Let $\mathcal{H}(\mathbb{U})$ denote the class of analytic functions in the open unit disk $\mathbb{U}=\{z\in\mathbb{C}: |z|<1\}$. For $a\in\mathbb{C}$ and $n\in\mathbb{N}=\{1, 2, \cdots\}$, let

$\mathcal{H}[a, n]=\left\{f\in\mathcal{H}(\mathbb{U}):f(z)=a+a_nz^n+a_{n+1}z^{n+1}+\cdots\right\}.$

Let $f$ and $g$ be two members of $\mathcal{H}(\mathbb{U})$. The function $f$ is said to be subordinate to $g$, or $g$ is said to be superordinate to $f$, if there exists a Schwarz function $\omega$, analytic in $\mathbb{U}$ with $\omega(0)=0$ and $|\omega(z)|<1 (z\in\mathbb{U})$, such that $f(z)=g(\omega(z)) (z\in\mathbb{U})$. In such a case, we write $f\prec g$ or $f(z)\prec g(z) (z\in\mathbb{U})$. Furthermore, if the function $g$ is univalent in $\mathbb{U}$, then we have (see [8] and [21])

$f\prec g (z\in\mathbb{U})\Longleftrightarrow f(0)=g(0)\;\mathrm{and}\;f(\mathbb{U})\subset g(\mathbb{U}).$

Definition 1.1 (see [8]) Let $\phi: {\mathbb{C}}^2\rightarrow\mathbb{C}$ and let $h$ be univalent in $\mathbb{U}$. If $\mathfrak{p}$ is analytic in $\mathbb{U}$ and satisfies the following differential subordination

$\phi(\mathfrak{p}(z), z{\mathfrak{p}}'(z))\prec h(z) (z\in\mathbb{U}), $

(1.1)

then $\mathfrak{p}$ is called a solution of the differential subordination (1.1). The univalent function $\mathfrak{q}$ is called a dominant of the solutions of the differential subordination (1.1), if $\mathfrak{p}\prec\mathfrak{q}$ for all $\mathfrak{p}$ satisfying (1.1). A dominant $\tilde{\mathfrak{q}}$ that satisfies $\tilde{\mathfrak{q}}\prec\mathfrak{q}$ for all dominants $\mathfrak{q}$ of (1.1) is said to be the best dominant.

Definition 1.2 (see [9]) Let $\varphi: {\mathbb{C}}^2\rightarrow\mathbb{C}$ and let $h$ be univalent in $\mathbb{U}$. If $\mathfrak{p}$ and $\varphi(\mathfrak{p}(z), z{\mathfrak{p}}'(z))$ are univalent in $\mathbb{U}$ and satisfy the following differential superordination

$h(z)\prec\varphi(\mathfrak{p}(z), z{\mathfrak{p}}'(z)) (z\in\mathbb{U}), $

(1.2)

then $\mathfrak{p}$ is called a solution of the differential superordination (1.2). An analytic function $\mathfrak{q}$ is called a subordination of the solutions of the differential superordination (1.2), if $\mathfrak{q}\prec\mathfrak{p}$ for all $\mathfrak{p}$ satisfying (1.2). A univalent subordination $\tilde{\mathfrak{q}}$ that satisfies $\mathfrak{q}\prec\tilde{\mathfrak{q}}$ for all subordinations $\mathfrak{q}$ of (1.2) is said to be the best subordination.

Definition 1.3 (see [9]) We denote by $\mathcal{Q}$ the class of functions f that are analytic and injective on $\overline{\mathbb{U}}\setminus E(f)$, where

$E(f)=\left\{\xi: \xi\in\partial\mathbb{U}\;\mathrm{and}\;\lim\limits_{z\rightarrow\xi}f(z)=\infty\right\}, $

and are such that $f'(\xi)\neq0 (\xi\in\partial\mathbb{U}\setminus E(f)).$

Let $\mathcal{A}(p)$ denote the class of all analytic functions of the form

$f(z)=z^p+\sum\limits_{n=1}^{\infty}a_{p+n}z^{p+n} (p\in\mathbb{N}; z\in\mathbb{U}).$

(1.3)

