Some subordination and superordination results of generalized Srivastava-Attiya operator

Abstract

In this article, we obtain some subordination and superordination-preserving results of the generalized Srivastava-Attyia operator. Sandwich-type result is also obtained.

Mathematics Subject Classification 2000: 30C45.

1 Introduction

Let H(U) be the class of functions analytic in $U=\left\{z\in ℂ:\left|z\right|<1\right\}$ and H[a, n] be the subclass of H(U) consisting of functions of the form f(z) = a + a _n zⁿ + a _{n +1}zⁿ⁺¹ + ..., with H₀ = H[0, 1] and H = H[1, 1]. Denote A(p) by the class of all analytic functions of the form

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

(1.1)

and let A (1) = A. For f, F ∈ H(U), the function f(z) is said to be subordinate to F(z), or F(z) is superordinate to f(z), if there exists a function ω(z) analytic in U with ω(0) = 0 and |ω(z)| < 1(z ∈ U), such that f(z) = F(ω(z)). In such a case we write f(z) ≺ F(z). If F is univalent, then f(z) ≺ F(z) if and only if f(0) = F(0) and f(U) ⊂ F(U) (see [1, 2]).

Let $\varphi :{ℂ}^2\times U\to ℂ$ and h(z) be univalent in U. If p(z) is analytic in U and satisfies the first order differential subordination:

$$\varphi \left(p\left(z\right),z{p}^{\prime }\left(z\right);z\right)\prec h\left(z\right),$$

(1.2)

then p (z) is a solution of the differential subordination (1.2). The univalent function q (z) is called a dominant of the solutions of the differential subordination (1.2) if p(z) ≺ q(z) for all p(z) satisfying (1.2). A univalent dominant $\tilde{q}$ that satisfies $\tilde{q}\prec q$ for all dominants of (1.2) is called the best dominant. If p(z) and ϕ(p(z), zp' (z) ; z) are univalent in U and if p(z) satisfies the first order differential superordination:

$$h\left(z\right)\prec \varphi \left(p\left(z\right),z{p}^{\prime }\left(z\right);z\right),$$

(1.3)

then p(z) is a solution of the differential superordination (1.3). An analytic function q(z) is called a subordinant of the solutions of the differential superordination (1.3) if q(z) ≺ p(z) for all p(z) satisfying (1.3). A univalent subordinant $\tilde{q}$ that satisfies $q\prec \tilde{q}$ for all subordinants of (1.3) is called the best subordinant (see [1, 2]).

The general Hurwitz-Lerch Zeta function Φ(z, s, a) is defined by:

$$\Phi \left(z,s,a\right)=\sum _{n=0}^{\infty }\frac{z^n}{{\left(n+a\right)}^s},$$

(1.4)

$(a\in ℂ\backslash {\mathbb{Z}}_0^{-};{\mathbb{Z}}_0^{-}=\left\{0,-1,-2,\dots \right\}$; $s\in ℂ$ when |z| < 1; R{s} > 1 when |z| = 1).

For further interesting properties and characteristics of the Hurwitz-Lerch Zeta function Φ(z, s, a) (see [3–7]).

Recently, Srivastava and Attiya [8] introduced the linear operator L _s,b : A → A, defined in terms of the Hadamard product by

$$L_{s,b}\left(f\right)\left(z\right)=G_{s,b}\left(z\right)*f\left(z\right)\left(z\in U;b\in ℂ\backslash {\mathbb{Z}}_0^{-};s\in ℂ\right),$$

(1.5)

where

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

(1.6)

The Srivastava-Attiya operator L _s,b contains among its special cases, the integral operators introduced and investigated by Alexander [9], Libera [10] and Jung et al. [11].

