1 Introduction

Let H be the class of functions analytic in the unit disk U and H[a,n] be the subclass of H consisting of functions of the form \(f(z) = a + a_n z^n + a_{n+1} z^{n+1} + \dots\) Let A be the subclass of H consisting of functions of the form

\[f(z) = z + \sum_{n=2}^{\infty} a_n z^n, \ z \in U.\] (1)

The following differential operator is defined in [1] and studied in [2] \(D_{\lambda,\delta}^k:A\to A\) by

\[\mathsf{D}_{\lambda,\delta}^{k} f(z) = z + \sum_{n=2}^{\infty} [1 + (n-1)\lambda]^{k} C(\delta, n) a_{n} z^{n}, k \in \mathbb{N} \cup \{0\}, \lambda \ge 0, \delta \ge 0,\] (2)

where

\[C(\delta,n) = {n+\delta-1 \choose \delta} = \frac{\Gamma(n+\delta)}{\Gamma(n)\Gamma(\delta+1)}.\]

Remark 1.1. When \(\lambda = 1\), \(\delta = 0\) we get \(S \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\)

Given two functions \(f,g \in A, f(z) = z + \sum_{n=2}^{\infty} a_n z^n\) and \(g(z) = z + \sum_{n=2}^{\infty} b_n z^n\) their convolution or Hadamard product f(z) * g(z) is defined by

\[f(z) * g(z) = z + \sum_{n=2}^{\infty} a_n b_n z^n, z \in U.\]

And for several functions \(f_1(z),...,f_m(z) \in A\)

\[f_1(z)*...*f_m(z) = z + \sum_{n=2}^{\infty} (a_{1n}...a_{nm})z^n, z \in U.\]

Analogous to \(\mathsf{D}^k_{\lambda,\delta}f(z), z{\in}U\) we define an integral operator \(\mathsf{J}^k_{\lambda,\delta}:\mathsf{A}\to\mathsf{A}\) as follows.

Let

\[\phi(z) := \frac{z}{1-z} + \frac{\lambda z}{(1-z)^2} - \frac{\lambda z}{1-z}, \lambda \ge 0.\]

\[F_{k}(z) = \underbrace{\phi(z) * ... * \phi(z)}_{k-times} * [\frac{z}{(1-z)^{\delta+1}}]\]\[= z + \sum_{n=2}^{\infty} [1 + (n-1)\lambda]^{k} C(\delta, n) z^{n}, k \in \mathbb{N}_{0}.\]

And let \(F_k^{(-1)}\) be defined such that

\[F_k(z) * F_k^{(-1)} = \frac{z}{1-z}\]
= \(z + \sum_{n=2}^{\infty} z^n\).

Then

\[\mathbf{J}_{\lambda,\delta}^{k} f(z) = F_{k}^{(-1)} * f(z) = [\underbrace{\phi(z) * \dots * \phi(z)}_{k-times} * \frac{z}{(1-z)^{\delta+1}}]^{(-1)} * f(z) = z + \sum_{n=2}^{\infty} \frac{a_{n}}{[1+(n-1)\lambda]^{k} C(\delta,n)} z^{n}, k \in \mathbb{N}_{0}, \lambda \geq 0, \delta \geq 0 \ z \in U.\] (3)

Remark 1.2. When \(\lambda = 1, \delta = 0\) we get the integral operator [3], also k = 0 gives Noor integral operator [7,8].

Some of relations for this integral operator are discussed in the next lemma.

Lemma 1.1. Let \(f \in A\). Then

(i) \[J_{\lambda,0}^{0} f(z) = f(z)\],
(ii) \(J_{1,0}^{1} f(z) = \int_{0}^{z} \frac{f(t)}{t} dt\).

Proof.