Motivated essentially by Jung et al.[4], Liu and Owa [5] introduced the integral operator $Q_{\beta, p}^\alpha: \mathcal{A}(p)\longrightarrow\mathcal{A}(p)$ as follows:

$Q_{\beta, p}^\alpha f(z)=\left( \begin{array}{ll} p+\alpha+\beta-1\\\\ p+\beta-1\\ \end{array} \right)\frac{\alpha}{z^\beta}\int_0^z\left(1-\frac{t}{z}\right)^{\alpha-1}t^{\beta-1}f(t)dt (\alpha>0;\beta>-1;p\in\mathbb{N}), $

(1.4)

and

$Q_{\beta, p}^0 f(z)=f(z) (\alpha=0; \beta>-1).$

If $f\in\mathcal{A}(p)$ given by (1.3), then from (1.4), we deduce that

$Q_{\beta, p}^\alpha f(z)=z^p+\frac{\Gamma(\alpha+\beta+p)}{\Gamma(\beta+p)}\sum\limits_{n=1}^{\infty} \frac{\Gamma(\beta+p+n)}{\Gamma(\alpha+\beta+p+n)}a_{p+n}z^{p+n} (\alpha>0;\beta>-1;p\in\mathbb{N}).$

(1.5)

It is easily verified from the definition (1.5) that (see [5])

$z\left(Q_{\beta, p}^\alpha f(z)\right)'=(\alpha+\beta+p-1)Q_{\beta, p}^{\alpha-1} f(z)-(\alpha+\beta-1)Q_{\beta, p}^\alpha f(z).$

(1.6)

We note that, for $p=1$, we obtain the operator $Q_{\beta, 1}^\alpha=Q_{\beta}^\alpha$ defined by Jung et al.[4], and studied by Aouf [16] and Gao et al.[6]. On the other hand, if we set $\alpha=1, \beta=c$ in (1.5), we obtain the generalized Libera operator $J_c (c>-p)$ defined by (see [1, 13]; also [19, 20])

$Q_{c, p}^1 f(z)=J_c(f)(z)=\frac{c+p}{z^c}\int_0^zt^{c-1}f(t)dt (c>-p; z\in\mathbb{U}).$

(1.7)

With the help of the principle of subordination, various subordination preserving properties involving certain integral operators for analytic functions in $\mathbb{U}$ were investigated by Bulboc$\breve{a}$ [2], Miller et al.[10], and Owa and Srivastava [14]. Moreover, Miller and Mocanu [9] considered differential superordinations, as the dual problem of differential subordinations (see also [3]). In the present paper, we investigate some subordination and superordination preserving properties of the integral operator $Q_{\beta, p}^\alpha$ defined by (1.4). Also, we obtain several sandwich-type results for these multivalent functions.

In order to establish our main results, we shall require the following lemmas.

Lemma 1.1 (see [11]) Suppose that the function $H: \mathbb{C}^2\longrightarrow\mathbb{C}$ satisfies the following condition

$\mathrm{Re}\left\{H(is, t)\right\}\leq0$

for all real $s$ and $t\leq-\frac{n(1+s^2)}{2} (n\in\mathbb{N})$. If the function $\mathfrak{p}(z)=1+\mathfrak{p}_nz^n+\cdots$ is analytic in $\mathbb{U}$ and

$\mathrm{Re}\left\{H(\mathfrak{p}(z), z\mathfrak{p}'(z))\right\}>0\;(z\in\mathbb{U}), $

then $\mathrm{Re}\{\mathfrak{p}(z)\}>0$ for $z\in\mathbb{U}.$

Lemma 1.2 (see [12]) Let $\kappa, \gamma\in\mathbb{C}$ with $\kappa\neq0$ and let $h\in\mathcal{H}(\mathbb{U})$ with $h(0)=b$. If $\mathrm{Re}\left\{\kappa h(z)+\gamma\right\}>0 (z\in\mathbb{U}), $ then the solution of the following differential equation

$q(z)+\frac{zq'(z)}{\kappa q(z)+\gamma}=h(z)\;(z\in\mathbb{U}; q(0)=b)$

is analytic in $\mathbb{U}$ and satisfies the inequality given by $\mathrm{Re}\left\{\kappa q(z)+\gamma\right\}>0$ for $z\in\mathbb{U}.$

