Remarks on some starlike functions

Abstract

Let $\mathcal{A}$ be the class of functions that are analytic in the unit disk $\mathbb{D}=\{z\in \mathbb{C}:|z|<1\}$ and normalized by $f(0)=f^{\prime }(0)-1=0$. In this work we investigate conditions under which $|z{f}^{\prime }(z)/f(z)-\delta |<\delta$. Next we also estimate $|Arg\{f^{\prime }(z)/z\}|$, $|Arg\{f(z)/z^2\}|$ and $|Arg\{z{f}^{\prime }(z)/f(z)\}|$ for functions of the form $f(z)=z^2+a_3{z}^3+\cdots$ in the unit disc $|z|<1$, which satisfy $|f^{″}(z)-2|<2$. Furthermore, some geometric consequences of these results are given.

MSC: Primary 30C45; secondary 30C80.

1 Introduction

Let $\mathcal{A}$ be the class of functions that are analytic in the unit disk $\mathbb{D}=\{z\in \mathbb{C}:|z|<1\}$ and normalized by $f(0)=f^{\prime }(0)-1=0$. The subclasses of $\mathcal{A}$ consisting of functions that are univalent in $\mathbb{D}$, starlike with respect to the origin and convex will be denoted by $\mathcal{S}$, ${\mathcal{S}}^{\ast }$ and $\mathcal{C}$, respectively. The class ${\mathcal{S}}_{\alpha }^{\ast }$ of starlike functions of order $\alpha <1$ may be defined as

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

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

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

were introduced by Robertson in [1]. If $\alpha \in [0;1)$, then a function in either of these sets is univalent. The convexity in one direction (it implies the univalence) of functions convex of negative order $-1/2$ was proved by Ozaki [2]. In [3] Pfaltzgraff et al. established that the constant $-1/2$ is, in a certain sense, the best possible. A lot of the other equivalent/sufficient conditions for univalence or for the starlikeness, or more, for the convexity in one direction, one can find in [3]. In this work we consider a similar problem, namely find α, β such that

$$\mathfrak{Re}(1+\frac{z{f}^{″}(z)}{f^{\prime }(z)})<\alpha \Rightarrow |\frac{z{f}^{\prime }(z)}{f(z)}-\beta |<\beta .$$

If $\beta \in (0,1]$, it implies also the starlikeness of f.

2 Preliminaries

The following lemma is a simple generalization of Nunokawa’s lemma [4], which together with the lemma from [5] has a surprising number of important applications in the theory of univalent functions.

Lemma 2.1 [6]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

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

and

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

where $|p(z_0)|=a$.

3 Main results

Theorem 3.1 Assume that $\delta \ge 3/4$ and m is a positive integer such that $m>4\delta -1$. If $f(z)=z+{\sum }_{n=m}^{\mathrm{\infty }}{a}_n{z}^n$, and $z{f}^{\prime }(z)/f(z)$ are analytic in the unit disc $\mathbb{D}$ with $z{f}^{\prime }(z)\ne 2\delta f(z)$, $f^{\prime }(z)\ne 0$, $z\in \mathbb{D}$ and

$$\begin{array}{c}\mathfrak{Re}\{1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}\}\hfill \\ <\{\begin{array}{ll}2\delta +(\delta -1/2)(m-1)& \mathit{\text{for }}\delta \in [3/4,1)\mathit{\text{ and }}m\ge \delta /(1-\delta ),m\in \mathbb{N},\\ \frac{m-1}{2(2\delta -1)}& \mathit{\text{for }}\delta \ge 1\mathit{\text{ and }}m>4\delta -1,m\in \mathbb{N},\\ \frac{m-1}{2(2\delta -1)}& \mathit{\text{for }}\delta \in [3/4,1)\mathit{\text{ and }}4\delta -1<m<\delta /(1-\delta ),m\in \mathbb{N},\end{array}\hfill \end{array}$$

(3.1)

then we have

$$|\frac{z{f}^{\prime }(z)}{f(z)}-\delta |<\delta \mathit{\text{for }}|z|<1.$$

Proof The function $z{f}^{\prime }(z)/f(z)$ is analytic in $\mathbb{D}$, thus we can define the function p by

$$\frac{z{f}^{\prime }(z)}{f(z)}-\delta =\delta \frac{p(z)+1-2\delta }{p(z)-1+2\delta }\text{for }|z|<1,$$

(3.2)

where $p(0)=1$, and $p(z)=1+p_{m-1}{z}^{m-1}+p_m{z}^m+\cdots$ , $z\in \mathbb{D}$.