(i) \[J_{\lambda,0}^0 f(z) = z + \sum_{n=2}^{\infty} a_n z^n = f(z),\]

\[(ii) \int_0^z \frac{f(t)}{t} dt = \int_0^z [1 + \sum_{n=2}^\infty a_n t^{n-1}] dt\]\[= z + \sum_{n=2}^\infty \frac{a_n}{n} z^n\]\[= J_{1,0}^1 f(z).\]

In the following definitions, we introduce new classes of analytic functions containing the integral operator (3):

Definition 1.1. Let \(f(z) \in A\). Then \(f(z) \in S_{\lambda,\delta}^k(\mu)\) if and only if

\[\Re\{\frac{z[\mathsf{J}_{\lambda,\delta}^k f(z)]'}{\mathsf{J}_{\lambda,\delta}^k f(z)}\} > \mu, 0 \le \mu < 1, z \in U.\]

Definition 1.2. Let \(f(z) \in A\). Then \(f(z) \in C_{\lambda,\delta}^k(\mu)\) if and only if

\[\Re\{\frac{[z(\mathsf{J}_{\lambda,\delta}^k f(z))']'}{(\mathsf{J}_{\lambda,\delta}^k f(z))'}\} > \mu, 0 \le \mu < 1, z \in U.\]

Let F and G be analytic functions in the unit disk U. The function F is subordinate to G, written \(F \prec G\), if G is univalent, F(0) = G(0) and \(F(U) \subset G(U)\). In general, given two functions F(z) and G(z), which are analytic in U, the function F(z) is said to be subordination to G(z) in U if there exists a function h(z), analytic in U with

\[h(0) = 0\] and \(|h(z)| < 1\) for all \(z \in U\)

such that

\[F(z) = G(h(z))\] for all \(z \in U\).

Let \(\phi: \mathbb{C}^2 \to \mathbb{C}\) and let h be univalent in U. If p is analytic in U and satisfies the differential subordination \(\phi(p(z)), zp'(z)) \prec h(z)\) then p is called a solution of the differential subordination. The univalent function q is called a dominant of the solutions of the differential subordination, if \(p \prec q\). If p and \(\phi(p(z)), zp'(z)\) are univalent in U and satisfy the differential superordination \(h(z) \prec \phi(p(z)), zp'(z)\) then p is called a solution of the differential superordination. An analytic function q is called subordinant of the solution of the differential superordination if \(q \prec p\). Let \(\Phi\) be an analytic function in a domain containing f(U), \(\Phi(0) = 0\) and \(\Phi'(0) > 0\).

The function \(f \in A\) is called \(\Phi\) – like if

\[\Re\{\frac{zf'(z)}{\Phi(f(z))}\} > 0, z \in U.\]

This concept was introduced by Brickman [9] and established that a function \(f \in A\) is univalent if and only if f is \(\Phi\)-like for some \(\Phi\).

Definition 1.3. Let \(\Phi\) be analytic function in a domain containing \(f(U), \Phi(0) = 0, \Phi'(0) = 1\) and \(\Phi(\omega) \neq 0\) for \(\omega \in f(U) - 0\). Let q(z) be a fixed analytic function in U, q(0) = 1. The function \(f \in A\) is called \(\Phi\)-like with respect to q if

\[\frac{zf'(z)}{\Phi(f(z))} \prec q(z), z \in U.\]

The paper is organized as follows: Section 2 discuses the characterization properties for functions belonging to the classes \(S_k(\mu)\), \(C_k(\mu)\) and Section 3, gives the subordination and superordination results involving the integral operator \(J_{\lambda,\delta}^k f(z)\). For this purpose we need to the following lemmas in the sequel.

Definition 1.4. [10] Denote by Q the set of all functions f(z) that are analytic and injective on \(\overline{U} - E(f)\) where \(E(f) := \{ \zeta \in \partial U : \lim_{z \to \zeta} f(z) = \infty \}\) and are such that \(f'(\zeta) \neq 0\) for \(\zeta \in \partial U - E(f)\).

