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

$$f(z)=z+\sum _{n=2}^{\mathrm{\infty }}{a}_n{z}^n,$$

(1.1)

which are analytic in the open unit disc $E=\{z:|z|<1\}$. Let S, K, $S^{\ast }$, 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 [1–25] 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 γ contained in E, with center ξ also in E, the image curve $f(\gamma )$ is a convex arc. The class of uniformly convex functions is denoted by $UCV$. The corresponding class $UST$ is defined by the relation that $f\in UCV$ if, and only if, $z{f}^{\prime }\in UST$. It is well known [13] that $f\in UCV$ if, and only if

$$\left|\frac{z{f}^{″}(z)}{f^{\prime }(z)}\right|<\mathrm{\Re }\{1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}\}(z\in E).$$

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

For $0\le \beta <1$, the class $P(\beta )$ consists of functions $p(z)$ analytic in E with $p(0)=1$ such that $\mathrm{\Re }p(z)>\beta$ for $z\in E$, and, with $\beta =0$, we obtain the well-known class P of Carathéodory functions with positive real part.

For $k\in [0,\mathrm{\infty })$, the conic regions ${\mathrm{\Omega }}_k$ are defined as follows, see [5]:

$${\mathrm{\Omega }}_k=\{u+iv:u>k\sqrt{{(u-1)}^2+v^2}\}.$$

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

We shall consider the case when $k\in [0,1]$. Related to the domain ${\mathrm{\Omega }}_k$, the following functions $p_k(z)$, $k\in [0,1]$, play the role of extremal functions mapping in E onto ${\mathrm{\Omega }}_k$:

$$p_k(z)=\{\begin{array}{cc}\frac{1+z}{1-z}\hfill & (k=0),\hfill \\ 1+\frac{2}{{\pi }^2}{(log\frac{1+\sqrt{z}}{1-\sqrt{z}})}^2\hfill & (k=1),\hfill \\ 1+\frac{2}{1-k^2}{sinh}^2[(\frac{2}{\pi }arccosk)arctanh\sqrt{z}]\hfill & (0<k<1).\hfill \end{array}$$

(1.2)

These functions are univalent in E and belong to the class P. Using the subordination concept, we define the class $P(p_k)$ as follows.

Let $p(z)$ be analytic in E with $p(0)=1$. Then $p\in P(p_k)$ if, and only if, $p\prec p_k$ in E and $p_k(z)$ are given by (1.2).

The conic domains ${\mathrm{\Omega }}_k$ can be generalized as given by

$${\mathrm{\Omega }}_{k,\beta }=(1-\beta ){\mathrm{\Omega }}_k+\beta ,$$

with the corresponding extremal function

$$p_{k,\beta }(z)=(1-\beta )p_k+\beta (0\le \beta <1,k\in [0,1]).$$

It can easily be seen that the analytic function $p(z)$, with $p(0)=1$, belongs to the class $P(p_{k,\beta })$ if $p(z)\prec p_{k,\beta }(z)$ in E.

It is easy to verify that $P(p_{k,\beta })$ is a convex set. It is known [6] that

$$P(p_k)\subset P\left(\frac{k}{k+1}\right)\subset P,$$

and, for $p\in P(p_k)$, we have

$$|argp(z)|\le \sigma \frac{\pi }{2},$$

where

$$\sigma =\frac{2}{\pi }arctan\frac{1}{k}.$$

(1.3)

So we can write $p(z)=h^{\sigma }(z)$, $h\in P$.

Also

$$P(p_{k,\beta })\subset P\left(\frac{k+\beta }{k+1}\right)\subset P.$$

Sakaguchi [24] introduced and studied the class $S_s^{\ast }$ of starlike functions with respect to symmetrical points. The class $S_s^{\ast }$ 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 $f\in S_s^{\ast }$ to be univalent and starlike with respect to symmetrical points in E is that

$$\left(\frac{2z{f}^{\prime }(z)}{f(z)-f(-z)}\right)\in P,z\in E.$$

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 $f\in C_s$ is that

$$\frac{2{(z{f}^{\prime }(z))}^{\prime }}{{(f(z)-f(-z))}^{\prime }}\in P,z\in E.$$

It is also well known [2] that $f\in C_s$ if, and only if, $z{f}^{\prime }\in S_s^{\ast }$.

We now define the following.

Definition 1.1 Let $f\in A$. The f is said to be in the class $k-S{T}_s(\beta )$ if, and only if,

$$\frac{2z{f}^{\prime }(z)}{(f(z)-f(-z))}\in P(p_{k,\beta }),z\in E.$$

It can easily be seen that

$$k-S{T}_s(\beta )\subset S_s^{\ast }\subset S_s^{\ast },{\beta }_1=\frac{k+\beta }{k+1}.$$

Also, for $\beta =0=k$, the class $k-S{T}_s(\beta )$ reduces to $S_s^{\ast }$.

