On some differential inequalities in the unit disk with applications

Abstract

In this paper we obtain a number of interesting relations associated with some differential inequalities in the open unit disk, $\mathbb{U}=\{z:|z|<1\}$. Some applications of the main results are also obtained.

MSC:30C45, 30C80.

1 Introduction

Let A denote the class of functions of the form

$$f(z)=z+\sum _{n=2}^{\mathrm{\infty }}{a}_n{z}^n$$

(1.1)

which are analytic in the unit disc $\mathbb{U}=\{z:|z|<1\}$. Also, we denote by K the class of functions $f(z)\in A$ that are convex in .

A function $f(z)$ in the class A is said to be in the class $S^{\ast }(\alpha )$ of starlike functions of order α ($0\le \alpha <1$) if it satisfies

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

(1.2)

for some α ($0\le \alpha <1$). Also, we write $S(0)=S^{\ast }$, the class of starlike functions in .

A function $f(z)\in A$ is in $S^{\lambda }$ ($|\lambda |<\frac{\pi }{2}$), the class of λ-spiral-like functions, if it satisfies

$$Re\left(e^{i\lambda }\frac{z{f}^{\prime }(z)}{f(z)}\right)>0(z\in \mathbb{U}).$$

(1.3)

Definition 1.1 Let $f(z)$ and $F(z)$ be analytic functions. The function $f(z)$ is said to be subordinate to $F(z)$, written $f(z)\prec F(z)$, if there exists a function $w(z)$ analytic in , with $w(0)=0$ and $|w(z)|\le 1$, and such that $f(z)=F(w(z))$. If $F(z)$ is univalent, then $f(z)\prec F(z)$ if and only if $f(0)=F(0)$ and $f(\mathbb{U})\subset F(\mathbb{U})$.

Let be the set of analytic functions $q(z)$ injective on $\overline{\mathbb{U}}\mathrm{\setminus }E(q)$, where

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

and $q^{\prime }(\zeta )\ne 0$ for $\zeta \in \partial \mathbb{U}\mathrm{\setminus }E(q)$. Further, let ${\mathbb{D}}_a=\{q(z)\in \mathbb{D}:q(0)=a\}$.

In this paper we obtain some interesting relations associated with some differential inequalities in . These relations extend and generalize the Carathéodory functions in which have been studied by many authors e.g., see [1–14].

2 Main results

To prove our results, we need the following lemma due to Miller and Mocanu [[15], p.24].

Lemma 2.1 Let $q(z)\in {\mathbb{D}}_a$ and let

$$p(z)=b+b_n{z}^n+\cdots$$

be analytic in with $p(z)\ne b$. If $p(z)\nprec q(z)$, then there exist points $z_0\in \mathbb{U}$ and ${\zeta }_0\in \partial \mathbb{U}\mathrm{\setminus }E(q)$ and on $m\ge n\ge 1$ for which

• (i) $p(z_0)=q({\zeta }_0)$,

• (ii) $z_0{p}^{\prime }(z_0)=m{\zeta }_0{q}^{\prime }({\zeta }_0)$.

Theorem 2.1 Let

$$P:\mathbb{U}\to \mathbb{C}$$

with

$$Re(\overline{a}P(z))>0(a\in \mathbb{C}).$$

If p is a function analytic in with $p(0)=1$ and

$$Re(p(z)+P(z)z{p}^{\prime }(z))>\frac{E}{2|a{|}^{2}Re(\overline{a}P(z))},$$

(2.1)

then

$$Re(ap(z))>\alpha ,$$

where

$$\begin{array}{rcl}E& =& -(Re(a)-\alpha ){(Re(\overline{a}P(z)))}^2\\ +2Re(\overline{a}P(z))[{(Im(a))}^2+2\alpha Re(a)]\\ +(Re(a)-\alpha ){(Im(a))}^2,\end{array}$$

(2.2)

with $Re(a)>\alpha$.

Proof Let us define $q(z)$ and $h(z)$ as follows:

$$q(z)=ap(z)$$

and

$$h(z)=\frac{a-(2\alpha -\overline{a})z}{1-z}(Re(a)>\alpha ).$$

The functions q and h are analytic in with $q(0)=h(0)=a\in \mathbb{C}$ with

$$h(\mathbb{U})=\{w:Re(w)>\alpha \}.$$

Now, we suppose that $q(z)\nprec h(z)$. Therefore, by using Lemma 2.1, there exist points

$$z_0\in \mathbb{U}\text{and}{\zeta }_0\in \partial \mathbb{U}\mathrm{\setminus }\{1\}$$

such that $q(z_0)=h({\zeta }_0)$ and $z_0{q}^{\prime }(z_0)=m{\zeta }_0{h}^{\prime }({\zeta }_0)$, $m\ge n\ge 1$.