Then it follows that

$$1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}=\frac{2\delta p(z)}{p(z)-1+2\delta }+\frac{2\delta -1}{p(z)-1+2\delta }\frac{z{p}^{\prime }(z)}{p(z)}.$$

(3.3)

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

$$|\frac{z{f}^{\prime }(z)}{f(z)}-\delta |<\delta \text{for }|z|<|z_0|$$

and

$$|\frac{z_0{f}^{\prime }(z_0)}{f(z_0)}-\delta |=\delta ,$$

then by (3.2)

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

and

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

and $p(z_0)\ne 0$ by (3.3). Then applying Lemma 2.1, we have

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

where

$$k\ge \frac{(m-1)(a^2+1)}{2a}\text{when }Arg\{p(z_0)\}=\frac{\pi }{2}$$

(3.4)

and

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

and where $p(z_0)=\pm ia$ and $0<a$. For the case $Arg\{p(z_0)\}=\pi /2$, $p(z_0)=ia$ and $0<a$ it follows from (3.3) that

$$\begin{array}{c}\mathfrak{Re}\{1+\frac{z_0{f}^{″}(z_0)}{f^{\prime }(z_0)}\}\hfill \\ =\mathfrak{Re}\frac{2\delta ia}{ia-1+2\delta }+\mathfrak{Re}\frac{(2\delta -1)ik}{ia-1+2\delta }\hfill \\ =\frac{2a^2\delta }{a^2+{(2\delta -1)}^2}+\frac{(2\delta -1)ak}{a^2+{(2\delta -1)}^2}\hfill \\ =\frac{2a^2\delta +(2\delta -1)ak}{a^2+{(2\delta -1)}^2}.\hfill \end{array}$$

Therefore, we have from (3.4)

$$\begin{array}{c}\mathfrak{Re}\{1+\frac{z_0{f}^{″}(z_0)}{f^{\prime }(z_0)}\}\hfill \\ \ge \frac{2a^2\delta +(2\delta -1)a\frac{m-1}{2}\frac{a^2+1}{a}}{a^2+{(2\delta -1)}^2}\hfill \\ =\frac{4a^2\delta +(2\delta -1)(m-1)(a^2+1)}{2(a^2+{(2\delta -1)}^2)}\hfill \\ =2\delta +(\delta -1/2)(m-1)+2\delta (2\delta -1)\frac{(m-1)(1-\delta )-(2\delta -1)}{a^2+{(2\delta -1)}^2}\text{for }a>0.\hfill \end{array}$$

In the last expression, the numerator $(m-1)(1-\delta )-(2\delta -1)$ is nonnegative if and only if $\delta \in [3/4,1)$ and $m\ge \delta /(1-\delta )$ but this expression tends to 0⁺ when $a\to \mathrm{\infty }$. Therefore, in this case we have

$$\mathfrak{Re}\{1+\frac{z_0{f}^{″}(z_0)}{f^{\prime }(z_0)}\}\ge 2\delta +(\delta -1/2)(m-1)\text{for }\delta \in [3/4,1)\text{ and }m>\delta /(1-\delta ).$$

(3.5)

Furthermore, the numerator $(m-1)(1-\delta )-(2\delta -1)$ is negative if and only if $\delta \ge 1$ and $m\in \mathbb{N}$ or $\delta \in [3/4,1)$ and $m<\delta /(1-\delta )$. In this case the quotient decreases when $a\to 0^{+}$. Therefore, in this case we have

$$\begin{array}{c}\mathfrak{Re}\{1+\frac{z_0{f}^{″}(z_0)}{f^{\prime }(z_0)}\}\hfill \\ \ge 2\delta +(\delta -1/2)(m-1)+\frac{2\delta \{(m-1)(1-\delta )-(2\delta -1)\}}{2\delta -1}\hfill \\ =\frac{m-1}{2(2\delta -1)}.\hfill \end{array}$$

(3.6)

We have assumed that $m>4\delta -1$ to have the right-hand side in (3.1) greater to 1. So in this case we have

$$4\delta -1<m<\frac{\delta }{1-\delta }\text{for }\delta \in [3/4,1).$$

(3.7)

Therefore, we can write (3.6) in the form

$$\begin{array}{c}\mathfrak{Re}\{1+\frac{z_0{f}^{″}(z_0)}{f^{\prime }(z_0)}\}\hfill \\ \ge \frac{m-1}{2(2\delta -1)}\{\begin{array}{ll}\text{either for}& \delta \ge 1\text{ and }m>4\delta -1,m\in \mathbb{N},\\ \text{or for}& \delta \in [3/4,1)\text{ and }4\delta -1<m<\delta /(1-\delta ).\end{array}\hfill \end{array}$$

