Applications of differential subordinations for certain classes of p-valent functions associated with generalized Srivastava-Attiya operator

Abstract

The object of the present paper is to investigate some inclusion relations and other interesting properties for certain classes of p-valent functions involving generalized Srivastava-Attiya operator by using the principle of differential subordination.

MSC:30C45.

1 Introduction

Let $A(p)$ be the class of functions which are analytic and p-valent in the unit disc $U=\{z\in \mathbb{C}:|z|<1\}$ of the form

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

(1.1)

Let also $A(1)=A_1$. For $g(z)\in A(p)$, given by $g(z)=z^p+{\sum }_{k=1}^{\mathrm{\infty }}{b}_{k+p}{z}^{k+p}$, the Hadamard product (or convolution) of $f(z)$ and $g(z)$ is defined by

$$(f\ast g)(z)=z^p+\sum _{k=1}^{\mathrm{\infty }}{a}_{k+p}{b}_{k+p}{z}^{k+p}=(g\ast f)(z).$$

(1.2)

Next, in the usual notation, let $\mathrm{\Phi }(z,s,a)$ denote the Hurwitz-Lerch Zeta function defined as follows:

(1.3)

For further interesting properties and characteristics of the Hurwitz-Lerch Zeta function $\mathrm{\Phi }(z,s,a)$ see [2, 5, 8, 9, 11], and [21].

Recently, Srivastava and Attiya [20] have introduced the linear operator $L_{s,b}:A_1\to A_1$, defined in terms of the Hadamard product by

$$L_{s,b}(f)(z)=G_{s,b}(z)\ast f(z)(z\in U;b\in \mathbb{C}\mathrm{\setminus }{\mathbb{Z}}_0^{-};s\in \mathbb{C}),$$

(1.4)

where

$$G_{s,b}={(1+b)}^s[\mathrm{\Phi }(z,s,b)-b^{-s}](z\in U).$$

(1.5)

The Srivastava-Attiya operator $L_{s,b}$ contains, among its special cases, the integral operators introduced and investigated by Alexander [1], Libera [7] and Jung et al. [6].

Analogous to $L_{s,b}$, Liu [10] defined the operator $J_{p,s,b}:A(p)\to A(p)$ by

$$J_{p,s,b}(f)(z)=G_{p,s,b}(z)\ast f(z)(z\in U;b\in \mathbb{C}\mathrm{\setminus }{\mathbb{Z}}_0^{-};s\in \mathbb{C};p\in \mathbb{N}),$$

(1.6)

where

$$G_{p,s,b}={(1+b)}^s[{\mathrm{\Phi }}_p(z,s,b)-b^{-s}]$$

and

$${\mathrm{\Phi }}_p(z,s,b)=\frac{1}{b^s}+\sum _{k=0}^{\mathrm{\infty }}\frac{z^{k+p}}{{(k+1+b)}^s}.$$

(1.7)

It is easy to observe from (1.6) and (1.7) that

$$J_{p,s,b}(f)(z)=z^p+\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1+b}{k+1+b}\right)}^s{a}_{k+p}{z}^{k+p}.$$

(1.8)

We note that

• (i) $J_{p,0,b}(f)(z)=f(z)$;

• (ii) $J_{1,1,0}(f)(z)=Lf(z)={\int }_0^z\frac{f(t)}{t}dt$ ($f\in A_1$), where the operator L was introduced by Alexander [1];

• (iii) $J_{1,s,b}(f)(z)=L_{s,b}f(z)$ ($s\in \mathbb{C}$$b\in \mathbb{C}\mathrm{\setminus }{\mathbb{Z}}_0^{-}$), where the operator $L_{s,b}$ was introduced by Srivastava-Attiya [20];

• (iv) $J_{p,1,\mu +p-1}(f)(z)=F_{\mu ,p}(f)(z)$ ($\mu >-p$$p\in \mathbb{N}$), where the operator $F_{\mu ,p}$ was introduced by Choi et al. [3];

• (v) $J_{p,\alpha ,p}(f)(z)=I_p^{\alpha }f(z)$ ($\alpha >0$$p\in \mathbb{N}$), where the operator $I_p^{\alpha }$ was introduced by Shams et al. [18];

• (vi) $J_{p,\gamma ,p-1}(f)(z)=J_p^{\gamma }f(z)$ ($\gamma \in {\mathbb{N}}_0=\mathbb{N}\cup \{0\}$$p\in \mathbb{N}$), where the operator $J_p^{\gamma }$ was introduced by El-Ashwah and Aouf [4];

• (vii) $J_{p,\gamma ,p+l-1}(f)(z)=J_p^{\gamma }(l)f(z)$ ($\gamma \in {\mathbb{N}}_0$$p\in \mathbb{N}$$l\ge 0$), where the operator $J_p^{\gamma }(l)$ was introduced by El-Ashwah and Aouf [4].

