Properties of multivalent functions associated with the integral operator defined by the hypergeometric function

Abstract

In this paper, we introduce a new class of multivalent functions by using a generalized integral operator defined by the hypergeometric function. Some properties such as inclusion, radius problem and integral preserving are considered.

MSC:30C45, 30C50.

1 Introduction and preliminaries

Let $A_p$ denote the class of functions $f(z)$ of the form

$$f(z)=z^p+\sum _{n=p+1}^{\mathrm{\infty }}{a}_n{z}^n(p\in \mathbb{N}=\{1,2,3,\dots \}),$$

(1.1)

which are analytic in the open unit disc E. Also $A_1=A$, the usual class of analytic functions defined in the open unit disc $E=\{z:|z|<1\}$. A function $f\in A_p$ is a p-valent starlike function of order ρ if and only if

$$Re\frac{z{f}^{\prime }(z)}{f(z)}>\rho ,0\le \rho <p,z\in E.$$

This class of functions is denoted by $S_p^{\ast }(\rho )$. It is noted that $S_p^{\ast }(0)=S_p^{\ast }$. Let $f(z)$ and $g(z)$ be analytic in E, we say $f(z)$ is subordinate to $g(z)$, written $f\prec g$ or $f(z)\prec g(z)$ if there exists a Schwarz function $w(z)$, $w(0)=0$ and $|w(z)|<1$ in E, then $f(z)=g(w(z))$. In particular, if g is univalent in E, then we have the following equivalence

$$f(z)\prec g(z)⟺f(0)=g(0)\text{and}f(E)\subset g(E).$$

For any two analytic functions $f(z)$ and $g(z)$ with

$$f(z)=\sum _{n=0}^{\mathrm{\infty }}{b}_n{z}^{n+1}\text{and}g(z)=\sum _{n=0}^{\mathrm{\infty }}{c}_n{z}^{n+1},z\in E,$$

the convolution (the Hadamard product) is given by

$$(f\ast g)(z)=\sum _{n=0}^{\mathrm{\infty }}{b}_n{c}_n{z}^{n+1},z\in E.$$

A function $f\in A$ is said to be in the class, denoted by $\mathit{SD}(k,\delta )$ ($0\le \delta <1$), if and only if

$$Re\left\{\frac{z{f}^{\prime }(z)}{f(z)}\right\}>k|\frac{z{f}^{\prime }(z)}{f(z)}-1|+\delta ,k\ge 0,z\in E.$$

(1.2)

Similarly, a function $f\in A$ is said to be in the class, denoted by $\mathit{CD}(k,\delta )$ of k-uniformly convex of order δ ($0\le \delta <1$), if

$$Re\{1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}\}>k\left|\frac{z{f}^{″}(z)}{f^{\prime }(z)}\right|+\delta ,k\ge 0,z\in E.$$

(1.3)

Geometric interpretation The functions $f\in \mathit{SD}(k,\delta )$ and $f\in \mathit{CD}(k,\delta )$ if and only if $\frac{z{f}^{\prime }(z)}{f(z)}$ and $\frac{z{f}^{″}(z)}{f(z)}+1$, respectively, take all the values in the conic domain ${\mathrm{\Omega }}_{k,\delta }$ defined by

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

with $p(z)=\frac{z{f}^{\prime }(z)}{f(z)}$ or $p(z)=\frac{z{f}^{″}(z)}{f^{\prime }(z)}+1$ and considering the functions which map E onto the conic domain ${\mathrm{\Omega }}_{k,\delta }$ such that $1\in {\mathrm{\Omega }}_{k,\delta }$. One may rewrite the conditions (1.2) or (1.3) in the form