(3.8)

Inequalities (3.5) and (3.8) contradict the hypothesis of Theorem 3.1, and therefore we have

$$\mathfrak{Re}\{p(z)\}>0\text{for }|z|<1.$$

(3.9)

Furthermore, from (3.2) and (3.9) we obtain

$$|\frac{z{f}^{\prime }(z)}{f(z)}-\delta |=\left|\delta \frac{p(z)+1-2\delta }{p(z)-1+2\delta }\right|<\delta \text{for }|z|<1.$$

(3.10)

For the case $Arg\{p(z_0)\}=-\pi /2$, $p(z_0)=-ia$ and $0<a$, applying the same method as above, we also have (3.9). Therefore, we get (3.10), which completes the proof of Theorem 3.1. □

Substituting $\delta =1$ in Theorem 3.1 leads to the following corollary.

Corollary 3.2 If $f(z)=z+{\sum }_{n=m}^{\mathrm{\infty }}{a}_n{z}^n$ is analytic in the unit disc $\mathbb{D}$ and

$$\mathfrak{Re}\{1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}\}<\frac{m-1}{2},$$

then we have

$$|\frac{z{f}^{\prime }(z)}{f(z)}-1|<1\mathit{\text{for }}|z|<1.$$

Substituting $\delta =3/4$, $m=3$ in Theorem 3.1 gives the following corollary.

Corollary 3.3 If $f(z)=z+{\sum }_{n=3}^{\mathrm{\infty }}{a}_n{z}^n$ is analytic in the unit disc $\mathbb{D}$ and

$$\mathfrak{Re}\{1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}\}<\frac{13}{8},$$

then we have

$$|\frac{z{f}^{\prime }(z)}{f(z)}-\frac{3}{4}|<\frac{3}{4}\mathit{\text{for }}|z|<1.$$

Substituting $\delta =4/5$, $m=3$ in Theorem 3.1 gives the following corollary.

Corollary 3.4 If $f(z)=z+{\sum }_{n=3}^{\mathrm{\infty }}{a}_n{z}^n$ is analytic in the unit disc $\mathbb{D}$ and

$$\mathfrak{Re}\{1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}\}<\frac{5}{2},$$

then we have

$$|\frac{z{f}^{\prime }(z)}{f(z)}-\frac{4}{5}|<\frac{4}{5}\mathit{\text{for }}|z|<1.$$

As a supplement to the above results recall here the known result [[7], p.61] that if $f(z)=z+{\sum }_{n=2}^{\mathrm{\infty }}{a}_n{z}^n$ is analytic in the unit disc $\mathbb{D}$ and

$$\mathfrak{Re}\{1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}\}\prec R_{1,1}(z)=\frac{1+z}{1-z}+\frac{2z}{1-z^2},$$

then

$$\mathfrak{Re}\frac{z{f}^{\prime }(z)}{f(z)}>0\text{for }|z|<1.$$

Note that the open door function $R_{1,1}(z)$ maps $\mathbb{D}$ onto the complex plane with slits along the half-lines $\mathfrak{Re}\{w\}=0$, and $|\mathfrak{Im}\{w\}|\ge \sqrt{3}$. Next we give the bounds for $|Arg\{z{f}^{\prime }(z)/f(z)\}|$.

Theorem 3.5 Let $f(z)=z^2+{\sum }_{n=3}^{\mathrm{\infty }}{a}_n{z}^n$ be analytic in the unit disc $\mathbb{D}$. If

$$|f^{″}(z)-2|<2\mathit{\text{for }}|z|<1,$$

(3.11)

then

$$|\frac{f^{\prime }(z)}{z}-2|<1\mathit{\text{for }}|z|<1.$$

(3.12)

Proof By the Schwarz lemma we have

$$|f^{″}\left(t{e}^{i\phi }\right)-2|\le 2t,t\in [0,1).$$

Let $z=r{e}^{i\phi }$, $r\in [0,1)$, and let φ be fixed. Using this we obtain

$$\begin{array}{rcl}|\frac{f^{\prime }(z)}{z}-2|& =& \frac{|f^{\prime }(z)-2z|}{|z|}\\ =& \frac{|{\int }_0^z(f^{″}(u)-2)\mathrm{d}u|}{|z|}\\ =& \frac{|{\int }_0^r(f^{″}(t{e}^{i\phi })-2)\mathrm{d}(t{e}^{i\phi })|}{|r{e}^{i\phi }|}\\ =& \frac{|{\int }_0^r{e}^{i\phi }(f^{″}(t{e}^{i\phi })-2)\mathrm{d}t|}{|r{e}^{i\phi }|}\\ \le & \frac{{\int }_0^r|e^{i\phi }(f^{″}(t{e}^{i\phi })-2)|\mathrm{d}t}{|r{e}^{i\phi }|}\\ \le & \frac{{\int }_0^{r}2t\mathrm{d}t}{r}\\ =& \frac{r^2}{r}=r<1.\end{array}$$