The class $k-UC{V}_s(\beta )$ is defined as follows.

Definition 1.2 Let $f\in A$. Then $f\in k-UC{V}_s(\beta )$ if, and only if $z{f}^{\prime }\in k-S{T}_s(\beta )$ for $z\in E$.

We note that

$$k-UC{V}_s(\beta )\subset C_s({\beta }_1)\subset C_s,{\beta }_1=\frac{k+\beta }{k+1}.$$

Definition 1.3 Let $f\in A$. Then $f\in k-U{K}_s(\beta )$ if, and only if, there exists $g\in k-S{T}_s(\beta )$ such that

$$\left(\frac{2z{f}^{\prime }(z)}{g(z)-g(-z)}\right)\in P(p_{k,\beta }),z\in E.$$

Since $P(p_{k,\beta })\subset P({\beta }_1)\subset P$, ${\beta }_1=\frac{k+\beta }{k+1}$, and $k-S{T}_s(\beta )\subset S_s^{\ast }$, we note that

$$k-U{K}_s(\beta )\subset K_s\subset K,$$

where $k_S$ consists of close-to-convex functions with respect to symmetrical starlike functions.

From the definition, it is clear that $k-U{K}_s(\beta )$ consists of univalent functions.

For $k=0$, $\beta =0$ and $f(z)=g(z)$, $k-U{K}_s(\beta )$ reduces to the class $S_s^{\ast }$.

2 Preliminary results

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

Lemma 2.1 [15]

Let $q(z)$ be a convex function in E with $q(0)=1$ and let another function $h:E\to \mathbb{C}$ be with $\mathrm{\Re }h(z)>0$. Let $p(z)$ be analytic in E with $p(0)=1$ such that

$$(p(z)+h(z)z{p}^{\prime }(z))\prec q(z),z\in E.$$

Then $p(z)\prec q(z)$, $z\in E$.

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

$$N(0)=0=D(z)$$

and let $D\in S^{\ast }$ for $z\in E$. Then $\frac{N^{\prime }(z)}{D^{\prime }(z)}\in P(p_{k,\beta })$ implies that $\frac{N(z)}{D(z)}\in P(p_{k,\beta })$ for $z\in E$.

Proof Let

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

Then

$$\frac{N^{\prime }(z)}{D^{\prime }(z)}=p(z)+h(z)(z{p}^{\prime }(z)),h(z)=\frac{1}{h_0(z)},$$

where

$$h_0(z)=\frac{z{D}^{\prime }(z)}{D(z)}\in P.$$

Since $\frac{N^{\prime }(z)}{D^{\prime }(z)}\in P(p_{k,\beta })$, we have

$$\frac{N^{\prime }(z)}{D^{\prime }(z)}=(p(z)+h(z)(z{p}^{\prime }(z)))\prec p_{k,\beta }(z),z\in E.$$

We now use Lemma 2.1 and this implies that

$$\frac{N(z)}{D(z)}=p(z)\prec p_{k,\beta }(z)\text{in }E.$$

This proves that $\frac{N(z)}{D(z)}\in P(p_{k,\beta })$ for $z\in E$. □

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

Lemma 2.3 Let $k\in [0,\mathrm{\infty })$ and ${\gamma }_1$, ${\delta }_1$ be any complex numbers with ${\gamma }_1\ne 0$ and let $\mathrm{\Re }\{\frac{{\gamma }_{1}k}{k+1}+{\delta }_1\}>\beta$. If $h(z)$ is analytic in E, $h(0)=1$ and it satisfies

$$(h(z)+\frac{z{h}^{\prime }(z)}{{\gamma }_{1}h(z)+{\delta }_1})\prec p_{k,\beta }(z),$$

(2.1)

and $q_{k,\beta }(z)$ is an analytic solution of

$$(q_{k,\beta }(z)+\frac{z{q}_{k,\beta }^{\prime }(z)}{{\gamma }_1{q}_{k,\beta }(z)+{\delta }_1})=p_{k,\beta }(z),$$

then $q_{k,\beta }$ is univalent and

$$h(z)\prec q_{k,\beta }(z)\prec p_{k,\beta }(z),$$

and $q_{k,\beta }(z)$ is the best dominant of (2.1).

3 The class $k-S{T}_s(\beta )$

In this section, we shall study some basic properties of the class $k-S{T}_s(\beta )$.

Theorem 3.1 Let $f\in k-S{T}_s(\beta )$. Then the odd function

$$\mathrm{\Psi }(z)=\frac{1}{2}[f(z)-f(-z)],$$

(3.1)

belongs to $k-ST(\beta )$ in E.

