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# Conditions for starlikeness of the Libera operator

Journal of Inequalities and Applications20142014:135

https://doi.org/10.1186/1029-242X-2014-135

• Received: 31 December 2013
• Accepted: 7 March 2014
• Published:

## Abstract

Let denote the class of functions f that are analytic in the unit disc and normalized by $f\left(0\right)={f}^{\prime }\left(0\right)-1=0$. In this paper some conditions are determined for starlikeness of the Libera integral operator $F\left(z\right)=\frac{2}{z}{\int }_{0}^{z}f\left(t\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t$.

MSC:30C45, 30C80.

## Keywords

• Nunokawa’s lemma
• starlike functions
• strongly starlike functions

## 1 Introduction

Let be the class of functions analytic in the unit disk $\mathbb{D}=\left\{z\in \mathbb{C}:|z|<1\right\}$, and let us denote by ${\mathcal{A}}_{n}$ the class of functions $f\in \mathcal{H}$ with the normalization of the form
$f\left(z\right)=z+{a}_{n+1}{z}^{n+1}+{a}_{n+2}{z}^{n+2}+\cdots ,\phantom{\rule{1em}{0ex}}z\in \mathbb{D},$

with ${\mathcal{A}}_{1}=\mathcal{A}$.

Let ${\mathcal{SS}}^{\ast }\left(\beta \right)$ denote the class of strongly starlike functions of order β, $0<\beta \le 1$,
${\mathcal{SS}}^{\ast }\left(\beta \right)=\left\{f\in \mathcal{A}:|arg\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}|<\frac{\beta \pi }{2},z\in \mathbb{D}\right\},$
which was introduced in  and , and ${\mathcal{SS}}^{\ast }\left(1\right)\equiv {\mathcal{S}}^{\ast }$ is the well-known class of starlike functions in . Functions in the class
$\mathcal{R}\left(\beta \right)=\left\{f\in \mathcal{A}:\mathfrak{Re}\left\{{f}^{\prime }\left(z\right)\right\}>\beta ,z\in \mathbb{D}\right\},$
where $\beta <1$ are called functions with bounded turning. The Libera transform $L:\mathcal{A}\to \mathcal{A}$, $L\left[f\right]=F$, where
$F\left(z\right)=\frac{2}{z}{\int }_{0}^{z}f\left(t\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t,$
(1.1)

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

Theorem 1.1 

If $f\left(z\right)$ is analytic and $\mathfrak{Re}\left\{{f}^{\prime }\left(z\right)\right\}>0$ 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:
$L\left[\mathcal{R}\left(0\right)\right]\subset {\mathcal{S}}^{\ast }={\mathcal{SS}}^{\ast }\left(1\right),$
(1.2)
where $L\left[\mathcal{R}\left(0\right)\right]=\left\{L\left[f\right]:f\in \mathcal{R}\left(0\right)\right\}$. In 1995 Mocanu  improved (1.2) by showing that
$L\left[\mathcal{R}\left(0\right)\right]\subset {\mathcal{SS}}^{\ast }\left(8/9\right).$
(1.3)
In 2002 Miller and Mocanu  showed that a subcase of this last result can be sharpened to
$L\left[\mathcal{R}\left(0\right)\cap {\mathcal{A}}_{2}\right]\subset {\mathcal{SS}}^{\ast }\left(2/3\right).$
The problem of strongly starlikeness of $L\left[f\right]$ for $f\in \mathcal{R}\left(0\right)$ was consider also in  where it is shown that
$L\left[\mathcal{R}\left(0\right)\cap {\mathcal{A}}_{2}\right]\subset {\mathcal{SS}}^{\ast }\left(3/5\right).$
The above inclusion relationship is equivalent to the following differential implication:
$f\in {\mathcal{A}}_{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathfrak{Re}\left\{{f}^{\prime }\left(z\right)\right\}>0\phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}|arg\left\{\frac{z{F}^{\prime }\left(z\right)}{F\left(z\right)}\right\}|<\frac{3\pi }{10}$
or equivalently
$F\in {\mathcal{A}}_{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathfrak{Re}\left\{{F}^{\prime }\left(z\right)+\frac{1}{2}z{F}^{″}\left(z\right)\right\}>0\phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}|arg\left\{\frac{z{F}^{\prime }\left(z\right)}{F\left(z\right)}\right\}|<\frac{3\pi }{10},$

where F is given by (1.1).