It follows from (1.8) that

$$z{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }}=(b+1)J_{p,s-1,b}(f)(z)-(b+1-p)J_{p,s,b}(f)(z).$$

(1.9)

For two analytic functions $f,g\in A(p)$, we say that f is subordinate to g, written $f(z)\prec g(z)$ if there exists a Schwarz function $w(z)$, which (by definition) is analytic in U with $w(0)=0$ and $|w(z)|<1$ for all $z\in U$, such that $f(z)=g(w(z))$$z\in U$. Furthermore, if the function $g(z)$ is univalent in U, then we have the following equivalence (see [14]):

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

Definition 1 For fixed parameters A and B, with $-1\le B<A\le 1$, we say that $f\in A(p)$ is in the class $S_p^{s,b}(A,B)$ if it satisfies the following subordination condition:

$$\frac{{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}\prec \frac{1+Az}{1+Bz}(p\in \mathbb{N}).$$

(1.10)

In view of the definition of subordination (1.10) is equivalent to the following condition:

$$\left|\frac{\frac{{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}-1}{B\frac{{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}-A}\right|<1(z\in U).$$

For convenience, we write $S_p^{s,b}(1-\frac{2\eta }{p},-1)=S_p^{s,b}(\eta )$, where $S_p^{s,b}(\eta )$ denotes the class of functions in $A(p)$ satisfying the inequality

$$Re\left(\frac{{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}\right)>\eta (0\le \eta <1;p\in \mathbb{N};z\in U).$$

In the present paper, we investigate some inclusion relations and other interesting properties for certain classes of p-valent functions involving an integral operator.

2 Preliminaries

To establish our main results, we need the following lemmas.

Lemma 1 ([13, 14])

Let h be analytic and convex (univalent) in U with$h(0)=1$. Suppose also that the function φ given by

$$\phi (z)=1+c_m{z}^m+c_{m+1}{z}^{m+1}+\cdots ,$$

(2.1)

is analytic in U, where m is a positive integer. If

$$\phi (z)+\frac{z{\phi }^{\mathrm{\prime }}(z)}{\varrho }\prec h(z)(Re\{\varrho \}\ge 0;\varrho \ne 0),$$

(2.2)

then

$$\phi (z)\prec \psi (z)=\frac{\varrho }{m}{z}^{-\frac{\varrho }{m}}{\int }_0^z{t}^{\frac{\varrho }{m}-1}h(t)dt\prec h(z)$$

(2.3)

and$\psi (z)$is the best dominant of (2.2).

We denote by $H(\varrho )$ the class of functions $\mathrm{\Phi }(z)$ given by

$$\mathrm{\Phi }(z)=1+c_{1}z+c_2{z}^2+\cdots ,$$

(2.4)

which are analytic in U and satisfy the following inequality:

$$Re\{\mathrm{\Phi }(z)\}>\varrho (0\le \varrho <1;z\in U).$$

Lemma 2 ([17])

Let the function$\mathrm{\Phi }(z)\in H(\varrho )$, where$\mathrm{\Phi }(z)$given by (2.4). Then

$$Re\{\mathrm{\Phi }(\varrho )\}\ge 2\varrho -1+\frac{2(1-\varrho )}{1+|z|}(0\le \varrho <1;z\in U).$$

Lemma 3 ([22])

For$0\le {\varrho }_1$, ${\varrho }_2<1$,

$$H({\varrho }_1)\ast H({\varrho }_2)\subset H({\varrho }_3),{\varrho }_3=1-2(1-{\varrho }_1)(1-{\varrho }_2).$$

The result is best possible.

Lemma 4 ([24])

Let μ be a positive measure on the unit interval$[0,1]$. Let$g(z,t)$be a complex valued function defined on$U\times [0,1]$such that$g(0,t)$is analytic in U for each$t\in [0,1]$and such that$g(z,0)$is μ integrable on$[0,1]$for all$z\in U$. In addition, suppose that$Re\{g(z,t)\}>0$, $g(-r,t)$is real and

$$Re\left\{\frac{1}{g(z,t)}\right\}\ge \frac{1}{g(-r,t)}(|z|\le r<1;t\in [0,1]).$$

If G is defined by

$$G(z)={\int }_0^{1}g(z,t)d\mu (t),$$

then

$$Re\left\{\frac{1}{G(z)}\right\}\ge \frac{1}{G(-r)}(|z|\le r<1).$$

Lemma 5 ([19])

Let the function g be analytic in U with$g(0)=1$and$Re\{g(z)\}>\frac{1}{2}$ ($z\in U$). Then, for any function F analytic in U, $(g\ast F)(U)$is contained in the convex hull of$F(U)$.

Lemma 6 ([16])

Let φ be analytic in U with$\phi (0)=1$and$\phi (z)=0$for$0<|z|<1$and let$A,B\in \mathbb{C}$with$A\ne B$, $|B|\le 1$.