Lemma 1.2. [11] Let q(z) be univalent in the unit disk U and \(\theta\) and \(\phi\) be analytic in a domain D containing q(U) with \(\phi(w) \neq 0\) when \(w \in q(U)\). Set \(Q(z) := zq'(z)\phi(q(z)), h(z) := \theta(q(z)) + Q(z)\). Suppose that 1. Q(z) is starlike univalent in U, and 2. \[\Re \frac{zh'(z)}{Q(z)} > 0\] for \(z \in U\).

If

\[\theta(p(z)) + zp'(z)\phi(p(z)) \prec \theta(q(z)) + zq'(z)\phi(q(z))\] then

\[p(z) \prec q(z)\] and q(z) is the best dominant.

Lemma 1.3. [12] Let q(z) be convex univalent in the unit disk U and \(\mathcal{G}\) and \(\varphi\) be analytic in a domain D containing q(U). Suppose that 1. \(zq'(z)\varphi(q(z))\) is starlike univalent in U, and 2. \[\Re{\frac{\mathcal{G}'(q(z))}{\varphi(q(z))}} > 0 \text{ for } z \in U.\]

If \(p(z) \in H[q(0),1] \cap Q\), with \(p(U) \subseteq D\) and \(\mathcal{G}(p(z)) + zp'(z)\varphi(z)\) is univalent in U and

\[\mathcal{G}(q(z)) + zq'(z)\varphi(q(z)) \prec \mathcal{G}(p(z)) + zp'(z)\varphi(p(z))\] then \(q(z) \prec p(z)\) and q(z) is the best subordinant.

2 General Properties of \(J_{\lambda,\delta}^k\)

In this section we study the characterization properties for the function \(f(z) \in A\) to belong to the classes \(S_{\lambda,\delta}^k(\mu)\) and \(C_{\lambda,\delta}^k(\mu)\) by obtaining the coefficient bounds.

Theorem 2.1. Let \(f(z) \in A\). If

\[\sum_{n=2}^{\infty} \frac{(n-\mu) |a_n|}{[1+(n-1)\lambda]^k C(\delta,n)} \le 1-\mu, \ 0 \le \mu < 1, \quad (4)\] then \(f(z) \in S_{\lambda}^k(\mu)\). The result (4) is sharp.

Proof. Suppose that (4) holds. Since

\[1-\mu \geq \sum_{n=2}^{\infty} \frac{|a_n || \mu - n|}{[1 + (n-1)\lambda]^k}\] \[\geq \mu \sum_{n=2}^{\infty} \frac{|a_n ||}{[1 + (n-1)\lambda]^k} - \sum_{n=2}^{\infty} \frac{n |a_n ||}{[1 + (n-1)\lambda]^k}\] then this implies that

\[\frac{1 + \sum_{n=2}^{\infty} \frac{n |a_n|}{[1 + (n-1)\lambda]^k}}{1 + \sum_{n=2}^{\infty} \frac{|a_n|}{[1 + (n-1)\lambda]^k}} > \mu,\]

hence

\[\Re\{\frac{z[\mathsf{J}_{\lambda,\delta}^k f(z)]'}{\mathsf{J}_{\lambda,\delta}^k f(z)}\} > \mu.\]

We also note that the assertion (4) is sharp and the extremal function is given by

\[f(z) = z + \sum_{n=2}^{\infty} \frac{[1 + (n-1)\lambda]^k C(\delta, n)(1-\mu)}{(n-\mu)} z^n.\]

Corollary 2.1. Let the assumption of Theorem 2.1. Then

\[|a_n| \le \frac{[1+(n-1)\lambda]^k C(\delta,n)(1-\mu)}{(n-\mu)}, \forall n \ge 2.\]

Corollary 2.2. Let the assumption of Theorem 2.1. Then for \(\mu = \delta = 0\) and \(\lambda = 1\)

\[|a_n| \le n^{k-1}, \forall n \ge 2, k \in \mathbb{N}_0.\]

In the same way we can verify the following results:

Theorem 2.2. Let \(f(z) \in A\). If

\[\sum_{n=2}^{\infty} \frac{n |a_n| |\mu+1-n|}{C(\delta,n)[1+(n-1)\lambda]^k} \le 1-\mu, \ 0 \le \mu < 1, \ (5)\] then \(f(z) \in \mathbf{C}_{\lambda,\delta}^k(\mu)\). The result (5) is sharp.

