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# Some subordination and superordination results of generalized Srivastava-Attiya operator

Journal of Inequalities and Applications20122012:115

https://doi.org/10.1186/1029-242X-2012-115

• Received: 8 December 2011
• Accepted: 24 May 2012
• Published:

## 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.

## Keywords

• analytic function
• differential subordination
• superordination

## 1 Introduction

Let H(U) be the class of functions analytic in $U=\left\{z\in ℂ:|z|<1\right\}$ and H[a, n] be the subclass of H(U) consisting of functions of the form f(z) = a + a n z n + a n +1zn+1 + ..., with H0 = H[0, 1] and H = H[1, 1]. Denote A(p) by the class of all analytic functions of the form
$f\left(z\right)={z}^{p}+\sum _{n=1}^{\infty }{a}_{p+n}{z}^{p+n}\left(p\in ℕ=\left\{1,2,3,\phantom{\rule{2.77695pt}{0ex}}\dots \right\};z\in U\right)$
(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 (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 $\varphi :{ℂ}^{2}×U\to ℂ$ 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 $\stackrel{̃}{q}$ that satisfies $\stackrel{̃}{q}\phantom{\rule{0.3em}{0ex}}\prec q$ 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 $\stackrel{̃}{q}$ that satisfies $q\prec \stackrel{̃}{q}\phantom{\rule{0.3em}{0ex}}$ 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:
$\Phi \left(z,\phantom{\rule{2.77695pt}{0ex}}s,\phantom{\rule{2.77695pt}{0ex}}a\right)=\sum _{n=0}^{\infty }\frac{{z}^{n}}{{\left(n+a\right)}^{s}},$
(1.4)

$\left(a\in ℂ\{ℤ}_{0}^{-};\phantom{\rule{0.3em}{0ex}}{ℤ}_{0}^{-}=\left\{0,-1,-2,\dots \right\}$; $s\in ℂ$ 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 ).

Recently, Srivastava and Attiya  introduced the linear operator L s,b : AA, defined in terms of the Hadamard product by
${L}_{s,b}\left(f\right)\left(z\right)={G}_{s,b}\left(z\right)*f\left(z\right)\left(z\in U;b\in ℂ\{ℤ}_{0}^{-};s\in ℂ\right),$
(1.5)
where
${G}_{s,b}={\left(1+b\right)}^{s}\left[\Phi \left(z,\phantom{\rule{2.77695pt}{0ex}}s,\phantom{\rule{2.77695pt}{0ex}}b\right)-{b}^{-s}\right]\left(z\in U\right).$
(1.6)

The Srivastava-Attiya operator L s,b contains among its special cases, the integral operators introduced and investigated by Alexander , Libera  and Jung et al. .

Analogous to L s,b , Liu  defined the operator J p,s,b : A(p) → A(p) by
${J}_{p,s,b}\left(f\right)\left(z\right)={G}_{p,s,b}\left(z\right)*f\left(z\right)\phantom{\rule{2.77695pt}{0ex}}\left(z\in U;b\in ℂ/{ℤ}_{0}^{-};s\in ℂ;p\in ℕ\right),$
(1.7)
where
${G}_{p,s,b}={\left(1+b\right)}^{s}\left[{\Phi }_{p}\left(z,\phantom{\rule{2.77695pt}{0ex}}s,\phantom{\rule{2.77695pt}{0ex}}b\right)-{b}^{-s}\right]$
and
${\Phi }_{p}\left(z,\phantom{\rule{2.77695pt}{0ex}}s,\phantom{\rule{2.77695pt}{0ex}}b\right)=\frac{1}{{b}^{s}}+\sum _{n=0}^{\infty }\frac{{z}^{n+p}}{{\left(n+1+b\right)}^{s}}.$
(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
1. (i)

J p ,0, b (f)(z) = f (z);

2. (ii)

${J}_{1,s,b}\left(f\right)\left(z\right)={L}_{s,b}f\left(z\right)\phantom{\rule{2.77695pt}{0ex}}\left(s\in ℂ,b\in ℂ\{ℤ}_{0}^{-}\right)$, where the operator L s,b was introduced by Srivastava and Attiya ;

3. (iii)

${{J}_{p}}_{,\mathsf{\text{1}},v+p-\mathsf{\text{1}}}\left(f\right)\left(z\right)={F}_{v,p}\left(f\left(z\right)\right)\left(v>-p,p\in ℕ\right)$, where the operator F v,p was introduced by Choi et al. ;

4. (iv)