• (i) Let$B\ne 0$and$\gamma \in {\mathbb{C}}^{\ast }=\mathbb{C}\mathrm{\setminus }\{0\}$satisfy either$|\frac{\gamma (A-B)}{B}-1|\le 1$or$|\frac{\gamma (A-B)}{B}+1|\le 1$. If φ satisfies

  $$1+\frac{z{\phi }^{\mathrm{\prime }}(z)}{\gamma \phi (z)}\prec \frac{1+Az}{1+Bz},$$

  (2.5)

then

$$\phi (z)\prec {(1+Bz)}^{\gamma (\frac{A-B}{B})},$$

and this is best dominant.

• (ii) Let$B=0$and$\gamma \in {\mathbb{C}}^{\ast }$be such that$|\gamma A|<\pi$. If φ satisfies (2.5), then

  $$\phi (z)\prec e^{\gamma Az}$$

and this is the best dominant.

For real or complex numbers a n and c ($c\notin {\mathbb{Z}}_0^{-}$) and $z\in U$, the Gaussian hypergeometric function defined by

$$\begin{array}{rl}{}_{2}F_1(a,n;c;z)& =1+\frac{an}{c}\cdot \frac{z}{1!}+\frac{a(a+1)n(n+1)}{c(c+1)}\cdot \frac{z^2}{2!}+\cdots \\ =\sum _{k=0}^{\mathrm{\infty }}\frac{{(a)}_k{(n)}_k}{{(c)}_k}\frac{z^k}{k!},\end{array}$$

(2.6)

where ${(d)}_k=d(d+1)\cdots (d+k-1)$ and ${(d)}_0=1$. We note that the series defined by (2.6) converges absolutely for $z\in U$, and hence, ${}_{2}F_1$ represents an analytic function in U (see, for details, [23], Ch.14]).

Lemma 7 ([23])

For real or complex numbers a, n and c ($c\notin {\mathbb{Z}}_0^{-}$)

(2.7)

(2.8)

and

$${}_{2}F_1(a,n;c;z){=}_2{F}_1(n,a;c;z).$$

(2.9)

3 Main results

Unless otherwise mentioned, we assume throughout this paper that $-1\le B<A\le 1$, $s\in \mathbb{C}$, $b\in \mathbb{C}\mathrm{\setminus }{\mathbb{Z}}_0^{-}$, $p\in \mathbb{N}\mathrm{\setminus }\{1\}$, $0<\alpha \le 1$, m is a positive integer and the powers are understood as principle values.

Theorem 1 Let f given by (1.1) satisfy the following subordination condition:

$$(1-\alpha )\frac{{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}+\alpha \frac{{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }\mathrm{\prime }}}{p(p-1)z^{p-2}}\prec \frac{1+Az}{1+Bz}.$$

(3.1)

Then

$$\frac{{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}\prec \mathrm{\Psi }(z)\prec \frac{1+Az}{1+Bz},$$

(3.2)

where

$$\mathrm{\Psi }(z)=\{\begin{array}{cc}\frac{A}{B}+(1-\frac{A}{B}){{(1+Bz)}^{-1}}_2{F}_1(1,1;\frac{p-1}{m\alpha }+1;\frac{Bz}{1+Bz})\hfill & \text{for}B\ne 0,\hfill \\ 1+\frac{p-1}{m\alpha +p-1}Az\hfill & \text{for}B=0,\hfill \end{array}$$

(3.3)

is the best dominant of (3.2). Furthermore,

$$f\in S_p^{s,b}(\beta ),$$

(3.4)

where

$$\beta =\{\begin{array}{cc}\frac{A}{B}+(1-\frac{A}{B})(1-B)^{-1}{}_2{F}_1(1,1;\frac{p-1}{m\alpha }+1;\frac{B}{B-1})\hfill & \text{for}B\ne 0,\hfill \\ 1-\frac{p-1}{m\alpha +p-1}A\hfill & \text{for}B=0.\hfill \end{array}$$

(3.5)

The estimate (3.4) is best possible.

Proof Let

$$\theta (z)=\frac{{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}(z\in U),$$

(3.6)

where θ is of the form (2.1) and is analytic in U. Differentiating (3.6) with respect to z, we get

$$(1-\alpha )\frac{{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}+\alpha \frac{{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }\mathrm{\prime }}}{p(p-1)z^{p-2}}=\theta (z)+\frac{\alpha }{p-1}z{\theta }^{\mathrm{\prime }}(z)\prec \frac{1+Az}{1+Bz}.$$

Applying Lemma 1 for $\varrho =\frac{p-1}{\alpha }$ and Lemma 7, we have

$$\begin{array}{rcl}\frac{{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}& \prec & \mathrm{\Psi }(z)=\frac{p-1}{m\alpha }{z}^{-\frac{p-1}{m\alpha }}{\int }_0^z{t}^{\frac{p-1}{m\alpha }-1}\left(\frac{1+At}{1+Bt}\right)dt\\ =& \{\begin{array}{cc}\frac{A}{B}+(1-\frac{A}{B})(1+Bz)^{-1}{}_2{F}_1(1,1;\frac{p-1}{m\alpha }+1;\frac{Bz}{1+Bz})\hfill & \text{for}B\ne 0,\hfill \\ 1+\frac{p-1}{m\alpha +p-1}Az\hfill & \text{for}B=0.\hfill \end{array}\end{array}$$

