1-z},(z\in \mathbb{U}).$$

(65)

Thus, $p\left(z\right)$ satisfies the subordination (52) for $\alpha =0.$ For such $p\left(z\right),$ we have that

$$Re\left(p\left(z\right)\right)>\frac{1}{2}>0,(z\in \mathbb{U}).$$

(66)

Corollary 8. 

If the function $p\left(z\right)$ in the class ${\mathcal{P}}_n$ satisfies

$$\left|arg\left(\frac{p\left(z\right)-\alpha }{1-\alpha }+\frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)-\alpha }\right)\right|<\frac{\pi }{2},(z\in \mathbb{U})$$

(67)

for some real $\alpha (0\le \alpha <1),$ then

$$Re\left(p\right(z\left)\right)>\alpha ,(z\in \mathbb{U}).$$

(68)

Corollary 9. 

If the function $f\left(z\right)$ in the class $\mathcal{A}$ satisfies

$$\frac{f^{\prime }\left(z\right)-\alpha }{1-\alpha }+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)-\alpha }\prec \frac{1+z}{1-z};(z\in \mathbb{U})$$

(69)

for some real α$(0\le \alpha <1),$ then

$${Ref}^{\prime }\left(z\right)>\alpha ;(z\in \mathbb{U}).$$

(70)

Corollary 10. 

If the function $f\left(z\right)$ in the class $\mathcal{A}$ satisfies

$$\frac{f\left(z\right)}{z}+\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-1\prec \frac{1+z}{1-z};(z\in \mathbb{U})$$

(71)

for some real α$(0\le \alpha <1),$ then

$$Re\left(\frac{f\left(z\right)}{z}\right)>0;(z\in \mathbb{U}).$$

(72)

3. Applications of Miller–Mocanu Lemma

In this section, we would like to apply the Miller–Mocanu lemma [1,6] (also from Jack [7]).

Lemma 3. 

Let $w\left(z\right)$ be analytic in $\mathbb{U}$ with $w\left(0\right)=0.$ Then, if $\left|w\right(z\left)\right|$ attains its maximum value on the circle $\left|z\right|=r<1$ at a point $z_0\in \mathbb{U},$ then we have

$$z_0{w}^{\prime }\left(z_0\right)=mw\left(z_0\right)$$

(73)

and

$$Re\left(1+\frac{z_0{w}^{″}\left(z_0\right)}{w^{\prime }\left(z_0\right)}\right)\ge m$$

(74)

where $m\ge 1.$

Theorem 4. 

If the function $f\left(z\right)$ in the class $\mathcal{A}$ satisfies

$$\frac{f\left(z\right)}{z}\prec \frac{\alpha (1+z)}{\alpha +(2-\alpha )z};(z\in \mathbb{U})$$

(75)

for some real α$(\alpha >1),$ then

$$\left|\frac{f\left(z\right)}{z}-\frac{\alpha }{2}\right|<\frac{\alpha }{2};(z\in \mathbb{U}).$$

(76)

Proof. 

Let us define a function $w\left(z\right)$ using

$$\frac{f\left(z\right)}{z}=\frac{\alpha (1+w(z\left)\right)}{\alpha +(2-\alpha )w\left(z\right)};(z\in \mathbb{U}).$$

(77)

Then, $w\left(z\right)$ is analytic in $\mathbb{U}$ with $w\left(0\right)=0$ and $\left|w\right(z\left)\right|<1(z\in \mathbb{U}).$ Letting

$$g\left(z\right)=\frac{f\left(z\right)}{z},$$

(78)

we see that

$$\left|w\left(z\right)\right|=\left|\frac{\alpha \left(g\right(z)-1)}{\alpha -(2-\alpha )g\left(z\right)}\right|<1;(z\in \mathbb{U}).$$

(79)

It follows from (79) that

$${2\left|g\left(z\right)\right|}^2-\alpha (g\left(z\right)+\overline{g\left(z\right)})<0;(z\in \mathbb{U})$$

(80)

and that

$$\left|g\left(z\right)-\frac{\alpha }{2}\right|<\frac{\alpha }{2};(z\in \mathbb{U}).$$

(81)

□

Next, we have the following theorem.

Theorem 5. 

If the function $f\left(z\right)$ in the class $\mathcal{A}$ satisfies

$$f^{\prime }\left(z\right)\prec \frac{\alpha (1+z)}{\alpha +(2-\alpha )z};(z\in \mathbb{U})$$

(82)

for some real α$(\alpha >1),$ then

$$\left|f^{\prime }\left(z\right)-\frac{\alpha }{2}\right|<\frac{\alpha }{2};(z\in \mathbb{U}).$$

(83)

Proof. 