${J}_{p,\alpha ,p}\left(f\right)\left(z\right)={I}_{p}^{\alpha }f\left(z\right)\phantom{\rule{2.77695pt}{0ex}}\left(\alpha \ge 0,\phantom{\rule{2.77695pt}{0ex}}p\in ℕ\right)$, where the operator ${I}_{p}^{\alpha }$ was introduced by Shams et al. ;

5. (v)

${J}_{p,m,p-1}\left(f\right)\left(z\right)={J}_{p}^{m}f\left(z\right)\phantom{\rule{2.77695pt}{0ex}}\left(m\in {ℕ}_{0}=ℕ\cup \left\{0\right\},\phantom{\rule{2.77695pt}{0ex}}p\in ℕ\right)$, where the operator ${J}_{p}^{m}$was introduced by El-Ashwah and Aouf ;

6. (vi)

${J}_{p,m,p+l-1}\left(f\right)\left(z\right)={J}_{p}^{m}\left(l\right)f\left(z\right)\phantom{\rule{2.77695pt}{0ex}}\left(m\in {ℕ}_{0},\phantom{\rule{2.77695pt}{0ex}}p\in ℕ,\phantom{\rule{2.77695pt}{0ex}}l\ge 0\right)$, where the operator ${J}_{p}^{m}\left(l\right)$ was introduced by El-Ashwah and Aouf .

It follows from (1.9) that:
(1.10)

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

Definition 1Denote by the set of all functions q(z) that are analytic and injective on$Ū\E\left(q\right)$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 2A 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 ≤ t1t2.

Lemma 1The function$L\left(z,t\right):U×\left[0;1\right)\to ℂ$of the form
$L\left(z,\phantom{\rule{2.77695pt}{0ex}}t\right)={a}_{1}\left(t\right)z+{a}_{2}\left(t\right){z}^{2}+\cdots \phantom{\rule{2.77695pt}{0ex}}\left({a}_{1}\left(t\right)\ne 0;t\ge 0\right)$
and $\underset{t\to \infty }{\text{lim}}|{a}_{1}\left(t\right)|=\infty$ is a subordination chain if and only if
$\mathsf{\text{Re}}\left\{\frac{z\partial L\left(z,t\right)/\partial z}{\partial L\left(z,t\right)/\partial t}\right\}>0\phantom{\rule{2.77695pt}{0ex}}\left(z\in U,\phantom{\rule{2.77695pt}{0ex}}t\ge 0\right).$
Lemma 2Suppose that the function$\mathcal{H}:{ℂ}^{2}\to ℂ$satisfies the condition
$\mathsf{\text{Re}}\left\{\phantom{\rule{2.77695pt}{0ex}}\mathcal{H}\phantom{\rule{2.77695pt}{0ex}}\left(is;\phantom{\rule{2.77695pt}{0ex}}t\right)\right\}\phantom{\rule{2.77695pt}{0ex}}\le 0$
for all real s and for all t-n (1 + s2) / 2, $n\in ℕ$. If the function p(z) = 1+p n z n +p n +1zn+1+ ...is analytic in U and
$\mathsf{\text{Re}}\left\{\mathcal{H}\left(p\left(z\right);z{p}^{\prime }\left(z\right)\right)\right\}>0\phantom{\rule{2.77695pt}{0ex}}\left(z\in U\right),$

then Re {p(z)} > 0 for z U.

Lemma 3Let 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:
$q\left(z\right)+\frac{z{q}^{\prime }\left(z\right)}{\kappa q\left(z\right)+\gamma }=h\left(z\right)\phantom{\rule{2.77695pt}{0ex}}\left(z\in U;q\left(0\right)=c\right)$

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

Lemma 4Let$p\in \mathcal{F}\left(a\right)$and let q(z) = a + a n z n + an+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 U and ζ0 δU\E(q) such that
$q\left({U}_{{r}_{0}}\right)\subset p\left(U\right);\phantom{\rule{0.3em}{0ex}}q\left({z}_{0}\right)=p\left({\zeta }_{0}\right)\phantom{\rule{0.3em}{0ex}}and\phantom{\rule{0.3em}{0ex}}{z}_{0}{p}^{\prime }\left({z}_{0}\right)=m{\zeta }_{0}{p}^{\prime }\left({\zeta }_{0}\right)\left(m\ge n\right).$
Lemma 5Let 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
$h\left(z\right)\prec \phi \left(q\left(z\right),\phantom{\rule{2.77695pt}{0ex}}z{q}^{\prime }\left(z\right)\right),$

