# On uniformly univalent functions with respect to symmetrical points

## Abstract

In this paper, we define and study some new subclasses of starlike and close-to-convex functions with respect to symmetrical points. These functions map the open unit disc onto certain conic regions in the right half plane. Some basic properties, a necessary condition, and coefficient and arc length problems are investigated. The mapping properties of the functions in these classes are studied under a certain linear operator.

MSC:30C45, 30C50.

## 1 Introduction

Let A be the class of functions of the form

$f\left(z\right)=z+\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}{a}_{n}{z}^{n},$
(1.1)

which are analytic in the open unit disc $E=\left\{z:|z|<1\right\}$. Let S, K, ${S}^{âˆ—}$, and C be the subclasses of A which consist of univalent, close-to-convex, starlike (with respect to origin), and convex functions, respectively. For recent developments, extensions, and applications, see [1â€“25] and the references therein.

A function f in A is said to be uniformly convex in E if f is a univalent convex function along with the property that, for every circular arc Î³ contained in E, with center Î¾ also in E, the image curve $f\left(\mathrm{Î³}\right)$ is a convex arc. The class of uniformly convex functions is denoted by $UCV$. The corresponding class $UST$ is defined by the relation that $fâˆˆUCV$ if, and only if, $z{f}^{â€²}âˆˆUST$. It is well known [13] that $fâˆˆUCV$ if, and only if

$|\frac{z{f}^{â€³}\left(z\right)}{{f}^{â€²}\left(z\right)}|<\mathrm{â„œ}\left\{1+\frac{z{f}^{â€³}\left(z\right)}{{f}^{â€²}\left(z\right)}\right\}\phantom{\rule{1em}{0ex}}\left(zâˆˆE\right).$

Uniformly starlike and convex functions were first introduced by Goodman [3] and then studied by various other authors. If $f,gâˆˆA$, we say f is subordinate to g in E, written as $fâ‰ºg$ or $f\left(z\right)â‰ºg\left(z\right)$, if there exists a Schwarz function $w\left(z\right)$ such that $f\left(z\right)=g\left(w\left(z\right)\right)$ for $zâˆˆE$.

For $0â‰¤\mathrm{Î²}<1$, the class $P\left(\mathrm{Î²}\right)$ consists of functions $p\left(z\right)$ analytic in E with $p\left(0\right)=1$ such that $\mathrm{â„œ}p\left(z\right)>\mathrm{Î²}$ for $zâˆˆE$, and, with $\mathrm{Î²}=0$, we obtain the well-known class P of CarathÃ©odory functions with positive real part.

For $kâˆˆ\left[0,\mathrm{âˆž}\right)$, the conic regions ${\mathrm{Î©}}_{k}$ are defined as follows, see [5]:

${\mathrm{Î©}}_{k}=\left\{u+iv:u>k\sqrt{{\left(uâˆ’1\right)}^{2}+{v}^{2}}\right\}.$

For fixed k, ${\mathrm{Î©}}_{k}$ represents the conic regions bounded, successively, by the imaginary axis ($k=0$), the right branch of a hyperbolic ($0) and a parabola ${v}^{2}=2uâˆ’1$ ($k=1$). When $k>1$, the domain becomes a bounded domain being the interior of the ellipse.

We shall consider the case when $kâˆˆ\left[0,1\right]$. Related to the domain ${\mathrm{Î©}}_{k}$, the following functions ${p}_{k}\left(z\right)$, $kâˆˆ\left[0,1\right]$, play the role of extremal functions mapping in E onto ${\mathrm{Î©}}_{k}$:

${p}_{k}\left(z\right)=\left\{\begin{array}{cc}\frac{1+z}{1âˆ’z}\hfill & \left(k=0\right),\hfill \\ 1+\frac{2}{{\mathrm{Ï€}}^{2}}{\left(log\frac{1+\sqrt{z}}{1âˆ’\sqrt{z}}\right)}^{2}\hfill & \left(k=1\right),\hfill \\ 1+\frac{2}{1âˆ’{k}^{2}}{sinh}^{2}\left[\left(\frac{2}{\mathrm{Ï€}}arccosk\right)arctanh\sqrt{z}\right]\hfill & \left(0
(1.2)

These functions are univalent in E and belong to the class P. Using the subordination concept, we define the class $P\left({p}_{k}\right)$ as follows.

Let $p\left(z\right)$ be analytic in E with $p\left(0\right)=1$. Then $pâˆˆP\left({p}_{k}\right)$ if, and only if, $pâ‰º{p}_{k}$ in E and ${p}_{k}\left(z\right)$ are given by (1.2).

The conic domains ${\mathrm{Î©}}_{k}$ can be generalized as given by

${\mathrm{Î©}}_{k,\mathrm{Î²}}=\left(1âˆ’\mathrm{Î²}\right){\mathrm{Î©}}_{k}+\mathrm{Î²},$

with the corresponding extremal function

${p}_{k,\mathrm{Î²}}\left(z\right)=\left(1âˆ’\mathrm{Î²}\right){p}_{k}+\mathrm{Î²}\phantom{\rule{1em}{0ex}}\left(0â‰¤\mathrm{Î²}<1,kâˆˆ\left[0,1\right]\right).$

It can easily be seen that the analytic function $p\left(z\right)$, with $p\left(0\right)=1$, belongs to the class $P\left({p}_{k,\mathrm{Î²}}\right)$ if $p\left(z\right)â‰º{p}_{k,\mathrm{Î²}}\left(z\right)$ in E.

It is easy to verify that $P\left({p}_{k,\mathrm{Î²}}\right)$ is a convex set. It is known [6] that

$P\left({p}_{k}\right)âŠ‚P\left(\frac{k}{k+1}\right)âŠ‚P,$

and, for $pâˆˆP\left({p}_{k}\right)$, we have

$|argp\left(z\right)|â‰¤\mathrm{Ïƒ}\frac{\mathrm{Ï€}}{2},$

where

$\mathrm{Ïƒ}=\frac{2}{\mathrm{Ï€}}arctan\frac{1}{k}.$
(1.3)

So we can write $p\left(z\right)={h}^{\mathrm{Ïƒ}}\left(z\right)$, $hâˆˆP$.

Also

$P\left({p}_{k,\mathrm{Î²}}\right)âŠ‚P\left(\frac{k+\mathrm{Î²}}{k+1}\right)âŠ‚P.$

Sakaguchi [24] introduced and studied the class ${S}_{s}^{âˆ—}$ of starlike functions with respect to symmetrical points. The class ${S}_{s}^{âˆ—}$ includes the classes of convex and odd starlike functions with respect to the origin. It was shown [24] that a necessary and sufficient condition for $fâˆˆ{S}_{s}^{âˆ—}$ to be univalent and starlike with respect to symmetrical points in E is that

$\left(\frac{2z{f}^{â€²}\left(z\right)}{f\left(z\right)âˆ’f\left(âˆ’z\right)}\right)âˆˆP,\phantom{\rule{1em}{0ex}}zâˆˆE.$

Das and Singh [2] defined the classes ${C}_{s}$ of convex functions with respect to symmetrical points and showed that a necessary and sufficient condition for $fâˆˆ{C}_{s}$ is that

$\frac{2{\left(z{f}^{â€²}\left(z\right)\right)}^{â€²}}{{\left(f\left(z\right)âˆ’f\left(âˆ’z\right)\right)}^{â€²}}âˆˆP,\phantom{\rule{1em}{0ex}}zâˆˆE.$

It is also well known [2] that $fâˆˆ{C}_{s}$ if, and only if, $z{f}^{â€²}âˆˆ{S}_{s}^{âˆ—}$.

We now define the following.