Analogous to L _s,b , Liu [12] defined the operator J _p,s,b : A(p) → A(p) by

$$J_{p,s,b}\left(f\right)\left(z\right)=G_{p,s,b}\left(z\right)*f\left(z\right)\left(z\in U;b\in ℂ/{\mathbb{Z}}_0^{-};s\in ℂ;p\in \mathbb{N}\right),$$

(1.7)

where

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

and

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

(1.8)

It is easy to observe from (1.7) and (1.8) that

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

(1.9)

We note that

• (i) J _{p ,0, b} (f)(z) = f (z);

• (ii) $J_{1,s,b}\left(f\right)\left(z\right)=L_{s,b}f\left(z\right)\left(s\in ℂ,b\in ℂ\backslash {\mathbb{Z}}_0^{-}\right)$, where the operator L _s,b was introduced by Srivastava and Attiya [8];

• (iii) ${J_p}_{,\mathsf{\text{1}},v+p-\mathsf{\text{1}}}\left(f\right)\left(z\right)=F_{v,p}\left(f\left(z\right)\right)\left(v>-p,p\in \mathbb{N}\right)$, where the operator F _v,p was introduced by Choi et al. [13];

• (iv) $J_{p,\alpha ,p}\left(f\right)\left(z\right)=I_p^{\alpha }f\left(z\right)\left(\alpha \ge 0,p\in \mathbb{N}\right)$, where the operator $I_p^{\alpha }$ was introduced by Shams et al. [14];

• (v) $J_{p,m,p-1}\left(f\right)\left(z\right)=J_p^{m}f\left(z\right)\left(m\in {\mathbb{N}}_0=\mathbb{N}\cup \left\{0\right\},p\in \mathbb{N}\right)$, where the operator $J_p^m$was introduced by El-Ashwah and Aouf [15];

• (vi) $J_{p,m,p+l-1}\left(f\right)\left(z\right)=J_p^m\left(l\right)f\left(z\right)\left(m\in {\mathbb{N}}_0,p\in \mathbb{N},l\ge 0\right)$, where the operator $J_p^m\left(l\right)$ was introduced by El-Ashwah and Aouf [15].

It follows from (1.9) that:

$$z\mathsf{\text{(}}{J_{p\mathsf{\text{,}}s\mathsf{\text{ + 1}}}}_{\mathsf{\text{,}}b}\mathsf{\text{(}}f\mathsf{\text{)(}}z\mathsf{\text{))' = (}}b\mathsf{\text{ + 1)}}{J}_{p\mathsf{\text{,}}s\mathsf{\text{,}}b}\mathsf{\text{(}}f\mathsf{\text{)(}}z\mathsf{\text{) { (}}b\mathsf{\text{ + 1 { }}p\mathsf{\text{)}}{J_{p\mathsf{\text{,}}s\mathsf{\text{ + 1}}}}_{\mathsf{\text{,}}b}\mathsf{\text{(}}f\mathsf{\text{)(}}z\mathsf{\text{)}}\mathsf{\text{.}}$$

(1.10)

To prove our results, we need the following definitions and lemmas.

Definition 1[1]Denote bythe set of all functions q(z) that are analytic and injective on$Ū\backslash E\left(q\right)$where

$$E\left(q\right)=\left\{\zeta \in \partial U:\underset{z\to \zeta }{\text{lim}}q\left(z\right)=\infty \right\}$$

and are such that q'(ζ) ≠ 0 for ζ ∈ δU\E(q). Further let the subclass of for which q(0) = a be denoted by$\mathcal{F}\left(a\right)$, $\mathcal{F}\left(0\right)\equiv {\mathcal{F}}_0$and$\mathcal{F}\left(1\right)\equiv {\mathcal{F}}_1$.

Definition 2[2]A function L (z, t) (z ∈ U, t ≥ 0) is said to be a subordination chain if L (0, t) is analytic and univalent in U for all t ≥ 0, L (z, 0) is continuously differentiable on [0; 1) for all z ∈ U and L (z, t₁) ≺ L (z, t₂) for all 0 ≤ t₁ ≤ t₂.