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

The function $q_{k,\delta }(z)$ plays the role of extremal for these classes and is given by

$$q_{k,\delta }(z)=\{\begin{array}{cc}\frac{1+(1-2\delta )z}{1-z},\hfill & k=0,\hfill \\ 1+\frac{2\delta \gamma }{{\pi }^2}{(log\frac{1+\sqrt{z}}{1-\sqrt{z}})}^2,\hfill & k=1,\hfill \\ 1+\frac{2\delta }{1-k^2}{sinh}^2[(\frac{2}{\pi }arccosk)arctanh\sqrt{z}],\hfill & 0<k<1,\hfill \\ 1+\frac{\delta }{k^2-1}sin(\frac{\pi }{2R(t)}{\int }_0^{\frac{u(z)}{\sqrt{t}}}\frac{1}{\sqrt{1-x^2}\sqrt{1-{(tx)}^2}}dx)+\frac{\delta }{k^2-1},\hfill & k>1.\hfill \end{array}$$

(1.4)

For $f(z)$ in $A_p$, the operator $D^{\mu +p-1}:A_p⟶A_p$ is defined by

$$D^{\mu +p-1}f(z)=\frac{z^p}{{(1-z)}^{\mu +p}}\ast f(z)(\mu >-p),$$

or equivalently

$$D^{\mu +p-1}f(z)=\frac{z^p{(z^{\mu -1}f(z))}^{\mu +p-1}}{(\mu +p-1)!},$$

(1.5)

where μ is any integer greater than −p. If $f(z)$ is given by (1.1), then it follows that

$$D^{\mu +p-1}f(z)=z^p+\sum _{n=p+1}^{\mathrm{\infty }}\frac{(\mu +n-1)!}{(n-p)!(\mu +p-1)!}{a}_n{z}^n.$$

The symbol $D^{\mu +p-1}$ when $p=1$, was introduced by Ruscheweyh [1] and $D^{\mu +p-1}$ is called the $(\mu +p-1)$th order Ruscheweyh derivative. We now introduce a function ${(z_2^p{F}_1(a,b,c;z))}^{-1}$ given by

$$(z^p{}_2{F}_1(a,b,c;z))\ast {(z^p{}_2{F}_1(a,b,c;z))}^{-1}=\frac{z^p}{{(1-z)}^{\mu +p}}(\mu >-p),$$

and the following linear operator

$$I_{\mu ,p}(a,b,c)f(z)={(z^p{}_2{F}_1(a,b,c;z))}^{-1}\ast f(z),$$

(1.6)

where a, b, c are real or complex numbers other than $0,-1,-2,\dots$ , $\mu >-p$, $z\in E$ and $f(z)\in A_p$. This operator was recently introduced in [2]. In particular, for $p=1$, this operator is studied by Noor [3]. For $b=1$, this operator reduces to the well-known Cho-Kwon-Srivastava operator $I_{\mu ,p}(a,c)$, which was studied by Cho et al. [4], and for $\mu =1$, $b=c$, $a=n+p$, see [5]. For $a=n+p$, $b=c=1$, this operator was investigated by Liu [6] and Liu and Noor [7].

Simple computations yield

$$I_{\mu ,p}(a,b,c)f(z)=z^p+\sum _{n=p+1}^{\mathrm{\infty }}\frac{{(c)}_n{(\mu +p)}_n}{{(a)}_n{(b)}_n}{a}_n{z}^n.$$

From (1.6), we note that

$$\begin{array}{r}{I}_{\mu ,1}(a,b,c)f(z)=I_{\mu }(a,b,c)f(z)\text{(see [3])},\\ I_{0,p}(a,p,a)f(z)=f(z),I_{1,p}(a,p,a)f(z)=\frac{z{f}^{\prime }(z)}{p}.\end{array}$$

Also, it can be easily seen that

$$z{(I_{\mu ,p}(a,b,c)f(z))}^{\prime }=(\mu +p)I_{\mu +1,p}(a,b,c)f(z)-\mu I_{\mu ,p}(a,b,c)f(z),$$

(1.7)

and

$$z{(I_{\mu ,p}(a+1,b,c)f(z))}^{\prime }=a{I}_{\mu ,p}(a,b,c)f(z)-(a-p)I_{\mu ,p}(a,b,c)f(z).$$

We define the following class of multivalent analytic functions by using the operator $I_{\mu ,p}(a,b,c)f(z)$ above.

