On some sufficient conditions for univalence and starlikeness

Abstract

In this work, the conditions for univalence, starlikeness and convexity are discussed.

MSC:30C45, 30C80.

1 Introduction

We shall consider the set ℋ of all analytic functions in the open unit disc

$$\mathbb{D}=\{z:|z|<1\}$$

on the complex plane ℂ and

$$\mathcal{A}=\{f\in \mathcal{H}:f(z)=z+a_2{z}^2+\cdots \}.$$

The class ${\mathcal{S}}_{\alpha }^{\ast }$ of starlike functions of order $\alpha <1$ may be defined as

$${\mathcal{S}}_{\alpha }^{\ast }=\{f\in \mathcal{A}:\mathfrak{Re}\frac{z{f}^{\prime }(z)}{f(z)}>\alpha ,z\in \mathbb{D}\}.$$

The class ${\mathcal{S}}_{\alpha }^{\ast }$ and the class ${\mathcal{K}}_{\alpha }$ of convex functions of order $\alpha <1$

$$\begin{array}{rcl}{\mathcal{K}}_{\alpha }& :=& \{f\in \mathcal{A}:\mathfrak{Re}(1+\frac{z{f}^{″}(z)}{f^{\prime }(z)})>\alpha ,z\in \mathbb{D}\}\\ =& \{f\in \mathcal{A}:z{f}^{\prime }\in {\mathcal{S}}_{\alpha }^{\ast }\}\end{array}$$

were introduced by Robertson in [1]. If $\alpha \in [0;1)$, then a function in either of these sets is univalent. In particular, we denote by ${\mathcal{S}}_0^{\ast }={\mathcal{S}}^{\ast }$, ${\mathcal{K}}_0=\mathcal{K}$ the classes of starlike and convex functions, respectively.

2 Preliminaries

Lemma 2.1 Let $O=0$, $P=\alpha +ia\alpha$, $Q=x+iax$ and $A\in (0,+\mathrm{\infty })$ be the points on the complex plane, where $0<\alpha \le A/2$, $\alpha <x$ and $-\mathrm{\infty }<a<\mathrm{\infty }$. Then we have

$$\frac{|A-Q|}{|O-Q|}<\frac{|A-P|}{|O-P|}\le \frac{A-\alpha }{\alpha }.$$

(2.1)

Proof For $a=0$, the assertion is obvious. For $a\ne 0$, consider the triangles $OAP$ and $OAQ$, see Figure 1. Both of them have the same angle φ at the point O. Let the first have the angle ψ at the point A and the second have the angle ${\psi }^{\prime }$ at the point A. Then the hypothesis $0<\alpha \le A/2$ implies $cos\phi \le cos\psi$. Further we have

$$cos\phi =\frac{\alpha }{|P|},cos\psi =\frac{A-\alpha }{|A-P|}.$$

This gives the second inequality of the assertion. The first one follows immediately from $sin{\psi }^{\prime }>sin\psi$ and

$$\frac{|A-P|}{|P|}=\frac{sin\phi }{sin\psi },\frac{|A-Q|}{|Q|}=\frac{sin\phi }{sin{\psi }^{\prime }}.$$

 □

Figure 1

The lines and the points.

3 Main result

Theorem 3.1 Let $p(z)=1+{\sum }_{n=k\ge 1}^{\mathrm{\infty }}{c}_n{z}^n$ be analytic in the unit disc $\mathbb{D}$ and α be a positive real number $0<\alpha \le 1/2$. Then suppose that there exists a point $z_0$, $|z_0|<1$ such that

$$\mathfrak{Re}\{p(z)\}>\alpha \mathit{\text{for }}|z|<|z_0|$$

(3.1)

and

$$\mathfrak{Re}\{p(z_0)\}=p(z_0)=\alpha .$$

(3.2)

Then we have

$$\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\le -k(1-\alpha ).$$

(3.3)

Proof Let us put

$$q(z)=\frac{p(z)-1}{p(z)},q(0)=0.$$

Then the function q is analytic in $|z|\le |z_0|<1$ and from the hypothesis of Theorem 3.1 and Lemma 2.1, with $A=1$, we have

$$|q(z)|<\frac{1-\alpha }{\alpha }$$

for $|z|\le |z_0|$ and

$$|q(z_0)|=\frac{1-\alpha }{\alpha }.$$

This shows that $|q(z)|$ takes its maximum at $z=z_0$ on the circle $|z|=|z_0|$. Then from Fukui-Sakaguchi [2] and Jack’s [3] lemmas, there exists a real number $k\ge 1$ such that

$$\begin{array}{rcl}\frac{z_0{q}^{\prime }(z_0)}{q(z_0)}& =& \frac{z_0{p}^{\prime }(z_0)}{p(z_0)-1}-\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\\ =& z_0{p}^{\prime }(z_0)(\frac{1}{\alpha -1}-\frac{1}{\alpha })\\ =& \frac{z_0{p}^{\prime }(z_0)}{\alpha (\alpha -1)}\\ \ge & k.\end{array}$$