Corollary 2.3. Let the assumption of Theorem 2.2. Then

\[|a_n| \le \frac{[1+(n-1)\lambda]^k C(\delta,n)(1-\mu)}{n |\mu-n+1|}, \forall n \ge 2.\]

Also we have the following inclusion results

Theorem 2.3. Let \(0 \le \mu_1 \le \mu_2 < 1\). Then \(S^k_{\lambda,\delta}(\mu_1) \supseteq S^k_{\lambda,\delta}(\mu_2)\).

Proof. By Theorem 2.1.

Theorem 2.4. Let \(0 \le \mu_1 \le \mu_2 < 1\). Then \(\mathbf{C}_{\lambda,\delta}^k(\mu_1) \supseteq \mathbf{C}_{\lambda,\delta}^k(\mu_2)\).

Proof. By Theorem 2.2.

Theorem 2.5. Let \(0 \le \lambda_1 \le \lambda_2\). Then \(S_{\lambda_1,\delta}^k(\mu) \supseteq S_{\lambda_2,\delta}^k(\mu)\).

Proof. By Theorem 2.1.

Theorem 2.6. Let \(0 \le \lambda_1 \le \lambda_2\). Then \(C_{\lambda_1,\delta}^k(\mu) \supseteq C_{\lambda_2,\delta}^k(\mu)\).

Proof. By Theorem 2.2.

Moreover, we introduce the following distortion theorems.

Theorem 2.7. Let \(f \in A\) and satisfies (4). Then for \(z \in U\) and \(0 \le \mu < 1\)

\[|\mathbf{J}_{\lambda,\delta}^k f(z)| \ge |z| - \frac{(1-\mu)}{(2-\mu)} |z|^2\] and

\[|\mathbf{J}_{\lambda,\delta}^{k}f(z)| \le |z| + \frac{(1-\mu)}{(2-\mu)}|z|^{2}\].

Proof. By using Theorem 2.1, one can verify that

\[(2-\mu)\sum_{n=2}^{\infty}\frac{|a_n|}{[1+(n-1)\lambda]^kC(\delta,n)} \leq \sum_{n=2}^{\infty}\frac{(n-\mu)|a_n|}{[1+(n-1)\lambda]^kC(\delta,n)} \leq 1-\mu\] then

\[\sum_{n=2}^{\infty} \frac{|a_n|}{[1+(n-1)\lambda]^k C(\delta,n)} \le \frac{1-\mu}{2-\mu}.\]

Thus we obtain

\[|\mathbf{J}_{\lambda,\delta}^{k} f(z)| = |z + \sum_{n=2}^{\infty} \frac{a_{n}}{[1 + (n-1)\lambda]^{k}} z^{n} |\] \[\leq |z| + \sum_{n=2}^{\infty} \frac{|a_{n}|}{[1 + (n-1)\lambda]^{k}} |z|^{2}\] \[\leq |z| + [\frac{1-\mu}{2-\mu}] |z|^{2}\]

The other assertion can be proved as follows

\[|J_{\lambda,\delta}^{k}f(z)| = |z + \sum_{n=2}^{\infty} \frac{a_{n}}{[1 + (n-1)\lambda]^{k}C(\delta, n)} z^{n} |\] \[\geq |z - \sum_{n=2}^{\infty} \frac{a_{n}}{[1 + (n-1)\lambda]^{k}C(\delta, n)} z^{n} |\] \[\geq |z| - \sum_{n=2}^{\infty} \frac{|a_{n}|}{[1 + (n-1)\lambda]^{k}C(\delta, n)} |z|^{2}\] \[\geq |z| - [\frac{1 - \mu}{2 - \mu}] |z|^{2}.\]

This complete the proof.