In particular $\mathrm{\Psi }(z)$ is an odd starlike function of order ${\beta }_1=\frac{k+\beta }{k+1}$ in E.

Proof Logarithmic differentiation of (3.1) and simple computation yield

$$\begin{array}{rl}\frac{z{\mathrm{\Psi }}^{\prime }(z)}{\mathrm{\Psi }(z)}& =\frac{1}{2}[\frac{2z{f}^{\prime }(z)}{f(z)-f(-z)}+\frac{2(-z)f^{\prime }(-z)}{f(-z)-f(z)}]\\ =\frac{1}{2}[p_1(z)+p_2(z)],\text{for }z\in E,p_1,p_2\in P(p_{k,\beta }).\end{array}$$

Since $P(p_{k,\beta })$ is a convex set, it follows that $\frac{z{\mathrm{\Psi }}^{\prime }(z)}{\mathrm{\Psi }(z)}\in P(p_{k,\beta })$ and thus $\mathrm{\Psi }\in k-ST(\beta )$ in E. □

As a special case, we note that, for $k=0=\beta$, $\frac{1}{2}[f(z)-f(-z)]=\mathrm{\Psi }(z)\in S^{\ast }$ in E, and hence $\frac{z{f}^{\prime }}{\mathrm{\Psi }}\in P$. We now discuss a geometric property for $f\in k-S{T}_s(\beta )$. Here we investigate the behavior of the inclusion of the tangent at a point $w(\theta )=f(r{e}^{i\theta })$ to the image ${\mathrm{\Gamma }}_r$ of the circle $C_r=\{z:|z|=r\}$, $0\le r<1$, $\theta \in [0,2\pi ]$, under the mapping by means of a function f from the class $f\in k-S{T}_s(\beta )$.

Let

$$\mathrm{\Phi }(\theta )=\frac{\pi }{2}+\theta +arg{f}^{\prime }\left(r{e}^{i\theta }\right)=arg\frac{\partial }{\partial \theta }f\left(r{e}^{i\theta }\right),$$

and, for ${\theta }_2>{\theta }_1$, ${\theta }_1,{\theta }_2\in [0,2\pi ]$,

$$\mathrm{\Phi }({\theta }_2)-\mathrm{\Phi }({\theta }_1)={\theta }_2+arg{f}^{\prime }\left(r{e}^{i{\theta }_2}\right)-{\theta }_1-arg{f}^{\prime }\left(r{e}^{i{\theta }_1}\right).$$

Now, since

$$\theta +arg{f}^{\prime }\left(r{e}^{i\theta }\right)=\theta +\mathrm{\Re }\{-iln{f}^{\prime }\left(r{e}^{i\theta }\right)\},$$

then

$$\frac{\partial }{\partial \theta }(\theta +arg{f}^{\prime }\left(r{e}^{i\theta }\right))=\mathrm{\Re }\{1+\frac{r{e}^{i\theta }{f}^{″}(r{e}^{i\theta })}{f^{\prime }(r{e}^{i\theta })}\}.$$

Hence

$${\int }_{{\theta }_1}^{{\theta }_2}\frac{\partial }{\partial \theta }(\theta +arg{f}^{\prime }\left(r{e}^{i\theta }\right))d\theta ={\int }_{{\theta }_1}^{{\theta }_2}\mathrm{\Re }\{1+\frac{r{e}^{i\theta }{f}^{″}(r{e}^{i\theta })}{f^{\prime }(r{e}^{i\theta })}\}d\theta .$$

Also, on the other hand,

$$\begin{array}{rl}{\int }_{{\theta }_1}^{{\theta }_2}\frac{\partial }{\partial \theta }(\theta +arg{f}^{\prime }\left(r{e}^{i\theta }\right))d\theta & ={\theta }_2+arg{f}^{\prime }\left(r{e}^{i{\theta }_2}\right)-{\theta }_1-arg{f}^{\prime }\left(r{e}^{i{\theta }_1}\right)\\ =\mathrm{\Phi }({\theta }_2)-\mathrm{\Phi }({\theta }_1).\end{array}$$

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 ${\mathrm{\Gamma }}_r$ between the points $w({\theta }_2)$ and $w({\theta }_1)$ for ${\theta }_2>{\theta }_1$.

We have the following necessary condition for $f\in k-S{T}_s(\beta )$.

Theorem 3.2 Let $f\in k-S{T}_s(\beta )$. Then, with $z=r{e}^{i\theta }$ and $0\le {\theta }_1<{\theta }_2\le 2\pi$, $0\le \beta <1$ and $0\le k\le 1$, we have

$${\int }_{{\theta }_1}^{{\theta }_2}\mathrm{\Re }\left\{\frac{{(z{f}^{\prime }(z))}^{\prime }}{f^{\prime }(z)}\right\}d\theta >-\sigma \pi +2{cos}^{-1}\left\{\frac{2(1-\beta )}{1-(1-2\beta )r^2}\right\}+{\beta }_1({\theta }_2-{\theta }_1),$$

where σ is given by (1.3) and ${\beta }_1=\frac{k+\beta }{k+1}$.