This shows that $z_0{p}^{\prime }(z_0)$ is a negative real number and

$$\frac{z_0{p}^{\prime }(z_0)}{\alpha }=\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\le -k(1-\alpha ).$$

This completes the proof of Theorem 3.1. □

Theorem 3.1 is, in a certain sense, the supplement of Nunokawa’s lemma [4]. From Theorem 3.1 we have the following corollaries.

Corollary 3.2 Let $p(z)=1+{\sum }_{n=1}^{\mathrm{\infty }}{c}_n{z}^n$ be analytic in the unit disc $\mathbb{D}$ and α be a positive real number $0<\alpha \le 1/2$. Suppose also that for arbitrary r, $0<r<1$, p satisfies the condition

$$\underset{|z|\le r}{min}\mathfrak{Re}\{p(z)\}=\underset{|z|\le r}{min}|p(z)|$$

(3.4)

and

$$\mathfrak{Re}\{p(z)+\frac{z{p}^{\prime }(z)}{p(z)}\}>2\alpha -1\mathit{\text{for }}|z|<1.$$

(3.5)

Then we have

$$\mathfrak{Re}\{p(z)\}>\alpha \mathit{\text{for }}|z|<1.$$

(3.6)

Proof If there exists a point $z_0$, $|z_0|<1$, such that

$$\mathfrak{Re}\{p(z)\}>\alpha \text{for }|z|<|z_0|$$

and

$$\mathfrak{Re}\{p(z_0)\}=\alpha ,0<\alpha \le 1/2,$$

then from the hypothesis of Corollary 3.2, we have

$$\mathfrak{Re}\{p(z_0)\}=p(z_0)=\alpha .$$

Then from Theorem 3.1, we have

$$\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\le \alpha -1,$$

and therefore we have

$$\mathfrak{Re}\{p(z_0)+\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\}\le 2\alpha -1.$$

This contradicts the hypothesis of Corollary 3.2 and it completes the proof of Corollary 3.2. □

Corollary 3.3 Let $f(z)=z+{\sum }_{n=2}^{\mathrm{\infty }}{a}_n{z}^n$ be analytic in the unit disc $\mathbb{D}$ and α be a positive real number $0<\alpha \le 1/2$. Suppose that for arbitrary r, $0<r<1$, f satisfies the condition

$$\underset{|z|\le r}{min}\mathfrak{Re}\left\{\frac{z{f}^{\prime }(z)}{f(z)}\right\}=\underset{|z|\le r}{min}\left|\frac{z{f}^{\prime }(z)}{f(z)}\right|$$

(3.7)

and

$$\mathfrak{Re}\{1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}\}>2\alpha -1\mathit{\text{for }}|z|<1,$$

(3.8)

where

$$-1<2\alpha -1\le 0.$$

Then we have

$$\mathfrak{Re}\left\{\frac{z{f}^{\prime }(z)}{f(z)}\right\}>\alpha \mathit{\text{for }}|z|<1,$$

(3.9)

or f is starlike of order α.

Proof Putting

$$p(z)=\frac{z{f}^{\prime }(z)}{f(z)},$$

it follows that

$$p(z)+\frac{z{p}^{\prime }(z)}{p(z)}=1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}.$$

Then from Corollary 3.2, we have (3.9). □

Theorem 3.4 Let $p(z)=1+{\sum }_{n=k\ge 1}^{\mathrm{\infty }}{c}_n{z}^n$ be analytic in the unit disc $\mathbb{D}$ and α be a positive real number $1/2<\alpha <1$. Then suppose that there exists a point $z_0\in \mathbb{D}$ such that

$$\mathfrak{Re}\{p(z)\}>\alpha \mathit{\text{for }}|z|<|z_0|$$

(3.10)

and

$$\mathfrak{Re}\{p(z_0)\}=p(z_0)=\alpha .$$

(3.11)

