Conditions for starlikeness of the Libera operator

Abstract

Let denote the class of functions f that are analytic in the unit disc and normalized by $f(0)=f^{\prime }(0)-1=0$. In this paper some conditions are determined for starlikeness of the Libera integral operator $F(z)=\frac{2}{z}{\int }_0^{z}f(t)\mathrm{d}t$.

MSC:30C45, 30C80.

1 Introduction

Let ℋ be the class of functions analytic in the unit disk $\mathbb{D}=\{z\in \mathbb{C}:|z|<1\}$, and let us denote by ${\mathcal{A}}_n$ the class of functions $f\in \mathcal{H}$ with the normalization of the form

$$f(z)=z+a_{n+1}{z}^{n+1}+a_{n+2}{z}^{n+2}+\cdots ,z\in \mathbb{D},$$

with ${\mathcal{A}}_1=\mathcal{A}$.

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

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

which was introduced in [1] and [2], and ${\mathcal{SS}}^{\ast }(1)\equiv {\mathcal{S}}^{\ast }$ is the well-known class of starlike functions in . Functions in the class

$$\mathcal{R}(\beta )=\{f\in \mathcal{A}:\mathfrak{Re}\{f^{\prime }(z)\}>\beta ,z\in \mathbb{D}\},$$

where $\beta <1$ are called functions with bounded turning. The Libera transform $L:\mathcal{A}\to \mathcal{A}$, $L[f]=F$, where

$$F(z)=\frac{2}{z}{\int }_0^{z}f(t)\mathrm{d}t,$$

(1.1)

is the Libera integral operator, which has been studied by several authors on different counts. In [3] Mocanu considered the problem of starlikeness of F and proved the following result.

Theorem 1.1 [3]

If $f(z)$ is analytic and $\mathfrak{Re}\{f^{\prime }(z)\}>0$ in the unit disc and if the function F is given in (1.1), then $F\in {\mathcal{S}}^{\ast }$.

This result may be written briefly as follows:

$$L[\mathcal{R}(0)]\subset {\mathcal{S}}^{\ast }={\mathcal{SS}}^{\ast }(1),$$

(1.2)

where $L[\mathcal{R}(0)]=\{L[f]:f\in \mathcal{R}(0)\}$. In 1995 Mocanu [4] improved (1.2) by showing that

$$L[\mathcal{R}(0)]\subset {\mathcal{SS}}^{\ast }(8/9).$$

(1.3)

In 2002 Miller and Mocanu [5] showed that a subcase of this last result can be sharpened to

$$L[\mathcal{R}(0)\cap {\mathcal{A}}_2]\subset {\mathcal{SS}}^{\ast }(2/3).$$

The problem of strongly starlikeness of $L[f]$ for $f\in \mathcal{R}(0)$ was consider also in [6] where it is shown that

$$L[\mathcal{R}(0)\cap {\mathcal{A}}_2]\subset {\mathcal{SS}}^{\ast }(3/5).$$

The above inclusion relationship is equivalent to the following differential implication:

$$f\in {\mathcal{A}}_2\text{and}\mathfrak{Re}\{f^{\prime }(z)\}>0⟹|arg\left\{\frac{z{F}^{\prime }(z)}{F(z)}\right\}|<\frac{3\pi }{10}$$

or equivalently

$$F\in {\mathcal{A}}_2\text{and}\mathfrak{Re}\{F^{\prime }(z)+\frac{1}{2}z{F}^{″}(z)\}>0⟹|arg\left\{\frac{z{F}^{\prime }(z)}{F(z)}\right\}|<\frac{3\pi }{10},$$

where F is given by (1.1).

In [7] Ponnusamy improved (1.2) by showing that

$$L[\mathcal{R}(-\varrho )]\subset {\mathcal{S}}^{\ast },\varrho =0.09032572\dots .$$

(1.4)

On the order of starlikeness of convex functions was considered also in the recent paper [8].

2 Main result

In this paper we go back to the problem of starlikeness of Libera transform. We need the following lemmas.