Definition 1.1 Let $f\in A_p$ for $p\in \mathbb{N}$. Then $f\in U{B}_{\mu ,p}^{\alpha }(a,b,c;\gamma ,k,\delta )$ for $a,b,c\in \mathbb{R}\mathrm{\setminus }{\mathbb{Z}}_0^{-}$, $\mu >-p$, $\alpha >0$, $k\ge 0$, $0\le \delta <1$ and $\gamma >0$ if and only if

$$\begin{array}{r}(1-\gamma ){\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)}^{\alpha }+\gamma \frac{I_{\mu +1,p}(a,b,c)f(z)}{I_{\mu +1,p}(a,b,c)g(z)}{\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)}^{\alpha -1}\\ \prec q_{k,\delta }(z),\end{array}$$

(1.8)

where $g\in A_p$ is such that

$$q(z)=\frac{I_{\mu +1,p}(a,b,c)g(z)}{I_{\mu ,p}(a,b,c)g(z)}\in P(\rho ),\rho =\frac{k+\delta }{k+1},z\in E.$$

(1.9)

Furthermore, for different choices of parameters being involved, we obtain many other well-known subclasses of the class $A_p$ and A as special cases.

• (i) $a=c$, $b=1$, $k=0$, $\mu =m\in {\mathbb{N}}_0$, we have $B_{m,p}^{\alpha }(\gamma ,\delta )$ studied in [8].

• (ii) $a=c=b=p=\gamma =1$, $k=\mu =0$, $g(z)=z$, the class $U{B}_{\mu ,p}^{\alpha }(a,b,c;\gamma ,k,\delta )$ reduces to the class

  $$B^{\alpha }(\delta )=\{f\in A(1):\frac{z{f}^{\prime }(z)}{f(z)}{\left(\frac{f(z)}{z}\right)}^{\alpha }\in P(\delta )\}$$

  studied in [9].

• (iii) $a=c=b=p=\gamma =1$, $k=\mu =0$, $g(z)=z$, the $U{B}_{\mu ,p}^{\alpha }(a,b,c;\gamma ,k,\delta )$ reduces to $B(\alpha )$ is the class of Bazilevich functions investigated by Singh [10].

• (iv) $a=c=b=p=\alpha =1$, $\gamma =0$, $k=\mu =0$, $g(z)=z$, the class $U{B}_{\mu ,p}^{\alpha }(a,b,c;\gamma ,k,\delta )$ reduces to the class

  $$B_{\delta }=\{f\in A(1):\frac{f(z)}{z}\in P(\delta )\},$$

the class studied by Chen [11].

Let $f\in A_{p\cdot }$ and $F_{\eta ,p}:A_{p\cdot }\to A_{p\cdot }$ be defined by

$$F_{\eta ,p}(z)=\frac{(\eta +p)}{z^{\eta }}{\int }_0^z{t}^{\eta -1}f(t)dt,\eta >-p.$$

(1.10)

We need the following lemmas which will be used in our main results.

Lemma 1.2 [12]

Let $u=u_1+i{u}_2$ and $v=v_1+i{v}_2$, and let $\psi :D\subset {\mathbb{C}}^2\to \mathbb{C}$ be a complex-valued function satisfying the conditions:

• (i) $\psi (u,v)$ is continuous in a domain $D\subset {\mathbb{C}}^2$,

• (ii) $(1,0)\in D$ and $\psi (1,0)>0$,

• (iii) $Re\psi (i{u}_2,v_1)\le 0$, whenever $(i{u}_2,v_1)\in D$ and $v_1\le -\frac{1}{2}(1+u_2^2)$.

If $h(z)=1+c_{1}z+c_2{z}^2+\cdots$ is analytic in E such that $(h,z{h}^{\prime })\in D$ and $Re\psi (h(z),z{h}^{\prime }(z))>0$ for $z\in E$, then $Reh(z)>0$.