Proof Since $\frac{f^{\prime }(z)}{{\mathrm{\Psi }}^{\prime }(z)}\in P(p_{k,\beta })$, $\mathrm{\Psi }(z)=\frac{1}{2}[f(z)-f(-z)]$ and $\mathrm{\Psi }\in k-UCV(\beta )\subset C(\beta )$.

We can write

$$f^{\prime }(z)={({\mathrm{\Psi }}_1^{\prime }(z))}^{1-{\beta }_1}{h}^{\sigma }(z),{\mathrm{\Psi }}_1\in C,h\in P(\beta ),$$

and this gives us, with $z=r{e}^{i\theta }$, $0\le r<1$, $0\le {\theta }_1<{\theta }_2\le 2\pi$,

$$\begin{array}{rl}{\int }_{{\theta }_1}^{{\theta }_2}\mathrm{\Re }\left\{\frac{{(z{f}^{\prime }(z))}^{\prime }}{f^{\prime }(z)}\right\}d\theta =& (1-{\beta }_1){\int }_{{\theta }_1}^{{\theta }_2}\mathrm{\Re }\left\{\frac{{(z{\mathrm{\Psi }}_1^{\prime }(z))}^{\prime }}{{\mathrm{\Psi }}_1^{\prime }(z)}\right\}d\theta \\ +\sigma {\int }_{{\theta }_1}^{{\theta }_2}\mathrm{\Re }\frac{2h^{\prime }(z)}{h(z)}d\theta +{\beta }_1({\theta }_2-{\theta }_1).\end{array}$$

(3.2)

For $h\in P(\beta )$, we observe that

$$\begin{array}{rl}\frac{\partial }{\partial \theta }argh\left(r{e}^{i\theta }\right)& =\frac{\partial }{\partial \theta }\mathrm{\Re }\{-ilnh\left(r{e}^{i\theta }\right)\}\\ =\mathrm{\Re }\left\{r{e}^{i\theta }\frac{h^{\prime }(r{e}^{i\theta })}{h(r{e}^{i\theta })}\right\}.\end{array}$$

Therefore

$${\int }_{{\theta }_1}^{{\theta }_2}\mathrm{\Re }\left\{\frac{r{e}^{i\theta }{h}^{\prime }(r{e}^{i\theta })}{h(r{e}^{i\theta })}\right\}d\theta =argh\left(r{e}^{i{\theta }_2}\right)-argh\left(r{e}^{i{\theta }_1}\right),$$

and

$$\underset{h\in P(\beta )}{max}\left|{\int }_{{\theta }_1}^{{\theta }_2}\mathrm{\Re }\left\{\frac{r{e}^{i\theta }{h}^{\prime }(r{e}^{i\theta })}{h(r{e}^{i\theta })}\right\}d\theta \right|=\underset{h\in P(\beta )}{max}|argh\left(r{e}^{i{\theta }_2}\right)-argh\left(r{e}^{i{\theta }_1}\right)|.$$

We can write

$$\frac{1}{1-\beta }[h(z)-\beta ]=p(z),p\in P,$$

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

$$|p(z)-\frac{1+r^2}{1-r^2}|\le \frac{2r}{1-r^2}.$$

From this, we have

$$|h(z)-\frac{1+(1-2\beta )r^2}{1-r^2}|\le \frac{2(1-\beta )r}{1-r^2}.$$

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

$$argh(z)=\pm {sin}^{-1}\left(\frac{2(1-\beta )r}{1-(1-2\beta )r^2}\right).$$

(3.3)

From (3.3), we observe that

$$\begin{array}{rl}\underset{h\in P(\beta )}{max}\left|{\int }_{{\theta }_1}^{{\theta }_2}\mathrm{\Re }\left\{r{e}^{i\theta }\frac{h^{\prime }(r{e}^{i\theta })}{h(r{e}^{i\theta })}\right\}d\theta \right|& \le 2{sin}^{-1}\left(\frac{2(1-\beta )r}{1-(1-2\beta )r^2}\right)\\ =\pi -2{cos}^{-1}\left(\frac{2(1-\beta )r}{1-(1-2\beta )r^2}\right).\end{array}$$

(3.4)

Also, for ${\mathrm{\Psi }}_1\in C$,

$${\int }_{{\theta }_1}^{{\theta }_2}\mathrm{\Re }\{1+r{e}^{i\theta }\frac{{\mathrm{\Psi }}_1^{″}(r{e}^{i\theta })}{{\mathrm{\Psi }}_1^{\prime }(r{e}^{i\theta })}\}d\theta \ge 0.$$

(3.5)

Using (3.4) and (3.5) in (3.2), we obtain the required result. □

We note the following special cases:

• 1. For $k=0$, $0\le {\theta }_1<{\theta }_2\le 2\pi$, $z=r{e}^{i\theta }$, it follows from Theorem 3.2 that

  $${\int }_{{\theta }_1}^{{\theta }_2}\mathrm{\Re }\{1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}\}d\theta >-\pi (z\in E).$$

This is a necessary and sufficient condition for f to be close-to-convex (hence univalent) in E; see [7]. This also shows that $S{T}_s(\beta )\subset K$.