Lemma 1[16]The function$L\left(z,t\right):U\times \left[0;1\right)\to ℂ$of the form

$$L\left(z,t\right)=a_1\left(t\right)z+a_2\left(t\right)z^2+\cdots \left(a_1\left(t\right)\ne 0;t\ge 0\right)$$

and $\underset{t\to \infty }{\text{lim}}\left|a_1\left(t\right)\right|=\infty$ is a subordination chain if and only if

$$\mathsf{\text{Re}}\left\{\frac{z\partial L\left(z,t\right)/\partial z}{\partial L\left(z,t\right)/\partial t}\right\}>0\left(z\in U,t\ge 0\right).$$

Lemma 2[17]Suppose that the function$\mathcal{H}:{ℂ}^2\to ℂ$satisfies the condition

$$\mathsf{\text{Re}}\left\{\mathcal{H}\left(is;t\right)\right\}\le 0$$

for all real s and for all t ≤ -n (1 + s²) / 2, $n\in \mathbb{N}$. If the function p(z) = 1+p _n zⁿ +p _{n +1}zⁿ⁺¹+ ...is analytic in U and

$$\mathsf{\text{Re}}\left\{\mathcal{H}\left(p\left(z\right);z{p}^{\prime }\left(z\right)\right)\right\}>0\left(z\in U\right),$$

then Re {p(z)} > 0 for z ∈ U.

Lemma 3[18]Let κ, $\gamma \in ℂ$with κ ≠ 0 and let h ∈ H(U) with h(0) = c. If Re {κh(z) + γ} > 0 (z ∈ U), then the solution of the following differential equation:

$$q\left(z\right)+\frac{z{q}^{\prime }\left(z\right)}{\kappa q\left(z\right)+\gamma }=h\left(z\right)\left(z\in U;q\left(0\right)=c\right)$$

is analytic in U and satisfies Re {κq(z) + γ} > 0 for z ∈ U.

Lemma 4[1]Let$p\in \mathcal{F}\left(a\right)$and let q(z) = a + a _n zⁿ + a_n+1zⁿ⁺¹ + ...be analytic in U with q (z) ≠ a and n ≥ 1. If q is not subordinate to p, then there exists two points z₀ = r₀e^iθ ∈ U and ζ₀ ∈ δU\E(q) such that

$$q\left(U_{r_0}\right)\subset p\left(U\right);q\left(z_0\right)=p\left({\zeta }_0\right)and{z}_0{p}^{\prime }\left(z_0\right)=m{\zeta }_0{p}^{\prime }\left({\zeta }_0\right)\left(m\ge n\right).$$

Lemma 5[2]Let q ∈ H[a; 1] and $\varphi :{ℂ}^2\to ℂ$. Also set φ(q(z), zq'(z)) = h(z). If L(z, t) = φ (q (z), tzq'(z)) is a subordination chain and$q\in H\left[a;1\right]\cap \mathcal{F}\left(a\right)$, then

$$h\left(z\right)\prec \phi \left(q\left(z\right),z{q}^{\prime }\left(z\right)\right),$$

implies that q(z) ≺ p(z). Furthermore, if φ(q(z), zq'(z)) = h(z) has a univalent solution$q\in \mathcal{F}\left(a\right)$, then q is the best subordinant.

In the present article, we aim to prove some subordination-preserving and superordination-preserving properties associated with the integral operator J _p,s,b . Sandwich-type result involving this operator is also derived.

2 Main results

Unless otherwise mentioned, we assume throughout this section that $b\in ℂ\backslash {\mathbb{Z}}_0^{-}$, $s\in ℂ$, Re {b}, µ > 0, $p\in \mathbb{N}$, $z\in U$ and the powers are understood as principle values.

Theorem 1. Let f, g ∈ A (p) and

$$\mathsf{\text{Re}}\left\{1+\frac{z{\varphi }^{″}\left(z\right)}{{\varphi }^{\prime }\left(z\right)}\right\}>-\delta \left(\varphi \left(z\right)=\left(\frac{J_{p,s-1,b}\left(g\right)\left(z\right)}{J_{p,s,b}\left(g\right)\left(z\right)}\right){\left(\frac{J_{p,s,b}\left(g\right)\left(z\right)}{z^p}\right)}^{\mu };z\in U\right),$$