Lemma 1.3 [13]

Let h be convex in the unit disc E, and let $A\ge 0$. Suppose that $B(z)$ is analytic in E with $ReB(z)\ge A$. If g is analytic in E and $g(0)=h(0)$. Then

$$A{z}^2{g}^{″}(z)+B(z)z{g}^{\prime }(z)+g(z)\prec h(z)\mathit{\text{implies that}}g(z)\prec h(z).$$

Lemma 1.4 [14]

Let F be analytic and convex in E. If $f,g\in A_p$ and $f,g\prec F$. Then

$$\sigma f+(1-\sigma )g\prec F,0\le \sigma \le 1.$$

Lemma 1.5 [15]

Let h be convex in E with $h(0)=a$ and $\beta \in \mathbb{C}$ such that $Re\beta \ge 0$. If $p\in H[a,n]$ and

$$p(z)+\frac{z{p}^{\prime }(z)}{\beta }\prec h(z),$$

then $p(z)\prec q(z)\prec h(z)$, where

$$q(z)=\frac{\beta }{n{z}^{\beta /n}}{\int }_0^{z}h(t)t^{\beta /n-1}dt$$

and $q(z)$ is the best dominant.

2 Main results

Theorem 2.1 Let $f\in U{B}_{\mu ,p}^{\alpha }(a,b,c;\gamma ,k,\delta )$ for $a,b,c\in \mathbb{R}\mathrm{\setminus }{\mathbb{Z}}_0^{-}$, $\mu >-p$, $p\in \mathbb{N}$, $\alpha >0$, $k\ge 0$, $0\le \delta <1$ and $\gamma >0$. Then $f\in U{B}_{\mu ,p}^{\alpha }(a,b,c;0,k,\delta )$.

Proof Consider

$$h(z)={\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)}^{\alpha },$$

(2.1)

where h is analytic in E with $h(0)=1$, and $g\in A_p$ satisfies condition (1.9). Differentiating (2.1) logarithmically and using (1.7), we have

$$\frac{z{h}^{\prime }(z)}{h(z)}=\alpha (\mu +p)\left\{(\frac{I_{\mu +1,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)f(z)}-\frac{I_{\mu +1,p}(a,b,c)g(z)}{I_{\mu ,p}(a,b,c)g(z)})\right\}.$$

Using (2.1) and simplifying, we obtain

$$\begin{array}{rcl}h(z)+\frac{\gamma z{h}^{\prime }(z)}{\alpha (\mu +p)q(z)}& =& (1-\gamma ){\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)}^{\alpha }\\ +\gamma \frac{I_{\mu +1,p}(a,b,c)f(z)}{I_{\mu +1,p}(a,b,c)g(z)}{\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)}^{\alpha -1}.\end{array}$$

Since $f\in U{B}_{\mu ,p}^{\alpha }(a,b,c;\gamma ,k,\delta )$, therefore, we can write

$$h(z)+\frac{\gamma z{h}^{\prime }(z)}{\alpha (\mu +p)q(z)}\prec q_{k,\delta }(z),z\in E.$$

Now using Lemma 1.3 for $A=0$ and $B(z)=\frac{\gamma }{\alpha (\mu +p)q(z)}$ with $Req(z)>0$, we have $ReB(z)\ge 0$, therefore, $h(z)\prec q_{k,\delta }(z)$. Hence $f\in U{B}_{\mu ,p}^{\alpha }(a,b,c;0,k,\delta )$. □

Theorem 2.2 Let $f\in U{B}_{\mu ,p}^{\alpha }(a,b,c;\gamma ,0,\delta )$ for $a,b,c\in \mathbb{R}\mathrm{\setminus }{\mathbb{Z}}_0^{-}$, $\mu >-p$, $p\in \mathbb{N}$. Then $f\in U{B}_{\mu ,p}^{\alpha }(a,b,c;0,0,{\delta }_1)$, where

$${\delta }_1=\frac{2\alpha \delta (p+\mu )|q(z){|}^2+\gamma \rho }{2\alpha (p+\mu )|q(z){|}^2+\gamma \rho }.$$