Considering a function $w\left(z\right)$ such that

$$f^{\prime }\left(z\right)=\frac{\alpha (1+w(z\left)\right)}{\alpha +(2-\alpha )w\left(z\right)};(z\in \mathbb{U}),$$

(84)

we prove the theorem. □

Remark 2. 

The inequality (76) implies that

$$0<Re\left(\frac{f\left(z\right)}{z}\right)<\alpha ;(z\in \mathbb{U})$$

(85)

and the inequality (83) implies that

$$0<{Ref}^{\prime }\left(z\right)<\alpha ;(z\in \mathbb{U}).$$

(86)

The following theorem is our next result.

Theorem 6. 

If the function $f\left(z\right)$ in the class $\mathcal{A}$ satisfies

$$Re\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)<1+\frac{\alpha -1}{2\delta };(z\in \mathbb{U})$$

(87)

for some real α$(1<\alpha \le 2)$ or

$$Re\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)<1+\frac{1}{2\delta (\alpha -1)};(z\in \mathbb{U})$$

(88)

for some real α$(\alpha >2),$ then

$$\left|{\left(\frac{f\left(z\right)}{z}\right)}^{\delta }-\frac{\alpha }{2}\right|<\frac{\alpha }{2};(z\in \mathbb{U})$$

(89)

where $0<\delta \le 1.$

Proof. 

We define a function $w\left(z\right)$ provided by

$${\left(\frac{f\left(z\right)}{z}\right)}^{\delta }=\frac{\alpha (1+w(z\left)\right)}{\alpha +(2-\alpha )w\left(z\right)};(z\in \mathbb{U})$$

(90)

for $0<\delta \le 1.$ Then, $w\left(z\right)$ is analytic in $\mathbb{U}$ with $w\left(0\right)=0.$ This function $w\left(z\right)$ satisfies

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

(91)

Suppose that there exists a point $z_0\in \mathbb{U}$ such that

$${\max }_{\left|z\right|\le |z_0|}\left|w\left(z\right)\right|=\left|w\right(z_0|=1.$$

(92)

Then, Lemma 3 shows us that

$$z_0{w}^{\prime }\left(z_0\right)=mw\left(z_0\right)(m\ge 1)$$

(93)

and $w\left(z_0\right)=e^{i\theta }(0\le \theta <2\pi ).$ It follows from the above that

$$\begin{array}{cc}\hfill \mathrm{Re}\left(\frac{z_0{f}^{\prime }\left(z_0\right)}{f\left(z_0\right)}\right)& =\mathrm{Re}\left(1+\frac{m}{\delta }\left(\frac{e^{i\theta }}{1+e^{i\theta }}-\frac{(2-\alpha )e^{i\theta }}{\alpha +(2-\alpha )e^{i\theta }}\right)\right)\hfill \\ \hfill & =1+\frac{m}{\delta }\left(\frac{1}{2}+\frac{(\alpha -2)(2-\alpha +\alpha cos\theta )}{{\alpha }^2+{(2-\alpha )}^2+2\alpha (2-\alpha )cos\theta }\right).\hfill \end{array}$$

(94)

We consider a function $g\left(t\right)$ provided by

$$g\left(t\right)=\frac{2-\alpha +\alpha t}{{\alpha }^2+{(2-\alpha )}^2+2\alpha (2-\alpha )t}(t=cos\theta ).$$

(95)

It follows from (95) that

$$g^{\prime }\left(t\right)=\frac{4\alpha (\alpha -1)}{{({\alpha }^2+{(2-\alpha )}^2+2\alpha (2-\alpha )t)}^2}>0$$

(96)

for $\alpha >1.$ Since $g\left(t\right)$ is increasing for $t=cos\theta ,$ we have

$$\begin{array}{cc}\hfill \mathrm{Re}\left(\frac{z_0{f}^{\prime }\left(z_0\right)}{f\left(z_0\right)}\right)& \ge 1+\frac{m}{\delta }\left(\frac{1}{2}+\frac{\alpha -2}{2}\right)\ge 1+\frac{\alpha -1}{2\delta }\hfill \end{array}$$