Lemma 1.3 (see [8]) Let $\mathfrak{p}\in\mathcal{Q}$ with $\phi(0)=a$ and let the function $q(z)=a+a_nz^n+\cdots$ be analytic in $\mathbb{U}$ with $q(z)\neq a$ and $n\in\mathbb{N}$. If $q$ is not subordinate to $\mathfrak{p}$, then there exist points

$z_0=r_0e^{i\theta}\in\mathbb{U}\;\mathrm{and}\;\xi_0\in\partial\mathbb{U}\setminus E(f), $

for which

$q(\mathbb{U}_{r_0})\subset\mathfrak{p}(\mathbb{U}), q(z_0)=\mathfrak{p}(z_0)\;\mathrm{and}\;z_0q'(z_0)=m\xi_0\mathfrak{p}'(\xi_0)\;(m\geq n), $

where $\mathbb{U}_{r_0}=\{z\in\mathbb{C}: |z|<r_0\}$,

A function $L(z, t)$ defined on $\mathbb{U}\times[0, \infty)$ is the subordination chain (or L$\ddot{o}$wner chain) if $L(\cdot, t)$ is analytic and univalent in $\mathbb{U}$ for all $t\in[0, \infty)$, $L(\cdot, t)$ is continuously differentiable on $[0, \infty)$ for all $z\in\mathbb{U}$ and $L(z, t_1)\prec L(z, t_2) (z\in\mathbb{U}; 0\leq t_1\leq t_2).$

Lemma 1.4 (see [9]) Let $q\in\mathcal{H}[a, 1]$ and $\varphi: \mathbb{C}^2\longrightarrow\mathbb{C}$. Also let

$\varphi(q(z), zq'(z))=h(z)\;(z\in\mathbb{U}).$

If $L(z, t)=\varphi(q(z), tzq'(z))$ is a subordination chain and $\mathfrak{p}\in\mathcal{H}[a, 1]\cap\mathcal{Q}$, then

$h(z)\prec\varphi(\mathfrak{p}(z), z\mathfrak{p}'(z))\;(z\in\mathbb{U}), $

implies that $q(z)\prec\mathfrak{p}(z)$. Furthermore, if $\varphi(q(z), zq'(z))=h(z)$ has a univalent solution $q\in\mathcal{Q}$, then $q$ is the best subordinant.

Lemma 1.5 (see [15]) The function $L(z, t)=a_1(t)z+a_2(t)z^2+\cdots$ with $a_1(t)\neq0$ and $\lim\limits_{t\rightarrow\infty}|a_1(t)|=\infty$ is a subordination chain if and only if

$\mathrm{Re}\left\{\frac{z\partial L(z, t)/\partial z}{\partial L(z, t)/\partial t}\right\}>0\;(z\in\mathbb{U}; 0\leq t<\infty).$

First of all, we begin by proving the following subordination theorem involving the operator $Q_{\beta, p}^\alpha$ defined by (1.4). Unless otherwise mentioned, we assume throughout this paper that $\alpha\geq1, \beta>-1, 0<\lambda\leq1, \mu>0, p\in\mathbb{N}$ and $z\in\mathbb{U}$.

Theorem 2.1 Let $f, g\in\mathcal{A}(p)$ and suppose that

$\mathrm{Re}\left\{1+\frac{z\phi''(z)}{\phi'(z)}\right\}>-\delta \left(\phi(z)=(1-\lambda)\left(\frac{Q_{\beta, p}^\alpha g(z)}{z^p}\right)^\mu+\lambda\left(\frac{Q_{\beta, p}^{\alpha-1} g(z)}{Q_{\beta, p}^\alpha g(z)}\right)\left(\frac{Q_{\beta, p}^\alpha g(z)}{z^p}\right)^\mu\right), $

(2.1)

where

$\delta=\frac{\lambda^2+\mu^2(\alpha+\beta+p-1)^2-|\lambda^2-\mu^2(\alpha+\beta+p-1)^2|} {4\lambda\mu(\alpha+\beta+p-1)}.$

(2.2)

Then the following subordination condition

$(1-\lambda)\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu+\lambda\left(\frac{Q_{\beta, p}^{\alpha-1} f(z)}{Q_{\beta, p}^\alpha f(z)}\right)\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu\prec \phi(z)$

(2.3)

implies that

$\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu\prec\left(\frac{Q_{\beta, p}^\alpha g(z)}{z^p}\right)^\mu.$

Moreover, the function $\left(\frac{Q_{\beta, p}^\alpha g(z)}{z^p}\right)^\mu$ is the best dominant.