In the same way we can get the following results.

Theorem 2.8. Let \(f(z) \in A\) and satisfies (5). Then for \(z \in U\) and \(0 \le \mu < 1\)

\[|\mathbf{J}_{\lambda,\delta}^{k}f(z)| \ge |z| - \frac{(1-\mu)}{2(2-\mu)}|z|^{2}\] and

\[|\mathbf{J}_{\lambda,\delta}^k f(z)| \le |z| + \frac{(1-\mu)}{2(2-\mu)} |z|^2\].

Also, we have the following distortion results

Theorem 2.9. Let \(f(z) \in A\) and and satisfies (4). Then for \(m \ge [1+(n-1)\lambda]^k C(\delta,n), z \in U\) and \(0 \le \mu < 1\)

\[|f(z)| \ge |z| - \frac{m(1-\mu)}{(2-\mu)} |z|^2\] and

\[|f(z)| \le |z| + \frac{m(1-\mu)}{(2-\mu)} |z|^2\].

Proof. By using Theorem 2.1, one can show that

\[(2-\mu)\sum_{n=2}^{\infty} |a_n| \le \sum_{n=2}^{\infty} (n-\mu) |a_n| \le m \sum_{n=2}^{\infty} \frac{(n-\mu) |a_n|}{[1+(n-1)\lambda]^k C(\delta, n)} \le m(1-\mu)\] then

\[\sum_{n=2}^{\infty} |a_n| \leq \frac{m(1-\mu)}{2-\mu}.\]

Thus we obtain

\[|f(z)| = |z + \sum_{n=2}^{\infty} a_n z^n|\] \[\leq |z| + \sum_{n=2}^{\infty} |a_n| |z|^2\] \[\leq |z| + \frac{m(1-\mu)}{2-\mu} |z|^2\]

The other assertion can be proved as follows

\[| f(z) | \ge | z - \sum_{n=2}^{\infty} a_n z^n |\] \[\ge | z | - \sum_{n=2}^{\infty} | a_n | | z |^2\] \[\ge | z | - \frac{m(1-\mu)}{2-\mu} | z |^2.\]

This completes the proof.

In the same way we can get the following results.

Theorem 2.10. Let \(f(z) \in A\) and and satisfies (5). Then for \(z \in U\) and \(0 \le \mu < 1\)

\[|f(z)| \ge |z| - \frac{m(1-\mu)}{2(2-\mu)} |z|^2\] and

\[|f(z)| \le |z| + \frac{m(1-\mu)}{2(2-\mu)} |z|^2\].

3 Sandwich Result.

By making use of lemmas 1.2 and 1.3, we prove the following subordination and superordination results involving the integral operator (3).

Theorem 3.1. Let \(q \neq 0\) be univalent in U such that \(\frac{zq'(z)}{q(z)}\) is starlike univalent in U and

\[\Re\{1+(\frac{\alpha}{\gamma}+z)\frac{q''(z)}{q'(z)}-(\frac{\alpha}{\gamma}+z)\frac{q'(z)}{q(z)}\}>0, \alpha, \gamma \in \mathbb{C}, \gamma \neq 0.\] (6)

If \(f \in A\) satisfies the subordination

\[(\alpha + \gamma z) \left\{ \frac{1}{z} + \frac{\left[ \mathbf{J}_{\lambda,\delta}^{k} f(z) \right]''}{\left[ \mathbf{J}_{\lambda,\delta}^{k} f(z) \right]'} - \frac{\Phi'[\mathbf{J}_{\lambda,\delta}^{k} f(z)]}{\Phi[\mathbf{J}_{\lambda,\delta}^{k} f(z)]} \right\} \prec (\alpha + \gamma z) \frac{q'(z)}{q(z)},\] then

\[\frac{z[\mathsf{J}_{\lambda,\delta}^k f(z)]'}{\Phi[\mathsf{J}_{\lambda,\delta}^k f(z)]} \prec q(z) \tag{7}\] and q is the best dominant.