Definition 1.1 Let $fâˆˆA$. The f is said to be in the class $kâˆ’S{T}_{s}\left(\mathrm{Î²}\right)$ if, and only if,

$\frac{2z{f}^{â€²}\left(z\right)}{\left(f\left(z\right)âˆ’f\left(âˆ’z\right)\right)}âˆˆP\left({p}_{k,\mathrm{Î²}}\right),\phantom{\rule{1em}{0ex}}zâˆˆE.$

It can easily be seen that

$kâˆ’S{T}_{s}\left(\mathrm{Î²}\right)âŠ‚{S}_{s}^{âˆ—}âŠ‚{S}_{s}^{âˆ—},\phantom{\rule{2em}{0ex}}{\mathrm{Î²}}_{1}=\frac{k+\mathrm{Î²}}{k+1}.$

Also, for $\mathrm{Î²}=0=k$, the class $kâˆ’S{T}_{s}\left(\mathrm{Î²}\right)$ reduces to ${S}_{s}^{âˆ—}$.

The class $kâˆ’UC{V}_{s}\left(\mathrm{Î²}\right)$ is defined as follows.

Definition 1.2 Let $fâˆˆA$. Then $fâˆˆkâˆ’UC{V}_{s}\left(\mathrm{Î²}\right)$ if, and only if $z{f}^{â€²}âˆˆkâˆ’S{T}_{s}\left(\mathrm{Î²}\right)$ for $zâˆˆE$.

We note that

$kâˆ’UC{V}_{s}\left(\mathrm{Î²}\right)âŠ‚{C}_{s}\left({\mathrm{Î²}}_{1}\right)âŠ‚{C}_{s},\phantom{\rule{2em}{0ex}}{\mathrm{Î²}}_{1}=\frac{k+\mathrm{Î²}}{k+1}.$

Definition 1.3 Let $fâˆˆA$. Then $fâˆˆkâˆ’U{K}_{s}\left(\mathrm{Î²}\right)$ if, and only if, there exists $gâˆˆkâˆ’S{T}_{s}\left(\mathrm{Î²}\right)$ such that

$\left(\frac{2z{f}^{â€²}\left(z\right)}{g\left(z\right)âˆ’g\left(âˆ’z\right)}\right)âˆˆP\left({p}_{k,\mathrm{Î²}}\right),\phantom{\rule{1em}{0ex}}zâˆˆE.$

Since $P\left({p}_{k,\mathrm{Î²}}\right)âŠ‚P\left({\mathrm{Î²}}_{1}\right)âŠ‚P$, ${\mathrm{Î²}}_{1}=\frac{k+\mathrm{Î²}}{k+1}$, and $kâˆ’S{T}_{s}\left(\mathrm{Î²}\right)âŠ‚{S}_{s}^{âˆ—}$, we note that

$kâˆ’U{K}_{s}\left(\mathrm{Î²}\right)âŠ‚{K}_{s}âŠ‚K,$

where ${k}_{S}$ consists of close-to-convex functions with respect to symmetrical starlike functions.

From the definition, it is clear that $kâˆ’U{K}_{s}\left(\mathrm{Î²}\right)$ consists of univalent functions.

For $k=0$, $\mathrm{Î²}=0$ and $f\left(z\right)=g\left(z\right)$, $kâˆ’U{K}_{s}\left(\mathrm{Î²}\right)$ reduces to the class ${S}_{s}^{âˆ—}$.

## 2 Preliminary results

We shall need the following lemmas to prove our main results.

Lemma 2.1 [15]

Let $q\left(z\right)$ be a convex function in E with $q\left(0\right)=1$ and let another function $h:Eâ†’\mathbb{C}$ be with $\mathrm{â„œ}h\left(z\right)>0$. Let $p\left(z\right)$ be analytic in E with $p\left(0\right)=1$ such that

$\left(p\left(z\right)+h\left(z\right)z{p}^{â€²}\left(z\right)\right)â‰ºq\left(z\right),\phantom{\rule{1em}{0ex}}zâˆˆE.$

Then $p\left(z\right)â‰ºq\left(z\right)$, $zâˆˆE$.

Lemma 2.2 Let $N\left(z\right)$, $D\left(z\right)$ be analytic in E with

$N\left(0\right)=0=D\left(z\right)$

and let $Dâˆˆ{S}^{âˆ—}$ for $zâˆˆE$. Then $\frac{{N}^{â€²}\left(z\right)}{{D}^{â€²}\left(z\right)}âˆˆP\left({p}_{k,\mathrm{Î²}}\right)$ implies that $\frac{N\left(z\right)}{D\left(z\right)}âˆˆP\left({p}_{k,\mathrm{Î²}}\right)$ for $zâˆˆE$.

Proof Let

$\frac{N\left(z\right)}{D\left(z\right)}=p\left(z\right).$

Then

$\frac{{N}^{â€²}\left(z\right)}{{D}^{â€²}\left(z\right)}=p\left(z\right)+h\left(z\right)\left(z{p}^{â€²}\left(z\right)\right),\phantom{\rule{1em}{0ex}}h\left(z\right)=\frac{1}{{h}_{0}\left(z\right)},$

where

${h}_{0}\left(z\right)=\frac{z{D}^{â€²}\left(z\right)}{D\left(z\right)}âˆˆP.$

Since $\frac{{N}^{â€²}\left(z\right)}{{D}^{â€²}\left(z\right)}âˆˆP\left({p}_{k,\mathrm{Î²}}\right)$, we have

$\frac{{N}^{â€²}\left(z\right)}{{D}^{â€²}\left(z\right)}=\left(p\left(z\right)+h\left(z\right)\left(z{p}^{â€²}\left(z\right)\right)\right)â‰º{p}_{k,\mathrm{Î²}}\left(z\right),\phantom{\rule{1em}{0ex}}zâˆˆE.$

We now use Lemma 2.1 and this implies that

This proves that $\frac{N\left(z\right)}{D\left(z\right)}âˆˆP\left({p}_{k,\mathrm{Î²}}\right)$ for $zâˆˆE$.â€ƒâ–¡

The following lemma is an easy extension of a result proved in [5].

Lemma 2.3 Let $kâˆˆ\left[0,\mathrm{âˆž}\right)$ and ${\mathrm{Î³}}_{1}$, ${\mathrm{Î´}}_{1}$ be any complex numbers with and let $\mathrm{â„œ}\left\{\frac{{\mathrm{Î³}}_{1}k}{k+1}+{\mathrm{Î´}}_{1}\right\}>\mathrm{Î²}$. If $h\left(z\right)$ is analytic in E, $h\left(0\right)=1$ and it satisfies

$\left(h\left(z\right)+\frac{z{h}^{â€²}\left(z\right)}{{\mathrm{Î³}}_{1}h\left(z\right)+{\mathrm{Î´}}_{1}}\right)â‰º{p}_{k,\mathrm{Î²}}\left(z\right),$
(2.1)

and ${q}_{k,\mathrm{Î²}}\left(z\right)$ is an analytic solution of

$\left({q}_{k,\mathrm{Î²}}\left(z\right)+\frac{z{q}_{k,\mathrm{Î²}}^{â€²}\left(z\right)}{{\mathrm{Î³}}_{1}{q}_{k,\mathrm{Î²}}\left(z\right)+{\mathrm{Î´}}_{1}}\right)={p}_{k,\mathrm{Î²}}\left(z\right),$

then ${q}_{k,\mathrm{Î²}}$ is univalent and

$h\left(z\right)â‰º{q}_{k,\mathrm{Î²}}\left(z\right)â‰º{p}_{k,\mathrm{Î²}}\left(z\right),$

and ${q}_{k,\mathrm{Î²}}\left(z\right)$ is the best dominant of (2.1).

## 3 The class $kâˆ’S{T}_{s}\left(\mathrm{Î²}\right)$

In this section, we shall study some basic properties of the class $kâˆ’S{T}_{s}\left(\mathrm{Î²}\right)$.