In  Ponnusamy improved (1.2) by showing that
$L\left[\mathcal{R}\left(-\varrho \right)\right]\subset {\mathcal{S}}^{\ast },\phantom{\rule{1em}{0ex}}\varrho =0.09032572\dots .$
(1.4)

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

## 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 [, p.73]

Let n be a positive integer, $\lambda >0$ and let ${\beta }_{0}={\beta }_{0}\left(\lambda ,n\right)$ be the positive root of the equation
$\beta \pi =3\pi /2-{tan}^{-1}\left(n\lambda \beta \right).$
(2.1)
In addition, let
$\alpha =\alpha \left(\beta ,\lambda ,n\right)=\beta +\left(2/\pi \right){tan}^{-1}\left(n\lambda \beta \right)$
(2.2)
for $0<\beta \le {\beta }_{0}$. If $p\left(z\right)=1+{p}_{n}{z}^{n}+{p}_{n+1}{z}^{n+1}+\cdots$ is analytic in , then
$p\left(z\right)+\lambda z{p}^{\prime }\left(z\right)\prec {\left(\frac{1+z}{1-z}\right)}^{\alpha },\phantom{\rule{1em}{0ex}}z\in \mathbb{D},$
(2.3)
implies the following subordination:
$p\left(z\right)\prec {\left(\frac{1+z}{1-z}\right)}^{\beta },\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$
(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 $\pi \alpha /2=\pi /2+{tan}^{-1}\left(1/2\right)=2.03\dots$ .

Corollary 2.2 Assume that $f\left(z\right)\in {\mathcal{A}}_{1}$. If
$|arg\left\{{F}^{\prime }\left(z\right)+\left(1/2\right)z{F}^{″}\left(z\right)\right\}|<\pi /2+{tan}^{-1}\left(1/2\right)=2.03\dots ,\phantom{\rule{1em}{0ex}}z\in \mathbb{D},$
then
$\mathfrak{Re}\left\{{F}^{\prime }\left(z\right)\right\}>0,\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$

Note that if $F\left(z\right)\in {\mathcal{A}}_{2}$, then a sufficient condition for $F\in \mathcal{R}\left(0\right)$ is $|arg\left\{{f}^{\prime }\left(z\right)\right\}|<3\pi /4=2.356\dots$ ; see [, p.96].

Lemma 2.3 

Let $p\left(z\right)$ be of the form
$p\left(z\right)=1+\sum _{n=m\ge 1}^{\mathrm{\infty }}{a}_{n}{z}^{n},\phantom{\rule{1em}{0ex}}{a}_{m}\ne 0\phantom{\rule{0.25em}{0ex}}\left(z\in \mathbb{D}\right),$
(2.5)
with $p\left(z\right)\ne 0$ in . If there exists a point ${z}_{0}$, $|{z}_{0}|<1$, such that
for some $\alpha >0$, then we have
$\frac{{z}_{0}{p}^{\prime }\left({z}_{0}\right)}{p\left({z}_{0}\right)}=ik\alpha ,$
where
(2.6)
and
(2.7)
where
${\left\{p\left({z}_{0}\right)\right\}}^{1/\alpha }=±ia,\phantom{\rule{1em}{0ex}}a>0.$

Lemma 2.4 [, p.75], 

Let $p\left(z\right)=1+{\sum }_{n=1}^{\mathrm{\infty }}{a}_{n}{z}^{n}$ be analytic in the unit disc . If
$p\left(z\right)+z{p}^{\prime }\left(z\right)\prec \frac{1+z}{1-z},\phantom{\rule{1em}{0ex}}z\in \mathbb{D},$
then
$p\left(z\right)\prec q\left(z\right)=\frac{2}{z}log\frac{1}{1-z}-1$
and
$|arg\left\{p\left(z\right)\right\}|<{\theta }_{0}=\underset{|z|=1}{max}|arg\left\{q\left(z\right)\right\}|=0.9110\dots ,\phantom{\rule{1em}{0ex}}z\in \mathbb{D},$
(2.8)

where ${\theta }_{0}$ lies between 0.911621904 and 0.911621907.