• 2. For $k=1$ ${\int }_{{\theta }_1}^{{\theta }_2}\mathrm{\Re }\{1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}\}d\theta >-\frac{\pi }{2}$.

• 3. When $k\in [0,1]$, it is obvious that $\sigma \in (0,1]$. In this case, the class $k-S{T}_s(\beta )$ consists of strongly close-to-convex functions of order σ in the sense of Pommerenke [20, 21].

Theorem 3.3 (Integral representation)

Let $f\in k-S{T}_s(\beta )$. Then

$$f^{\prime }(z)=\frac{1}{2}p(z)exp{\int }_0^z\frac{1}{t}[p(t)+p(-t)-2]dt,$$

where $p\in P(p_{k,\beta })$, $z\in E$.

Proof Since $f\in k-S{T}_s(\beta )$, we can write

$$\frac{2z{f}^{\prime }(z)}{f(z)-f(-z)}=p(z),p\in P(p_{k,\beta }).$$

This gives us

$$\frac{2{[f(z)-f(-z)]}^{\prime }}{f(z)-f(-z)}-\frac{1}{z}=\frac{1}{2}[p(z)-p(-z)-2]$$

and the result follows when we integrate. □

When $k=0$, $\beta =0$, we obtain the result for the class $S_s^{\ast }$ given in [5].

We now study the class $k-S{T}_s(\beta )$ under a certain integral operator.

Theorem 3.4 Let $g\in k-S{T}_s(\beta )$ and let for $m=1,2,3,\dots ,G$ be defined by

$$G(z)=\frac{m+1}{2z^m}{\int }_0^z{t}^{m-1}\{g(t)-g(-t)\}dt.$$

(3.6)

Then $G(z)$ belongs to $k-S{T}_s(\beta )$ in E.

Proof Let

$$J(z)={\int }_0^z{t}^{m-1}\frac{g(t)-g(-t)}{2}dt.$$

Since $g\in k-S{T}_s(\beta )$, $\frac{1}{2}\{g(z)-g(-z)\}\in k-ST(\beta )\subset S^{\ast }({\beta }_1)\subset S^{\ast }$, and ${\beta }_1=\frac{k+\beta }{k+1}$. Therefore it can easily be verified that $J(z)$ is $(m+1)$-valently starlike in E.

We can write (3.6) as

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

and, differentiating logarithmically, we have

$$\frac{z{G}^{\prime }(z)}{G(z)}=\frac{z{J}^{\prime }(z)-mJ(z)}{J(z)}=\frac{N(z)}{D(z)},$$

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

Let

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

Then

$$\begin{array}{rl}\frac{N^{\prime }(z)}{D^{\prime }(z)}& =h(z)+\frac{z{h}^{\prime }(z)}{h_0(z)},h_0(z)=\frac{z{D}^{\prime }(z)}{D(z)}\in P\\ =h(z)+H_0(z)(z{h}^{\prime }(z)),H_0=\frac{1}{h_0}\in P.\end{array}$$

(3.7)

Since

$$\begin{array}{rl}\frac{N^{\prime }(z)}{D^{\prime }(z)}& =\frac{{(z{h}^{\prime }(z))}^{\prime }-m{J}^{\prime }(z)}{J^{\prime }(z)}\\ =\{\frac{{(z{J}^{\prime }(z))}^{\prime }}{J^{\prime }(z)}-m\}\in P(p_{k,\beta }).\end{array}$$

We now apply Lemma 2.2 to obtain

$$\frac{N(z)}{D(z)}=\frac{z{G}^{\prime }(z)}{G(z)}\in P(p_{k,\beta }),z\in E.$$

This proves that $G\in k-ST(\beta )$ in E. □

Theorem 3.5 Let $f,g\in k-S{T}_s(\beta )$ and let F be defined by the following integral operator:

$$F(z)=(\gamma +\frac{1}{\delta })z^{1-\frac{1}{\delta }}{\int }_0^z{t}^{\frac{1}{\delta }-2}{\left[\frac{f(t)-f(-t)}{2}\right]}^{\frac{1}{1+\gamma }}\left[\frac{g(t)-g(-t)}{2}\right]dt,$$

(3.8)

where $z\in E$, $\delta >0$, $\gamma \ge 0$ and $[\frac{k(1+\gamma )}{k+1}+(\frac{1}{\delta }-1)]>\beta$. Then $F(z)$ belongs to $k-ST(\beta )$ for $z\in E$.