Then we have

$$\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}=\mathfrak{Re}\left\{\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\right\}\le -\frac{k(2-\alpha )}{2}.$$

(3.12)

Proof Let us put

$$q(z)=\frac{2-p(z)}{p(z)}-1,q(0)=0.$$

Then from Lemma 2.1, with $A=2$, we have that $|q(z)+1|$ takes its maximum value at $z=z_0$ on the circle $|z|=|z_0|$ or

$$\underset{|z|=|z_0|}{max}|q(z)+1|=\underset{|z|=|z_0|}{max}\left|\frac{2-p(z)}{p(z)}\right|=\frac{2-\alpha }{\alpha }.$$

Applying Jack [3], Miller-Mocanu [5] and Fukui-Sakaguchi’s [2] lemmas, there exists a real number $m\ge k$ such that

$$\begin{array}{rcl}\frac{z_0{q}^{\prime }(z_0)}{q(z_0)}& =& -\frac{z_0{p}^{\prime }(z_0)}{2-p(z_0)}-\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\\ =& -z_0{p}^{\prime }(z_0)(\frac{1}{2-\alpha }+\frac{1}{\alpha })\\ =& -\frac{2z_0{p}^{\prime }(z_0)}{\alpha (2-\alpha )}\\ =& -\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\left(\frac{2}{2-\alpha }\right)\\ \ge & m\ge k.\end{array}$$

This shows that

$$\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\le -\frac{k(2-\alpha )}{2}.$$

This completes the proof of Theorem 3.4. □

Corollary 3.5 Let $f(z)=z+{\sum }_{n=2}^{\mathrm{\infty }}{a}_n{z}^n$ be analytic in the unit disc $\mathbb{D}$ and α be a positive real number $1/2<\alpha <1$. Suppose that for arbitrary r, $0<r<1$, f satisfies the condition

$$\underset{|z|\le r}{min}\mathfrak{Re}\left\{\frac{z{f}^{\prime }(z)}{f(z)}\right\}=\underset{|z|\le r}{min}\left|\frac{z{f}^{\prime }(z)}{f(z)}\right|$$

(3.13)

and

$$\mathfrak{Re}\{1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}\}>\frac{3\alpha -2}{2}\mathit{\text{for }}|z|<1.$$

(3.14)

Then we have

$$\mathfrak{Re}\left\{\frac{z{f}^{\prime }(z)}{f(z)}\right\}>\alpha \mathit{\text{for }}|z|<1.$$

(3.15)

Proof Applying the same method as in the proof of Corollary 3.3 and in the proof of Lemma 2.1, we can obtain Corollary 3.5. □

Corollary 3.6 Let $F(z)=1/z+{\sum }_{n=1}^{\mathrm{\infty }}{b}_n{z}^n$ be analytic and not vanishing in the punctured unit disc $0<|z|<1$ and let α be a positive real number $0<\alpha <1/2$. Suppose also that for arbitrary r, $0<r<1$, F satisfies the following condition:

$$\underset{|z|\le r}{min}\mathfrak{Re}\{-\frac{z{F}^{\prime }(z)}{F(z)}\}=\underset{|z|\le r}{min}\left|\frac{z{F}^{\prime }(z)}{F(z)}\right|$$

(3.16)

and

$$-\mathfrak{Re}\{1+\frac{z{F}^{″}(z)}{F^{\prime }(z)}\}<2-\alpha \mathit{\text{for }}|z|<1.$$

(3.17)

Then we have

$$\mathfrak{Re}\{-\frac{z{F}^{\prime }(z)}{F(z)}\}>\alpha \mathit{\text{for }}|z|<1.$$

(3.18)

Proof Putting

$$p(z)=-\frac{z{F}^{\prime }(z)}{F(z)}=-\frac{z(-1/z^2+{\sum }_{n=1}^{\mathrm{\infty }}n{b}_n{z}^{n-1})}{1/z+{\sum }_{n=1}^{\mathrm{\infty }}{b}_n{z}^n},$$

then we have

$$p(z)=1+\sum _{n=2}^{\mathrm{\infty }}{c}_n{z}^n,p(0)=1$$

and it follows that

$$p(z)-\frac{z{p}^{\prime }(z)}{p(z)}=-(1+\frac{z{F}^{″}(z)}{F^{\prime }(z)}).$$

If there exists a point $z_0\in \mathbb{D}$ such that

$$\mathfrak{Re}\{p(z)\}>\alpha \text{for }|z|<|z_0|$$

and

$$\mathfrak{Re}\{p(z_0)\}=p(z_0)=\alpha ,$$

then from the hypothesis, we have $p(z_0)=\alpha$. Applying Theorem 3.1, we have

$$\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\le -2(1-\alpha )$$

and therefore we have

$$\begin{array}{rcl}\mathfrak{Re}\{p(z_0)-\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\}& =& -\mathfrak{Re}\{1+\frac{z_0{F}^{″}(z_0)}{F^{\prime }(z_0)}\}\\ \ge & \alpha +2(1-\alpha )\\ =& 2-\alpha .\end{array}$$