Proof. Our aim is to apply Lemma 1.2. Setting

\[p(z) := \frac{z[\mathsf{J}_{\lambda,\delta}^k f(z)]'}{\Phi[\mathsf{J}_{\lambda,\delta}^k f(z)]}.\]

By computation shows that

\[\frac{zp'(z)}{p(z)} = 1 + \frac{z[J_{\lambda}^{k}f(z)]''}{[J_{\lambda,\delta}^{k}f(z)]'} - \frac{z\Phi'[J_{\lambda,\delta}^{k}f(z)]}{\Phi[J_{\lambda,\delta}^{k}f(z)]}\] which yields the following subordination

\[(\alpha + \gamma z) \frac{p'(z)}{p(z)} \prec (\alpha + \gamma z) \frac{q'(z)}{q(z)}, \alpha, \gamma \in \mathbb{C}.\]

By setting

\[\theta(\omega) := \frac{\alpha \omega'}{\omega}\] and \(\phi(\omega) := \frac{\gamma}{\omega}, \gamma \neq 0\), it can be easily observed that \(\theta(\omega), \phi(\omega)\) are analytic in \(\mathbb{C}\setminus\{0\}\) and that \(\phi(\omega) \neq 0\) when \(\omega \in \mathbb{C}\setminus\{0\}\). Also, by letting

\[Q(z) = zq'(z)\phi(q(z)) = \gamma z \frac{q'(z)}{q(z)}\] and

\[h(z) = \theta(q(z)) + Q(z) = \frac{\alpha q'(z)}{q(z)} + \gamma z \frac{q'(z)}{q(z)} = (\alpha + \gamma z) \frac{q'(z)}{q(z)},\]

we find that Q(z) is starlike univalent in U and that

\[\Re\left\{\frac{zh'(z)}{Q(z)}\right\} = \Re\left\{1 + \left(\frac{\alpha}{\gamma} + z\right)\frac{q''(z)}{q'(z)} - \left(\frac{\alpha}{\gamma} + z\right)\frac{q'(z)}{q(z)}\right\} > 0, \alpha, \gamma \in \mathbb{C}, \gamma \neq 0.\]

Then the relation (7) follows by an application of Lemma 1.2.

Corollary 3.1. Let the assumptions of Theorem 3.1 hold. Then the subordination

\[1 + \frac{z[\mathsf{J}_{\lambda,\delta}^k f(z)]''}{[\mathsf{J}_{\lambda,\delta}^k f(z)]'} - \frac{z[\mathsf{J}_{\lambda,\delta}^k f(z)]'}{[\mathsf{J}_{\lambda,\delta}^k f(z)]} \prec \frac{zq'(z)}{q(z)},\]

implies

\[\frac{z[\mathsf{J}_{\lambda,\delta}^k f(z)]'}{[\mathsf{J}_{\lambda,\delta}^k f(z)]} \prec q(z) \tag{8}\] and q is the best dominant.

Proof. By letting \(\alpha = 0, \gamma = 1, \Phi(\omega) := \omega\).

Corollary 3.2. If \(f \in A\) and assume that (7) holds then

\[1 + \frac{z[\mathsf{J}_{\lambda,\delta}^k f(z)]''}{[\mathsf{J}_{\lambda,\delta}^k f(z)]'} - \frac{z[\mathsf{J}_{\lambda,\delta}^k f(z)]'}{[\mathsf{J}_{\lambda,\delta}^k f(z)]} \prec \frac{(A-B)z}{(1+Az)(1+Bz)}\]

implies

\[\frac{z[\mathbf{J}_{\lambda,\delta}^k f(z)]'}{[\mathbf{J}_{\lambda,\delta}^k f(z)]} \prec \frac{1+Az}{1+Bz}, -1 \le B < A \le 1\] and \(\frac{1+Az}{1+Bz}\) is the best dominant.