(2.1)

where δ is given by

$$\delta =\frac{1+{\mu }^2|b+1{|}^2-|1-{\mu }^2{\left(b+1\right)}^2|}{4\mu \left[1+\mathsf{\text{Re}}\left\{b\right\}\right]}\left(z\in U\right).$$

(2.2)

Then the subordination condition

$$\left(\frac{J_{p,s-1,b}\left(f\right)\left(z\right)}{J_{p,s,b}\left(f\right)\left(z\right)}\right){\left(\frac{J_{p,s,b}\left(f\right)\left(z\right)}{z^p}\right)}^{\mu }\prec \left(\frac{J_{p,s-1,b}\left(g\right)\left(z\right)}{J_{p,s,b}\left(g\right)\left(z\right)}\right){\left(\frac{J_{p,s,b}\left(g\right)\left(z\right)}{z^p}\right)}^{\mu },$$

(2.3)

implies that

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

(2.4)

where${\left(\frac{J_{p,s,b}\left(g\right)\left(z\right)}{z^p}\right)}^{\mu }$is the best dominant.

Proof. Let us define the functions F(z) and G(z) in U by

$$F\left(z\right)={\left(\frac{J_{p,s,b}\left(f\right)\left(z\right)}{z^p}\right)}^{\mu }\mathsf{\text{and}}G\left(z\right)={\left(\frac{J_{p,s,b}\left(g\right)\left(z\right)}{z^p}\right)}^{\mu }\left(z\in U\right)$$

(2.5)

and without loss of generality we assume that G(z) is analytic, univalent on and

$$G^{\prime }\left(\zeta \right)\ne 0\left(\left|\zeta \right|=1\right).$$

If not, then we replace F(z) and G(z) by F(ρz) and G(ρz), respectively, with 0 < ρ < 1. These new functions have the desired properties on , so we can use them in the proof of our result and the results would follow by letting ρ → 1.

We first show that, if

$$q\left(z\right)=1+\frac{z{G}^{″}\left(z\right)}{G^{\prime }\left(z\right)}\left(z\in U\right),$$

(2.6)

then

$$\mathsf{\text{Re}}\left\{q\left(z\right)\right\}>0\left(z\in U\right).$$

From (1.10) and the definition of the functions G, ϕ, we obtain that

$$\varphi \left(z\right)=G\left(z\right)+\frac{z{G}^{\prime }\left(z\right)}{\mu \left(b+1\right)}.$$

(2.7)

Differentiating both sides of (2.7) with respect to z yields

$${\varphi }^{\prime }\left(z\right)=\left(1+\frac{1}{\mu \left(b+1\right)}\right)G^{\prime }\left(z\right)+\frac{z{G}^{″}\left(z\right)}{\mu \left(b+1\right)}.$$

(2.8)

Combining (2.6) and (2.8), we easily get

$$1+\frac{z{\varphi }^{″}\left(z\right)}{{\varphi }^{\prime }\left(z\right)}=q\left(z\right)+\frac{z{q}^{\prime }\left(z\right)}{q\left(z\right)+\mu \left(b+1\right)}=h\left(z\right)\left(z\in U\right).$$

(2.9)

It follows from (2.1) and (2.9) that

$$\mathsf{\text{Re}}\left\{h\left(z\right)+\mu \left(b+1\right)\right\}>0\left(z\in U\right).$$

(2.10)

Moreover, by using Lemma 3, we conclude that the differential Equation (2.9) has a solution q(z) ∈ H(U) with h(0) = q(0) = 1. Let

$$\mathcal{H}\left(u,v\right)=u+\frac{v}{u+\mu \left(b+1\right)}+\delta ,$$

Where δ is given by (2.2). From (2.9) and (2.10), we obtain $\mathsf{\text{Re}}\left\{\mathcal{H}\left(q\left(z\right);\right)z{q}^{\prime }\left(z\right)\right)\}>0\left(z\in U\right)$.