Theorem 2.5 Let $q\left(z\right)$ be analytic in and suppose that
$|arg\left\{q\left(z\right)\right\}|<\frac{\beta \pi }{2},\phantom{\rule{1em}{0ex}}z\in \mathbb{D}$
for some $\beta \in \left(0,1\right]$. If $p\left(z\right)$ is analytic and $p\left(z\right)\ne 0$ in with $p\left(0\right)=1$ and such that
$|arg\left\{q\left(z\right)\left(z{p}^{\prime }\left(z\right)+{p}^{2}\left(z\right)+p\left(z\right)\right)\right\}|<{tan}^{-1}\beta ,\phantom{\rule{1em}{0ex}}z\in \mathbb{D},$
(2.9)
then we have
$|arg\left\{p\left(z\right)\right\}|<\frac{\beta \pi }{2},\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$
Proof If there exists a point ${z}_{0}$, $|{z}_{0}|<1$, for which
$|arg\left\{p\left(z\right)\right\}|<\pi \beta /2\phantom{\rule{1em}{0ex}}\left(|z|<|{z}_{0}|\right)$
and
$|arg\left\{p\left({z}_{0}\right)\right\}|=\pi \beta /2,\phantom{\rule{1em}{0ex}}p\left({z}_{0}\right)={\left(±ia\right)}^{\beta },$
then from Nunokawa’s Lemma 2.3, we have
$\frac{{z}_{0}{p}^{\prime }\left({z}_{0}\right)}{p\left({z}_{0}\right)}=ik\beta ,$
where
and
For the case $arg\left\{p\left({z}_{0}\right)\right\}=\beta \pi /2$, we have
$\begin{array}{c}|arg\left\{q\left({z}_{0}\right)\left[{z}_{0}{p}^{\prime }\left({z}_{0}\right)+{p}^{2}\left({z}_{0}\right)+p\left({z}_{0}\right)\right]\right\}|\hfill \\ \phantom{\rule{1em}{0ex}}=|arg\left\{q\left({z}_{0}\right)p\left({z}_{0}\right)\left[1+p\left({z}_{0}\right)+{z}_{0}{p}^{\prime }\left({z}_{0}\right)/p\left({z}_{0}\right)\right]\right\}|\hfill \\ \phantom{\rule{1em}{0ex}}=|\frac{\pi \beta }{2}+arg\left\{q\left({z}_{0}\right)\right\}+arg\left\{1+p\left({z}_{0}\right)+{z}_{0}{p}^{\prime }\left({z}_{0}\right)/p\left({z}_{0}\right)\right\}|\hfill \\ \phantom{\rule{1em}{0ex}}=|\frac{\pi \beta }{2}+arg\left\{q\left({z}_{0}\right)\right\}+{tan}^{-1}\left\{\frac{\beta k+{a}^{\beta }sin\left(\pi \beta /2\right)}{1+{a}^{\beta }cos\left(\pi \beta /2\right)}\right\}|,\hfill \end{array}$
(2.10)
where $p\left({z}_{0}\right)={\left(ia\right)}^{\beta }$, $0 and
$k\ge \frac{{a}^{2}+1}{2a}\ge 1.$
Let us put
$g\left(a\right)=\frac{k\beta +{a}^{\beta }sin\left(\pi \beta /2\right)}{1+{a}^{\beta }cos\left(\pi \beta /2\right)},\phantom{\rule{1em}{0ex}}0
then it is easy to see that
$g\left(a\right)\ge \frac{\beta +{a}^{\beta }sin\left(\pi \beta /2\right)}{1+{a}^{\beta }cos\left(\pi \beta /2\right)},\phantom{\rule{1em}{0ex}}0
(2.11)
Putting
$h\left(x\right)=\frac{\beta +xsin\left(\pi \beta /2\right)}{1+xcos\left(\pi \beta /2\right)},\phantom{\rule{1em}{0ex}}0\le x,$
we have
${h}^{\prime }\left(x\right)=\frac{sin\left(\pi \beta /2\right)-\beta cos\left(\pi \beta /2\right)}{{\left(1+xcos\left(\pi \beta /2\right)\right)}^{2}}>0,\phantom{\rule{1em}{0ex}}0\le x,$
because $tan\left(\pi \beta /2\right)>\beta$. Therefore, for $x>0$ we get $h\left(x\right)>h\left(0\right)=\beta$, so from (2.11) we have
$g\left(a\right)>\beta ,$
and so
${tan}^{-1}\left\{\frac{k\beta +{a}^{\beta }sin\left(\pi \beta /2\right)}{1+{a}^{\beta }cos\left(\pi \beta /2\right)}\right\}>{tan}^{-1}\beta ,\phantom{\rule{1em}{0ex}}0
Therefore, we have the following inequality from (2.10):
$\begin{array}{c}|arg\left\{q\left({z}_{0}\right)\left[{z}_{0}{p}^{\prime }\left({z}_{0}\right)+{p}^{2}\left({z}_{0}\right)+p\left({z}_{0}\right)\right]\right\}|\hfill \\ \phantom{\rule{1em}{0ex}}\ge \frac{\pi \beta }{2}+{tan}^{-1}\frac{k+{a}^{\beta }sin\left(\pi \beta /2\right)}{1+{a}^{\beta }cos\left(\pi \beta /2\right)}-|arg\left\{q\left({z}_{0}\right)\right\}|\hfill \end{array}$
(2.12)
$\phantom{\rule{1em}{0ex}}>{tan}^{-1}\beta .$
(2.13)