When $g(z)=z$, $\gamma =0$, we obtain a generalized form of the Bernardi operator; see [1]. Also for $g(z)=z$, $\gamma =0$, and $\delta =\frac{1}{2}$, 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 $\frac{f(z)-f(-z)}{2}={\mathrm{\Psi }}_1(z)$, $\frac{g(z)-g(-z)}{2}={\mathrm{\Psi }}_2(z)$. Then ${\mathrm{\Psi }}_1,{\mathrm{\Psi }}_2\in k-ST(\beta )$ in E. We can write (3.8) as

$$F(z)=(\gamma +\frac{1}{\delta })z^{1-\frac{1}{\delta }}{\int }_0^z{t}^{\frac{1}{\delta }-2}{({\mathrm{\Psi }}_1(t))}^{\frac{1}{1+\gamma }}({\mathrm{\Psi }}_2(t))dt.$$

(3.9)

Differentiating (3.9) logarithmically, and with $p(z)=\frac{z{F}^{\prime }(z)}{F(z)}$, we have

$$\frac{\gamma }{1+\gamma }\frac{z{\mathrm{\Psi }}_1^{\prime }}{{\mathrm{\Psi }}_1(z)}+\frac{1}{1+\gamma }\frac{z{\mathrm{\Psi }}_2^{\prime }}{{\mathrm{\Psi }}_2(z)}=p(z)+\frac{z{p}^{\prime }(z)}{(1+\gamma )p(z)+(\frac{1}{\delta }-1)}.$$

(3.10)

Since, for $i=1,2$, ${\mathrm{\Psi }}_i\in k-ST(\beta )$, $\frac{z{\mathrm{\Psi }}_1^{\prime }(z)}{{\mathrm{\Psi }}_1}=h_1(z)$, $\frac{z{\mathrm{\Psi }}_2^{\prime }(z)}{{\mathrm{\Psi }}_2}=h_2(z)$ both belong to $P(p_{k,\beta })$ in E, and $P(p_{k,\beta })$ is a convex set. Therefore

$$(\frac{\gamma }{1+\gamma }{h}_1(z)+\frac{1}{1+\gamma }{h}_2(z))\in P(p_{k,\beta }),z\in E.$$

(3.11)

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

$$(p(z)+\frac{z{p}^{\prime }(z)}{(1+\gamma )p(z)+(\frac{1}{\delta }-1)})\prec p_{k,\beta }(z).$$

We now apply Lemma 2.3 which gives us

$$p(z)\prec q_{k,\beta }(z)\prec p_{k,\beta }(z).$$

Thus $F\in k-ST(\beta )$ and the proof is complete. □

4 The class $k-U{K}_s(\beta )$

Here we shall study some properties of the class $k-U{K}_s(\beta )$ which consists of k-uniformly close-to-convex functions.

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

Theorem 4.1 Let $f\in k-U{K}_s(\beta )$. Then, for $0<r<1$, $k\in [0,1]$,

$$L(r,f)=O(1){\left(\frac{1}{1-r}\right)}^{\sigma -{\beta }_1},{\beta }_1<\frac{\sigma }{2},$$

where ${\beta }_1=\frac{k+\beta }{k+1}$ and σ is given by (1.3), and $O(1)$ is a constant depending only on k, β.

Proof For $f\in k-U{K}_s(\beta )$, we can write

$$z{f}^{\prime }(z)=\mathrm{\Psi }(z)h^{\sigma }(z),h\in P,\mathrm{\Psi }\in S^{\ast }({\beta }_1),$$

(4.1)

and $\mathrm{\Psi }(z)=\{g(z)-g(-z)\}$, $g\in k-S{T}_s(\beta )$.

Since $\mathrm{\Psi }\in S^{\ast }({\beta }_1)$ and is odd, there exists an odd starlike function ${\mathrm{\Psi }}_1(z)$ such that

$$\mathrm{\Psi }(z)=z{\left(\frac{{\mathrm{\Psi }}_1(z)}{z}\right)}^{1-{\beta }_1}=z{\left(\frac{{\mathrm{\Psi }}_1(z)}{z}\right)}^{\frac{1-{\beta }_1}{k+1}}.$$