Lemma 2.1 [[9], p.73]

Let n be a positive integer, $\lambda >0$ and let ${\beta }_0={\beta }_0(\lambda ,n)$ be the positive root of the equation

$$\beta \pi =3\pi /2-{tan}^{-1}(n\lambda \beta ).$$

(2.1)

In addition, let

$$\alpha =\alpha (\beta ,\lambda ,n)=\beta +(2/\pi ){tan}^{-1}(n\lambda \beta )$$

(2.2)

for $0<\beta \le {\beta }_0$. If $p(z)=1+p_n{z}^n+p_{n+1}{z}^{n+1}+\cdots$ is analytic in , then

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

(2.3)

implies the following subordination:

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

(2.4)

If in Lemma 2.1 we put $n=1$, $\lambda =1/2$, then the solution ${\beta }_0$ of (2.1) satisfies ${\beta }_0>1$, so we may take $\beta =1$, which gives $\pi \alpha /2=\pi /2+{tan}^{-1}(1/2)=2.03\dots$ .

Corollary 2.2 Assume that $f(z)\in {\mathcal{A}}_1$. If

$$|arg\{F^{\prime }(z)+(1/2)z{F}^{″}(z)\}|<\pi /2+{tan}^{-1}(1/2)=2.03\dots ,z\in \mathbb{D},$$

then

$$\mathfrak{Re}\{F^{\prime }(z)\}>0,z\in \mathbb{D}.$$

Note that if $F(z)\in {\mathcal{A}}_2$, then a sufficient condition for $F\in \mathcal{R}(0)$ is $|arg\{f^{\prime }(z)\}|<3\pi /4=2.356\dots$ ; see [[5], p.96].

Lemma 2.3 [10]

Let $p(z)$ be of the form

$$p(z)=1+\sum _{n=m\ge 1}^{\mathrm{\infty }}{a}_n{z}^n,a_m\ne 0(z\in \mathbb{D}),$$

(2.5)

with $p(z)\ne 0$ in . If there exists a point $z_0$, $|z_0|<1$, such that

$$|arg\{p(z)\}|<\pi \alpha /2\mathit{\text{in }}|z|<|z_0|\mathit{\text{and}}|arg\{p(z_0)\}|=\pi \alpha /2$$

for some $\alpha >0$, then we have

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

where

$$k\ge m(a^2+1)/(2a),\mathit{\text{when }}arg\{p(z_0)\}=\pi \alpha /2$$

(2.6)

and

$$k\le -m(a^2+1)/(2a),\mathit{\text{when }}arg\{p(z_0)\}=-\pi \alpha /2,$$

(2.7)

where

$${\{p(z_0)\}}^{1/\alpha }=\pm ia,a>0.$$

Lemma 2.4 [[9], p.75], [11]

Let $p(z)=1+{\sum }_{n=1}^{\mathrm{\infty }}{a}_n{z}^n$ be analytic in the unit disc . If

$$p(z)+z{p}^{\prime }(z)\prec \frac{1+z}{1-z},z\in \mathbb{D},$$

then

$$p(z)\prec q(z)=\frac{2}{z}log\frac{1}{1-z}-1$$

and

$$|arg\{p(z)\}|<{\theta }_0=\underset{|z|=1}{max}|arg\{q(z)\}|=0.9110\dots ,z\in \mathbb{D},$$

(2.8)

where ${\theta }_0$ lies between 0.911621904 and 0.911621907.

Theorem 2.5 Let $q(z)$ be analytic in and suppose that

$$|arg\{q(z)\}|<\frac{\beta \pi }{2},z\in \mathbb{D}$$

for some $\beta \in (0,1]$. If $p(z)$ is analytic and $p(z)\ne 0$ in with $p(0)=1$ and such that

$$|arg\{q(z)(z{p}^{\prime }(z)+p^2(z)+p(z))\}|<{tan}^{-1}\beta ,z\in \mathbb{D},$$

(2.9)