This contradicts the hypothesis and for the case $arg\left\{p\left({z}_{0}\right)\right\}=-\beta \pi /2$, 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
$|arg\left\{{f}^{\prime }\left(z\right)\right\}|<{tan}^{-1}\beta ,\phantom{\rule{1em}{0ex}}z\in \mathbb{D}$
(2.14)
and
$|arg\left\{F\left(z\right)/z\right\}|<\frac{\beta \pi }{2},\phantom{\rule{1em}{0ex}}z\in \mathbb{D}$
(2.15)
for some $\beta \in \left(0,1\right]$, where $F\left(z\right)$ is given in (1.1). Then we have
$|arg\left\{\frac{z{F}^{\prime }\left(z\right)}{F\left(z\right)}\right\}|<\frac{\beta \pi }{2},\phantom{\rule{1em}{0ex}}z\in \mathbb{D},$

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

Proof If we set
$p\left(z\right)=\frac{z{F}^{\prime }\left(z\right)}{F\left(z\right)},$
then
${f}^{\prime }\left(z\right)={F}^{\prime }\left(z\right)+\frac{1}{2}z{F}^{″}\left(z\right)=\frac{1}{2}\left(\frac{F\left(z\right)}{z}\right)\left(z{p}^{\prime }\left(z\right)+{p}^{2}\left(z\right)+p\left(z\right)\right).$
If we let $q\left(z\right)=F\left(z\right)/z$, then by (2.14) and (2.15) the assumptions of Theorem 2.5 are satisfied. Therefore,
$|arg\left\{p\left(z\right)\right\}|<\frac{\beta \pi }{2},\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$

□

Theorem 2.7 Let $q\left(z\right)$ be analytic in , with $q\left(0\right)=1$ and satisfy
$\mathfrak{Re}\left\{z{q}^{\prime }\left(z\right)+q\left(z\right)\right\}>0,\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$
If $p\left(z\right)$ is analytic in , with $p\left(0\right)=1$ and if
$|arg\left\{q\left(z\right)\left(z{p}^{\prime }\left(z\right)+{p}^{2}\left(z\right)+p\left(z\right)\right)\right\}|<\frac{5\pi }{6}-{\theta }_{0}=1.706\dots ,\phantom{\rule{1em}{0ex}}z\in \mathbb{D},$
where ${\theta }_{0}$ is given in (2.8), then we have
$\mathfrak{Re}\left\{p\left(z\right)\right\}>0,\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$
Proof By Lemma 2.4, we have
$|arg\left\{q\left(z\right)\right\}|<{\theta }_{0}=0.911\dots ,\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$
(2.16)
If there exists a point ${z}_{0}$, $|{z}_{0}|<1$, such that
$|arg\left\{p\left(z\right)\right\}|<\pi /2\phantom{\rule{1em}{0ex}}\left(|z|<|{z}_{0}|\right)$
and
$|arg\left\{p\left({z}_{0}\right)\right\}|=\pi /2,\phantom{\rule{1em}{0ex}}p\left({z}_{0}\right)=±ia,0
then from Nunokawa’s Lemma 2.3, we have
$\frac{{z}_{0}{p}^{\prime }\left({z}_{0}\right)}{p\left({z}_{0}\right)}=ik,$
where
and
For the case $arg\left\{p\left({z}_{0}\right)\right\}=\pi /2$, we have
$\begin{array}{rcl}arg\left\{1+ia+ik\right\}& \ge & arg\left\{1+ia+i\frac{{a}^{2}+1}{2a}\right\}\\ =& {tan}^{-1}\frac{\mathfrak{Im}\left\{1+ia+i\frac{{a}^{2}+1}{2a}\right\}}{\mathfrak{Re}\left\{1+ia+i\frac{{a}^{2}+1}{2a}\right\}}\\ =& {tan}^{-1}\left\{\frac{3{a}^{2}+1}{2a}\right\}\\ \ge & {tan}^{-1}\left\{\sqrt{3}\right\}\\ =& \frac{\pi }{3}.\end{array}$
Therefore, for the case $arg\left\{p\left({z}_{0}\right)\right\}=\pi /2$, we have
$\frac{\pi }{3}\le arg\left\{1+ia+ik\right\}<\frac{\pi }{2}.$
Moreover, by (2.16)
$arg\left\{q\left({z}_{0}\right)\right\}<{\theta }_{0}.$
Therefore, we can write
$\begin{array}{c}|arg\left\{q\left({z}_{0}\right)\left({z}_{0}{p}^{\prime }\left({z}_{0}\right)+{p}^{2}\left({z}_{0}\right)+p\left({z}_{0}\right)\right)\right\}|\hfill \\ \phantom{\rule{1em}{0ex}}=|arg\left\{p\left({z}_{0}\right)\left[1+p\left({z}_{0}\right)+{z}_{0}{p}^{\prime }\left({z}_{0}\right)/p\left({z}_{0}\right)\right]q\left({z}_{0}\right)\right\}|\hfill \\ \phantom{\rule{1em}{0ex}}\ge |arg\left\{p\left({z}_{0}\right)\left(1+ia+ik\right)\right\}|-|arg\left\{q\left({z}_{0}\right)\right\}|\hfill \\ \phantom{\rule{1em}{0ex}}\ge \frac{\pi }{2}+\frac{\pi }{3}-|arg\left\{q\left({z}_{0}\right)\right\}|\hfill \\ \phantom{\rule{1em}{0ex}}\ge \frac{5\pi }{6}-{\theta }_{0}.\hfill \end{array}$
(2.17)
This contradicts the hypothesis and for the case $arg\left\{p\left({z}_{0}\right)\right\}=-\pi /2$, applying the same method as above, we have
$|arg\left\{q\left({z}_{0}\right)\left({z}_{0}{p}^{\prime }\left({z}_{0}\right)+{p}^{2}\left({z}_{0}\right)+p\left({z}_{0}\right)\right)\right\}|\ge \frac{5\pi }{6}-{\theta }_{0}.$