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 $b\in ℂ\{ℤ}_{0}^{-}$, $s\in ℂ$, Re {b}, µ > 0, $p\in ℕ$, $z\in U$ 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
$F\left(z\right)={\left(\frac{{J}_{p,s,b}\left(f\right)\left(z\right)}{{z}^{p}}\right)}^{\mu }\mathsf{\text{and}}\phantom{\rule{0.3em}{0ex}}G\left(z\right)={\left(\frac{{J}_{p,s,b}\left(g\right)\left(z\right)}{{z}^{p}}\right)}^{\mu }\left(z\in U\right)$
(2.5)
and without loss of generality we assume that G(z) is analytic, univalent on and
${G}^{\prime }\left(\zeta \right)\ne 0\phantom{\rule{2.77695pt}{0ex}}\left(|\zeta |=1\right).\phantom{\rule{0.3em}{0ex}}$

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
$\mathsf{\text{Re}}\left\{q\left(z\right)\right\}>0\phantom{\rule{2.77695pt}{0ex}}\left(z\in U\right).$
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
$1+\frac{z{\varphi }^{″}\left(z\right)}{{\varphi }^{\prime }\left(z\right)}=q\left(z\right)+\frac{z{q}^{\prime }\left(z\right)}{q\left(z\right)+\mu \left(b+1\right)}=h\left(z\right)\phantom{\rule{0.3em}{0ex}}\left(z\in U\right).$
(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
$\mathcal{H}\left(u,\phantom{\rule{2.77695pt}{0ex}}v\right)=u+\frac{v}{u+\mu \left(b+1\right)}+\delta ,$

Where δ is given by (2.2). From (2.9) and (2.10), we obtain $\mathsf{\text{Re}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\left\{\mathcal{H}\left(q\left(z\right);\right)z{q}^{\prime }\left(z\right)\right)\right\}>0\left(z\in U\right)$.

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:
$\begin{array}{ll}\hfill \mathsf{\text{Re}}\left\{\mathcal{H}\left(i\vartheta ;t\right)\right\}\phantom{\rule{2.77695pt}{0ex}}& =\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{Re}}\left\{i\vartheta +\frac{t}{\mu \left(b+1\right)+i\vartheta }+\delta \right\}=\frac{t\mu \left(1+\mathsf{\text{Re}}\left(b\right)\right)}{|\mu \left(b+1\right)+i\vartheta {|}^{2}}+\delta \phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}-\frac{\Upsilon \left(b,\vartheta ,\delta \right)}{2|\mu \left(b+1\right)+i\vartheta {|}^{2}},\phantom{\rule{2em}{0ex}}\end{array}$
where
$\Upsilon \left(b,\phantom{\rule{2.77695pt}{0ex}}\vartheta ,\phantom{\rule{2.77695pt}{0ex}}\delta \right)=\left[\mu \left(1+\mathsf{\text{Re}}\left(b\right)\right)-2\delta \right]{\vartheta }^{2}-4\delta \mu \mathsf{\text{Im}}\left(b\right)\vartheta -2\delta |\mu \left(b+1\right){|}^{2}+\mu \left(1+\mathsf{\text{Re}}\left\{b\right\}\right).\phantom{\rule{2.77695pt}{0ex}}$
(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
$\mu \left(1+\mathsf{\text{Re}}\left(b\right)\right)={c}_{1}\phantom{\rule{0.3em}{0ex}}\mathsf{\text{and}}\phantom{\rule{0.3em}{0ex}}\mu \mathsf{\text{Im}}\left(b\right)={c}_{2}.$
We thus have to verify that
${c}_{1}-2\delta \ge 0,$
or
${c}_{1}\ge 2\delta =\frac{1+|c{|}^{2}-|1-{c}^{2}|}{2{c}_{1}}.$
This inequality will hold true if
$2{c}_{1}^{2}+|1-{c}^{2}|\ge 1+|c{|}^{2}=1+{c}_{1}^{2}+{c}_{2}^{2},$
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
$\mathsf{\text{Re}}\phantom{\rule{2.77695pt}{0ex}}\left\{q\left(z\right)\right\}>0\phantom{\rule{2.77695pt}{0ex}}\left(z\in U\right),$
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
$L\left(z,\phantom{\rule{2.77695pt}{0ex}}t\right)=G\left(z\right)+\frac{\left(1+t\right)z{G}^{\prime }\left(z\right)}{\mu \left(b+1\right)}\left(0\le t<\infty ;z\in U\right).$
(2.13)
We note that
This show that the function
$L\left(z,\phantom{\rule{2.77695pt}{0ex}}t\right)={a}_{1}\left(t\right)z+\cdots \phantom{\rule{0.3em}{0ex}}$
satisfies the condition a1 (t) ≠ 0 (0 ≤ t < ∞). Further, we have
$\mathsf{\text{Re}}\left\{\frac{z\partial L\left(z,t\right)/\partial z}{\partial L\left(z,t\right)/\partial t}\right\}=\mathsf{\text{Re}}\left\{\mu \left(b+1\right)+\left(1+t\right)q\left(z\right)\right\}>0\phantom{\rule{0.3em}{0ex}}\left(0\le t<\infty ;z\in U\right).$
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
$\varphi \left(z\right)=G\left(z\right)+\frac{z{G}^{\prime }\left(z\right)}{\mu \left(b+1\right)}=L\left(z,\phantom{\rule{2.77695pt}{0ex}}0\right)$
and
$L\left(z,\phantom{\rule{2.77695pt}{0ex}}0\right)\prec L\left(z,\phantom{\rule{2.77695pt}{0ex}}t\right)\phantom{\rule{2.77695pt}{0ex}}\left(0\le t<\infty \right),$
which implies that
$L\left(\zeta ,\phantom{\rule{2.77695pt}{0ex}}t\right)\notin L\left(U,\phantom{\rule{2.77695pt}{0ex}}0\right)=\varphi \left(U\right)\phantom{\rule{0.3em}{0ex}}\left(0\le t<\infty ;\zeta \in \partial U\right).$
(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
$F\left({z}_{0}\right)=G\left({\zeta }_{0}\right)\phantom{\rule{0.3em}{0ex}}and\phantom{\rule{0.3em}{0ex}}{z}_{0}{F}^{\prime }\left({z}_{0}\right)=\left(1+t\right){\zeta }_{0}{G}^{\prime }\left({\zeta }_{0}\right)\phantom{\rule{0.3em}{0ex}}\left(0\le t<\infty \right).$
(2.15)
Hence, by using (2.5), (2.13), (2.15) and (2.3), we have
$\begin{array}{ll}\hfill L\left({\zeta }_{0},\phantom{\rule{2.77695pt}{0ex}}t\right)\phantom{\rule{2.77695pt}{0ex}}& =\phantom{\rule{2.77695pt}{0ex}}G\left({\zeta }_{0}\right)+\frac{\left(1+t\right){\zeta }_{0}{G}^{\prime }\left({\zeta }_{0}\right)}{\mu \left(b+1\right)}=F\left({z}_{0}\right)+\frac{{z}_{0}{F}^{\prime }\left({z}_{0}\right)}{\mu \left(b+1\right)}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{2.77695pt}{0ex}}\left(\frac{{J}_{p,s-1,b}\left(f\right)\left(z\right)}{{J}_{p,s,b}\left(f\right)\left(z\right)}\right){\left(\frac{{J}_{p,s,b}\left(f\right)\left(z\right)}{{z}^{p}}\right)}^{\mu }\in \varphi \left(U\right).\phantom{\rule{2em}{0ex}}\end{array}$