This contradicts the hypothesis and therefore we have

$$\mathfrak{Re}\{p(z)\}=\mathfrak{Re}\{-\frac{z{F}^{\prime }(z)}{F(z)}\}>\alpha \text{for }|z|<1,$$

and this shows that F is meromorphic starlike of order α in the punctured unit disc $0<|z|<1$. □

We note the following interesting result which was published in a minor journal and so it was not well known in the public of univalent function theory but is strongly connected with the previous Corollaries 3.3, 3.5 and 3.6.

Lemma 3.7 ([6])

Let $f(z)=z+a_2{z}^2+\cdots$ be an analytic function in $|z|<1$ with $f(z)f^{\prime }(z)/z\ne 0$ in $|z|<1$. Then, for each α, $-1/2<\alpha <0$, there exists a function f which satisfies

$$1+\mathfrak{Re}\frac{z{f}^{″}(z)}{f^{\prime }(z)}>\alpha \mathit{\text{in }}|z|<1,$$

(3.19)

but f is not starlike in $|z|<1$.

In [6] the authors pointed out that the function

$$f(z)=\frac{{(1-z)}^{2\alpha -1}-1}{1-2\alpha }$$

(3.20)

satisfies the above conditions.

Lemma 3.8 Let $p(z)=1+{\sum }_{n=m\ge 2}^{\mathrm{\infty }}{c}_n{z}^n$ be an analytic function in $\mathbb{D}$. Suppose also that there exists a point $z_0\in \mathbb{D}$ such that

$$\mathfrak{Re}\{p(z)\}>0\mathit{\text{for }}|z|<|z_0|$$

and

$$\mathfrak{Re}\{p(z_0)\}=0\mathit{\text{and}}p(z_0)\ne 0.$$

Then we have

$$\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}=ik,$$

where k is a real number and

$$k\ge \frac{m}{2}(a+\frac{1}{a})\ge m\ge 2\mathit{\text{when }}argp(z_0)=\frac{\pi }{2}$$

and

$$k\le -\frac{m}{2}(a+\frac{1}{a})\le -m\le -2\mathit{\text{when }}argp(z_0)=-\frac{\pi }{2},$$

where

$$p(z_0)=\pm ia\mathit{\text{and}}a>0.$$

Proof Let us put

$$\varphi (z)=\frac{1-p(z)}{1+p(z)},|z|<1.$$

Then we have $\varphi (0)={\varphi }^{\prime }(0)=\cdots ={\varphi }^{(m-1)}(0)=0$, $|\varphi (z)|<1$, for $|z|<|z_0|$ and $|\varphi (z_0)|=1$. From [2, 3] and [5], we obtain

$$\frac{z_0{\varphi }^{\prime }(z_0)}{\varphi (z_0)}=-\frac{2z_0{p}^{\prime }(z_0)}{1-p^2(z_0)}=-\frac{2z_0{p}^{\prime }(z_0)}{1+{|p(z_0)|}^2}\ge m.$$

This shows that

$$-z_0{p}^{\prime }(z_0)\ge \frac{m}{2}(1+{|p(z_0)|}^2)$$

and $z_0{p}^{\prime }(z_0)$ is a negative real number. For the case $argp(z_0)=\pi /2$, $p(z_0)=ia$ and $0<a$, we have

$$\mathfrak{Re}\left\{\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\right\}=\mathfrak{Re}\left\{\frac{z_0{p}^{\prime }(z_0)}{ia}\right\}=\mathfrak{Re}\{-\frac{i{z}_0{p}^{\prime }(z_0)}{a}\}=0$$

and

$$\mathfrak{Im}\left\{\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\right\}=\mathfrak{Im}\{-\frac{i{z}_0{p}^{\prime }(z_0)}{a}\}\ge \frac{m}{2}\left(\frac{1+{|p(z_0)|}^2}{a}\right)=\frac{m}{2}(a+\frac{1}{a})\ge m.$$