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

Corollary 2.8 Assume that
$|arg\left\{{f}^{\prime }\left(z\right)\right\}|<\frac{5\pi }{6}-{\theta }_{0}=1.706\dots ,\phantom{\rule{1em}{0ex}}z\in \mathbb{D},$
(2.18)
then we have
$\mathfrak{Re}\left\{\frac{z{F}^{\prime }\left(z\right)}{F\left(z\right)}\right\}>0,\phantom{\rule{1em}{0ex}}z\in \mathbb{D},$
(2.19)

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

Proof Because
${f}^{\prime }\left(z\right)={F}^{\prime }\left(z\right)+\frac{1}{2}z{F}^{″}\left(z\right),$
by Corollary 2.2 and by (2.18) we obtain
$\mathfrak{Re}\left\{{F}^{\prime }\left(z\right)\right\}>0,\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$
Therefore, if we let $q\left(z\right)=F\left(z\right)/z$, then
$\mathfrak{Re}\left\{z{q}^{\prime }\left(z\right)+q\left(z\right)\right\}=\mathfrak{Re}\left\{{F}^{\prime }\left(z\right)\right\}>0,\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$
If we set
$p\left(z\right)=\frac{z{F}^{\prime }\left(z\right)}{F\left(z\right)},$
then
${f}^{\prime }\left(z\right)={F}^{\prime }\left(z\right)+\frac{1}{2}z{F}^{″}\left(z\right)=\frac{1}{2}\left(\frac{F\left(z\right)}{z}\right)\left(z{p}^{\prime }\left(z\right)+{p}^{2}\left(z\right)+p\left(z\right)\right).$

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  because in (2.18) we have $|arg\left\{{f}^{\prime }\left(z\right)\right\}|<1.706\dots$ , while in (1.2) we have the stronger assumption that $|arg\left\{{f}^{\prime }\left(z\right)\right\}|<\pi /2=1.57\dots$ .

## Declarations

### Acknowledgements

The authors would like to express their thanks to the referees for valuable advice regarding a previous version of this paper. This research was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (No. 2011-0007037).

## Authors’ Affiliations

(1)
University of Gunma, Hoshikuki-cho 798-8, Chuou-Ward, Chiba 260-0808, Japan
(2)
Department of Mathematics, Rzeszów University of Technology, Al. Powstańców Warszawy 12, Rzeszów, 35-959, Poland
(3)
Department of Applied Mathematics, College of Natural Sciences, Pukyong National University, Pusan, 608-737, Korea
(4)
Department of Mathematics, Kyungsung University, Busan, 608-736, Korea

## References

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