Proof Consider

$$h(z)=\frac{1}{(1-{\delta }_1)}\{{\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)}^{\alpha }-{\delta }_1\},$$

(2.2)

where h is analytic in E with $h(0)=1$, and $g\in A_p$ satisfies condition (1.9). Differentiating (2.2), we have

$$\begin{array}{rcl}\frac{(1-{\delta }_1)}{\alpha }{h}^{\prime }(z)& =& {\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)}^{\alpha -1}\\ \times \{\frac{{(I_{\mu ,p}(a,b,c)f(z))}^{\prime }}{I_{\mu ,p}(a,b,c)g(z)}-\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\frac{{(I_{\mu ,p}(a,b,c)g(z))}^{\prime }}{I_{\mu ,p}(a,b,c)g(z)}\}.\end{array}$$

Using (1.7) and simplifying, we obtain

$$\begin{array}{r}(1-\gamma ){\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)}^{\alpha }+\gamma \frac{I_{\mu +1,p}(a,b,c)f(z)}{I_{\mu +1,p}(a,b,c)g(z)}{\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)}^{\alpha -1}\\ =(1-{\delta }_1)h(z)+{\delta }_1+\frac{\gamma (1-{\delta }_1)z{h}^{\prime }(z)}{\alpha (p+\mu )q(z)}.\end{array}$$

Since $f\in U{B}_{\mu ,p}^{\alpha }(a,b,c;\gamma ,0,\delta )$, therefore we have

$$(1-{\delta }_1)h(z)+{\delta }_1+\frac{\gamma (1-{\delta }_1)z{h}^{\prime }(z)}{\alpha (p+\mu )q(z)}\prec \frac{1+(1-2\delta )z}{1-z},0\le \delta <1,z\in E.$$

This implies that

$$\frac{1}{1-\delta }\{(1-{\delta }_1)h(z)+{\delta }_1-\delta +\frac{\gamma (1-{\delta }_1)z{h}^{\prime }(z)}{\alpha (p+\mu )q(z)}\}\in q_{0,0}(E)=P.$$

To obtain our desired result, we show that $h\in P$, for $z\in E$. Let $u=u_1+i{u}_2$, $v=v_1+i{v}_2$, and let $\mathrm{\Psi }:D\subset {\mathbb{C}}^2\to \mathbb{C}$ be a complex-valued function such that $u=h(z)$, $v=z{h}^{\prime }(z)$. Then

$$\mathrm{\Psi }(u,v)=(1-{\delta }_1)u+{\delta }_1-\delta +\frac{\gamma (1-{\delta }_1)v}{\alpha (p+\mu )q(z)}.$$

The first two conditions of Lemma 1.2 are easily verified. To verify the third condition, we consider

$$\begin{array}{r}Re\mathrm{\Psi }(i{u}_2,v_1)\\ =Re\{(1-{\delta }_1)i{u}_2+{\delta }_1-\delta +\frac{\gamma (1-{\delta }_1)v_1}{\alpha (p+\mu )q(z)}\}\\ ={\delta }_1-\delta +Re\frac{\gamma (1-{\delta }_1)v_1}{\alpha (p+\mu )q(z)}\\ \le {\delta }_1-\delta -Re\frac{\gamma (1-{\delta }_1)(1+u_2^2)\overline{q(z)}}{2\alpha (p+\mu )|q(z){|}^2}\\ \le {\delta }_1-\delta -\frac{\gamma (1-{\delta }_1)(1+u_2^2)\rho }{2\alpha (p+\mu )|q(z){|}^2}=\frac{A+B{u}^2}{C},\end{array}$$

where $A=2\alpha (p+\mu )({\delta }_1-\delta )|q(z){|}^2-\gamma \rho (1-{\delta }_1)$, $B=-\gamma \rho (1-{\delta }_1)\le 0$ if $0\le {\delta }_1<1$ and $C=2\alpha (p+\mu )|q(z){|}^2>0$. From the relation ${\delta }_1=\frac{2\alpha \delta (p+\mu )|q(z){|}^2+\gamma \rho }{2\alpha (p+\mu )|q(z){|}^2+\gamma \rho }$, we have $A\le 0$. This implies that $Re\mathrm{\Psi }(i{u}_2,v_1)\le 0$. Using Lemma 1.2, we have $h\in P$ for $z\in E$. This completes the proof. □