To verify the condition

$$\mathsf{\text{Re}}\left\{\mathcal{H}\left(i\vartheta ;t\right)\right\}\le 0\left(\vartheta \in ℝ;t\le -\frac{1+{\vartheta }^2}{2}\right),$$

(2.11)

we proceed as follows:

$$\begin{array}{ll}\hfill \mathsf{\text{Re}}\left\{\mathcal{H}\left(i\vartheta ;t\right)\right\}& =\mathsf{\text{Re}}\left\{i\vartheta +\frac{t}{\mu \left(b+1\right)+i\vartheta }+\delta \right\}=\frac{t\mu \left(1+\mathsf{\text{Re}}\left(b\right)\right)}{|\mu \left(b+1\right)+i\vartheta {|}^2}+\delta \\ \le -\frac{\Upsilon \left(b,\vartheta ,\delta \right)}{2|\mu \left(b+1\right)+i\vartheta {|}^2},\end{array}$$

where

$$\Upsilon \left(b,\vartheta ,\delta \right)=\left[\mu \left(1+\mathsf{\text{Re}}\left(b\right)\right)-2\delta \right]{\vartheta }^2-4\delta \mu \mathsf{\text{Im}}\left(b\right)\vartheta -2\delta |\mu \left(b+1\right){|}^2+\mu \left(1+\mathsf{\text{Re}}\left\{b\right\}\right).$$

(2.12)

For δ given by (2.2), the coefficient of ϑ² in the quadratic expression ϒ(b, ϑ, δ) given by (2.12) is positive or equal to zero. To check this, put µ(b + 1) = c, so that

$$\mu \left(1+\mathsf{\text{Re}}\left(b\right)\right)=c_1\mathsf{\text{and}}\mu \mathsf{\text{Im}}\left(b\right)=c_2.$$

We thus have to verify that

$$c_1-2\delta \ge 0,$$

or

$$c_1\ge 2\delta =\frac{1+|c{|}^2-|1-c^2|}{2c_1}.$$

This inequality will hold true if

$$2c_1^2+|1-c^2|\ge 1+|c{|}^2=1+c_1^2+c_2^2,$$

that is, if

$$|1-c^2|\ge 1-\mathsf{\text{Re}}\left(c^2\right),$$

which is obviously true. Moreover, the quadratic expression ϒ(b, ϑ, δ)by ϑ in (2.12) is a perfect square for the assumed value of δ given by (2.2). Hence we see that (2.11) holds. Thus, by Lemma 2, we conclude that

$$\mathsf{\text{Re}}\left\{q\left(z\right)\right\}>0\left(z\in U\right),$$

that is, that G defined by (2.5) is convex (univalent) in U. Next, we prove that the subordination condition (2.3) implies that

$$F\left(z\right)\prec G\left(z\right),$$

for the functions F and G defined by (2.5). Consider the function L(z, t) given by

$$L\left(z,t\right)=G\left(z\right)+\frac{\left(1+t\right)z{G}^{\prime }\left(z\right)}{\mu \left(b+1\right)}\left(0\le t<\infty ;z\in U\right).$$

(2.13)

We note that

$${\left.\frac{\partial L\left(z,t\right)}{\partial z}\right|}_{z=0}=G^{\prime }\left(0\right)\left(1+\frac{1+t}{\mu \left(b+1\right)}\right)\ne 0\left(0\le t<\infty ;z\in U;\mathsf{\text{Re}}\left\{\mu \left(b+1\right)\right\}>0\right).$$

This show that the function

$$L\left(z,t\right)=a_1\left(t\right)z+\cdots$$

satisfies the condition a₁ (t) ≠ 0 (0 ≤ t < ∞). Further, we have

$$\mathsf{\text{Re}}\left\{\frac{z\partial L\left(z,t\right)/\partial z}{\partial L\left(z,t\right)/\partial t}\right\}=\mathsf{\text{Re}}\left\{\mu \left(b+1\right)+\left(1+t\right)q\left(z\right)\right\}>0\left(0\le t<\infty ;z\in U\right).$$