Thus, with $z=r{e}^{i\theta }$,

$$L(r,f)={\int }_0^{2\pi }|z{f}^{\prime }(z)|d\theta ={\int }_0^{2\pi }|z^{{\beta }_1}{({\mathrm{\Psi }}_1(z))}^{1-{\beta }_1}{h}^{\sigma }(z)|d\theta ,$$

and using Hölder’s inequality, we have

$$L(r,f)\le 2\pi r^{{\beta }_1}{\left(\frac{1}{2\pi }{\int }_0^{2\pi }\right|{\mathrm{\Psi }}_1(z){|}^{(1-\beta )(\frac{z}{z-\sigma })}d\theta )}^{\frac{2-\sigma }{z}}{\left(\frac{1}{2\pi }{\int }_0^{2\pi }\right|h(z){|}^{2}d\theta )}^{\frac{\sigma }{2}}.$$

(4.2)

For $h\in P$, it is well known [20] that

$$\frac{1}{2\pi }{\int }_0^{2\pi }|h(z){|}^{2}d\theta \le \frac{1+3r^2}{1-r^2}.$$

(4.3)

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

$$\begin{array}{rl}L(r,f)& \le C({\beta }_1,\sigma ){\left(\frac{1}{1-r^2}\right)}^{[(1-{\beta }_1)(\frac{2}{2-\sigma })-1]{[\frac{1+3r^2}{1-r}]}^{\frac{\sigma }{2}}}\\ =O(1){\left(\frac{1}{1-r}\right)}^{\sigma -{\beta }_1},\end{array}$$

where C and $O(1)$ are constants depending only on ${\beta }_1$ and σ. This completes the proof. □

We now discuss the growth rate of coefficients of $f\in k-U{K}_s(\beta )$.

Theorem 4.2 Let $f\in k-U{K}_s(\beta )$ and be given by (1.1). Then

$$a_n=O(1)n^{\sigma -{\beta }_1-1},n\ge 1,{\beta }_1<\frac{\sigma }{2},$$

where $O(1)$ is a constant depending only on σ and ${\beta }_1$ and σ, ${\beta }_1$ are as given in Theorem  4.1.

Proof For $z=r{e}^{i\theta }$, $n\ge 1$, Cauchy’s Theorem gives us

$$\begin{array}{rl}n|a_n|& =\frac{1}{2\pi r^{n+1}}|{\int }_0^{2\pi }z{f}^{\prime }(z)e^{-in\theta }d\theta |\\ \le \frac{1}{2\pi r^{n+1}}{\int }_0^{2\pi }|z{f}^{\prime }(z)|d\theta \\ =\frac{1}{2\pi r^n}L(r,f).\end{array}$$

With $r=(1-\frac{1}{n})$, we use Theorem 4.1 and obtain the required result. □

Theorem 4.3 Let $f\in k-U{K}_s(\beta )$ and let F be defined by

$$F(z)=\frac{m+1}{2z^m}{\int }_0^z{t}^{m-1}\{f(t)-f(-t)\}dt.$$

(4.4)

Then $F\in k-U{K}_s(\beta )$ in E. That is, the class $k-U{K}_s(\beta )$ is preserved under the integral operator (4.4).

Proof Since $f\in k-U{K}_s(\beta )$, we can write

$$\left\{\frac{2z{f}^{\prime }(z)}{g(z)-g(-z)}\right\}\in P(p_{k,\beta }),g\in k-S{T}_s(\beta )\subset S_S^{\ast }({\beta }_1).$$

Let $G(z)=\frac{1}{2}\{g_1(z)-g_1(-z)\}$ and be defined by (3.5). By Theorem 3.4, $g_1\in k-ST(\beta )$ and $G\in k-S_{s}T(\beta )\subset S_s^{\ast }({\beta }_1)$. Let $G=z{G}_1^{\prime }$. Then we can write

$$G_1^{\prime }(z)=\frac{1}{2}{[z{g}_1(z)-g_1(-z)]}^{\prime },G_1\in k-UC{V}_s(\beta ).$$

Thus, from (4.4) and $g=z{g}_1^{\prime }$, $g_1\in C_s({\beta }_1)$, we have

$$\begin{array}{rl}\frac{2F^{\prime }(z)}{{[g_1(z)-g_1(-z)]}^{\prime }}=& \frac{z^m\{f(z)-f(-z)\}-m{\int }_0^z{t}^{m-1}\{f(t)-f(-t)\}dt}{z^m\{g_1(z)-g_1(-z)\}-m{\int }_0^z{t}^{m-1}\{g_1(t)-g_1(-t)\}dt}\\ =& \frac{N(z)}{D(z)},\end{array}$$

say. We note that $N(0)=D(0)=0$, and for $g_1\in C_S({\beta }_1)$,

$$\begin{array}{rl}\frac{{(z{D}^{\prime }(z))}^{\prime }}{D^{\prime }(z)}& =m+\frac{{\{z{[g_1(z)-g_1(-z)]}^{\prime }\}}^{\prime }}{{\{g_1(z)-g_1(-z)\}}^{\prime }}\\ =m+h_1(z),h_1\in P({\beta }_1).\end{array}$$