then we have

$$|arg\{p(z)\}|<\frac{\beta \pi }{2},z\in \mathbb{D}.$$

Proof If there exists a point $z_0$, $|z_0|<1$, for which

$$|arg\{p(z)\}|<\pi \beta /2(|z|<|z_0|)$$

and

$$|arg\{p(z_0)\}|=\pi \beta /2,p(z_0)={(\pm ia)}^{\beta },$$

then from Nunokawa’s Lemma 2.3, we have

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

where

$$k\ge \frac{a^2+1}{2a}\ge 1,\text{when }arg\{p(z_0)\}=\pi \beta /2$$

and

$$k\le -\frac{a^2+1}{2a}\le -1,\text{when }arg\{p(z_0)\}=-\pi \beta /2.$$

For the case $arg\{p(z_0)\}=\beta \pi /2$, we have

$$\begin{array}{c}|arg\{q(z_0)[z_0{p}^{\prime }(z_0)+p^2(z_0)+p(z_0)]\}|\hfill \\ =|arg\{q(z_0)p(z_0)[1+p(z_0)+z_0{p}^{\prime }(z_0)/p(z_0)]\}|\hfill \\ =|\frac{\pi \beta }{2}+arg\{q(z_0)\}+arg\{1+p(z_0)+z_0{p}^{\prime }(z_0)/p(z_0)\}|\hfill \\ =|\frac{\pi \beta }{2}+arg\{q(z_0)\}+{tan}^{-1}\left\{\frac{\beta k+a^{\beta }sin(\pi \beta /2)}{1+a^{\beta }cos(\pi \beta /2)}\right\}|,\hfill \end{array}$$

(2.10)

where $p(z_0)={(ia)}^{\beta }$, $0<a$ and

$$k\ge \frac{a^2+1}{2a}\ge 1.$$

Let us put

$$g(a)=\frac{k\beta +a^{\beta }sin(\pi \beta /2)}{1+a^{\beta }cos(\pi \beta /2)},0<a,$$

then it is easy to see that

$$g(a)\ge \frac{\beta +a^{\beta }sin(\pi \beta /2)}{1+a^{\beta }cos(\pi \beta /2)},0<a.$$

(2.11)

Putting

$$h(x)=\frac{\beta +xsin(\pi \beta /2)}{1+xcos(\pi \beta /2)},0\le x,$$

we have

$$h^{\prime }(x)=\frac{sin(\pi \beta /2)-\beta cos(\pi \beta /2)}{{(1+xcos(\pi \beta /2))}^2}>0,0\le x,$$

because $tan(\pi \beta /2)>\beta$. Therefore, for $x>0$ we get $h(x)>h(0)=\beta$, so from (2.11) we have

$$g(a)>\beta ,$$

and so

$${tan}^{-1}\left\{\frac{k\beta +a^{\beta }sin(\pi \beta /2)}{1+a^{\beta }cos(\pi \beta /2)}\right\}>{tan}^{-1}\beta ,0<a.$$

Therefore, we have the following inequality from (2.10):

$$\begin{array}{c}|arg\{q(z_0)[z_0{p}^{\prime }(z_0)+p^2(z_0)+p(z_0)]\}|\hfill \\ \ge \frac{\pi \beta }{2}+{tan}^{-1}\frac{k+a^{\beta }sin(\pi \beta /2)}{1+a^{\beta }cos(\pi \beta /2)}-|arg\{q(z_0)\}|\hfill \end{array}$$

(2.12)

$$>{tan}^{-1}\beta .$$

(2.13)

This contradicts the hypothesis and for the case $arg\{p(z_0)\}=-\beta \pi /2$, applying the same method as above, we also have (2.12). This is also a contradiction and it completes the proof. □

Corollary 2.6 Assume that

$$|arg\{f^{\prime }(z)\}|<{tan}^{-1}\beta ,z\in \mathbb{D}$$

(2.14)

and

$$|arg\{F(z)/z\}|<\frac{\beta \pi }{2},z\in \mathbb{D}$$

(2.15)

for some $\beta \in (0,1]$, where $F(z)$ is given in (1.1). Then we have

$$|arg\left\{\frac{z{F}^{\prime }(z)}{F(z)}\right\}|<\frac{\beta \pi }{2},z\in \mathbb{D},$$

hence $F(z)$ is strongly starlike of order β.

Proof If we set

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

then

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

If we let $q(z)=F(z)/z$, then by (2.14) and (2.15) the assumptions of Theorem 2.5 are satisfied. Therefore,

$$|arg\{p(z)\}|<\frac{\beta \pi }{2},z\in \mathbb{D}.$$

 □

Theorem 2.7 Let $q(z)$ be analytic in , with $q(0)=1$ and satisfy

$$\mathfrak{Re}\{z{q}^{\prime }(z)+q(z)\}>0,z\in \mathbb{D}.$$

If $p(z)$ is analytic in , with $p(0)=1$ and if

$$|arg\{q(z)(z{p}^{\prime }(z)+p^2(z)+p(z))\}|<\frac{5\pi }{6}-{\theta }_0=1.706\dots ,z\in \mathbb{D},$$

where ${\theta }_0$ is given in (2.8), then we have

$$\mathfrak{Re}\{p(z)\}>0,z\in \mathbb{D}.$$

Proof By Lemma 2.4, we have

$$|arg\{q(z)\}|<{\theta }_0=0.911\dots ,z\in \mathbb{D}.$$

(2.16)