Since G(z) is convex and Re {µ(b + 1)} > 0. Therefore, by using Lemma 1, we deduce that L(z, t) is a subordination chain. It follows from the definition of subordination chain that

$$\varphi \left(z\right)=G\left(z\right)+\frac{z{G}^{\prime }\left(z\right)}{\mu \left(b+1\right)}=L\left(z,0\right)$$

and

$$L\left(z,0\right)\prec L\left(z,t\right)\left(0\le t<\infty \right),$$

which implies that

$$L\left(\zeta ,t\right)\notin L\left(U,0\right)=\varphi \left(U\right)\left(0\le t<\infty ;\zeta \in \partial U\right).$$

(2.14)

If F is not subordinate to G, by using Lemma 4, we know that there exist two points z₀ ∈ U and ζ₀ ∈ ∂U such that

$$F\left(z_0\right)=G\left({\zeta }_0\right)and{z}_0{F}^{\prime }\left(z_0\right)=\left(1+t\right){\zeta }_0{G}^{\prime }\left({\zeta }_0\right)\left(0\le t<\infty \right).$$

(2.15)

Hence, by using (2.5), (2.13), (2.15) and (2.3), we have

$$\begin{array}{ll}\hfill L\left({\zeta }_0,t\right)& =G\left({\zeta }_0\right)+\frac{\left(1+t\right){\zeta }_0{G}^{\prime }\left({\zeta }_0\right)}{\mu \left(b+1\right)}=F\left(z_0\right)+\frac{z_0{F}^{\prime }\left(z_0\right)}{\mu \left(b+1\right)}\\ =\left(\frac{J_{p,s-1,b}\left(f\right)\left(z\right)}{J_{p,s,b}\left(f\right)\left(z\right)}\right){\left(\frac{J_{p,s,b}\left(f\right)\left(z\right)}{z^p}\right)}^{\mu }\in \varphi \left(U\right).\end{array}$$

This contradicts (2.14). Thus, we deduce that F ≺ G. Considering F = G, we see that the function G is the best dominant. This completes the proof of Theorem 1.

We now derive the following superordination result.

Theorem 2. Let f, g ∈ A (p) and

$$\mathsf{\text{Re}}\left\{1+\frac{z{\varphi }^{″}\left(z\right)}{{\varphi }^{\prime }\left(z\right)}\right\}>-\delta \left(\varphi \left(z\right)=\left(\frac{J_{p,s-1,b}\left(g\right)\left(z\right)}{J_{p,s,b}\left(g\right)\left(z\right)}\right){\left(\frac{J_{p,s,b}\left(g\right)\left(z\right)}{z^p}\right)}^{\mu };z\in U\right),$$

(2.16)

where δ is given by (2.2) . If the function$\left(\frac{J_{p,s-1,b}\left(f\right)\left(z\right)}{J_{p,s,b}\left(f\right)\left(z\right)}\right){\left(\frac{J_{p,s,b}\left(f\right)\left(z\right)}{z^p}\right)}^{\mu }$is univalent in U and ${\left(\frac{J_{p,s,b}\left(f\right)\left(z\right)}{z^p}\right)}^{\mu }\in \mathcal{F}$, then the superordination condition

$$\left(\frac{J_{p,s-1,b}\left(g\right)\left(z\right)}{J_{p,s,b}\left(g\right)\left(z\right)}\right){\left(\frac{J_{p,s,b}\left(g\right)\left(z\right)}{z^p}\right)}^{\mu }\prec \left(\frac{J_{p,s-1,b}\left(f\right)\left(z\right)}{J_{p,s,b}\left(f\right)\left(z\right)}\right){\left(\frac{J_{p,s,b}\left(f\right)\left(z\right)}{z^p}\right)}^{\mu },$$

(2.17)

implies that

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

(2.18)

where ${\left(\frac{J_{p,s,b}\left(f\right)\left(z\right)}{z^p}\right)}^{\mu }$ is the best subordinant.