Therefore, we obtain (3.12). □

For the function $f(z)=z^3/3+z^2$, condition (3.11) is satisfied while (3.12) becomes $|z|<1$ in the unit disc, which shows that the constant 1 in (3.12) cannot be replaced by a smaller one. A simple geometric observation yields the following corollary.

Corollary 3.6 Let $f(z)=z^2+{\sum }_{n=3}^{\mathrm{\infty }}{a}_n{z}^n$ be analytic in the unit disc $\mathbb{D}$. If

$$|f^{″}(z)-2|<2\mathit{\text{for }}|z|<1,$$

(3.13)

then

$$|Arg\left\{\frac{f^{\prime }(z)}{z}\right\}|<\frac{\pi }{6}\mathit{\text{for }}|z|<1.$$

(3.14)

Using the same method as in the proof of Theorem 3.5, we can obtain the following result.

Theorem 3.7 Let $f(z)=z^2+{\sum }_{n=3}^{\mathrm{\infty }}{a}_n{z}^n$ be analytic in the unit disc $\mathbb{D}$. If

$$|f^{″}(z)-2|<2\mathit{\text{for }}|z|<1,$$

(3.15)

then

$$|\frac{f(z)}{z^2}-1|<\frac{1}{3}\mathit{\text{for }}|z|<1.$$

(3.16)

For the function $f(z)=z^3/3+z^2$, condition (3.15) is satisfied while (3.16) becomes $|z/3|<1/3$ in the unit disc, which shows that the constant $1/3$ in (3.16) cannot be replaced by a smaller one. A simple geometric observation yields the following corollary.

Corollary 3.8 Let $f(z)=z^2+{\sum }_{n=3}^{\mathrm{\infty }}{a}_n{z}^n$ be analytic in the unit disc $\mathbb{D}$. If

$$|f^{″}(z)-2|<2\mathit{\text{for }}|z|<1,$$

(3.17)

then

$$|Arg\left\{\frac{f(z)}{z^2}\right\}|<{sin}^{-1}\frac{1}{3}\mathit{\text{for }}|z|<1.$$

(3.18)

Using Corollaries 3.6 and 3.8 together, we obtain the next one.

Corollary 3.9 Let $f(z)=z^2+{\sum }_{n=3}^{\mathrm{\infty }}{a}_n{z}^n$ be analytic in the unit disc $\mathbb{D}$. If

$$|f^{″}(z)-2|<2\mathit{\text{for }}|z|<1,$$

(3.19)

then

$$|Arg\left\{\frac{z{f}^{\prime }(z)}{f(z)}\right\}|<\frac{\pi }{6}+{sin}^{-1}\frac{1}{3}\approx 0.8634\mathit{\text{for }}|z|<1.$$

(3.20)

Proof From (3.12) and from (3.16), we have

$$\begin{array}{rcl}|Arg\left\{\frac{z{f}^{\prime }(z)}{f(z)}\right\}|& =& |Arg\left\{\frac{f^{\prime }(z)}{z}\frac{z^2}{f(z)}\right\}|\\ \le & |Arg\left\{\frac{f^{\prime }(z)}{z}\right\}|+|Arg\left\{\frac{f(z)}{z^2}\right\}|\\ <& \frac{\pi }{6}+{sin}^{-1}\frac{1}{3}\\ \approx & 0.8634.\end{array}$$

 □

Recall the class ${\mathcal{SS}}^{\ast }(\beta )$ of strongly starlike functions of order β, $0<\beta \le 1$,

$${\mathcal{SS}}^{\ast }(\beta ):=\{f\in \mathcal{A}:|Arg\frac{z{f}^{\prime }(z)}{f(z)}|<\frac{\beta \pi }{2},z\in \mathbb{U}\},$$

which was introduced in [8] and [9]. Therefore, Corollary 3.9 says that if f satisfies the assumptions, then it is 2-valently strongly starlike of order at least 0.8634.