(97)

for $1<\alpha \le 2$ and

$$\begin{array}{cc}\hfill \mathrm{Re}\left(\frac{z_0{f}^{\prime }\left(z_0\right)}{f\left(z_0\right)}\right)& \ge 1+\frac{m}{\delta }\left(\frac{1}{2}-\frac{\alpha -2}{2(\alpha -1)}\right)\ge 1+\frac{1}{2\delta (\alpha -1)}\hfill \end{array}$$

(98)

for $\alpha >2.$ Thus, inequalities (97) and (98) contradict the conditions (87) and (88). Therefore, we say that there is no $w\left(z\right)$ such that $w\left(0\right)=0$ and $\left|w\right(z_0\left)\right|=1$ for $z_0\in \mathbb{U}.$ This implies that $\left|w\right(z\left)\right|<1$ for all $z\in \mathbb{U},$ that is

$$\begin{array}{c}\hfill \left|w\left(z\right)\right|=\left|\frac{\alpha \left({\left(\frac{f\left(z\right)}{z}\right)}^{\delta }-1\right)}{\alpha -(2-\alpha ){\left(\frac{f\left(z\right)}{z}\right)}^{\delta }}\right|<1;(z\in \mathbb{U}).\end{array}$$

(99)

This completes the proof of the theorem. □

Using $\delta =1$ in Theorem 6, we have the following corollary.

Corollary 11. 

If the function $f\left(z\right)$ in the class $\mathcal{A}$ satisfies

$$Re\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)<\frac{\alpha +1}{2};(z\in \mathbb{U})$$

(100)

for some real α$(1<\alpha \le 2)$ or

$$Re\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)<\frac{2\alpha -1}{2(\alpha -1)};(z\in \mathbb{U})$$

(101)

for some real α$(\alpha >2),$ then

$$\left|\frac{f\left(z\right)}{z}-\frac{\alpha }{2}\right|<\frac{\alpha }{2};(z\in \mathbb{U}).$$

(102)

Letting $\delta =\frac{1}{2},$ we have the following corollary.

Corollary 12. 

If the function $f\left(z\right)$ in the class $\mathcal{A}$ satisfies

$$Re\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)<\alpha ;(z\in \mathbb{U})$$

(103)

for some real α$(1<\alpha \le 2)$ or

$$Re\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)<\frac{\alpha }{\alpha -1};(z\in \mathbb{U})$$

(104)

for some real α$(\alpha >2),$ then

$$\left|\sqrt{\frac{f\left(z\right)}{z}}-\frac{\alpha }{2}\right|<\frac{\alpha }{2};(z\in \mathbb{U}).$$

(105)

Theorem 7. 

If the function $f\left(z\right)$ in the class $\mathcal{A}$ satisfies

$$Re\left(\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)<\frac{\alpha -1}{2\delta };(z\in \mathbb{U})$$

(106)

for some real α$(1<\alpha \le 2)$ or

$$Re\left(\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)<\frac{1}{2\delta (\alpha -1)};(z\in \mathbb{U})$$

(107)

for some real α$(\alpha >2),$ then

$$\left|{\left(f^{\prime }\left(z\right)\right)}^{\delta }-\frac{\alpha }{2}\right|<\frac{\alpha }{2};(z\in \mathbb{U})$$

(108)

where $0<\delta \le 1.$

Proof. 

Let us consider a function $w\left(z\right)$ provided by

$${\left(f^{\prime }\left(z\right)\right)}^{\delta }=\frac{\alpha (1+w(z\left)\right)}{\alpha +(2-\alpha )w\left(z\right)};(z\in \mathbb{U})$$

(109)

for $0<\delta \le 1.$ Then, $w\left(z\right)$ is analytic in $\mathbb{U}$ with $w\left(0\right)=0$ and satisfies

$$\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}=\frac{z{w}^{\prime }\left(z\right)}{\delta w\left(z\right)}\left(\frac{w\left(z\right)}{1+w\left(z\right)}-\frac{(2-\alpha )w\left(z\right)}{\alpha +(2-\alpha )w\left(z\right)}\right).$$

(110)

Therefore, applying Lemma 3 as the proof of Theorem 6, we prove the theorem. □

Using $\delta =1,$ we have the following corollary.

Corollary 13. 

If the function $f\left(z\right)$ in the class $\mathcal{A}$ satisfies

$$Re\left(\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)<\frac{\alpha -1}{2};(z\in \mathbb{U})$$

(111)

for some real α$(1<\alpha \le 2)$ or

$$Re\left(\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)<\frac{1}{2(\alpha -1)};(z\in \mathbb{U})$$

(112)

for some real α$(\alpha >2),$ then

$$\left|f^{\prime }\left(z\right)-\frac{\alpha }{2}\right|<\frac{\alpha }{2};(z\in \mathbb{U}).$$

(113)

Example 2. 