Since $P({\beta }_1)$ is a convex set, $D\in C_s({\beta }_1)\subset S^{\ast }$ in E. We thus have

$$\frac{N^{\prime }(z)}{D^{\prime }(z)}=\frac{1}{2}[\frac{2z{f}^{\prime }(z)}{{[g_1(z)-g_1(-z)]}^{\prime }}+\frac{2(-z)f^{\prime }(-z)}{{[g_1(-z)-g_1(z)]}^{\prime }}]\in P(p_{k,\beta }).$$

Now, using Lemma 2.2, it follows that

$$\frac{N(z)}{D(z)}=\frac{2F^{\prime }(z)}{{(g_1(z)-g_1(-z))}^{\prime }}\in P(p_{k,\beta })\text{for }z\in E.$$

This proves that $F\in k-U{K}_S(\beta )$ in E. □

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

Theorem 4.4 Let $(\frac{2z{f}^{\prime }(z)}{g(z)-g(-z)})\prec p_k(z)$ in E and let

$$F_1(z)=\frac{1}{1+m}{z}^{1-m}{(z^{m}f(z))}^{\prime },m=1,2,3,\dots .$$

(4.5)

Then $F_1\in K_s$ for $|z|<r_1$, where

$$r_1=\left\{\frac{m+1}{(2-{\beta }_1)+\sqrt{{(z-{\beta }_1)}^2+(m+1)(m-1+2{\beta }_1)}}\right\},{\beta }_1=\frac{k+\beta }{k+1}.$$

(4.6)

Proof We shall need the following well-known results for $p\in P(\alpha )$, $0\le \alpha <1$; see [4]:

$$\frac{1-(1-2\alpha )r}{1+r}\le |p(z)|\le \frac{1+(1-2\alpha )r}{1-r},$$

(4.7)

$$|p^{\prime }(z)|\le \frac{2[\mathrm{\Re }p(z)-\alpha ]}{1-r^2}.$$

(4.8)

Since $f\in k-U{K}_s(\beta )$, there exists $g\in S_s^{\ast }({\beta }_1)$ such that, for $z\in E$.

$$\left(\frac{2z{f}^{\prime }(z)}{g(z)-g(-z)}\right)=p(z),p\in P(p_k)\subset P(\alpha ),\alpha =\frac{k}{k+1}.$$

From (4.5), we have

$$F_1(z)=\frac{1}{1+m}[mf(z)+z{f}^{\prime }(z)],$$

and this gives us

$$\begin{array}{rl}\frac{2z{F}_1^{\prime }(z)}{g(z)-g(-z)}& =\frac{1}{m+1}[\frac{2m{f}^{\prime }(z)}{g(z)-g(-z)}+\frac{2z{(z{f}^{\prime }(z))}^{\prime }}{g(z)-g(-z)}]\\ =\frac{1}{m+1}[mp(z)+z{p}^{\prime }(z)+p(z)h(z)],\end{array}$$

where

$$h(z)=\frac{z{\mathrm{\Psi }}^{\prime }(z)}{\mathrm{\Psi }(z)}\in P({\beta }_1),\mathrm{\Psi }(z)=g(z)-g(-z).$$

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

$$\begin{array}{rl}\mathrm{\Re }\left\{\frac{2z{F}_1^{\prime }(z)}{g(z)-g(-z)}\right\}& \ge \frac{(\mathrm{\Re }p(z)-\alpha )}{1+m}\{m+\frac{1-(1-2{\beta }_1)r}{1+r}-\frac{2r}{1-r^2}\}\\ =\frac{\mathrm{\Re }p(z)-\alpha }{1+m}\left[\frac{T(r)}{1-r^2}\right],\end{array}$$

(4.9)

where

$$T(r)=(m+1)-2(2-{\beta }_1)r+(-m-2{\beta }_1+1)r^2.$$

We note that $T(0)=1+m>0$ and $T(1)=-3<0$. So there exists $r_1\in (0,1)$. The right hand side of (4.9) is positive for $|z|<r_1$, where $r_1$ is given by (4.6). This implies that $F\in K_s$ for $|z|<r_1$ and the proof is complete. □

We have the following special cases.

• 1. For $k=0=\beta$, $f\in K_s$. Then $F_1$, defined by (4.5) belongs to $K_s$ for $|z|<r_0=\frac{1+m}{2+\sqrt{3+m^2}}$.

• 2. When $m=1$ and ${\beta }_1=0$ (that is, $k=0=\beta$), then $F_1(z)=\frac{{(zf(z))}^{\prime }}{2}$ belongs to the same class for $|z|<\frac{1}{2}$. This result has been proved by Livingston [12] for convex and starlike functions.