We note that

$${\zeta }_0=h^{-1}(q(z_0))=\frac{q(z_0)-a}{q(z_0)-(2\alpha -\overline{a})}$$

(2.3)

and

$${\zeta }_0{h}^{\prime }({\zeta }_0)=\frac{-|q(z_0)-a{|}^2}{2Re(a-q(z_0))}.$$

(2.4)

We have $h({\zeta }_0)=\alpha +\rho i$ ($\alpha ,\rho \in \mathbb{R}$), therefore

$$\begin{array}{c}Re(p(z_0)+P(z_0)z_0{p}^{\prime }(z_0))\hfill \\ =Re(\frac{1}{a}h({\zeta }_0)+\frac{1}{a}P(z_0)m{\zeta }_0{h}^{\prime }({\zeta }_0))\hfill \\ =Re\left(\frac{\alpha +\rho i}{a}\right)-m\frac{|\alpha +\rho i-a{|}^2}{2Re(a-\alpha )}Re\left(\frac{P(z_0)}{a}\right)\hfill \\ \le Re\left(\frac{\alpha +\rho i}{a}\right)-\frac{|\alpha +\rho i-a{|}^2}{2Re(a-\alpha )}Re\left(\frac{P(z_0)}{a}\right)\hfill \\ =A{\rho }^2+B\rho +C\hfill \\ =g(\rho ),\hfill \end{array}$$

(2.5)

where

$$\begin{array}{c}A=-\frac{Re(\overline{a}P(z_0))}{2|a{|}^{2}Re(a-\alpha )},\hfill \\ B=\frac{Im(a)}{|a{|}^2}(1+\frac{Re(\overline{a}P(z_0))}{Re(a)-\alpha })\hfill \end{array}$$

and

$$C=\frac{1}{|a{|}^2}(\alpha Re(a)-\frac{{\alpha }^2+|a{|}^2-2\alpha Re(a)Re(\overline{a}P(z_0))}{2(Re(a)-\alpha )}).$$

We can see that the function $g(\rho )$ in (2.5) takes the maximum value at ${\rho }_1$ given by

$${\rho }_1=Im(a)(1+\frac{Re(a)-\alpha }{Re(\overline{a}P(z_0))}).$$

Hence, we have

$$\begin{array}{c}Re(p(z_0)+P(z_0)z{p}^{\prime }(z_0))\hfill \\ \le g({\rho }_1)\hfill \\ =\frac{E}{2|a{|}^{2}Re(\overline{a}P(z))},\hfill \end{array}$$

where E is defined by (2.2). This is in contradiction to (2.1). Then we obtain $Re(ap(z))>\alpha$. □

Theorem 2.2 Let $p(z)$ a nonzero analytic function in with $p(0)=1$. If

$$|p(z)+\frac{z{p}^{\prime }(z)}{p(z)}-1|<\frac{3Re(a-\alpha )}{2|a|}|p(z)|,$$

(2.6)

then

$$Re\left(\frac{a}{p(z)}\right)>\alpha ,$$

where $Re(a)>\alpha$.

Proof Let us define both $q(z)$ and $h(z)$ as follows:

$$q(z)=a/p(z)$$

and

$$h(z)=\frac{a-(2\alpha -\overline{a})z}{1-z}(Re(a)>\alpha ).$$

The functions q and h are analytic in with $q(0)=h(0)=a\in \mathbb{C}$ with

$$h(\mathbb{U})=\{w:Re(w)>\alpha \}.$$

Now, we suppose that $q(z)\nprec h(z)$. Therefore, by using Lemma 2.1, there exist points

$$z_0\in \mathbb{U}\text{and}{\zeta }_0\in \partial \mathbb{U}\mathrm{\setminus }\{1\}$$

such that $q(z_0)=h({\zeta }_0)$ and $z_0{q}^{\prime }(z_0)=m{\zeta }_0{h}^{\prime }({\zeta }_0)$, $m\ge n\ge 1$.

We note that

$${\zeta }_0{h}^{\prime }({\zeta }_0)=\frac{-|q(z_0)-a{|}^2}{2Re(a-q(z_0))}.$$

(2.7)

We have $h({\zeta }_0)=\alpha +\rho i$ ($\rho \in \mathbb{R}$); therefore,

$$\begin{array}{rcl}\frac{|p(z_0)+\frac{z{p}^{\prime }(z_0)}{p(z_0)}-1|}{|p(z_0)|}& =& |\frac{\alpha +\rho i}{a}-\frac{m}{a}\frac{|a-\alpha -i\rho {|}^2}{2Re(a-\alpha )}-1|\\ \ge & \frac{1}{|a|}|\frac{m|a-\alpha -i\rho {|}^2}{2Re(a-\alpha )}+Re(a-\alpha )|\\ \ge & \frac{1}{|a|}(\frac{|a-\alpha -i\rho {|}^2}{2Re(a-\alpha )}+Re(a-\alpha ))\\ \ge & \frac{1}{2|a|Re(a-\alpha )}(3{(Re(a-\alpha ))}^2+{(Im(a)-\rho )}^2)\\ \ge & \frac{3Re(a-\alpha )}{2|a|}.\end{array}$$

This is in contradiction to (2.6). Then we obtain $Re(\frac{a}{p(z)})>\alpha$. □

3 Applications and examples

Putting $P(z)=\beta$ ($\beta >0$; real) in Theorem 2.1 we have the following corollary.