This proves the assertion (3.2) of Theorem 1. Next, in order to prove the assertion (3.4) of Theorem 1, it suffices to show that

$$\underset{|z|<1}{inf}\{Re(\mathrm{\Psi }(z))\}=\mathrm{\Psi }(-1).$$

Indeed, we have

$$Re\left\{\frac{1+Az}{1+Bz}\right\}\ge \frac{1-Ar}{1-Br}(|z|\le r<1).$$

Setting

$$G(z,\zeta )=\frac{1+A\zeta z}{1+B\zeta z}\text{and}d\nu (\zeta )=\frac{p-1}{m\alpha }{\zeta }^{\frac{p-1}{m\alpha }-1}d\zeta (0\le \zeta \le 1),$$

which is a positive measure on the closed interval $[0,1]$, we get

$$\mathrm{\Psi }(z)={\int }_0^{1}G(z,\zeta )d\nu (\zeta ).$$

Then

$$Re\{\mathrm{\Psi }(z)\}\ge {\int }_0^1\frac{1-A\zeta r}{1-B\zeta r}d\nu (\zeta )=\mathrm{\Psi }(-r)(|z|\le r<1).$$

Letting $r\to 1^{-}$ in the above inequality, we obtain the assertion (3.4). Finally, the estimate (3.4) is best possible as Ψ is the best dominant of (3.2). This completes the proof of Theorem 1. □

Theorem 2 If$f\in S_p^{s,b}(\eta )$ ($0\le \eta <1$), then

$$Re\{(1-\alpha )\frac{{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}+\alpha \frac{{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }\mathrm{\prime }}}{p(p-1)z^{p-2}}\}>\eta (|z|<R),$$

where

$$R={\left\{\frac{\sqrt{{(p-1)}^2+{(m\alpha )}^2}-m\alpha }{p-1}\right\}}^{\frac{1}{m}}.$$

(3.7)

The result is best possible.

Proof Let $f\in S_p^{s,b}(\eta )$, then we write

$$\frac{{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}=\eta +(1-\eta )u(z)(z\in U),$$

(3.8)

where u is of the form (2.1), is analytic in U and has a positive real part in U. Differentiating (3.8) with respect to z, we have

$$\frac{1}{1-\eta }\{(1-\alpha )\frac{{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}+\alpha \frac{{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }\mathrm{\prime }}}{p(p-1)z^{p-2}}-\eta \}=u(z)+\frac{\alpha }{p-1}z{u}^{\mathrm{\prime }}(z).$$

(3.9)

Applying the following well-known estimate [12]:

$$\frac{|z{u}^{\mathrm{\prime }}(z)|}{Re\{u(z)\}}\le \frac{2m{r}^m}{1-r^{2m}}(|z|=r<1),$$

in (3.9), we have

(3.10)

such that the right-hand side of (3.10) is positive, if $r<R$, where R is given by (3.7).

In order to show that the bound R is best possible, we consider the function $f\in A(p)$ defined by

$$\frac{{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}=\eta +(1-\eta )\frac{1+z^m}{1-z^m}(0\le \eta <1;z\in U).$$

Note that

$$\frac{1}{1-\eta }\{(1-\alpha )\frac{{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}+\alpha \frac{{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }\mathrm{\prime }}}{p(p-1)z^{p-2}}-\eta \}=\frac{(p-1)(1-z^{2m})-2\alpha m{z}^m}{(p-1){(1-z^m)}^2}=0,$$

for $z=R^{\cdot }exp\{\frac{\mathbf{i}\mathit{\pi }}{\mathbf{m}}\}$. This completes the proof of Theorem 2. □

For a function $f\in A(p)$, the generalized Bernardi-Libera-Livingston integral operator $F_{\mu ,p}$ is defined by

$$\begin{array}{rl}{F}_{\mu ,p}(f)(z)& =\frac{\mu +p}{z^{\mu }}{\int }_0^z{t}^{\mu -1}f(t)dt=(z^p+\sum _{k=1}^{\mathrm{\infty }}\frac{\mu +p}{\mu +p+k}{z}^{k+p})\ast f(z)\\ =z^p{}_2{F}_1(1,\mu +p;\mu +p+1;z)\ast f(z)(\mu >-p;z\in U).\end{array}$$

(3.11)

From (1.8) and (3.11), we have

$$z{\left(J_{p,s,b}{\mathrm{F}}_{\mu ,p}(f(z))\right)}^{\mathrm{\prime }}=(\mu +p)J_{p,s,b}(f)(z)-\mu J_{p,s,b}{\mathrm{F}}_{\mu ,p}(f(z))(\mu >-p;z\in U)$$

(3.12)

and

$$J_{p,s,b}{\mathrm{F}}_{\mu ,p}(f(z))={\mathrm{F}}_{\mu ,p}(J_{p,s,b}(f)(z)).$$

Theorem 3 Let$f\in S_p^{s,b}(A,B)$and$F_{\mu ,p}$be defined by (3.11). Then

$$\frac{{(J_{p,s,b}{\mathrm{F}}_{\mu ,p}(f(z)))}^{\mathrm{\prime }}}{p{z}^{p-1}}\prec \mathrm{\Phi }(z)\prec \frac{1+Az}{1+Bz},$$