Theorem 2.3 Let $a,b,c\in \mathbb{R}\mathrm{\setminus }{\mathbb{Z}}_0^{-}$, $\mu >-p$, $p\in \mathbb{N}$, $\alpha >0$, $k\ge 0$ and $0\le \delta <1$. Then

$$U{B}_{\mu ,p}^{\alpha }(a,b,c;{\gamma }_2,k,\delta )\subset U{B}_{\mu ,p}^{\alpha }(a,b,c;{\gamma }_1,k,\delta ),0\le {\gamma }_1<{\gamma }_2,z\in E.$$

Proof Since $f\in U{B}_{\mu ,p}^{\alpha }(a,b,c;{\gamma }_2,k,\delta )$, therefore, we have

$$\begin{array}{r}(1-{\gamma }_2){\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)}^{\alpha }+{\gamma }_2\frac{I_{\mu +1,p}(a,b,c)f(z)}{I_{\mu +1,p}(a,b,c)g(z)}{\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)}^{\alpha -1}\\ =h_1(z)\prec q_{k,\delta }(z),\end{array}$$

(2.3)

where $g\in A_p$ satisfies condition (1.9). From Theorem 2.1, we write

$${\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)}^{\alpha }=h_2(z)\prec q_{k,\delta }(z),z\in E.$$

(2.4)

Now, for ${\gamma }_1\ge 0$, we obtain

$$\begin{array}{r}(1-{\gamma }_1){\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)}^{\alpha }+{\gamma }_1\frac{I_{\mu +1,p}(a,b,c)f(z)}{I_{\mu +1,p}(a,b,c)g(z)}{\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)}^{\alpha -1}\\ =(1-\frac{{\gamma }_1}{{\gamma }_2}){\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)}^{\alpha }\\ +\frac{{\gamma }_1}{{\gamma }_2}\{(1-{\gamma }_2){\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)}^{\alpha }+{\gamma }_2\frac{I_{\mu +1,p}(a,b,c)f(z)}{I_{\mu +1,p}(a,b,c)g(z)}{\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)}^{\alpha -1}\}\\ =\frac{{\gamma }_1}{{\gamma }_2}{h}_1(z)+(1-\frac{{\gamma }_1}{{\gamma }_2})h_2(z).\end{array}$$

Using the convexity of the class of the function $q_{k,\delta }(z)$ and Lemma 1.4, we write

$$\frac{{\gamma }_1}{{\gamma }_2}{h}_1(z)+(1-\frac{{\gamma }_1}{{\gamma }_2})h_2(z)\prec q_{k,\delta }(z),z\in E,$$

where $h_1$ and $h_2$ are given by (2.3) and (2.4), respectively. This implies that $f\in U{B}_{\mu ,p}^{\alpha }(a,b,c;{\gamma }_1,k,\delta )$. Hence the proof of the theorem is completed. □

Theorem 2.4 Let $f\in U{B}_{\mu ,p}^1(a,b,c;\gamma ,k,\delta )$, $a,b,c\in \mathbb{R}\mathrm{\setminus }{\mathbb{Z}}_0^{-}$, $\mu >-p$, $p\in \mathbb{N}$, $\alpha >0$, $k\ge 0$, $\gamma \ge 1$ and $0\le \delta <1$. Then $f\in U{B}_{\mu +1,p}^1(a,b,c;1,k,\delta )$.