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
$\mathsf{\text{Re}}\left\{q\left(z\right)\right\}>0\phantom{\rule{2.77695pt}{0ex}}\left(z\in U\right).$
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
$\varphi \left(z\right)=G\left(z\right)+\frac{z{G}^{\prime }\left(z\right)}{\mu \left(b+1\right)}=\phi \left(G\left(z\right),\phantom{\rule{2.77695pt}{0ex}}z{G}^{\prime }\left(z\right)\right)$

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
$\begin{array}{ll}\hfill \left(\frac{{J}_{p,s-1,b}\left({g}_{1}\right)\left(z\right)}{{J}_{p,s,b}\left({g}_{1}\right)\left(z\right)}\right){\left(\frac{{J}_{p,s,b}\left({g}_{1}\right)\left(z\right)}{{z}^{p}}\right)}^{\mu }& \prec \left(\frac{{J}_{p,s-1,b}\left(f\right)\left(z\right)}{{J}_{p,s,b}\left(f\right)\left(z\right)}\right){\left(\frac{{J}_{p,s,b}\left(f\right)\left(z\right)}{{z}^{p}}\right)}^{\mu }\phantom{\rule{2em}{0ex}}\\ \prec \left(\frac{{J}_{p,s-1,b}\left({g}_{2}\right)\left(z\right)}{{J}_{p,s,b}\left({g}_{2}\right)\left(z\right)}\right){\left(\frac{{J}_{p,s,b}\left({g}_{2}\right)\left(z\right)}{{z}^{p}}\right)}^{\mu },\phantom{\rule{2em}{0ex}}\end{array}$
(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, $p\in ℕ$) in our results of this article, we obtain the results obtained by Aouf and Seoudy;

(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.

## Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, Mansoura University, Mansoura, 35516, Egypt

## References 