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

$$|arg\{p(z)\}|<\pi /2(|z|<|z_0|)$$

and

$$|arg\{p(z_0)\}|=\pi /2,p(z_0)=\pm ia,0<a,$$

then from Nunokawa’s Lemma 2.3, we have

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

where

$$k\ge \frac{a^2+1}{2a}\ge 1,\text{when }arg\{p(z_0)\}=\pi /2$$

and

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

For the case $arg\{p(z_0)\}=\pi /2$, we have

$$\begin{array}{rcl}arg\{1+ia+ik\}& \ge & arg\{1+ia+i\frac{a^2+1}{2a}\}\\ =& {tan}^{-1}\frac{\mathfrak{Im}\{1+ia+i\frac{a^2+1}{2a}\}}{\mathfrak{Re}\{1+ia+i\frac{a^2+1}{2a}\}}\\ =& {tan}^{-1}\left\{\frac{3a^2+1}{2a}\right\}\\ \ge & {tan}^{-1}\{\sqrt{3}\}\\ =& \frac{\pi }{3}.\end{array}$$

Therefore, for the case $arg\{p(z_0)\}=\pi /2$, we have

$$\frac{\pi }{3}\le arg\{1+ia+ik\}<\frac{\pi }{2}.$$

Moreover, by (2.16)

$$arg\{q(z_0)\}<{\theta }_0.$$

Therefore, we can write

$$\begin{array}{c}|arg\{q(z_0)(z_0{p}^{\prime }(z_0)+p^2(z_0)+p(z_0))\}|\hfill \\ =|arg\{p(z_0)[1+p(z_0)+z_0{p}^{\prime }(z_0)/p(z_0)]q(z_0)\}|\hfill \\ \ge |arg\{p(z_0)(1+ia+ik)\}|-|arg\{q(z_0)\}|\hfill \\ \ge \frac{\pi }{2}+\frac{\pi }{3}-|arg\{q(z_0)\}|\hfill \\ \ge \frac{5\pi }{6}-{\theta }_0.\hfill \end{array}$$

(2.17)

This contradicts the hypothesis and for the case $arg\{p(z_0)\}=-\pi /2$, applying the same method as above, we have

$$|arg\{q(z_0)(z_0{p}^{\prime }(z_0)+p^2(z_0)+p(z_0))\}|\ge \frac{5\pi }{6}-{\theta }_0.$$

This is also a contradiction and it completes the proof. □

Corollary 2.8 Assume that

$$|arg\{f^{\prime }(z)\}|<\frac{5\pi }{6}-{\theta }_0=1.706\dots ,z\in \mathbb{D},$$

(2.18)

then we have

$$\mathfrak{Re}\left\{\frac{z{F}^{\prime }(z)}{F(z)}\right\}>0,z\in \mathbb{D},$$

(2.19)

where $F(z)$ is Libera integral given in (1.1).

Proof Because

$$f^{\prime }(z)=F^{\prime }(z)+\frac{1}{2}z{F}^{″}(z),$$

by Corollary 2.2 and by (2.18) we obtain

$$\mathfrak{Re}\{F^{\prime }(z)\}>0,z\in \mathbb{D}.$$

Therefore, if we let $q(z)=F(z)/z$, then

$$\mathfrak{Re}\{z{q}^{\prime }(z)+q(z)\}=\mathfrak{Re}\{F^{\prime }(z)\}>0,z\in \mathbb{D}.$$

If we set

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

then

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

The assumptions of Theorem 2.7 are satisfied. Therefore, (2.19) holds. □

Corollary 2.8 is an extension of Mocanu’s result (1.2) from the paper [3] because in (2.18) we have $|arg\{f^{\prime }(z)\}|<1.706\dots$ , while in (1.2) we have the stronger assumption that $|arg\{f^{\prime }(z)\}|<\pi /2=1.57\dots$ .