(3.13)

where

$$\mathrm{\Phi }(z)=\{\begin{array}{cc}\frac{A}{B}+(1-\frac{A}{B})(1+Bz)^{-1}{}_2{F}_1(1,1;\frac{\mu +p}{m}+1;\frac{Bz}{1+Bz})\hfill & \text{for}B\ne 0,\hfill \\ 1+\frac{\mu +p}{\mu +p+m}Az\hfill & \text{for}B=0,\hfill \end{array}$$

(3.14)

is the best dominant of (3.13). Furthermore,

$$Re\left\{\frac{{(J_{p,s,b}{\mathrm{F}}_{\mu ,p}(f(z)))}^{\mathrm{\prime }}}{p{z}^{p-1}}\right\}>\psi (z\in U),$$

where

$$\psi =\{\begin{array}{cc}\frac{A}{B}+(1-\frac{A}{B})(1-B)^{-1}{}_2{F}_1(1,1;\frac{\mu +p}{m}+1;\frac{B}{B-1})\hfill & \text{for}B\ne 0,\hfill \\ 1-\frac{\mu +p}{\mu +p+m}A\hfill & \text{for}B=0.\hfill \end{array}$$

The result is best possible.

Proof Let

$$K(z)=\frac{{(J_{p,s,b}{\mathrm{F}}_{\mu ,p}(f(z)))}^{\mathrm{\prime }}}{p{z}^{p-1}}(z\in U),$$

(3.15)

where K is of the form (2.1) and is analytic in U. Using (3.12) in (3.15) and differentiating the resulting equation with respect to z, we have

$$\frac{{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}=K(z)+\frac{z{K}^{\mathrm{\prime }}(z)}{p+\mu }\prec \frac{1+Az}{1+Bz}.$$

The remaining part of the proof is similar to that of Theorem 1, and so we omit it. □

We note that

$$\frac{{(J_{p,s,b}{\mathrm{F}}_{\mu ,p}(f(z)))}^{\mathrm{\prime }}}{p{z}^{p-1}}=\frac{p+\mu }{p{z}^{p+\mu }}{\int }_0^z{t}^{\mu }{(J_{p,s,b}(f)(t))}^{\mathrm{\prime }}dt(f\in A(p);z\in U).$$

(3.16)

Putting $A=1-\frac{2\delta }{p}$ ($0\le \delta <1$) and $B=-1$ in Theorem 3 and using (3.16), we obtain the following corollary.

Corollary 1 If$f\in A(p)$satisfies the following inequality:

$$Re\left\{\frac{{(J_{p,s,b}f(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}\right\}>\delta (0\le \delta <1;z\in U),$$

then

$$Re\left\{\frac{p+\mu }{p{z}^{p+\mu }}{\int }_0^z{t}^{\mu }{(J_{p,s,b}(f)(t))}^{\mathrm{\prime }}dt\right\}>\frac{\delta }{p}+(1-\frac{\delta }{p}){[}_2{F}_1(1,1;\frac{\mu +p}{m}+1;\frac{1}{2})-1](z\in U).$$

The result is best possible.

Theorem 4 Let$f,g\in A(p)$satisfy the following inequality:

$$Re\left\{\frac{J_{p,s,b}(g)(z)}{z^p}\right\}>0(z\in U).$$

If

$$|\frac{J_{p,s,b}(f)(z)}{J_{p,s,b}(g)(z)}-1|<1(z\in U),$$

then

where

(3.17)

Proof Let

$$q(z)=\frac{J_{p,s,b}(f)(z)}{J_{p,s,b}(g)(z)}-1=c_m{z}^m+c_{m+1}{z}^{m+1}+\cdots ,$$

(3.18)

where $q(z)$ is analytic in U with $q(0)=0$ and $|q(z)|\le |z{|}^m$. Then, by applying the familiar Schwartz Lemma [15], we have $q(z)=z^m\mathcal{X}(z)$, where $\mathcal{X}$ is analytic in U and $|\mathcal{X}(z)|\le 1$. Therefore (3.18) leads to

$$J_{p,s,b}(f)(z)=J_{p,s,b}(g)(z)(1+z^m\mathcal{X}(z))(z\in U).$$

(3.19)

Differentiating (3.19) logarithmically with respect to z, we have

$$\frac{z{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }}}{J_{p,s,b}(f)(z)}=\frac{z{(J_{p,s,b}(g)(z))}^{\mathrm{\prime }}}{J_{p,s,b}(g)(z)}+\frac{z^m\{m\mathcal{X}(z)+z{\mathcal{X}}^{\mathrm{\prime }}(z)\}}{1+z^m\mathcal{X}(z)}.$$

(3.20)