Theorem 3.1 Let $fâˆˆkâˆ’S{T}_{s}\left(\mathrm{Î²}\right)$. Then the odd function

$\mathrm{Î¨}\left(z\right)=\frac{1}{2}\left[f\left(z\right)âˆ’f\left(âˆ’z\right)\right],$
(3.1)

belongs to $kâˆ’ST\left(\mathrm{Î²}\right)$ in E.

In particular $\mathrm{Î¨}\left(z\right)$ is an odd starlike function of order ${\mathrm{Î²}}_{1}=\frac{k+\mathrm{Î²}}{k+1}$ in E.

Proof Logarithmic differentiation of (3.1) and simple computation yield

Since $P\left({p}_{k,\mathrm{Î²}}\right)$ is a convex set, it follows that $\frac{z{\mathrm{Î¨}}^{â€²}\left(z\right)}{\mathrm{Î¨}\left(z\right)}âˆˆP\left({p}_{k,\mathrm{Î²}}\right)$ and thus $\mathrm{Î¨}âˆˆkâˆ’ST\left(\mathrm{Î²}\right)$ in E.â€ƒâ–¡

As a special case, we note that, for $k=0=\mathrm{Î²}$, $\frac{1}{2}\left[f\left(z\right)âˆ’f\left(âˆ’z\right)\right]=\mathrm{Î¨}\left(z\right)âˆˆ{S}^{âˆ—}$ in E, and hence $\frac{z{f}^{â€²}}{\mathrm{Î¨}}âˆˆP$. We now discuss a geometric property for $fâˆˆkâˆ’S{T}_{s}\left(\mathrm{Î²}\right)$. Here we investigate the behavior of the inclusion of the tangent at a point $w\left(\mathrm{Î¸}\right)=f\left(r{e}^{i\mathrm{Î¸}}\right)$ to the image ${\mathrm{Î“}}_{r}$ of the circle ${C}_{r}=\left\{z:|z|=r\right\}$, $0â‰¤r<1$, $\mathrm{Î¸}âˆˆ\left[0,2\mathrm{Ï€}\right]$, under the mapping by means of a function f from the class $fâˆˆkâˆ’S{T}_{s}\left(\mathrm{Î²}\right)$.

Let

$\mathrm{Î¦}\left(\mathrm{Î¸}\right)=\frac{\mathrm{Ï€}}{2}+\mathrm{Î¸}+arg{f}^{â€²}\left(r{e}^{i\mathrm{Î¸}}\right)=arg\frac{\mathrm{âˆ‚}}{\mathrm{âˆ‚}\mathrm{Î¸}}f\left(r{e}^{i\mathrm{Î¸}}\right),$

and, for ${\mathrm{Î¸}}_{2}>{\mathrm{Î¸}}_{1}$, ${\mathrm{Î¸}}_{1},{\mathrm{Î¸}}_{2}âˆˆ\left[0,2\mathrm{Ï€}\right]$,

$\mathrm{Î¦}\left({\mathrm{Î¸}}_{2}\right)âˆ’\mathrm{Î¦}\left({\mathrm{Î¸}}_{1}\right)={\mathrm{Î¸}}_{2}+arg{f}^{â€²}\left(r{e}^{i{\mathrm{Î¸}}_{2}}\right)âˆ’{\mathrm{Î¸}}_{1}âˆ’arg{f}^{â€²}\left(r{e}^{i{\mathrm{Î¸}}_{1}}\right).$

Now, since

$\mathrm{Î¸}+arg{f}^{â€²}\left(r{e}^{i\mathrm{Î¸}}\right)=\mathrm{Î¸}+\mathrm{â„œ}\left\{âˆ’iln{f}^{â€²}\left(r{e}^{i\mathrm{Î¸}}\right)\right\},$

then

$\frac{\mathrm{âˆ‚}}{\mathrm{âˆ‚}\mathrm{Î¸}}\left(\mathrm{Î¸}+arg{f}^{â€²}\left(r{e}^{i\mathrm{Î¸}}\right)\right)=\mathrm{â„œ}\left\{1+\frac{r{e}^{i\mathrm{Î¸}}{f}^{â€³}\left(r{e}^{i\mathrm{Î¸}}\right)}{{f}^{â€²}\left(r{e}^{i\mathrm{Î¸}}\right)}\right\}.$

Hence

${âˆ«}_{{\mathrm{Î¸}}_{1}}^{{\mathrm{Î¸}}_{2}}\frac{\mathrm{âˆ‚}}{\mathrm{âˆ‚}\mathrm{Î¸}}\left(\mathrm{Î¸}+arg{f}^{â€²}\left(r{e}^{i\mathrm{Î¸}}\right)\right)\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}={âˆ«}_{{\mathrm{Î¸}}_{1}}^{{\mathrm{Î¸}}_{2}}\mathrm{â„œ}\left\{1+\frac{r{e}^{i\mathrm{Î¸}}{f}^{â€³}\left(r{e}^{i\mathrm{Î¸}}\right)}{{f}^{â€²}\left(r{e}^{i\mathrm{Î¸}}\right)}\right\}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}.$

Also, on the other hand,

$\begin{array}{rl}{âˆ«}_{{\mathrm{Î¸}}_{1}}^{{\mathrm{Î¸}}_{2}}\frac{\mathrm{âˆ‚}}{\mathrm{âˆ‚}\mathrm{Î¸}}\left(\mathrm{Î¸}+arg{f}^{â€²}\left(r{e}^{i\mathrm{Î¸}}\right)\right)\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}& ={\mathrm{Î¸}}_{2}+arg{f}^{â€²}\left(r{e}^{i{\mathrm{Î¸}}_{2}}\right)âˆ’{\mathrm{Î¸}}_{1}âˆ’arg{f}^{â€²}\left(r{e}^{i{\mathrm{Î¸}}_{1}}\right)\\ =\mathrm{Î¦}\left({\mathrm{Î¸}}_{2}\right)âˆ’\mathrm{Î¦}\left({\mathrm{Î¸}}_{1}\right).\end{array}$

So, the integral on the left side of the last inequality characterizes the increment of the angle of the inclination of the tangent to the curve ${\mathrm{Î“}}_{r}$ between the points $w\left({\mathrm{Î¸}}_{2}\right)$ and $w\left({\mathrm{Î¸}}_{1}\right)$ for ${\mathrm{Î¸}}_{2}>{\mathrm{Î¸}}_{1}$.

We have the following necessary condition for $fâˆˆkâˆ’S{T}_{s}\left(\mathrm{Î²}\right)$.

Theorem 3.2 Let $fâˆˆkâˆ’S{T}_{s}\left(\mathrm{Î²}\right)$. Then, with $z=r{e}^{i\mathrm{Î¸}}$ and $0â‰¤{\mathrm{Î¸}}_{1}<{\mathrm{Î¸}}_{2}â‰¤2\mathrm{Ï€}$, $0â‰¤\mathrm{Î²}<1$ and $0â‰¤kâ‰¤1$, we have

${âˆ«}_{{\mathrm{Î¸}}_{1}}^{{\mathrm{Î¸}}_{2}}\mathrm{â„œ}\left\{\frac{{\left(z{f}^{â€²}\left(z\right)\right)}^{â€²}}{{f}^{â€²}\left(z\right)}\right\}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}>âˆ’\mathrm{Ïƒ}\mathrm{Ï€}+2{cos}^{âˆ’1}\left\{\frac{2\left(1âˆ’\mathrm{Î²}\right)}{1âˆ’\left(1âˆ’2\mathrm{Î²}\right){r}^{2}}\right\}+{\mathrm{Î²}}_{1}\left({\mathrm{Î¸}}_{2}âˆ’{\mathrm{Î¸}}_{1}\right),$

where Ïƒ is given by (1.3) and ${\mathrm{Î²}}_{1}=\frac{k+\mathrm{Î²}}{k+1}$.