We consider a function $f\left(z\right)\in \mathcal{A}$ provided by

$$f\left(z\right)=\frac{\alpha }{2-\alpha }\left(z+\frac{2(1-\alpha )}{2-\alpha }log\left(1+\frac{2-\alpha }{\alpha }z\right)\right).$$

(114)

Then, we see

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

(115)

and

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

(116)

It follows from (116) that

$$\left|f^{\prime }\left(z\right)-\frac{\alpha }{2}\right|<\frac{\alpha }{2};(z\in \mathbb{U}).$$

(117)

On the other hand, $f\left(z\right)$ implies that

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

(118)

Thus, $f\left(z\right)$ satisfies the conditions (111) and (112) of Corollary 13.

Causing $\delta =\frac{1}{2}$ in Theorem 7, we have the following corollary.

Corollary 14. 

If the function $f\left(z\right)$ in the class $\mathcal{A}$ satisfies

$$Re\left(\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)<\alpha -1;(z\in \mathbb{U})$$

(119)

for some real α$(1<\alpha \le 2)$ or

$$Re\left(\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)<\frac{1}{\alpha -1};(z\in \mathbb{U})$$

(120)

for some real α$(\alpha >2),$ then

$$\left|\sqrt{f^{\prime }\left(z\right)}-\frac{\alpha }{2}\right|<\frac{\alpha }{2};(z\in \mathbb{U}).$$

(121)

Further, we obtain the following theorem.

Theorem 8. 

If the function $f\left(z\right)$ in the class $\mathcal{A}$ satisfies

$$Re\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}-\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)<\frac{\alpha -1}{2\delta };(z\in \mathbb{U})$$

(122)

for some real α$(1<\alpha \le 2)$ or

$$Re\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}-\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)<\frac{1}{2\delta (\alpha -1)};(z\in \mathbb{U})$$

(123)

for some real α$(\alpha >2),$ then

$$\left|{\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)}^{\delta }-\frac{\alpha }{2}\right|<\frac{\alpha }{2};(z\in \mathbb{U})$$

(124)

where $0<\delta \le 1.$

Letting $\delta =1,$ we obtain the following corollary.

Corollary 15. 

If the function $f\left(z\right)$ in the class $\mathcal{A}$ satisfies

$$Re\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}-\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)<\frac{\alpha -1}{2};(z\in \mathbb{U})$$

(125)

for some real α$(1<\alpha \le 2)$ or

$$Re\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}-\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)<\frac{1}{2(\alpha -1)};(z\in \mathbb{U})$$

(126)

for some real α$(\alpha >2),$ then

$$\left|\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-\frac{\alpha }{2}\right|<\frac{\alpha }{2};(z\in \mathbb{U}).$$

(127)

Example 3. 

We consider a function $f\left(z\right)\in \mathcal{A}$ provided by

$$f\left(z\right)=z{(\alpha +(2-\alpha )z)}^{\frac{2(\alpha -1)}{2-\alpha }},(\alpha \ne 2).$$

(128)

It follows from (128) that

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

(129)

and

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

(130)

With (130), we know that $f\left(z\right)$ satisfies the inequalities (125) and (126). Furthermore, by (129) we see that $f\left(z\right)$ satisfies the inequality (127).

Letting $\delta =\frac{1}{2}$ in Theorem 8, we have the following corollary.

Corollary 16. 

If the function $f\left(z\right)$ in the class $\mathcal{A}$ satisfies

$$Re\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}-\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)<\alpha -1;(z\in \mathbb{U})$$

(131)

for some real α$(1<\alpha \le 2)$ or

$$Re\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}-\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)<\frac{1}{\alpha -1};(z\in \mathbb{U})$$

(132)

for some real α$(\alpha >2),$ then

$$\left|\sqrt{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}-\frac{\alpha }{2}\right|<\frac{\alpha }{2};(z\in \mathbb{U}).$$

(133)

In addition to our results given above, we can add the following:

In Theorem 3, we prove that if $p\left(z\right)\in {\mathcal{P}}_n$ satisfies the subordination (52), then $p\left(z\right)$ satisfies the inequality (53). We know that

$$\mathrm{Re}\left(\frac{1+z}{1-z}\right)>0;(z\in \mathbb{U})$$

(134)

and

$$\frac{1+e^{i\theta }}{1-e^{i\theta }}=icot\frac{\theta }{2};(0\le \theta <2\pi ).$$