Letting

$$\omega (z)=\frac{J_{p,s,b}(g)(z)}{z^p}(z\in U),$$

where ω is in the form (2.1), is analytic in U$Re\{\omega (z)\}>0$ and

$$\frac{z{(J_{p,s,b}(g)(z))}^{\mathrm{\prime }}}{J_{p,s,b}(g)(z)}=\frac{z{\omega }^{\mathrm{\prime }}(z)}{\omega (z)}+p,$$

then we have

$$Re\left\{\frac{z{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }}}{J_{p,s,b}(f)(z)}\right\}\ge p-\left|\frac{z{\omega }^{\mathrm{\prime }}(z)}{\omega (z)}\right|-\left|\frac{z^m\{m\mathcal{X}(z)+z{\mathcal{X}}^{\mathrm{\prime }}(z)\}}{1+z^m\mathcal{X}(z)}\right|.$$

(3.21)

Using the following known estimates [12] (see also [15]):

$$\left|\frac{{\omega }^{\mathrm{\prime }}(z)}{\omega (z)}\right|\le \frac{2m{r}^{m-1}}{1-r^{2m}}\text{and}\left|\frac{m\mathcal{X}(z)+z{\mathcal{X}}^{\mathrm{\prime }}(z)}{1+z^m\mathcal{X}(z)}\right|\le \frac{m}{1-r^m}(|z|=r<1),$$

in (3.21), we have

$$Re\left\{\frac{z{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }}}{J_{p,s,b}(f)(z)}\right\}\ge \frac{p-3m{r}^m-(p+m)r^{2m}}{1-r^{2m}}(|z|=r<1),$$

which is certainly positive, provided that , where is given by (3.17). This completes the proof of Theorem 4. □

Theorem 5 Let$-1\le B_i<A_i\le 1$ ($i=1,2$) and$\tau <p$. If each of the functions$f_i\in A(p)$satisfies the following subordination condition:

$$(1-\alpha )\frac{{(J_{p,s,b}(f_i)(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}+\alpha \frac{{(J_{p,s,b}(f_i)(z))}^{\mathrm{\prime }\mathrm{\prime }}}{p(p-1)z^{p-2}}\prec \frac{1+A_{i}z}{1+B_{i}z}(i=1,2),$$

(3.22)

then

$$(1-\alpha )\frac{{(J_{p,s,b}(F)(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}+\alpha \frac{{(J_{p,s,b}(F)(z))}^{\mathrm{\prime }\mathrm{\prime }}}{p(p-1)z^{p-2}}\prec \frac{1+(1-\frac{2\tau }{p})z}{1-z},$$

(3.23)

where

$$F(z)=J_{p,s,b}(f_1\ast f_2)(z)$$

(3.24)

and

$$\tau =p-4p\frac{(A_1-B_1)(A_2-B_2)}{(1-B_1)(1-B_2)}[1-{\frac{1}{2}}_2{F}_1(1,1;\frac{p-1}{\alpha }+1;\frac{1}{2})].$$

(3.25)

The result is best possible when$B_1=B_2=-1$.

Proof Suppose that the functions $f_i\in A(p)$ ($i=1,2$) satisfy the condition (3.22). Then by setting

$$h_i(z)=(1-\alpha )\frac{{(J_{p,s,b}(f_i)(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}+\alpha \frac{{(J_{p,s,b}(f_i)(z))}^{\mathrm{\prime }\mathrm{\prime }}}{p(p-1)z^{p-2}}(i=1,2),$$

(3.26)

we have

$$h_i\in H({\varrho }_i),{\varrho }_i=\frac{1-A_{\mathbf{i}}}{1-B_{\mathbf{i}}}(i=1,2).$$

And

$${(J_{p,s,b}(f_{\mathbf{i}})(z))}^{\mathrm{\prime }}=\frac{p(p-1)}{\alpha }{z}^{\frac{(1-p)(1-\alpha )}{\alpha }}{\int }_0^z{t}^{\frac{p-1}{\mathit{\alpha }}-\mathbf{1}}{h}_i(t)dt(i=1,2),$$

(3.27)

from (3.24), (3.26) and (3.27), we have

$${(J_{p,s,b}(F)(z))}^{\mathrm{\prime }}=\frac{p(p-1)}{\alpha }{\mathbf{z}}^{\frac{(\mathbf{1}-\mathbf{p})(\mathbf{1}-\mathit{\alpha })}{\mathit{\alpha }}}{\int }_0^z{t}^{\frac{p-1}{\mathit{\alpha }}-1}H(t)dt(i=1,2).$$

(3.28)

For convenience,

$$\begin{array}{rl}H(z)& =(1-\alpha )\frac{{(J_{p,s,b}(F)(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}+\alpha \frac{{(J_{p,s,b}(F)(z))}^{\mathrm{\prime }\mathrm{\prime }}}{p(p-1)z^{p-2}}\\ =\frac{p(p-1)}{\alpha }{z}^{\frac{(1-p)}{\alpha }}{\int }_0^z{t}^{\frac{p-1}{\mathit{\alpha }}-1}(h_1\ast h_2)(t)dt.\end{array}$$

(3.29)