For the case $argp(z_0)=-\pi /2$, $p(z_0)=-ia$ and $0<a$, applying the same method as above, we have

$$\mathfrak{Re}\left\{\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\right\}=0$$

and

$$\mathfrak{Im}\left\{\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\right\}\le -\frac{m}{2}(a+\frac{1}{a})\le -m.$$

This completes the proof. □

Theorem 3.9 Let $f(z)=z+{\sum }_{n=m+1}^{\mathrm{\infty }}{a}_n{z}^n$ be analytic in the unit disc $\mathbb{D}$. Suppose also that

$$\mathfrak{Re}\{1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}\}<\frac{8+m}{6}\mathit{\text{for }}|z|<1$$

(3.21)

where $1\le m$. If $z{f}^{\prime }(z)/f(z)$ is analytic in $\mathbb{D}$ and omits $4/3$, then f is starlike in the unit disc $\mathbb{D}$.

Proof Let us put

$$\frac{z{f}^{\prime }(z)}{f(z)}=\frac{4p(z)}{1+3p(z)},$$

where $p(0)=1$, $p^{\prime }(0)=p^{″}(0)=\cdots =p^{(m-1)}(0)=0$. Then it follows that

$$1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}=\frac{z{f}^{\prime }(z)}{f(z)}+\frac{z{p}^{\prime }(z)}{p(z)}-\frac{3z{p}^{\prime }(z)}{1+3p(z)}.$$

If there exists a point $z_0\in \mathbb{D}$ such that

$$\mathfrak{Re}\{p(z)\}>0\text{for }|z|<|z_0|$$

and

$$\mathfrak{Re}\{p(z_0)\}=0$$

and $p(z_0)\ne 0$, because $p(z_0)=0$ contradicts the hypothesis (3.21), then from Lemma 3.8, we have

$$\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}=ik,$$

where k is a real number and $m\le |k|$. For the case $argp(z_0)=\pi /2$, $p(z_0)=ia$ and $0<a$, we have

$$\begin{array}{rcl}1+\frac{z_0{f}^{″}(z_0)}{f^{\prime }(z_0)}& =& \frac{z_0{f}^{\prime }(z_0)}{f(z_0)}+\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}-\frac{3z_0{p}^{\prime }(z_0)}{p(z_0)}\frac{p(z_0)}{1+3p(z_0)}\\ =& \frac{4ia}{1+3ia}+ik-3ik\frac{ia}{1+3ia}\\ =& \frac{4ia(1-3ia)}{1+9a^2}+ik+\frac{ak(1-3ia)}{1+9a^2}.\end{array}$$

Therefore, we have

$$\begin{array}{rcl}\mathfrak{Re}\{1+\frac{z_0{f}^{″}(z_0)}{f^{\prime }(z_0)}\}& =& \frac{12a^2}{1+9a^2}+\frac{3ak}{1+9a^2}\\ \ge & \frac{12a^2}{1+9a^2}+\frac{3m(1+a^2)}{2(1+9a^2)}\\ =& \frac{24a^2+3m{a}^2+3m}{2(1+9a^2)}.\end{array}$$

Putting

$$h(x)=\frac{(8+m)x^2+m}{1+9x^2},x>0,$$

we have

$$h^{\prime }(x)=\frac{16(1-m)x}{{(1+9x^2)}^2}<0,x>0.$$

This shows that

$$\mathfrak{Re}\{1+\frac{z_0{f}^{″}(z_0)}{f^{\prime }(z_0)}\}\ge \underset{a\to \mathrm{\infty }}{lim}\frac{24a^2+3m{a}^2+3m}{2(1+9a^2)}=\frac{24+3m}{18}.$$

(3.22)

This contradicts the hypothesis (3.21).

For the case $argp(z_0)=-\pi /2$, $p(z_0)=-ia$ and $0<a$, applying the same method as above, we also have (3.22). This is also a contradiction and therefore it completes the proof of Theorem 3.9. □

Remark 3.10 Singh and Singh obtained in [7] that if $f(z)=z+a_2{z}^2+\cdots$ is analytic in $\mathbb{D}$ and

$$\mathfrak{Re}\{1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}\}<\frac{3}{2}\text{for }|z|<1,$$

(3.23)

then f is starlike in $\mathbb{D}$. Earlier, Ozaki [8] proved the univalence of f in $\mathbb{D}$ under the same assumption (3.23).

For $m=1$, the inequality (3.21) becomes (3.23), so Theorem 3.9 is a generalization of the above result.