Proof. By setting \(\Phi(\omega) := \omega, \gamma = 1, \alpha = 0\) and \(q(z) := \frac{1 + Az}{1 + Bz}\) where \(-1 \le B < A \le 1\).

Corollary 3.3. If \(f \in A\) and assume that (7) holds then

\[1 + \frac{z[\mathsf{J}_{\lambda,\delta}^k f(z)]''}{[\mathsf{J}_{\lambda,\delta}^k f(z)]'} - \frac{z[\mathsf{J}_{\lambda,\delta}^k f(z)]'}{[\mathsf{J}_{\lambda,\delta}^k f(z)]} \prec \frac{2z}{1 - z^2}\]

implies

\[\frac{z[\mathbf{J}_{\lambda,\delta}^k f(z)]'}{\mathbf{J}_{\lambda,\delta}^k f(z)} \prec \frac{1+z}{1-z},\] and \(\frac{1+z}{1-z}\) is the best dominant.

Proof. By setting \(\Phi(\omega) := \omega\), \(\alpha = 0, \gamma = 1\), and \(q(z) := \frac{1+z}{1-z}\).

Corollary 3.4. If \(f \in A\) and assume that (7) holds then

\[1 + \frac{z[\mathbf{J}_{\lambda,\delta}^k f(z)]''}{[\mathbf{J}_{\lambda,\delta}^k f(z)]'} - \frac{z[\mathbf{J}_{\lambda,\delta}^k f(z)]'}{[\mathbf{J}_{\lambda,\delta}^k f(z)]} \prec Az\]

implies

\[\frac{z[\mathsf{J}_{\lambda,\delta}^k f(z)]'}{\mathsf{J}_{\lambda,\delta}^k f(z)} \prec e^{Az},\] and \(e^{Az}\) is the best dominant.

Proof. By setting \(\Phi(\omega) := \omega\), \(\alpha = 0, \gamma = 1\), and \(q(z) := e^{Az}\), \(|A| < \pi\).

Theorem 3.2. Let \(q(z) \neq 0\) be convex univalent in the unit disk U. Suppose that

\[\Re\{\frac{\alpha}{\gamma}q''(z) - \frac{\alpha}{\gamma}\frac{q'(z)}{q(z)}\} > 0, \alpha, \gamma \in \mathbb{C} \text{ for } z \in U\] (9)

and \(\frac{zq'(z)}{q(z)}\) is starlike univalent in U. If \(\frac{z[\mathbf{J}_{\lambda,\delta}^k f(z)]'}{\Phi[\mathbf{J}_{\lambda,\delta}^k f(z)]} \in \mathbf{H}[q(0),1] \cap Q\) where \(f \in \mathbf{A}\),

\[(\alpha + \gamma z) \left\{ \frac{1}{z} + \frac{\left[ \mathbf{J}_{\lambda,\delta}^{k} f(z) \right]''}{\left[ \mathbf{J}_{\lambda,\delta}^{k} f(z) \right]'} - \frac{\Phi'[\mathbf{J}_{\lambda,\delta}^{k} f(z)]}{\Phi[\mathbf{J}_{\lambda,\delta}^{k} f(z)]} \right\}\] is univalent is U and the subordination

\[(\alpha + \gamma z) \frac{q'(z)}{q(z)} \prec (\alpha + \gamma z) \{ \frac{1}{z} + \frac{[J_{\lambda,\delta}^k f(z)]''}{[J_{\lambda,\delta}^k f(z)]'} - \frac{\Phi'[J_{\lambda,\delta}^k f(z)]}{\Phi[J_{\lambda,\delta}^k f(z)]} \},\]

holds, then

\[q(z) \prec \frac{z[\mathsf{J}_{\lambda,\delta}^k f(z)]'}{\Phi[\mathsf{J}_{\lambda,\delta}^k f(z)]} \tag{10}\] and q is the best subordinant.