ProofLet us define the functions $F$ and $G$, respectively, by

$F(z)=\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu\;\mathrm{and}\;G(z)=\left(\frac{Q_{\beta, p}^\alpha g(z)}{z^p}\right)^\mu.$

(2.4)

We first prove that, if the function $q$ is defined by

$q(z)=1+\frac{zG''(z)}{G'(z)}\;(z\in\mathbb{U}), $

(2.5)

then $\mathrm{Re}\{q(z)\}>0$ for $z\in\mathbb{U}$.

Taking the logarithmic differentiation on both sides of the second equation in (2.4) and using (1.6) for $g\in\mathcal{A}(p)$, we have

$\phi(z)=G(z)+\frac{\lambda zG'(z)}{\mu(\alpha+\beta+p-1)}.$

(2.6)

Differentiating both sides of (2.6) with respect to $z$ yields

$\phi'(z)=\left(1+\frac{\lambda}{\mu(\alpha+\beta+p-1)}\right)G'(z)+\frac{\lambda zG''(z)}{\mu(\alpha+\beta+p-1)}.$

(2.7)

Combining (2.5) and (2.7), we easily get

$1+\frac{z\phi''(z)}{\phi'(z)}=q(z)+\frac{zq'(z)}{q(z)+\mu(\alpha+\beta+p-1)/\lambda}=h(z)\;(z\in\mathbb{U}).$

(2.8)

Thus, form (2.1) and (2.8), we see that

$\mathrm{Re}\left\{h(z)+\frac{\mu(\alpha+\beta+p-1)}{\lambda}\right\}>0\;(z\in\mathbb{U}).$

Also, in view of Lemma 1.2, we conclude that the differential equation (2.8) has a solution $q\in\mathcal{H}(\mathbb{U})$ with $q(0)=h(0)=1.$

Let us put

$H(u, v)=u+\frac{v}{u+\mu(\alpha+\beta+p-1)/\lambda}+\delta, $

(2.9)

where $\delta$ is given by (2.2). From (2.1), (2.8), together with (2.9), we obtain

$\mathrm{Re}\left\{H(q(z), zq'(z))\right\}>0\;(z\in\mathbb{U}).$

Now, we proceed to show that

$\mathrm{Re}\left\{H(is, t)\right\}\leq0\;\left(s\in\mathbb{R}; t\leq-\frac{1+s^2}{2}\right).$

(2.10)

In fact, from (2.9), we have

$\quad \mathrm{Re}\left\{H(is, t)\right\}=\mathrm{Re}\left\{is+\frac{t}{is+\mu(\alpha+\beta+p-1)/\lambda}+\delta\right\} \\ =\frac{t\lambda\mu(\alpha+\beta+p-1)}{\lambda^2s^2+\mu^2(\alpha+\beta+p-1)^2}+\delta \\ \leq-\frac{E_\delta(s)}{2[\lambda^2s^2+\mu^2(\alpha+\beta+p-1)^2]},$

where

$E_\delta(s)=[\lambda\mu(\alpha+\beta+p-1)-2\delta\lambda^2]s^2-2\delta\mu^2(\alpha+\beta+p-1)^2+\lambda\mu(\alpha+\beta+p-1). $

(2.11)

For $\delta$ given by (2.2), we can prove easily that the expression $E_\delta(s)$ in (2.11) is greater than or equal to zero, which implies that (2.10) holds true. Therefore, by using Lemma 1.1, we conclude that $\mathrm{Re}\{q(z)\}>0$ for $z\in\mathbb{U}$, that is, that the function $G$ defined by (2.4) is convex (univalent) in $\mathbb{U}$.