Proof Since $\frac{{f}^{â€²}\left(z\right)}{{\mathrm{Î¨}}^{â€²}\left(z\right)}âˆˆP\left({p}_{k,\mathrm{Î²}}\right)$, $\mathrm{Î¨}\left(z\right)=\frac{1}{2}\left[f\left(z\right)âˆ’f\left(âˆ’z\right)\right]$ and $\mathrm{Î¨}âˆˆkâˆ’UCV\left(\mathrm{Î²}\right)âŠ‚C\left(\mathrm{Î²}\right)$.

We can write

${f}^{â€²}\left(z\right)={\left({\mathrm{Î¨}}_{1}^{â€²}\left(z\right)\right)}^{1âˆ’{\mathrm{Î²}}_{1}}{h}^{\mathrm{Ïƒ}}\left(z\right),\phantom{\rule{1em}{0ex}}{\mathrm{Î¨}}_{1}âˆˆC,hâˆˆP\left(\mathrm{Î²}\right),$

and this gives us, with $z=r{e}^{i\mathrm{Î¸}}$, $0â‰¤r<1$, $0â‰¤{\mathrm{Î¸}}_{1}<{\mathrm{Î¸}}_{2}â‰¤2\mathrm{Ï€}$,

$\begin{array}{rl}{âˆ«}_{{\mathrm{Î¸}}_{1}}^{{\mathrm{Î¸}}_{2}}\mathrm{â„œ}\left\{\frac{{\left(z{f}^{â€²}\left(z\right)\right)}^{â€²}}{{f}^{â€²}\left(z\right)}\right\}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}=& \left(1âˆ’{\mathrm{Î²}}_{1}\right){âˆ«}_{{\mathrm{Î¸}}_{1}}^{{\mathrm{Î¸}}_{2}}\mathrm{â„œ}\left\{\frac{{\left(z{\mathrm{Î¨}}_{1}^{â€²}\left(z\right)\right)}^{â€²}}{{\mathrm{Î¨}}_{1}^{â€²}\left(z\right)}\right\}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\\ +\mathrm{Ïƒ}{âˆ«}_{{\mathrm{Î¸}}_{1}}^{{\mathrm{Î¸}}_{2}}\mathrm{â„œ}\frac{2{h}^{â€²}\left(z\right)}{h\left(z\right)}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}+{\mathrm{Î²}}_{1}\left({\mathrm{Î¸}}_{2}âˆ’{\mathrm{Î¸}}_{1}\right).\end{array}$
(3.2)

For $hâˆˆP\left(\mathrm{Î²}\right)$, we observe that

$\begin{array}{rl}\frac{\mathrm{âˆ‚}}{\mathrm{âˆ‚}\mathrm{Î¸}}argh\left(r{e}^{i\mathrm{Î¸}}\right)& =\frac{\mathrm{âˆ‚}}{\mathrm{âˆ‚}\mathrm{Î¸}}\mathrm{â„œ}\left\{âˆ’ilnh\left(r{e}^{i\mathrm{Î¸}}\right)\right\}\\ =\mathrm{â„œ}\left\{r{e}^{i\mathrm{Î¸}}\frac{{h}^{â€²}\left(r{e}^{i\mathrm{Î¸}}\right)}{h\left(r{e}^{i\mathrm{Î¸}}\right)}\right\}.\end{array}$

Therefore

${âˆ«}_{{\mathrm{Î¸}}_{1}}^{{\mathrm{Î¸}}_{2}}\mathrm{â„œ}\left\{\frac{r{e}^{i\mathrm{Î¸}}{h}^{â€²}\left(r{e}^{i\mathrm{Î¸}}\right)}{h\left(r{e}^{i\mathrm{Î¸}}\right)}\right\}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}=argh\left(r{e}^{i{\mathrm{Î¸}}_{2}}\right)âˆ’argh\left(r{e}^{i{\mathrm{Î¸}}_{1}}\right),$

and

$\underset{hâˆˆP\left(\mathrm{Î²}\right)}{max}|{âˆ«}_{{\mathrm{Î¸}}_{1}}^{{\mathrm{Î¸}}_{2}}\mathrm{â„œ}\left\{\frac{r{e}^{i\mathrm{Î¸}}{h}^{â€²}\left(r{e}^{i\mathrm{Î¸}}\right)}{h\left(r{e}^{i\mathrm{Î¸}}\right)}\right\}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}|=\underset{hâˆˆP\left(\mathrm{Î²}\right)}{max}|argh\left(r{e}^{i{\mathrm{Î¸}}_{2}}\right)âˆ’argh\left(r{e}^{i{\mathrm{Î¸}}_{1}}\right)|.$

We can write

$\frac{1}{1âˆ’\mathrm{Î²}}\left[h\left(z\right)âˆ’\mathrm{Î²}\right]=p\left(z\right),\phantom{\rule{1em}{0ex}}pâˆˆP,$

and for $|z|=r<1$, it is well known that

$|p\left(z\right)âˆ’\frac{1+{r}^{2}}{1âˆ’{r}^{2}}|â‰¤\frac{2r}{1âˆ’{r}^{2}}.$

From this, we have

$|h\left(z\right)âˆ’\frac{1+\left(1âˆ’2\mathrm{Î²}\right){r}^{2}}{1âˆ’{r}^{2}}|â‰¤\frac{2\left(1âˆ’\mathrm{Î²}\right)r}{1âˆ’{r}^{2}}.$

Thus the values of h are contained in the circle of Apollonius whose diameter is the line segment from $\frac{1âˆ’\left(1âˆ’2\mathrm{Î²}\right)r}{1+r}$ to $\frac{1+\left(1âˆ’2\mathrm{Î²}\right)r}{1âˆ’r}$ and has the radius $\frac{2\left(1âˆ’\mathrm{Î²}\right)r}{1âˆ’{r}^{2}}$. So $|argh\left(z\right)|$ attains its maximum at points where a ray from origin is tangent to the circle, that is, when

$argh\left(z\right)=Â±{sin}^{âˆ’1}\left(\frac{2\left(1âˆ’\mathrm{Î²}\right)r}{1âˆ’\left(1âˆ’2\mathrm{Î²}\right){r}^{2}}\right).$
(3.3)

From (3.3), we observe that

$\begin{array}{rl}\underset{hâˆˆP\left(\mathrm{Î²}\right)}{max}|{âˆ«}_{{\mathrm{Î¸}}_{1}}^{{\mathrm{Î¸}}_{2}}\mathrm{â„œ}\left\{r{e}^{i\mathrm{Î¸}}\frac{{h}^{â€²}\left(r{e}^{i\mathrm{Î¸}}\right)}{h\left(r{e}^{i\mathrm{Î¸}}\right)}\right\}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}|& â‰¤2{sin}^{âˆ’1}\left(\frac{2\left(1âˆ’\mathrm{Î²}\right)r}{1âˆ’\left(1âˆ’2\mathrm{Î²}\right){r}^{2}}\right)\\ =\mathrm{Ï€}âˆ’2{cos}^{âˆ’1}\left(\frac{2\left(1âˆ’\mathrm{Î²}\right)r}{1âˆ’\left(1âˆ’2\mathrm{Î²}\right){r}^{2}}\right).\end{array}$
(3.4)

Also, for ${\mathrm{Î¨}}_{1}âˆˆC$,

${âˆ«}_{{\mathrm{Î¸}}_{1}}^{{\mathrm{Î¸}}_{2}}\mathrm{â„œ}\left\{1+r{e}^{i\mathrm{Î¸}}\frac{{\mathrm{Î¨}}_{1}^{â€³}\left(r{e}^{i\mathrm{Î¸}}\right)}{{\mathrm{Î¨}}_{1}^{â€²}\left(r{e}^{i\mathrm{Î¸}}\right)}\right\}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}â‰¥0.$
(3.5)

Using (3.4) and (3.5) in (3.2), we obtain the required result.â€ƒâ–¡

We note the following special cases:

1. 1.

