Some applications of second-order differential subordination for a class of analytic function defined by the lambda operator

we denote a class of all convex functions defined in U and normalized by f (0) = 0 and f ′(0) = 1. Let f and F be elements of H(U). A function f is said to be subordinate to F, if there exists a Schwartz function w analytic in U with w(0) = 0 and |w(z)| < 1, z ∈ U, such that f (z) = F(w(z)). In this case, we write f (z) ≺ F(z) or f ≺ F. Furthermore, if the function F is univalent in U, then we get the following equivalence [1,2]: f (z) ≺ F(z)⇔ f (0) = F(0) and f (U) ≺ F(U).


Introduction
L et C be complex plane and let U = {z : z ∈ C and |z| < 1} = U \ {0} be an open unit disc in C. Also let H(U) be a class of analytic functions in U. For n ∈ N = {1, 2, 3, · · · , } and a ∈ C, let H[a, n] be a subclass of H(U) formed by the functions of the form f (z) = z + a n z n + a n+1 z n+1 + · · · with H 0 ≡ H[0, 1] and H ≡ H [1,1]. Suppose that A n is a class of all analytic functions of the form f (z) = z + ∞ ∑ k=n+1 a n z n (1) in the open unit disk U with A 1 = A. A function f ∈ H(U) is univalent if it is a one-to-one function in U. By S, we denote a subclass of A formed by functions univalent in U. If a function f ∈ A maps U onto a convex domain and f is univalent, then f is called a convex function. By we denote a class of all convex functions defined in U and normalized by f (0) = 0 and f (0) = 1. Let f and F be elements of H(U). A function f is said to be subordinate to F, if there exists a Schwartz function w analytic in U with w(0) = 0 and |w(z)| < 1, z ∈ U, such that f (z) = F(w(z)). In this case, we write f (z) ≺ F(z) or f ≺ F. Furthermore, if the function F is univalent in U, then we get the following equivalence [1,2]: The method of differential subordinations (also known as the method of admissible functions) was first introduced by Miller and Mocanu in 1978 [3], and the development of the theory was originated in 1981 [4]. All details can be found in the book by Miller and Mocanu [2]. In recent years, numerous authors studied the properties of differential subordinations (see [5][6][7][8], etc.).
Let Ψ : C 3 × U → C and let h be univalent in U. If p is analytic in U and satisfies the second-order differential subordination: then p is called the solution of differential subordination. The univalent function q is called a dominant of the solution of the differential subordination or, simply, a dominant if p ≺ q for all p satisfying (2). The dominant q 1 satisfying q 1 ≺ q for all dominants q of (2) is called the best dominant of (2). Let us recall lambda function [9] defined by: We now define zλ (−1) (z, s) as: where where (µ) k is the Pochhammer symbol defined in terms of the Gamma function by: .
The function q is convex and is the best dominant for subordination (5).
. Also, let h be an analytic function in U with h(0) = 1.
In the present paper, we use the subordination results from [10] to prove our main results.

Main results
Proof. Let f j (z) = z + ∞ ∑ k=2 a k,j z k , z ∈ U, j = 1, · · · , m be in the class L µ,s ( ). Then, by Definition 1, we get For any positive numbers ς 1 , ς 2 , ς 3 , · · · , ς m such that m ∑ j=1 ς j = 1, it is necessary to show that the function Thus, we have ς j a k,j z k .
If we differentiate (9) with respect to z, then we obtain ς j a k,j z k−1 .
Thus by using (8), we have Hence, inequality (7) is true and we arrive at the desired result.

Theorem 2.
Let q be convex function in U with q(0) = 1 and h(z) = q(z) implies that (I s µ ℵ(z)) ≺ q(z) and this result is sharp.
Proof. In view of equality (10), we can write Differentiating (12) with respect to z, we obtain (γ)ℵ(z) + zℵ (z) = (γ + 1) f (z). Further, by applying the operator I s µ to the last equation, we get If we differentiate (13) with respect to z, then we find By using the differential subordination given by (11) in equality (14), we obtain We define p(z) = (I s µ ℵ(z)) .
Hence, as a result of simple computations, we get By using (16) in subordination (15), we obtain If we use Lemma 2, then we write p(z) ≺ q(z). Thus, we obtained the desired result and q is the best dominant.  . Suppose that h is an analytic function in U with h(0) = 1 and that implies that (I s µ ℵ(z)) ≺ q(z), where q is the solution of the differential equation h(z) = q(z) + 1 γ+1 zq (z), q(0) = 1, given by q(z) = γ+1 z γ+1 z 0 t γ f (t)dt. Moreover, q is the best dominant for subordination (17).
Proof. If we choose n = 1 and µ = γ + 1 in Lemma 1, then the proof is obtained by means of the proof of Theorem 3.

Theorem 4.
Let be convex in U with h(0) = 1. If f ∈ A and verifies the differential subordination (I s µ f (z)) ≺ h(z), then (I s µ ℵ(z)) ≺ q(z) = (2 − 1) , where τ is given by the formula and ℵ is given by equation (10). The function q is convex and is the best dominant.
The function q is convex. Moreover, it is the best dominant. Hence the theorem is proved.