Corollary 3.1 If p is a function analytic in with $p(0)=1$ and

$$Re(p(z)+\beta z{p}^{\prime }(z))>\frac{E}{2\beta |a{|}^{2}Re(a)},$$

then

$$Re(ap(z))>\alpha ,$$

where

$$E=-(Re(a)-\alpha ){\beta }^2{(Re(a))}^2+2\beta Re(a)[{(Im(a))}^2+2\alpha Re(a)]+(Re(a)-\alpha ){(Im(a))}^2,$$

with $Re(a)>\alpha$ ($\alpha \ge 0$).

Putting $\beta =1$ in Corollary 3.1, we obtain the following corollary.

Corollary 3.2 If p is a function analytic in with $p(0)=1$ and

$$Re(p(z)+z{p}^{\prime }(z))>\frac{1}{2Re(a)}(3Re(a)-\alpha )-\frac{Re(a)}{|a{|}^2}(2Re(a)-3\alpha ),$$

then

$$Re(ap(z))>\alpha ,$$

with $Re(a)>\alpha$ ($\alpha \ge 0$).

Corollary 3.3 Let $f(z)\in A$, ${(g(z))}^a\in S^{\ast }$ and

$$Re\left(\frac{f^{\prime }(z)}{g^{\prime }(z)}\right)>\frac{E}{2|a{|}^{2}Re(\overline{a}\frac{g(z)}{z{g}^{\prime }(z)})},$$

then

$$Re\left(a\frac{f(z)}{g(z)}\right)>\alpha ,$$

where $Re(a)>\alpha$ ($\alpha \ge 0$) and E is defined by (2.2) with $P(z)=\frac{g(z)}{z{g}^{\prime }(z)}$.

Proof Putting $p(z)=\frac{f(z)}{g(z)}$ and $P(z)=\frac{g(z)}{z{g}^{\prime }(z)}$ in Theorem 2.1, we have

$$Re(p(z)+P(z)z{p}^{\prime }(z))=Re\left(\frac{f^{\prime }(z)}{g^{\prime }(z)}\right).$$

Since ${(g(z))}^a\in S^{\ast }$, which gives $Re(a\frac{z{g}^{\prime }(z)}{g(z)})>0$, therefore, $Re(\overline{a}P(z))>0$. This completes the proof of the corollary. □

Example 3.1 Let $f(z)\in A$ and

$$Re(f^{\prime }(z))>\frac{1}{2Re(a)}(3Re(a)-\alpha )-\frac{Re(a)}{|a{|}^2}(2Re(a)-3\alpha ),$$

then

$$Re\left(a\frac{f(z)}{z}\right)>\alpha ,$$

where $Re(a)>\alpha$.

Example 3.2 Let $f(z)\in A$ and

$$Re\left((2+\frac{z{f}^{″}(z)}{f^{\prime }(z)}-\frac{z{f}^{\prime }(z)}{f(z)})\frac{z{f}^{\prime }(z)}{f(z)}\right)>\frac{1}{2Re(a)}(3Re(a)-\alpha )-\frac{Re(a)}{|a{|}^2}(2Re(a)-3\alpha ),$$

then

$$Re\left(a\frac{z{f}^{\prime }(z)}{f(z)}\right)>\alpha ,$$

where $Re(a)>\alpha$.

• (1) Putting $a=e^{i\lambda }$ ($|\lambda |<\frac{\pi }{2}$) and $\alpha =0$ in Theorem 2.1, we have Theorem 1 due to Kim and Cho [4].

• (2) Putting $a=e^{i\lambda }$ ($|\lambda |<\frac{\pi }{2}$), $P(z)=\beta$ ($\beta >0$; real) and $\alpha =0$ in Theorem 2.1, we have Corollary 1 due to Kim and Cho [4].

• (3) Putting $a=\alpha =0$ and $P(z)=1$ in Theorem 2.1, we have the result due to Nunokawa et al. [16].

• (4) Putting $a=e^{i\lambda }$ ($|\lambda |<\frac{\pi }{2}$), $P(z)=1$ and $\alpha =0$ in Theorem 2.1, we have Corollary 2 due to Kim and Cho [4].

Putting $p(z)=\frac{z{f}^{\prime }(z)}{f(z)}$ in Theorem 2.2, we have the following corollary.

Corollary 3.4 Let $p(z)$ a nonzero analytic function in U with $p(0)=1$. If

$$\left|\frac{z{f}^{″}(z)}{f^{\prime }(z)}\right|<\frac{3Re(a-\alpha )}{2|a|}\left|\frac{z{f}^{\prime }(z)}{f(z)}\right|,$$

then

$$Re\left(\frac{1}{a}\frac{z{f}^{\prime }(z)}{f(z)}\right)>\alpha ,$$

where $Re(a)>\alpha$.

Remark

• (1) Putting $a=1$ and $\alpha =0$ in Corollary 3.4, we have the result due to Attiya and Nasr [1].

• (2) Putting $a=1$ and $\alpha =0$ in Corollary 3.4, we have the result due to Kim and Cho [4].