Next, we prove that $F\prec G (z\in\mathbb{U})$ holds for the functions $F$ and $G$ defined by (2.4). Without loss of generality, we assume that $G$ is analytic and univalent on $\overline{\mathbb{U}}$ and that $G'(\xi)\neq0$ for $|\xi|=1$.

Let us define the function $L(z, t)$ by

$L(z, t)=G(z)+\frac{\lambda(1+t)}{\mu(\alpha+\beta+p-1)}zG'(z)\;(0\leq t<\infty; z\in\mathbb{U}).$

Then

$\frac{\partial L(z, t)}{\partial z}\bigg |_{z=0}=G'(0)\left(1+\frac{\lambda(1+t)}{\mu(\alpha+\beta+p-1)}\right)=1+\frac{\lambda(1+t)}{\mu(\alpha+\beta+p-1)}\neq0\;(0\leq t<\infty; z\in\mathbb{U}), $

and this show that the function $L(z, t)=a_1(t)z+a_2(t)z^2+\cdots$ satisfies the conditions $a_1(t)\neq0$ for all $t\in[0, \infty)$ and $\lim\limits_{t\rightarrow\infty}|a_1(t)|=+\infty$.

Moreover, we have

$\mathrm{Re}\left\{\frac{z\partial L(z, t)/\partial z}{\partial L(z, t)/\partial t}\right\} =\mathrm{Re}\left\{\mu(\alpha+\beta+p-1)+\lambda(1+t)\left(1+\frac{zG''(z)}{G'(z)}\right)\right\}>0\;(0\leq t<\infty), $

because $G$ is convex in $\mathbb{U}$. Hence, by virtue of Lemma 1.5, we deduce that $L(z, t)$ is a subordination chain. We notice from the definition of subordination chain that

$\phi(z)=G(z)+\frac{\lambda zG'(z)}{\mu(\alpha+\beta+p-1)}=L(z, 0)$

and

$L(z, 0)\prec L(z, t) (0\leq t<\infty), $

which implies that

$L(\xi, t)\notin L(\mathbb{U}, 0)=\phi(\mathbb{U})\;(\xi\in\partial\mathbb{U}; 0\leq t<\infty).$

(2.12)

Now, we suppose that $F$ is not subordinate $G$, then by Lemma 1.3, there exist two points $z_0\in\mathbb{U}$ and $\xi_0\in\partial\mathbb{U}$, such that

$F(z_0)=G(\xi_0)\;\mathrm{and}\;z_0F'(z_0)=(1+t)\xi_0G'(\xi_0) (0\leq t<\infty).$

Thus, by means of the subordination condition (2.3), we have

$\quad L(\xi_0, t)=G(\xi_0)+\frac{\lambda(1+t)\xi_0G'(\xi_0)}{\mu(\alpha+\beta+p-1)} =F(z_0)+\frac{\lambda z_0F'(z_0)}{\mu(\alpha+\beta+p-1)} \\ =(1-\lambda)\left(\frac{Q_{\beta, p}^\alpha f(z_0)}{z_0^p}\right)^\mu+\lambda\left(\frac{Q_{\beta, p}^{\alpha-1} f(z_0)}{Q_{\beta, p}^\alpha f(z_0)}\right)\left(\frac{Q_{\beta, p}^\alpha f(z_0)}{z_0^p}\right)^\mu\in\phi(\mathbb{U}),$

which contradicts to (2.12). Hence, we deduce that $F\prec G$. Considering $F=G$, we know that the function $G$ is the best dominant. This completes the proof of Theorem 2.1.

We next derive a dual result of Theorem 2.1, in the sense that subordinations are replaced by superordinations.