For $k=0$, $0â‰¤{\mathrm{Î¸}}_{1}<{\mathrm{Î¸}}_{2}â‰¤2\mathrm{Ï€}$, $z=r{e}^{i\mathrm{Î¸}}$, it follows from Theorem 3.2 that

${âˆ«}_{{\mathrm{Î¸}}_{1}}^{{\mathrm{Î¸}}_{2}}\mathrm{â„œ}\left\{1+\frac{z{f}^{â€³}\left(z\right)}{{f}^{â€²}\left(z\right)}\right\}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}>âˆ’\mathrm{Ï€}\phantom{\rule{1em}{0ex}}\left(zâˆˆE\right).$

This is a necessary and sufficient condition for f to be close-to-convex (hence univalent) in E; see [7]. This also shows that $S{T}_{s}\left(\mathrm{Î²}\right)âŠ‚K$.

1. 2.

For $k=1$ ${âˆ«}_{{\mathrm{Î¸}}_{1}}^{{\mathrm{Î¸}}_{2}}\mathrm{â„œ}\left\{1+\frac{z{f}^{â€³}\left(z\right)}{{f}^{â€²}\left(z\right)}\right\}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}>âˆ’\frac{\mathrm{Ï€}}{2}$.

2. 3.

When $kâˆˆ\left[0,1\right]$, it is obvious that $\mathrm{Ïƒ}âˆˆ\left(0,1\right]$. In this case, the class $kâˆ’S{T}_{s}\left(\mathrm{Î²}\right)$ consists of strongly close-to-convex functions of order Ïƒ in the sense of Pommerenke [20, 21].

Theorem 3.3 (Integral representation)

Let $fâˆˆkâˆ’S{T}_{s}\left(\mathrm{Î²}\right)$. Then

${f}^{â€²}\left(z\right)=\frac{1}{2}p\left(z\right)exp{âˆ«}_{0}^{z}\frac{1}{t}\left[p\left(t\right)+p\left(âˆ’t\right)âˆ’2\right]\phantom{\rule{0.2em}{0ex}}dt,$

where $pâˆˆP\left({p}_{k,\mathrm{Î²}}\right)$, $zâˆˆE$.

Proof Since $fâˆˆkâˆ’S{T}_{s}\left(\mathrm{Î²}\right)$, we can write

$\frac{2z{f}^{â€²}\left(z\right)}{f\left(z\right)âˆ’f\left(âˆ’z\right)}=p\left(z\right),\phantom{\rule{1em}{0ex}}pâˆˆP\left({p}_{k,\mathrm{Î²}}\right).$

This gives us

$\frac{2{\left[f\left(z\right)âˆ’f\left(âˆ’z\right)\right]}^{â€²}}{f\left(z\right)âˆ’f\left(âˆ’z\right)}âˆ’\frac{1}{z}=\frac{1}{2}\left[p\left(z\right)âˆ’p\left(âˆ’z\right)âˆ’2\right]$

and the result follows when we integrate.â€ƒâ–¡

When $k=0$, $\mathrm{Î²}=0$, we obtain the result for the class ${S}_{s}^{âˆ—}$ given in [5].

We now study the class $kâˆ’S{T}_{s}\left(\mathrm{Î²}\right)$ under a certain integral operator.

Theorem 3.4 Let $gâˆˆkâˆ’S{T}_{s}\left(\mathrm{Î²}\right)$ and let for $m=1,2,3,â€¦,G$ be defined by

$G\left(z\right)=\frac{m+1}{2{z}^{m}}{âˆ«}_{0}^{z}{t}^{mâˆ’1}\left\{g\left(t\right)âˆ’g\left(âˆ’t\right)\right\}\phantom{\rule{0.2em}{0ex}}dt.$
(3.6)

Then $G\left(z\right)$ belongs to $kâˆ’S{T}_{s}\left(\mathrm{Î²}\right)$ in E.

Proof Let

$J\left(z\right)={âˆ«}_{0}^{z}{t}^{mâˆ’1}\frac{g\left(t\right)âˆ’g\left(âˆ’t\right)}{2}\phantom{\rule{0.2em}{0ex}}dt.$

Since $gâˆˆkâˆ’S{T}_{s}\left(\mathrm{Î²}\right)$, $\frac{1}{2}\left\{g\left(z\right)âˆ’g\left(âˆ’z\right)\right\}âˆˆkâˆ’ST\left(\mathrm{Î²}\right)âŠ‚{S}^{âˆ—}\left({\mathrm{Î²}}_{1}\right)âŠ‚{S}^{âˆ—}$, and ${\mathrm{Î²}}_{1}=\frac{k+\mathrm{Î²}}{k+1}$. Therefore it can easily be verified that $J\left(z\right)$ is $\left(m+1\right)$-valently starlike in E.

We can write (3.6) as

${z}^{m}G\left(z\right)=\left(m+1\right)J\left(z\right),$

and, differentiating logarithmically, we have

$\frac{z{G}^{â€²}\left(z\right)}{G\left(z\right)}=\frac{z{J}^{â€²}\left(z\right)âˆ’mJ\left(z\right)}{J\left(z\right)}=\frac{N\left(z\right)}{D\left(z\right)},$

say, where $N\left(0\right)=D\left(0\right)=0$ and D is $\left(m+1\right)$-valently starlike.

Let

$\frac{N\left(z\right)}{D\left(z\right)}=h\left(z\right).$

Then

$\begin{array}{rl}\frac{{N}^{â€²}\left(z\right)}{{D}^{â€²}\left(z\right)}& =h\left(z\right)+\frac{z{h}^{â€²}\left(z\right)}{{h}_{0}\left(z\right)},\phantom{\rule{1em}{0ex}}{h}_{0}\left(z\right)=\frac{z{D}^{â€²}\left(z\right)}{D\left(z\right)}âˆˆP\\ =h\left(z\right)+{H}_{0}\left(z\right)\left(z{h}^{â€²}\left(z\right)\right),\phantom{\rule{1em}{0ex}}{H}_{0}=\frac{1}{{h}_{0}}âˆˆP.\end{array}$
(3.7)

Since

$\begin{array}{rl}\frac{{N}^{â€²}\left(z\right)}{{D}^{â€²}\left(z\right)}& =\frac{{\left(z{h}^{â€²}\left(z\right)\right)}^{â€²}âˆ’m{J}^{â€²}\left(z\right)}{{J}^{â€²}\left(z\right)}\\ =\left\{\frac{{\left(z{J}^{â€²}\left(z\right)\right)}^{â€²}}{{J}^{â€²}\left(z\right)}âˆ’m\right\}âˆˆP\left({p}_{k,\mathrm{Î²}}\right).\end{array}$

We now apply Lemma 2.2 to obtain

$\frac{N\left(z\right)}{D\left(z\right)}=\frac{z{G}^{â€²}\left(z\right)}{G\left(z\right)}âˆˆP\left({p}_{k,\mathrm{Î²}}\right),\phantom{\rule{1em}{0ex}}zâˆˆE.$

This proves that $Gâˆˆkâˆ’ST\left(\mathrm{Î²}\right)$ in E.â€ƒâ–¡

Theorem 3.5 Let $f,gâˆˆkâˆ’S{T}_{s}\left(\mathrm{Î²}\right)$ and let F be defined by the following integral operator:

$F\left(z\right)=\left(\mathrm{Î³}+\frac{1}{\mathrm{Î´}}\right){z}^{1âˆ’\frac{1}{\mathrm{Î´}}}{âˆ«}_{0}^{z}{t}^{\frac{1}{\mathrm{Î´}}âˆ’2}{\left[\frac{f\left(t\right)âˆ’f\left(âˆ’t\right)}{2}\right]}^{\frac{1}{1+\mathrm{Î³}}}\left[\frac{g\left(t\right)âˆ’g\left(âˆ’t\right)}{2}\right]\phantom{\rule{0.2em}{0ex}}dt,$
(3.8)

where $zâˆˆE$, $\mathrm{Î´}>0$, $\mathrm{Î³}â‰¥0$ and $\left[\frac{k\left(1+\mathrm{Î³}\right)}{k+1}+\left(\frac{1}{\mathrm{Î´}}âˆ’1\right)\right]>\mathrm{Î²}$. Then $F\left(z\right)$ belongs to $kâˆ’ST\left(\mathrm{Î²}\right)$ for $zâˆˆE$.