Since $h_i\in H({\varrho }_i)$ ($i=1,2$), it follows from Lemma 3 that

$$(h_1\ast h_2)(z)\in H({\varrho }_3),{\varrho }_3=1-2(1-{\varrho }_1)(1-{\varrho }_2).$$

(3.30)

By using (3.30) in (3.29) and applying Lemmas 2 and 3, we have

$$\begin{array}{rcl}Re\{H(z)\}& =& \frac{p(p-1)}{\alpha }{\int }_0^1{s}^{\frac{p-1}{\alpha }-1}Re\{(h_1\ast h_2)(sz)\}ds\\ \ge & \frac{p(p-1)}{\alpha }{\int }_0^1{s}^{\frac{p-1}{\alpha }-1}(2{\varrho }_3-1+\frac{2(1-{\varrho }_3)}{1+s|z|})ds\\ >& \frac{p(p-1)}{\alpha }{\int }_0^1{s}^{\frac{p-1}{\alpha }-1}(2{\varrho }_3-1+\frac{2(1-{\varrho }_3)}{1+s})ds\\ =& p-4p\frac{(A_1-B_1)(A_2-B_2)}{(1-B_1)(1-B_2)}[1-\frac{p-1}{\alpha }{\int }_0^1{s}^{\frac{p-1}{\alpha }-1}{(1+s)}^{-1}ds]\\ =& p-4p\frac{(A_1-B_1)(A_2-B_2)}{(1-B_1)(1-B_2)}[1-{\frac{1}{2}}_2{F}_1(1,1;\frac{p-1}{\alpha }+1;\frac{1}{2})](\mathbf{z}\to -\mathbf{1})\\ =& \tau (z\in U).\end{array}$$

When $B_1=B_2=-1$, we consider $f_i\in A(p)$ ($i=1,2$) satisfy the condition (3.22) and are defined by

$${(J_{p,s,b}(f_i)(z))}^{\mathrm{\prime }}=\frac{p(p-1)}{\alpha }{z}^{\frac{(1-p)(1-\alpha )}{\alpha }}{\int }_0^z{t}^{\frac{p-1}{\alpha }-1}\left(\frac{1+A_{i}t}{1-t}\right)dt(i=1,2).$$

By using (3.29) and applying Lemma 3, we have

$$\begin{array}{rcl}H(z)& =& \frac{p(p-1)}{\alpha }{\int }_0^1{s}^{\frac{p-1}{\alpha }-1}[1-(1+A_1)(1+A_2)+\frac{(1+A_1)(1+A_2)}{1-sz}]ds\\ =& p-p(1+A_1)(1+A_2)+p(1+A_1)(1+A_2)(1-z)^{-1}{}_2{F}_1(1,1;\frac{p-1}{\alpha }+1;\frac{z}{z-1})\\ \to & p-p(1+A_1)(1+A_2)+\frac{p}{2}(1+A_1){(1+A_2)}_2{F}_1(1,1;\frac{p-1}{\alpha }+1;\frac{1}{2})(z\to -1).\end{array}$$

This completes the proof of Theorem 5. □

Remark 1 Putting $A_i=1-2{\theta }_i$ ($0\le {\theta }_i<1$) and $B_i=-1$ ($i=1,2$) in Theorem 5, we obtain the result obtained by Liu [10], Theorem 5].

Putting $A_i=1-2{\theta }_i$ ($0\le {\theta }_i<1$), $B_i=-1$ ($i=1,2$) and $s=0$ in Theorem 5, we obtain the following corollary.

Corollary 2 Let$\chi <p$and$f_i\in A(p)$satisfy the following inequality:

$$Re\{(1-\alpha )\frac{f_i^{\mathrm{\prime }}(z)}{p{z}^{p-1}}+\alpha \frac{f_i^{\mathrm{\prime }\mathrm{\prime }}(z)}{p(p-1)z^{p-2}}\}>{\theta }_i(0\le {\theta }_i<1;i=1,2),$$

then

$$Re\{(1-\alpha )\frac{{(f_1\ast f_2)}^{\mathrm{\prime }}(z)}{p{z}^{p-1}}+\alpha \frac{{(f_1\ast f_2)}^{\mathrm{\prime }\mathrm{\prime }}(z)}{p(p-1)z^{p-2}}\}>\frac{\chi }{p},$$

where

$$\chi =p-4p(1-{\theta }_1)(1-{\theta }_2)[1-{\frac{1}{2}}_2{F}_1(1,1;\frac{p-1}{\alpha }+1;\frac{1}{2})].$$

The result is best possible.