Theorem 2.2 Let $f, g\in\mathcal{A}(p)$ and suppose that

$\mathrm{Re}\left\{1+\frac{z\phi''(z)}{\phi'(z)}\right\}>-\delta \left(\phi(z)=(1-\lambda)\left(\frac{Q_{\beta, p}^\alpha g(z)}{z^p}\right)^\mu+\lambda\left(\frac{Q_{\beta, p}^{\alpha-1} g(z)}{Q_{\beta, p}^\alpha g(z)}\right)\left(\frac{Q_{\beta, p}^\alpha g(z)}{z^p}\right)^\mu\right), $

where $\delta$ is given by (2.2). If the function

$(1-\lambda)\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu+\lambda\left(\frac{Q_{\beta, p}^{\alpha-1} f(z)}{Q_{\beta, p}^\alpha f(z)}\right)\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu$

is univalent in $\mathbb{U}$ and $\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu\in\mathcal{H}[1,1]\cap\mathcal{Q}$. Then the following superordination condition

$\phi(z)\prec(1-\lambda)\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu+\lambda\left(\frac{Q_{\beta, p}^{\alpha-1} f(z)}{Q_{\beta, p}^\alpha f(z)}\right)\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu$

implies that

$\left(\frac{Q_{\beta, p}^\alpha g(z)}{z^p}\right)^\mu\prec\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu.$

Moreover, the function $\left(\frac{Q_{\beta, p}^\alpha g(z)}{z^p}\right)^\mu$ is the best subordination.

Proof Let us define the functions $F$ and $G$ just as (2.4). We first observe that, if the function $q$ is defined by (2.5), then we obtain from (2.6) that

$\phi(z)=G(z)+\frac{\lambda zG'(z)}{\mu(\alpha+\beta+p-1)}=\varphi\left(G(z), zG'(z)\right).$

(2.13)

By applying the same method as in the proof of Theorem 2.1, we can prove that $\mathrm{Re}\{q(z)\}>0$ for $z\in\mathbb{U}$. That is, the function $G$ defined by (2.4) is convex (univalent) in $\mathbb{U}$.

Next, we will show that $G\prec F$. For this purpose, we consider the function $L(z, t)$ defined by

$L(z, t)=G(z)+\frac{\lambda t}{\mu(\alpha+\beta+p-1)}zG'(z)\;(0\leq t<\infty; z\in\mathbb{U}).$

Since the function $G$ is convex in $\mathbb{U}$, we can prove easily that $L(z, t)$ is a subordination chain as in the proof of Theorem 2.1. Therefore, by Lemma 1.4, we conclude that $G\prec F$. Furthermore, since the differential equation (2.13) has the univalent solution $G$, it is the best subordination of the given differential superordination. We thus complete the proof of Theorem 2.2.

If we combine Theorems 2.1 and 2.2, then we get the following sandwich-type theorem.

Theorem 2.3 Let $f, g_j\in\mathcal{A}(p) (j=1, 2)$ and suppose that

$\mathrm{Re}\left\{1+\frac{z\phi_j''(z)}{\phi_j'(z)}\right\}>-\delta \left(\phi_j(z)=(1-\lambda)\left(\frac{Q_{\beta, p}^\alpha g_j(z)}{z^p}\right)^\mu+\lambda\left(\frac{Q_{\beta, p}^{\alpha-1}g_j(z)}{Q_{\beta, p}^\alpha g_j(z)}\right)\left(\frac{Q_{\beta, p}^\alpha g_j(z)}{z^p}\right)^\mu\right), $

(2.14)

where $\delta$ is given by (2.2). If the function

$(1-\lambda)\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu+\lambda\left(\frac{Q_{\beta, p}^{\alpha-1} f(z)}{Q_{\beta, p}^\alpha f(z)}\right)\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu$

is univalent in $\mathbb{U}$ and $\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu\in\mathcal{H}[1,1]\cap\mathcal{Q}$. Then the following subordination relationship

$\phi_1(z)\prec(1-\lambda)\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu+\lambda\left(\frac{Q_{\beta, p}^{\alpha-1} f(z)}{Q_{\beta, p}^\alpha f(z)}\right)\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu\prec\phi_2(z)$

implies that

$\left(\frac{Q_{\beta, p}^\alpha g_1(z)}{z^p}\right)^\mu\prec\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu\prec\left(\frac{Q_{\beta, p}^\alpha g_2(z)}{z^p}\right)^\mu.$

Moreover, the functions $\left(\frac{Q_{\beta, p}^\alpha g_1(z)}{z^p}\right)^\mu$ and $\left(\frac{Q_{\beta, p}^\alpha g_2(z)}{z^p}\right)^\mu$ are, respectively, the best subordination and the best dominant.