Proof. Our aim is to apply Lemma 1.3. Setting

\[p(z) := \frac{z[\mathsf{J}_{\lambda,\delta}^k f(z)]'}{\Phi[\mathsf{J}_{\lambda,\delta}^k f(z)]}.\]

By computation shows that

\[\frac{zp'(z)}{p(z)} = 1 + \frac{z[\mathbf{J}_{\lambda,\delta}^k f(z)]''}{[\mathbf{J}_{\lambda,\delta}^k f(z)]'} - \frac{z\Phi'[\mathbf{J}_{\lambda,\delta}^k f(z)]}{\Phi[\mathbf{J}_{\lambda,\delta}^k f(z)]}\] which yields the following subordination

\[(\alpha + \gamma z) \frac{q'(z)}{q(z)} \prec (\alpha + \gamma z) \frac{p'(z)}{p(z)}, \alpha, \gamma \in \mathbb{C}.\]

By setting

\[\vartheta(\omega) := \frac{\alpha \omega'}{\omega}\] and \(\varphi(\omega) := \frac{\gamma}{\omega}, \gamma \neq 0\), it can be easily observed that \(\mathcal{G}(\omega), \varphi(\omega)\) are analytic in \(\mathbb{C}\setminus\{0\}\) and that \(\varphi(\omega) \neq 0\) when \(\omega \in \mathbb{C}\setminus\{0\}\). Also, we obtain

\[\Re\left\{\frac{\mathcal{G}'(q(z))}{\varphi(q(z))}\right\} = \Re\left\{\frac{\alpha}{\gamma}q''(z) - \frac{\alpha}{\gamma}\frac{q'(z)}{q(z)}\right\} > 0.\]

Then (10) follows by an application of Lemma 1.3.

Combining Theorems 3.1 and 3.2 in order to get the following Sandwich theorems

Theorem 3.3 Let \(q_1(z) \neq 0, q_2(z) \neq 0\) be convex univalent in the unit disk U satisfy (9) and (6) respectively. Suppose that and \(\frac{zq_i'(z)}{q_i(z)}, i=1,2\) is starlike univalent in U. If \(f \in A\) and

\[\frac{z[\mathsf{J}_{\lambda,\delta}^k f(z)]'}{\Phi[\mathsf{J}_{\lambda,\delta}^k f(z)]} \in \mathsf{H}[q_1(0),1] \cap Q\]

\[(\alpha + \gamma z) \left\{ \frac{1}{z} + \frac{\left[ \mathbf{J}_{\lambda,\delta}^{k} f(z) \right]''}{\left[ \mathbf{J}_{\lambda,\delta}^{k} f(z) \right]'} - \frac{\Phi'[\mathbf{J}_{\lambda,\delta}^{k} f(z)]}{\Phi[\mathbf{J}_{\lambda,\delta}^{k} f(z)]} \right\}\] is univalent in U and the subordination

\[(\alpha + \gamma z) \frac{q_1'(z)}{q_1(z)} \prec (\alpha + \gamma z) \{ \frac{1}{z} + \frac{[J_{\lambda,\delta}^k f(z)]''}{[J_{\lambda,\delta}^k f(z)]'} - \frac{\Phi'[J_{\lambda,\delta}^k f(z)]}{\Phi[J_{\lambda,\delta}^k f(z)]} \} \prec (\alpha + \gamma z) \frac{q_2'(z)}{q_2(z)}\]

holds, then

\[q_1(z) \prec \frac{z[\mathbf{J}_{\lambda,\delta}^k f(z)]'}{\Phi[\mathbf{J}_{\lambda,\delta}^k f(z)]} \prec q_2(z)\] and \(q_1(z)\) is the best subordinant and \(q_2(z)\) is the best dominant.

Acknowledgement

The work here was supported by UKM-ST-06-FRGS0107-2009, MOHE Malaysia. The authors also would like to thank the anonymous referee for the informative and creative comments given to the article.