Proof. Suppose that the functions F, G and q are defined by (2.5) and (2.6), respectively. By applying similar method as in the proof of Theorem 1, we get

$$\mathsf{\text{Re}}\left\{q\left(z\right)\right\}>0\left(z\in U\right).$$

Next, to arrive at our desired result, we show that G ≺ F. For this, we suppose that the function L(z, t) be defined by (2.13). Since G is convex, by applying a similar method as in Theorem 1, we deduce that L(z, t) is subordination chain. Therefore, by using Lemma 5, we conclude that G ≺ F. Moreover, since the differential equation

$$\varphi \left(z\right)=G\left(z\right)+\frac{z{G}^{\prime }\left(z\right)}{\mu \left(b+1\right)}=\phi \left(G\left(z\right),z{G}^{\prime }\left(z\right)\right)$$

has a univalent solution G, it is the best subordinant. This completes the proof of Theorem 2.

Combining the above-mentioned subordination and superordination results involving the operator J _p,s,b , the following "sandwich-type result" is derived.

Theorem 3. Let f, g _j ∈ A (p) (j = 1, 2) and

$$\mathsf{\text{Re}}\left\{1+\frac{z{\varphi }_j^{″}\left(z\right)}{{\varphi }_j^{\prime }\left(z\right)}\right\}>-\delta \left({\varphi }_j\left(z\right)=\left(\frac{J_{p,s-1,b}\left(g_j\right)\left(z\right)}{J_{p,s,b}\left(g_j\right)\left(z\right)}\right){\left(\frac{J_{p,s,b}\left(g_j\right)\left(z\right)}{z^p}\right)}^{\mu }\left(j=1,2\right);z\in U\right),$$

where δ is given by (2.2) . If the function$\left(\frac{J_{p,s-1,b}\left(f\right)\left(z\right)}{J_{p,s,b}\left(f\right)\left(z\right)}\right){\left(\frac{J_{p,s,b}\left(f\right)\left(z\right)}{z^p}\right)}^{\mu }$is univalent in U and${\left(\frac{J_{p,s,b}\left(f\right)\left(z\right)}{z^p}\right)}^{\mu }\in \mathcal{F}$, then the condition

$$\begin{array}{ll}\hfill \left(\frac{J_{p,s-1,b}\left(g_1\right)\left(z\right)}{J_{p,s,b}\left(g_1\right)\left(z\right)}\right){\left(\frac{J_{p,s,b}\left(g_1\right)\left(z\right)}{z^p}\right)}^{\mu }& \prec \left(\frac{J_{p,s-1,b}\left(f\right)\left(z\right)}{J_{p,s,b}\left(f\right)\left(z\right)}\right){\left(\frac{J_{p,s,b}\left(f\right)\left(z\right)}{z^p}\right)}^{\mu }\\ \prec \left(\frac{J_{p,s-1,b}\left(g_2\right)\left(z\right)}{J_{p,s,b}\left(g_2\right)\left(z\right)}\right){\left(\frac{J_{p,s,b}\left(g_2\right)\left(z\right)}{z^p}\right)}^{\mu },\end{array}$$

(2.19)

implies that

$${\left(\frac{J_{p,s,b}\left(g_1\right)\left(z\right)}{z^p}\right)}^{\mu }\prec {\left(\frac{J_{p,s,b}\left(f\right)\left(z\right)}{z^p}\right)}^{\mu }\prec {\left(\frac{J_{p,s,b}\left(g_2\right)\left(z\right)}{z^p}\right)}^{\mu },$$

(2.20)

where${\left(\frac{J_{p,s,b}\left(g_1\right)\left(z\right)}{z^p}\right)}^{\mu }$and${\left(\frac{J_{p,s,b}\left(g_2\right)\left(z\right)}{z^p}\right)}^{\mu }$are, respectively, the best subordinant and the best dominant.

Remark. (i) Putting µ = 1, b = p and s = α(α = 0, $p\in \mathbb{N}$) in our results of this article, we obtain the results obtained by Aouf and Seoudy[19];

(ii) Specializing the parameters s and b in our results of this article, we obtain the results for the corresponding operators F _v,p , $I_p^{\alpha }$, $J_p^m$ and $J_p^m\left(l\right)$which are defined in the introduction.