Remark 2.1 By putting $\lambda=1$ in Theorems 2.1-2.3, we obtain the results obtained by Aouf and Seoudy [18].

Remark 2.2 By taking $\lambda=\mu=1$ in Theorems 2.1-2.3, we obtain the results obtained by Aouf and Seoudy [17].

Since the assumption of Theorem 2.3 of the preceding section that the functions

$(1-\lambda)\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu+\lambda\left(\frac{Q_{\beta, p}^{\alpha-1} f(z)}{Q_{\beta, p}^\alpha f(z)}\right)\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu\;\mathrm{and}\;\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu$

need to be univalent in $\mathbb{U}$, is not so easy to check, we will replace these conditions by another simple conditions in the following result.

Corollary 3.1 Let $f, g_j\in\mathcal{A}(p) (j=1, 2)$. Suppose that the condition (2.14) is satisfied and

$\mathrm{Re}\left\{1+\frac{z\psi''(z)}{\psi'(z)}\right\}>-\delta \left(\psi(z)=(1-\lambda)\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu+\lambda\left(\frac{Q_{\beta, p}^{\alpha-1} f(z)}{Q_{\beta, p}^\alpha f(z)}\right)\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu\right), $

(3.1)

where $\delta$ is given by (2.2). Then the following subordination relationship

$\phi_1(z)\prec(1-\lambda)\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu+\lambda\left(\frac{Q_{\beta, p}^{\alpha-1} f(z)}{Q_{\beta, p}^\alpha f(z)}\right)\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu\prec\phi_2(z)$

implies that

$\left(\frac{Q_{\beta, p}^\alpha g_1(z)}{z^p}\right)^\mu\prec\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu\prec\left(\frac{Q_{\beta, p}^\alpha g_2(z)}{z^p}\right)^\mu.$

Moreover, the functions $\left(\frac{Q_{\beta, p}^\alpha g_1(z)}{z^p}\right)^\mu$ and $\left(\frac{Q_{\beta, p}^\alpha g_2(z)}{z^p}\right)^\mu$ are, respectively, the best subordination and the best dominant.

Proof To prove our result, we have to show that the condition (3.1) implies the univalence of $\psi$ and $F(z)=\left(\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\right)^\mu$. Since $\delta$ given by (2.2) in Theorem 2.1 satisfies the inequality $0<\delta\leq\frac{1}{2}$, the condition (3.1) means that $\psi$ is a close-to-convex function in $\mathbb{U}$ (see [7]) and hence $\psi$ is univalent in $\mathbb{U}$. Also, by using the same techniques as in the proof of Theorem 2.1, we can prove that $F$ is convex (univalent) in $\mathbb{U}$, and so the details may be omitted. Therefore, by applying Theorem 2.3, we obtain the desired result.

Upon setting $\mu=1$ in Theorem 2.3, we are easily led to the following result.

Corollary 3.2 Let $f, g_j\in\mathcal{A}(p) (j=1, 2)$ and suppose that

$\mathrm{Re}\left\{1+\frac{z\phi_j''(z)}{\phi_j'(z)}\right\}>-\delta \left(\phi_j(z)=\frac{(1-\lambda)Q_{\beta, p}^\alpha g_j(z)+\lambda Q_{\beta, p}^{\alpha-1} g_j(z)}{z^p} (j=1, 2); z\in\mathbb{U}\right), $

where $\delta$ is given by (2.2) with $\mu=1$. If the function

$\frac{(1-\lambda)Q_{\beta, p}^\alpha f(z)+\lambda Q_{\beta, p}^{\alpha-1}f(z)}{z^p}$

is univalent in $\mathbb{U}$ and $\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\in\mathcal{H}[1,1]\cap\mathcal{Q}$. Then the following subordination relationship

$\phi_1(z)\prec\frac{(1-\lambda)Q_{\beta, p}^\alpha f(z)+\lambda Q_{\beta, p}^{\alpha-1}f(z)}{z^p}\prec\phi_2(z)$

implies that

$\frac{Q_{\beta, p}^\alpha g_1(z)}{z^p}\prec\frac{Q_{\beta, p}^\alpha f(z)}{z^p}\prec\frac{Q_{\beta, p}^\alpha g_2(z)}{z^p}.$

Moreover, the functions $\frac{Q_{\beta, p}^\alpha g_1(z)}{z^p}$ and $\frac{Q_{\beta, p}^\alpha g_2(z)}{z^p}$ are, respectively, the best subordination and the best dominant.