Proof Since $f\in U{B}_{\mu ,p}^1(a,b,c;\gamma ,k,\delta )$, therefore, we have

$$(1-\gamma )\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)+\gamma \frac{I_{\mu +1,p}(a,b,c)f(z)}{I_{\mu +1,p}(a,b,c)g(z)}\prec q_{k,\delta }(z).$$

Now, consider

$$\begin{array}{rcl}\gamma \frac{I_{\mu +1,p}(a,b,c)f(z)}{I_{\mu +1,p}(a,b,c)g(z)}& =& (1-\gamma )\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)+\gamma \frac{I_{\mu +1,p}(a,b,c)f(z)}{I_{\mu +1,p}(a,b,c)g(z)}\\ +(\gamma -1)\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right).\end{array}$$

This implies that

$$\begin{array}{rcl}\frac{I_{\mu +1,p}(a,b,c)f(z)}{I_{\mu +1,p}(a,b,c)g(z)}& =& \frac{1}{\gamma }\{(1-\gamma )\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)+\gamma \frac{I_{\mu +1,p}(a,b,c)f(z)}{I_{\mu +1,p}(a,b,c)g(z)}\}\\ +(1-\frac{1}{\gamma })\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right).\end{array}$$

Using Theorem 2.1, Lemma 1.4 and the convexity of $q_{k,\delta }(z)$, we have the required result. □

Now, using the operators $I_{\mu ,p}(a,b,c)$ and $F_{\eta ,p}$ defined by (1.7) and (1.10), respectively, we have

$$z{(I_p^{\mu }(a,b,c)F_{\eta ,p}(f)(z))}^{\prime }=(p+\eta )I_p^{\mu }(a,b,c)f(z)-\eta I_p^{\mu }(a,b,c)F_{\eta ,p}(f)(z),\eta >-p.$$

(2.5)

Theorem 2.5 Let $f\in A_p$ and $F_{\eta ,p}$ be given by (1.10). If

$$(1-\gamma )\frac{I_{\mu ,p}(a,b,c)F_{\eta ,p}(f)(z)}{z^p}+\gamma \frac{I_{\mu ,p}(a,b,c)(f(z))}{z^p}\prec q_{k,\delta }(z),z\in E,$$

(2.6)

with $a,b,c\in \mathbb{R}\mathrm{\setminus }{\mathbb{Z}}_0^{-}$, $\mu ,\eta >-p$, $p\in \mathbb{N}$, $\gamma >0$, then

$$\frac{I_{\mu ,p}(a,b,c)F_{\eta ,p}(f(z))}{z^p}\prec h(z)\prec q_{k,\delta }(z),z\in E,$$

where

$$h(z)=\frac{p+\eta }{\gamma z^{(p+\eta )/\gamma }}{\int }_0^z{q}_{k,\delta }(z)t^{{}^{(p+\eta )/\gamma }-1}dt.$$

Proof Let

$$\frac{I_{\mu ,p}(a,b,c)F_{\eta ,p}(f(z))}{z^p}=h_1(z),z\in E,$$

where $h_1$ is analytic in E with $h_1(0)=1$. Then

$$z{(I_{\mu ,p}(a,b,c)F_{\eta ,p}(f(z)))}^{\prime }=p{z}^p{h}_1(z)+z^{p+1}{h}_1^{\prime }(z).$$

Using (2.5), we have

$$\gamma (p+\eta )\frac{I_{\mu ,p}(a,b,c)f(z)}{z^p}-\gamma \eta \frac{I_{\mu ,p}(a,b,c)F_{\eta ,p}(f)(z)}{z^p}=p\gamma h_1(z)+\gamma z{h}_1^{\prime }(z).$$

Thus,

$$(1-\gamma )\frac{I_{\mu ,p}(a,b,c)F_{\eta ,p}(f(z))}{z^p}+\gamma \frac{I_{\mu ,p}(a,b,c)(f(z))}{z^p}=h_1(z)+\gamma \frac{z{h}_1^{\prime }(z)}{p+\eta }.$$