(135)

Furthermore, Equation (59) implies that

$$\mathrm{Re}\left\{\frac{p\left(z_0\right)-\alpha }{1-\alpha }+\frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)-\alpha }\right\}=0;(z\in \mathbb{U}).$$

(136)

Thus, we see that

$$\left|arg{\left\{\frac{p\left(z_0\right)-\alpha }{1-\alpha }+\frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)-\alpha }\right\}}^{\beta }\right|>\left|arg{\left(\frac{1+z}{1-z}\right)}^{\gamma }\right|$$

(137)

for some real $\beta$ and $\gamma$$(0<\beta \le \gamma \le 2).$

With the above comment, we derive the following theorem.

Theorem 9. 

If $p\left(z\right)\in {\mathcal{P}}_n$ satisfies

$${\left(\frac{p\left(z\right)-\alpha }{1-\alpha }+\frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)-\alpha }\right)}^{\beta }\prec {\left(\frac{1+z}{1-z}\right)}^{\gamma },(z\in \mathbb{U})$$

(138)

for some real $\alpha (0\le \alpha <1)$ and for some real β and γ$(0<\beta \le \gamma \le 2),$ then

$$Re\left(p\right(z\left)\right)>\alpha ,(z\in \mathbb{U}).$$

(139)

Corollary 17. 

If the function $p\left(z\right)$ in the class ${\mathcal{P}}_n$ satisfies

$$\left|arg{\left(\frac{p\left(z\right)-\alpha }{1-\alpha }+\frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)-\alpha }\right)}^{\beta }\right|<\frac{\pi }{2}\beta ,(z\in \mathbb{U})$$

(140)

for some real $\alpha (0\le \alpha <1)$ and $\beta (0<\beta \le 2),$ then

$$Re\left(p\right(z\left)\right)>\alpha ,(z\in \mathbb{U}).$$

(141)

Corollary 18. 

If the function $f\left(z\right)$ in the class $\mathcal{A}$ satisfies

$$\left|arg{\left(\frac{f^{\prime }\left(z\right)-\alpha }{1-\alpha }+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)-\alpha }\right)}^{\beta }\right|<\frac{\pi }{2}\beta ;(z\in \mathbb{U})$$

(142)

for some real $\alpha (0\le \alpha <1)$ and $\beta (0<\beta \le 2),$ then

$${Ref}^{\prime }\left(z\right)>\alpha ;(z\in \mathbb{U}).$$

(143)

Example 4. 

We consider a function $f\left(z\right)\in \mathcal{A}$ provided by

$$f\left(z\right)=log\left(\frac{1}{1-z}\right).$$

(144)

Then, we have that

$$\begin{array}{cc}\hfill Re\left(\frac{f^{\prime }\left(z\right)-\alpha }{1-\alpha }+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)-\alpha }\right)& =Re\left(\frac{1}{1-\alpha }\left(\frac{1}{1-z}-\alpha \right)\right)\hfill \\ \hfill & >\frac{1-2\alpha }{1-\alpha }\ge 0;(z\in \mathbb{U})\hfill \end{array}$$

(145)

with $\alpha (0\le \alpha \le \frac{1}{2}).$ This provides

$$\begin{array}{c}\hfill \left|arg{\left(\frac{f^{\prime }\left(z\right)-\alpha }{1-\alpha }+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)-\alpha }\right)}^{\beta }\right|<\frac{\pi }{2}\beta ;(z\in \mathbb{U})\end{array}$$

(146)

with $\alpha (0\le \alpha \le \frac{1}{2})$ and $\beta (0<\beta \le 2).$ Furthermore, we have that

$$Re{f}^{\prime }\left(z\right)=Re\left(\frac{1}{1-z}\right)>\frac{1}{2}\ge \alpha ;(z\in \mathbb{U}).$$

(147)

4. Conclusions

There are many interesting properties of functions $f\left(z\right)$ that are analytic in the open-unit disk concerning subordinations. In this paper, we consider many interesting properties of $f\left(z\right)$ that are analytic in the open-unit disk with subordinations by applying the three lemmas for $f\left(z\right)$ provided by Miller and Mocanu and by Nunokawa. Furthermore, we provide simple examples for our results since we think it is very important to consider examples of the obtained results.