When $g\left(z\right)=z$, $\mathrm{Î³}=0$, we obtain a generalized form of the Bernardi operator; see [1]. Also for $g\left(z\right)=z$, $\mathrm{Î³}=0$, and $\mathrm{Î´}=\frac{1}{2}$, we have the well-known integral operator studied by Libera [11] who showed that it preserves the geometric properties of convexity, starlikeness, and close-to-convexity.

Proof Let $\frac{f\left(z\right)âˆ’f\left(âˆ’z\right)}{2}={\mathrm{Î¨}}_{1}\left(z\right)$, $\frac{g\left(z\right)âˆ’g\left(âˆ’z\right)}{2}={\mathrm{Î¨}}_{2}\left(z\right)$. Then ${\mathrm{Î¨}}_{1},{\mathrm{Î¨}}_{2}âˆˆkâˆ’ST\left(\mathrm{Î²}\right)$ in E. We can write (3.8) as

$F\left(z\right)=\left(\mathrm{Î³}+\frac{1}{\mathrm{Î´}}\right){z}^{1âˆ’\frac{1}{\mathrm{Î´}}}{âˆ«}_{0}^{z}{t}^{\frac{1}{\mathrm{Î´}}âˆ’2}{\left({\mathrm{Î¨}}_{1}\left(t\right)\right)}^{\frac{1}{1+\mathrm{Î³}}}\left({\mathrm{Î¨}}_{2}\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt.$
(3.9)

Differentiating (3.9) logarithmically, and with $p\left(z\right)=\frac{z{F}^{â€²}\left(z\right)}{F\left(z\right)}$, we have

$\frac{\mathrm{Î³}}{1+\mathrm{Î³}}\frac{z{\mathrm{Î¨}}_{1}^{â€²}}{{\mathrm{Î¨}}_{1}\left(z\right)}+\frac{1}{1+\mathrm{Î³}}\frac{z{\mathrm{Î¨}}_{2}^{â€²}}{{\mathrm{Î¨}}_{2}\left(z\right)}=p\left(z\right)+\frac{z{p}^{â€²}\left(z\right)}{\left(1+\mathrm{Î³}\right)p\left(z\right)+\left(\frac{1}{\mathrm{Î´}}âˆ’1\right)}.$
(3.10)

Since, for $i=1,2$, ${\mathrm{Î¨}}_{i}âˆˆkâˆ’ST\left(\mathrm{Î²}\right)$, $\frac{z{\mathrm{Î¨}}_{1}^{â€²}\left(z\right)}{{\mathrm{Î¨}}_{1}}={h}_{1}\left(z\right)$, $\frac{z{\mathrm{Î¨}}_{2}^{â€²}\left(z\right)}{{\mathrm{Î¨}}_{2}}={h}_{2}\left(z\right)$ both belong to $P\left({p}_{k,\mathrm{Î²}}\right)$ in E, and $P\left({p}_{k,\mathrm{Î²}}\right)$ is a convex set. Therefore

$\left(\frac{\mathrm{Î³}}{1+\mathrm{Î³}}{h}_{1}\left(z\right)+\frac{1}{1+\mathrm{Î³}}{h}_{2}\left(z\right)\right)âˆˆP\left({p}_{k,\mathrm{Î²}}\right),\phantom{\rule{1em}{0ex}}zâˆˆE.$
(3.11)

From (3.10) and (3.11), it follows that

$\left(p\left(z\right)+\frac{z{p}^{â€²}\left(z\right)}{\left(1+\mathrm{Î³}\right)p\left(z\right)+\left(\frac{1}{\mathrm{Î´}}âˆ’1\right)}\right)â‰º{p}_{k,\mathrm{Î²}}\left(z\right).$

We now apply Lemma 2.3 which gives us

$p\left(z\right)â‰º{q}_{k,\mathrm{Î²}}\left(z\right)â‰º{p}_{k,\mathrm{Î²}}\left(z\right).$

Thus $Fâˆˆkâˆ’ST\left(\mathrm{Î²}\right)$ and the proof is complete.â€ƒâ–¡

## 4 The class $kâˆ’U{K}_{s}\left(\mathrm{Î²}\right)$

Here we shall study some properties of the class $kâˆ’U{K}_{s}\left(\mathrm{Î²}\right)$ which consists of k-uniformly close-to-convex functions.

Let $L\left(r,f\right)$ denote the length of the image of the circle $|z|=r$ under f. We prove the following.

Theorem 4.1 Let $fâˆˆkâˆ’U{K}_{s}\left(\mathrm{Î²}\right)$. Then, for $0, $kâˆˆ\left[0,1\right]$,

$L\left(r,f\right)=O\left(1\right){\left(\frac{1}{1âˆ’r}\right)}^{\mathrm{Ïƒ}âˆ’{\mathrm{Î²}}_{1}},\phantom{\rule{1em}{0ex}}{\mathrm{Î²}}_{1}<\frac{\mathrm{Ïƒ}}{2},$

where ${\mathrm{Î²}}_{1}=\frac{k+\mathrm{Î²}}{k+1}$ and Ïƒ is given by (1.3), and $O\left(1\right)$ is a constant depending only on k, Î².

Proof For $fâˆˆkâˆ’U{K}_{s}\left(\mathrm{Î²}\right)$, we can write

$z{f}^{â€²}\left(z\right)=\mathrm{Î¨}\left(z\right){h}^{\mathrm{Ïƒ}}\left(z\right),\phantom{\rule{1em}{0ex}}hâˆˆP,\mathrm{Î¨}âˆˆ{S}^{âˆ—}\left({\mathrm{Î²}}_{1}\right),$
(4.1)

and $\mathrm{Î¨}\left(z\right)=\left\{g\left(z\right)âˆ’g\left(âˆ’z\right)\right\}$, $gâˆˆkâˆ’S{T}_{s}\left(\mathrm{Î²}\right)$.

Since $\mathrm{Î¨}âˆˆ{S}^{âˆ—}\left({\mathrm{Î²}}_{1}\right)$ and is odd, there exists an odd starlike function ${\mathrm{Î¨}}_{1}\left(z\right)$ such that

$\mathrm{Î¨}\left(z\right)=z{\left(\frac{{\mathrm{Î¨}}_{1}\left(z\right)}{z}\right)}^{1âˆ’{\mathrm{Î²}}_{1}}=z{\left(\frac{{\mathrm{Î¨}}_{1}\left(z\right)}{z}\right)}^{\frac{1âˆ’{\mathrm{Î²}}_{1}}{k+1}}.$

Thus, with $z=r{e}^{i\mathrm{Î¸}}$,

$L\left(r,f\right)={âˆ«}_{0}^{2\mathrm{Ï€}}|z{f}^{â€²}\left(z\right)|\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}={âˆ«}_{0}^{2\mathrm{Ï€}}|{z}^{{\mathrm{Î²}}_{1}}{\left({\mathrm{Î¨}}_{1}\left(z\right)\right)}^{1âˆ’{\mathrm{Î²}}_{1}}{h}^{\mathrm{Ïƒ}}\left(z\right)|\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸},$

and using HÃ¶lderâ€™s inequality, we have

$L\left(r,f\right)â‰¤2\mathrm{Ï€}{r}^{{\mathrm{Î²}}_{1}}{\left(\frac{1}{2\mathrm{Ï€}}{âˆ«}_{0}^{2\mathrm{Ï€}}|{\mathrm{Î¨}}_{1}\left(z\right){|}^{\left(1âˆ’\mathrm{Î²}\right)\left(\frac{z}{zâˆ’\mathrm{Ïƒ}}\right)}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right)}^{\frac{2âˆ’\mathrm{Ïƒ}}{z}}{\left(\frac{1}{2\mathrm{Ï€}}{âˆ«}_{0}^{2\mathrm{Ï€}}|h\left(z\right){|}^{2}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\right)}^{\frac{\mathrm{Ïƒ}}{2}}.$
(4.2)