Theorem 6 Let$f\in S_p^{s,b}(A,B)$and$g\in A(p)$satisfy the following inequality:

$$Re\left\{\frac{g(z)}{z^p}\right\}>\frac{1}{2}(z\in U),$$

(3.31)

then

$$(f\ast g)(z)\in S_p^{s,b}(A,B).$$

Proof We have

$$\frac{{(J_{p,s,b}(f\ast g)(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}=\frac{{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}\ast \frac{g(z)}{z^p}(z\in U),$$

where $g(z)$ satisfies (3.31) and $\frac{1+Az}{1+Bz}$ is convex (univalent) in U. By using (1.10) and applying Lemma 5, we complete the proof of Theorem 6. □

Theorem 7 Let$\sigma >0$and$f\in A(p)$satisfy the following subordination condition:

$$(1-\alpha )\frac{J_{p,s,b}(f)(z)}{z^p}+\alpha \frac{{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}\prec \frac{1+Az}{1+Bz}.$$

(3.32)

Then

$$Re{\left\{\frac{J_{p,s,b}(f)(z)}{z^p}\right\}}^{\frac{1}{\sigma }}>{\gamma }^{\frac{1}{\sigma }},$$

where

$$\gamma =\{\begin{array}{cc}\frac{A}{B}+(1-\frac{A}{B})(1-B)^{-1}{}_2{F}_1(1,1;\frac{p}{m\alpha }+1;\frac{B}{B-1})\hfill & \text{for}B\ne 0,\hfill \\ 1-\frac{p}{m\alpha +p}A\hfill & \text{for}B=0.\hfill \end{array}$$

The result is best possible.

Proof Let

$$M(z)=\frac{J_{p,s,b}(f)(z)}{z^p}(z\in U),$$

(3.33)

where M is of the form (2.1) and is analytic in U. Differentiating (3.33) with respect to z, we have

$$(1-\alpha )\frac{J_{p,s,b}(f)(z)}{z^p}+\alpha \frac{{(J_{p,s,b}(f)(z))}^{\mathrm{\prime }}}{p{z}^{p-1}}=M(z)+\frac{\alpha }{p}z{M}^{\mathrm{\prime }}(z)\prec \frac{1+Az}{1+Bz}.$$

Now, by following steps similar to the proof of Theorem 1 and using the elementary inequality

$$Re\left\{{\mathrm{\Upsilon }}^{1/\varkappa }\right\}\ge {\{Re\mathrm{\Upsilon }\}}^{1/\varkappa }(Re\{\mathrm{\Upsilon }\}>0;\varkappa \in \mathbb{N}),$$

we obtain the result asserted by Theorem 7. □

Theorem 8 Let$\nu \in {\mathbb{C}}^{\ast }$and$A,B\in \mathbb{C}$with$A\ne B$and$|B|\le 1$. Suppose that

If$f\in A(p)$with$J_{p,s,b}(f)(z)\ne 0$for all$z\in U^{\ast }=U\mathrm{╲}\{0\}$, then

$$\frac{J_{p,s-1,b}(f)(z)}{J_{p,s,b}(f)(z)}\prec \frac{1+Az}{1+Bz},$$

implies

$${\left(\frac{J_{p,s,b}(f)(z)}{z^p}\right)}^{\nu }\prec q_1(z),$$

where

$$q_1(z)=\{\begin{array}{cc}{(1+Bz)}^{\nu (b+1)(A-B)/B},\hfill & \text{if}B\ne 0,\hfill \\ e^{\nu (b+1)Az},\hfill & \text{if}B=0,\hfill \end{array}$$

is the best dominant.

Proof Let us put

$$\phi (z)={\left(\frac{J_{p,s,b}(f)(z)}{z^p}\right)}^{\nu }(z\in U).$$

(3.34)

Then φ is analytic in U, $\phi (0)=1$ and $\phi (z)\ne 0$ for all $z\in U$. Taking the logarithmic derivatives in both sides of (3.34) and using the identity (1.9), we have

$$1+\frac{z{\phi }^{\mathrm{\prime }}(z)}{\nu (b+1)\phi (z)}=\frac{J_{p,s-1,b}(f)(z)}{J_{p,s,b}(f)(z)}\prec \frac{1+Az}{1+Bz}.$$

Now the assertions of Theorem 8 follow by using Lemma 6 for $\gamma =\nu (b+1)$. □

Putting $B=-1$ and $A=1-2\sigma$, $0\le \sigma <1$, in Theorem 8, we obtain the following corollary.

Corollary 3 Assume that$\nu \in {\mathbb{C}}^{\ast }$satisfies either$|2\nu (b+1)(1-\sigma )-1|\le 1$or$|2\nu (b+1)(1-\sigma )+1|\le 1$. If$f\in A(p)$with$J_{p,s,b}(f)(z)\ne 0$for$z\in U^{\ast }$, then

$$Re\left\{\frac{J_{p,s-1,b}(f)(z)}{J_{p,s,b}(f)(z)}\right\}>\sigma (z\in U),$$

implies

$${\left(\frac{J_{p,s,b}(f)(z)}{z^p}\right)}^{\nu }\prec q_2(z)={(1-z)}^{-2\nu (b+1)(1-\sigma )},$$

and$q_2$is the best dominant.

Remark 2 Specializing the parameters s and b in the above results of this paper, we obtain the results for the corresponding operators $F_{\mu ,p}$, $I_p^{\alpha }$, $J_p^{\gamma }$ and $J_p^{\gamma }(l)$ which are defined in the introduction.