By putting $\alpha=1$ and $\beta=c$ in Theorem 2.3, we can derive the following result involving the integral operator $J_c$ defined by (1.7).

Corollary 3.3 Let $f, g_j\in\mathcal{A}(p) (j=1, 2)$ and suppose that

$\mathrm{Re}\left\{1+\frac{z\phi_j''(z)}{\phi_j'(z)}\right\}>-\delta \left(\phi_j(z)=(1-\lambda)\left(\frac{J_{c}(g_j)(z)}{z^p}\right)^\mu +\lambda\left(\frac{g_j(z)}{J_{c}(g_j)(z)}\right) \left(\frac{J_{c}(g_j)(z)}{z^p}\right)^\mu\right), $

where

$\delta=\frac{\lambda^2+\mu^2(c+p)^2-|\lambda^2-\mu^2(c+p)^2|} {4\lambda\mu(c+p)} (c>-p).$

If the function

$(1-\lambda)\left(\frac{J_{c}(f)(z)}{z^p}\right)^\mu +\lambda\left(\frac{f(z)}{J_{c}(f)(z)}\right) \left(\frac{J_{c}(f)(z)}{z^p}\right)^\mu$

is univalent in $\mathbb{U}$ and $\left(\frac{J_{c}(f)(z)}{z^p}\right)^\mu\in\mathcal{H}[1,1]\cap\mathcal{Q}$. Then the following subordination relationship

$\phi_1(z)\prec(1-\lambda)\left(\frac{J_{c}(f)(z)}{z^p}\right)^\mu +\lambda\left(\frac{f(z)}{J_{c}(f)(z)}\right) \left(\frac{J_{c}(f)(z)}{z^p}\right)^\mu\prec\phi_2(z)$

implies that

$\left(\frac{J_{c}(g_1)(z)}{z^p}\right)^\mu\prec\left(\frac{J_{c}(f)(z)}{z^p}\right)^\mu \prec\left(\frac{J_{c}(g_2)(z)}{z^p}\right)^\mu.$

Moreover, the functions $\left(\frac{J_{c}(g_1)(z)}{z^p}\right)^\mu$ and $\left(\frac{J_{c}(g_2)(z)}{z^p}\right)^\mu$ are, respectively, the best subordination and the best dominant.

Further, setting $\lambda=\mu=1$ in Corollary 3.3, we have the following result.

Corollary 3.4 Let $f, g_j\in\mathcal{A}(p)\;(j=1, 2)$ and suppose that

$\mathrm{Re}\left\{1+\frac{z\phi_j''(z)}{\phi_j'(z)}\right\}>-\delta\;\left(\phi_j(z)=\frac{g_j(z)}{z^p}\;(j=1, 2); z\in\mathbb{U}\right), $

where

$\delta=\frac{1+(c+p)^2-|1-(c+p)^2|} {4(c+p)}\;(c>-p).$

If the function $\frac{f(z)}{z^p}$ is univalent in $\mathbb{U}$ and $\frac{J_{c}(f)(z)}{z^p}\in\mathcal{H}[1,1]\cap\mathcal{Q}$. Then the following subordination relationship

$\frac{g_1(z)}{z^p}\prec\frac{f(z)}{z^p}\prec\frac{g_2(z)}{z^p}$

implies that

$\frac{J_{c}(g_1)(z)}{z^p}\prec\frac{J_{c}(f)(z)}{z^p}\prec\frac{J_{c}(g_2)(z)}{z^p}.$

Moreover, the functions $\frac{J_{c}(g_1)(z)}{z^p}$ and $\frac{J_{c}(g_2)(z)}{z^p}$ are, respectively, the best subordination and the best dominant.