For $hâˆˆP$, it is well known [20] that

$\frac{1}{2\mathrm{Ï€}}{âˆ«}_{0}^{2\mathrm{Ï€}}|h\left(z\right){|}^{2}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}â‰¤\frac{1+3{r}^{2}}{1âˆ’{r}^{2}}.$
(4.3)

Using (4.3) and subordination for odd starlike functions in (4.2), it follows that

$\begin{array}{rl}L\left(r,f\right)& â‰¤C\left({\mathrm{Î²}}_{1},\mathrm{Ïƒ}\right){\left(\frac{1}{1âˆ’{r}^{2}}\right)}^{\left[\left(1âˆ’{\mathrm{Î²}}_{1}\right)\left(\frac{2}{2âˆ’\mathrm{Ïƒ}}\right)âˆ’1\right]{\left[\frac{1+3{r}^{2}}{1âˆ’r}\right]}^{\frac{\mathrm{Ïƒ}}{2}}}\\ =O\left(1\right){\left(\frac{1}{1âˆ’r}\right)}^{\mathrm{Ïƒ}âˆ’{\mathrm{Î²}}_{1}},\end{array}$

where C and $O\left(1\right)$ are constants depending only on ${\mathrm{Î²}}_{1}$ and Ïƒ. This completes the proof.â€ƒâ–¡

We now discuss the growth rate of coefficients of $fâˆˆkâˆ’U{K}_{s}\left(\mathrm{Î²}\right)$.

Theorem 4.2 Let $fâˆˆkâˆ’U{K}_{s}\left(\mathrm{Î²}\right)$ and be given by (1.1). Then

${a}_{n}=O\left(1\right){n}^{\mathrm{Ïƒ}âˆ’{\mathrm{Î²}}_{1}âˆ’1},\phantom{\rule{1em}{0ex}}nâ‰¥1,{\mathrm{Î²}}_{1}<\frac{\mathrm{Ïƒ}}{2},$

where $O\left(1\right)$ is a constant depending only on Ïƒ and ${\mathrm{Î²}}_{1}$ and Ïƒ, ${\mathrm{Î²}}_{1}$ are as given in Theorem  4.1.

Proof For $z=r{e}^{i\mathrm{Î¸}}$, $nâ‰¥1$, Cauchyâ€™s Theorem gives us

$\begin{array}{rl}n|{a}_{n}|& =\frac{1}{2\mathrm{Ï€}{r}^{n+1}}|{âˆ«}_{0}^{2\mathrm{Ï€}}z{f}^{â€²}\left(z\right){e}^{âˆ’in\mathrm{Î¸}}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}|\\ â‰¤\frac{1}{2\mathrm{Ï€}{r}^{n+1}}{âˆ«}_{0}^{2\mathrm{Ï€}}|z{f}^{â€²}\left(z\right)|\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¸}\\ =\frac{1}{2\mathrm{Ï€}{r}^{n}}L\left(r,f\right).\end{array}$

With $r=\left(1âˆ’\frac{1}{n}\right)$, we use Theorem 4.1 and obtain the required result.â€ƒâ–¡

Theorem 4.3 Let $fâˆˆkâˆ’U{K}_{s}\left(\mathrm{Î²}\right)$ and let F be defined by

$F\left(z\right)=\frac{m+1}{2{z}^{m}}{âˆ«}_{0}^{z}{t}^{mâˆ’1}\left\{f\left(t\right)âˆ’f\left(âˆ’t\right)\right\}\phantom{\rule{0.2em}{0ex}}dt.$
(4.4)

Then $Fâˆˆkâˆ’U{K}_{s}\left(\mathrm{Î²}\right)$ in E. That is, the class $kâˆ’U{K}_{s}\left(\mathrm{Î²}\right)$ is preserved under the integral operator (4.4).

Proof Since $fâˆˆkâˆ’U{K}_{s}\left(\mathrm{Î²}\right)$, we can write

$\left\{\frac{2z{f}^{â€²}\left(z\right)}{g\left(z\right)âˆ’g\left(âˆ’z\right)}\right\}âˆˆP\left({p}_{k,\mathrm{Î²}}\right),\phantom{\rule{1em}{0ex}}gâˆˆkâˆ’S{T}_{s}\left(\mathrm{Î²}\right)âŠ‚{S}_{S}^{âˆ—}\left({\mathrm{Î²}}_{1}\right).$

Let $G\left(z\right)=\frac{1}{2}\left\{{g}_{1}\left(z\right)âˆ’{g}_{1}\left(âˆ’z\right)\right\}$ and be defined by (3.5). By Theorem 3.4, ${g}_{1}âˆˆkâˆ’ST\left(\mathrm{Î²}\right)$ and $Gâˆˆkâˆ’{S}_{s}T\left(\mathrm{Î²}\right)âŠ‚{S}_{s}^{âˆ—}\left({\mathrm{Î²}}_{1}\right)$. Let $G=z{G}_{1}^{â€²}$. Then we can write

${G}_{1}^{â€²}\left(z\right)=\frac{1}{2}{\left[z{g}_{1}\left(z\right)âˆ’{g}_{1}\left(âˆ’z\right)\right]}^{â€²},\phantom{\rule{1em}{0ex}}{G}_{1}âˆˆkâˆ’UC{V}_{s}\left(\mathrm{Î²}\right).$

Thus, from (4.4) and $g=z{g}_{1}^{â€²}$, ${g}_{1}âˆˆ{C}_{s}\left({\mathrm{Î²}}_{1}\right)$, we have

$\begin{array}{rl}\frac{2{F}^{â€²}\left(z\right)}{{\left[{g}_{1}\left(z\right)âˆ’{g}_{1}\left(âˆ’z\right)\right]}^{â€²}}=& \frac{{z}^{m}\left\{f\left(z\right)âˆ’f\left(âˆ’z\right)\right\}âˆ’m{âˆ«}_{0}^{z}{t}^{mâˆ’1}\left\{f\left(t\right)âˆ’f\left(âˆ’t\right)\right\}\phantom{\rule{0.2em}{0ex}}dt}{{z}^{m}\left\{{g}_{1}\left(z\right)âˆ’{g}_{1}\left(âˆ’z\right)\right\}âˆ’m{âˆ«}_{0}^{z}{t}^{mâˆ’1}\left\{{g}_{1}\left(t\right)âˆ’{g}_{1}\left(âˆ’t\right)\right\}\phantom{\rule{0.2em}{0ex}}dt}\\ =& \frac{N\left(z\right)}{D\left(z\right)},\end{array}$

say. We note that $N\left(0\right)=D\left(0\right)=0$, and for ${g}_{1}âˆˆ{C}_{S}\left({\mathrm{Î²}}_{1}\right)$,

$\begin{array}{rl}\frac{{\left(z{D}^{â€²}\left(z\right)\right)}^{â€²}}{{D}^{â€²}\left(z\right)}& =m+\frac{{\left\{z{\left[{g}_{1}\left(z\right)âˆ’{g}_{1}\left(âˆ’z\right)\right]}^{â€²}\right\}}^{â€²}}{{\left\{{g}_{1}\left(z\right)âˆ’{g}_{1}\left(âˆ’z\right)\right\}}^{â€²}}\\ =m+{h}_{1}\left(z\right),\phantom{\rule{1em}{0ex}}{h}_{1}âˆˆP\left({\mathrm{Î²}}_{1}\right).\end{array}$