(2.7)

From (2.7), it follows that

$$h_1(z)+\gamma \frac{z{h}_1^{\prime }(z)}{p+\eta }\prec q_{k,\delta }(z),z\in E.$$

Using Lemma 1.5, for ${\beta }_1=\frac{p+\eta }{\gamma }$, $n=1$ and $a=1$, we obtain $h_1(z)\prec h(z)\prec q_{k,\delta }(z)$. That is, $\frac{I_{\mu ,p}(a,b,c)F_{\eta ,p}(f(z))}{z^p}\prec q_{k,\delta }(z)$. □

Theorem 2.6 Let $f\in U{B}_{\mu ,p}^{\alpha }(a,b,c;0,0,\delta )$ for $a,b,c\in \mathbb{R}\mathrm{\setminus }{\mathbb{Z}}_0^{-}$, $\mu >-p$, $p\in \mathbb{N}$, $\alpha ,\gamma >0$, $0\le \delta <1$. Then $f\in U{B}_{\mu ,p}^{\alpha }(a,b,c;\gamma ,0,\delta )$, for $|z|<r_0$, where

$$r_0=\frac{\alpha (p}{+}\mu )+\gamma -\sqrt{{\gamma }^2+2\alpha \gamma (p+\mu )}\alpha (p+\mu ).$$

(2.8)

Proof Let $f\in U{B}_{\mu ,p}^{\alpha }(a,b,c;0,0,\delta )$. Then we have

$${\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)}^{\alpha }=(1-\delta )h(z)+\delta ,$$

(2.9)

where $g\in A_p$ satisfies the condition

$$q(z)=\frac{I_{\mu +1,p}(a,b,c)g(z)}{I_{\mu ,p}(a,b,c)g(z)}\in P,z\in E$$

and $h\in P$. Differentiating (2.9) and then using (1.7), we obtain

$$\gamma \frac{I_{\mu +1,p}(a,b,c)f(z)}{I_{\mu +1,p}(a,b,c)g(z)}{\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)}^{\alpha -1}-\gamma {\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)}^{\alpha }=\frac{\gamma (p-\delta )z{h}^{\prime }(z)}{\alpha (p+\mu )q(z)}.$$

This implies that

$$\begin{array}{r}\frac{1}{1-\delta }\{(1-\gamma ){\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)}^{\alpha }+\gamma \frac{I_{\mu +1,p}(a,b,c)f(z)}{I_{\mu +1,p}(a,b,c)g(z)}{\left(\frac{I_{\mu ,p}(a,b,c)f(z)}{I_{\mu ,p}(a,b,c)g(z)}\right)}^{\alpha -1}-\delta \}\\ =h(z)+\frac{\gamma z{h}^{\prime }(z)}{\alpha (p+\mu )q(z)},z\in E.\end{array}$$

(2.10)

Now, using the well-known distortion result for class P, we have

$$|z{h}^{\prime }(z)|\le \frac{2rReh(z)}{1-r^2}\text{and}Reh(z)\ge \frac{1-r}{1+r},|z|<r<1,z\in E.$$

Thus, due to the applications of these inequalities, we have

$$\begin{array}{rl}Re(h(z)+\frac{\gamma z{h}^{\prime }(z)}{\alpha (p+\mu )q(z)})& \ge Reh(z)-\frac{\gamma |z{h}^{\prime }(z)|}{\alpha (p+\mu )|q(z)|}\\ \ge Reh(z)(1-\frac{2\gamma r}{\alpha (p+\mu ){(1-r)}^2})\\ =Reh(z)\left(\frac{\alpha (p+\mu ){(1-r)}^2-2\gamma r}{\alpha (p+\mu ){(1-r)}^2}\right).\end{array}$$

For $|z|<r_0$, where $r_0$ is given in (2.8), the inequality above is positive. Sharpness of the result follows by taking $h(z)=\frac{1+z}{1-z}$. Hence from (2.10), we have the required result. □