Since $P\left({\mathrm{Î²}}_{1}\right)$ is a convex set, $Dâˆˆ{C}_{s}\left({\mathrm{Î²}}_{1}\right)âŠ‚{S}^{âˆ—}$ in E. We thus have

$\frac{{N}^{â€²}\left(z\right)}{{D}^{â€²}\left(z\right)}=\frac{1}{2}\left[\frac{2z{f}^{â€²}\left(z\right)}{{\left[{g}_{1}\left(z\right)âˆ’{g}_{1}\left(âˆ’z\right)\right]}^{â€²}}+\frac{2\left(âˆ’z\right){f}^{â€²}\left(âˆ’z\right)}{{\left[{g}_{1}\left(âˆ’z\right)âˆ’{g}_{1}\left(z\right)\right]}^{â€²}}\right]âˆˆP\left({p}_{k,\mathrm{Î²}}\right).$

Now, using Lemma 2.2, it follows that

This proves that $Fâˆˆkâˆ’U{K}_{S}\left(\mathrm{Î²}\right)$ in E.â€ƒâ–¡

We study a partial converse of the above result as follows.

Theorem 4.4 Let $\left(\frac{2z{f}^{â€²}\left(z\right)}{g\left(z\right)âˆ’g\left(âˆ’z\right)}\right)â‰º{p}_{k}\left(z\right)$ in E and let

${F}_{1}\left(z\right)=\frac{1}{1+m}{z}^{1âˆ’m}{\left({z}^{m}f\left(z\right)\right)}^{â€²},\phantom{\rule{1em}{0ex}}m=1,2,3,â€¦.$
(4.5)

Then ${F}_{1}âˆˆ{K}_{s}$ for $|z|<{r}_{1}$, where

${r}_{1}=\left\{\frac{m+1}{\left(2âˆ’{\mathrm{Î²}}_{1}\right)+\sqrt{{\left(zâˆ’{\mathrm{Î²}}_{1}\right)}^{2}+\left(m+1\right)\left(mâˆ’1+2{\mathrm{Î²}}_{1}\right)}}\right\},\phantom{\rule{1em}{0ex}}{\mathrm{Î²}}_{1}=\frac{k+\mathrm{Î²}}{k+1}.$
(4.6)

Proof We shall need the following well-known results for $pâˆˆP\left(\mathrm{Î±}\right)$, $0â‰¤\mathrm{Î±}<1$; see [4]:

$\frac{1âˆ’\left(1âˆ’2\mathrm{Î±}\right)r}{1+r}â‰¤|p\left(z\right)|â‰¤\frac{1+\left(1âˆ’2\mathrm{Î±}\right)r}{1âˆ’r},$
(4.7)
$|{p}^{â€²}\left(z\right)|â‰¤\frac{2\left[\mathrm{â„œ}p\left(z\right)âˆ’\mathrm{Î±}\right]}{1âˆ’{r}^{2}}.$
(4.8)

Since $fâˆˆkâˆ’U{K}_{s}\left(\mathrm{Î²}\right)$, there exists $gâˆˆ{S}_{s}^{âˆ—}\left({\mathrm{Î²}}_{1}\right)$ such that, for $zâˆˆE$.

$\left(\frac{2z{f}^{â€²}\left(z\right)}{g\left(z\right)âˆ’g\left(âˆ’z\right)}\right)=p\left(z\right),\phantom{\rule{1em}{0ex}}pâˆˆP\left({p}_{k}\right)âŠ‚P\left(\mathrm{Î±}\right),\mathrm{Î±}=\frac{k}{k+1}.$

From (4.5), we have

${F}_{1}\left(z\right)=\frac{1}{1+m}\left[mf\left(z\right)+z{f}^{â€²}\left(z\right)\right],$

and this gives us

$\begin{array}{rl}\frac{2z{F}_{1}^{â€²}\left(z\right)}{g\left(z\right)âˆ’g\left(âˆ’z\right)}& =\frac{1}{m+1}\left[\frac{2m{f}^{â€²}\left(z\right)}{g\left(z\right)âˆ’g\left(âˆ’z\right)}+\frac{2z{\left(z{f}^{â€²}\left(z\right)\right)}^{â€²}}{g\left(z\right)âˆ’g\left(âˆ’z\right)}\right]\\ =\frac{1}{m+1}\left[mp\left(z\right)+z{p}^{â€²}\left(z\right)+p\left(z\right)h\left(z\right)\right],\end{array}$

where

$h\left(z\right)=\frac{z{\mathrm{Î¨}}^{â€²}\left(z\right)}{\mathrm{Î¨}\left(z\right)}âˆˆP\left({\mathrm{Î²}}_{1}\right),\phantom{\rule{1em}{0ex}}\mathrm{Î¨}\left(z\right)=g\left(z\right)âˆ’g\left(âˆ’z\right).$

Now, using (4.7) and (4.8), we have

$\begin{array}{rl}\mathrm{â„œ}\left\{\frac{2z{F}_{1}^{â€²}\left(z\right)}{g\left(z\right)âˆ’g\left(âˆ’z\right)}\right\}& â‰¥\frac{\left(\mathrm{â„œ}p\left(z\right)âˆ’\mathrm{Î±}\right)}{1+m}\left\{m+\frac{1âˆ’\left(1âˆ’2{\mathrm{Î²}}_{1}\right)r}{1+r}âˆ’\frac{2r}{1âˆ’{r}^{2}}\right\}\\ =\frac{\mathrm{â„œ}p\left(z\right)âˆ’\mathrm{Î±}}{1+m}\left[\frac{T\left(r\right)}{1âˆ’{r}^{2}}\right],\end{array}$
(4.9)

where

$T\left(r\right)=\left(m+1\right)âˆ’2\left(2âˆ’{\mathrm{Î²}}_{1}\right)r+\left(âˆ’mâˆ’2{\mathrm{Î²}}_{1}+1\right){r}^{2}.$

We note that $T\left(0\right)=1+m>0$ and $T\left(1\right)=âˆ’3<0$. So there exists ${r}_{1}âˆˆ\left(0,1\right)$. The right hand side of (4.9) is positive for $|z|<{r}_{1}$, where ${r}_{1}$ is given by (4.6). This implies that $Fâˆˆ{K}_{s}$ for $|z|<{r}_{1}$ and the proof is complete.â€ƒâ–¡

We have the following special cases.

1. 1.

For $k=0=\mathrm{Î²}$, $fâˆˆ{K}_{s}$. Then ${F}_{1}$, defined by (4.5) belongs to ${K}_{s}$ for $|z|<{r}_{0}=\frac{1+m}{2+\sqrt{3+{m}^{2}}}$.

2. 2.

When $m=1$ and ${\mathrm{Î²}}_{1}=0$ (that is, $k=0=\mathrm{Î²}$), then ${F}_{1}\left(z\right)=\frac{{\left(zf\left(z\right)\right)}^{â€²}}{2}$ belongs to the same class for $|z|<\frac{1}{2}$. This result has been proved by Livingston [12] for convex and starlike functions.

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## Acknowledgements

The author would like to thank editor and anonymous referee for their valuable suggestions. The author is grateful to Dr. SM Junaid Zaidi, Rector, COMSATS Institute of Information Technology, Pakistan for providing an excellent research and academic environment. This research is supported by HEC NRPU project No: 20-1966/R&D/11-2553, titled, Research unit od Academic Excellence in Geometric Function Theory and Applications.

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Noor, K.I. On uniformly univalent functions with respect to symmetrical points. J Inequal Appl 2014, 254 (2014). https://doi.org/10.1186/1029-242X-2014-254