E-mail address: [email protected] Received May 12, 2019
Available online at http://scik.org Eng. Math. Lett. 2019, 2019:12
https://doi.org/10.28919/eml/4118 ISSN: 2049-9337
SOME RESULTS FOR STRONG DIFFERENTIAL SUBORDINATION OF ANALYTIC FUNCTIONS DEFINED BY DIFFERENTIAL OPERATOR
ASRAA ABDUL JALEEL HUSIEN
Technical Institute, Diwaniya, Al-Furat Al-Awsat Technical University, Iraq
Copyright © 2019 the author(s). This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract: The purpose of this paper is to introduce some results associated with the strong differential subordinations of analytic functions defined in the open unit disk and closed unit disk of the complex plane.
Keywords: analytic functions; convex functions; strong differential subordinations; differential operator. 2010 AMS Subject Classification: 30C45.
1.
I
NTRODUCTIONDenote by ℋ(𝑈 × 𝑈̅) the family of all analytic functions in 𝑈 × 𝑈̅. Let 𝑈 = {𝑧 ∈ ℂ ∶ |𝑧| <
1} and 𝑈̅ = {𝑧 ∈ ℂ ∶ |𝑧| ≤ 1} indicate the open unit disk and the closed unit disk of the
complex plane, respectively. For 𝑛 ∈ ℕ = {1,2, … } and 𝑎 ∈ ℂ, let ℋ∗[𝑎, 𝑛, 𝜁] = {𝑓 ∈
ℋ(𝑈 × 𝑈̅): 𝑓(𝑧, 𝜁) = 𝑎 + 𝑎𝑛(𝜁)𝑧𝑛 + 𝑎𝑛+1(𝜁)𝑧𝑛+1+ ⋯ , 𝑧 ∈ 𝑈, 𝜁 ∈ 𝑈̅}, where 𝑎𝑘(𝜁) are
holomorphic functions in 𝑈̅ for 𝑘 ≥ 𝑛.
Also, let 𝒜𝑛 𝜁∗ = {𝑓 ∈ ℋ(𝑈 × 𝑈̅): 𝑓(𝑧, 𝜁) = 𝑧 + 𝑎
𝑛+1(𝜁)𝑧𝑛+1+ ⋯ , 𝑧 ∈ 𝑈, 𝜁 ∈ 𝑈̅}, where
𝑎𝑘(𝜁) are holomorphic functions in 𝑈̅ for 𝑘 ≥ 𝑛 + 1.
𝑅𝑒 {𝑧𝑓𝑧
′(𝑧, 𝜁)
𝑓(𝑧, 𝜁) } > 0 , ( 𝑧 ∈ 𝑈, 𝜁 ∈ 𝑈̅).
Denote the class of all starlike functions in 𝑈 × 𝑈̅ by 𝑆 𝜁∗.
Similar, 𝑓 ∈ ℋ∗[𝑎, 𝑛, 𝜁] is said to be convex in 𝑈 × 𝑈̅ if
𝑅𝑒 {𝑧𝑓𝑧2
′′(𝑧, 𝜁)
𝑓𝑧′(𝑧, 𝜁) + 1} > 0 , ( 𝑧 ∈ 𝑈, 𝜁 ∈ 𝑈̅).
Denote the class of all convex functions in 𝑈 × 𝑈̅ by 𝐾 𝜁∗.
Definition 1.1 (Oros and Gh Oros, [9]) Let 𝑓(𝑧, 𝜁), 𝑔(𝑧, 𝜁) analytic in 𝑈 × 𝑈̅. The function
𝑓(𝑧, 𝜁) is said to be strongly subordinate to 𝑔(𝑧, 𝜁), written 𝑓(𝑧, 𝜁) ≺≺ 𝐹(𝑧, 𝜁), 𝑧 ∈ 𝑈, 𝜁 ∈ 𝑈̅,
if there exists an analytic function 𝑤 in 𝑈 with 𝑤(0) = 0 and |𝑤(𝑧)| < 1, 𝑧 ∈ 𝑈 such that
𝑓(𝑧, 𝜁) = 𝑔(𝑤(𝑧), 𝜁) for all 𝜁 ∈ 𝑈̅.
Remark 1.1 (Oros and Gh Oros, [9])
1) Since 𝑓(𝑧, 𝜁) is analytic in 𝑈 × 𝑈̅, for all 𝜁 ∈ 𝑈̅ and univalent in 𝑈, for all 𝜁 ∈ 𝑈̅,
Definition1.1 is equivalent to 𝑓(0, 𝜁) = 𝑔(0, 𝜁) for all 𝜁 ∈ 𝑈̅ and 𝑓(𝑈 × 𝑈̅) ⊂ 𝑔(𝑈 × 𝑈̅).
2) If 𝑓(𝑧, 𝜁) = 𝑓(𝑧) and 𝑔(𝑧, 𝜁) = 𝑔(𝑧), the strong subordination becomes the usual notion
of subordination.
Let 𝒜 𝜁∗ denote the subclass of the functions 𝑓(𝑧, 𝜁) ∈ ℋ(𝑈 × 𝑈̅) of the form:
𝑓(𝑧, 𝜁) = 𝑧 + ∑ 𝑎𝑘(𝜁)𝑧𝑘 ∞
𝑘=2
, 𝑧 ∈ 𝑈, 𝜁 ∈ 𝑈̅ (1.1)
which are analytic and univalent in 𝑈 × 𝑈̅.
For 𝜂 ∈ ℕ0 = ℕ ∪ {0} 𝛼, 𝜏 ≥ 0, 𝜇, 𝜆, 𝛽 > 0 and 𝛼 ≠ 𝜆, we consider the differential operator
𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽) ∶ 𝒜 𝜁∗ ⟶ 𝒜
𝜁∗, introduced by Amourah and Darus [2], where
𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧) = 𝑧 + ∑ [1 +(𝑛 − 1)[(𝜆 − 𝛼)𝛽 + 𝑛𝜏]
𝜇 + 𝜆 ]
𝜂
𝑎𝑘(𝜁)𝑧𝑘 ∞
𝑘=2
. (1.2)
It is readily verified from (1.2) that
𝑧 (𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁))
= 𝜇 + 𝜆
(𝜆 − 𝛼)𝛽 + 𝑛𝜏𝐴𝜇,𝜆,𝜏
𝜂+1(𝛼, 𝛽)𝑓(𝑧, 𝜁) − (1 − 𝜇 + 𝜆
(𝜆 − 𝛼)𝛽 + 𝑛𝜏) 𝐴𝜇,𝜆,𝜏
𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁). (1.3)
Some of the special cases of the operator defined by (1.2) can be found in [1,4,10,11].
In recent years, many authors obtained various interesting results associated with strong
differential subordination and superordination for example (see [3,5,6,12,13,14]).
In order to derive our main results, we need the following Lemmas.
Lemma 1.1 (Miller and Mocanu, [8]) Let ℎ(𝑧, 𝜁) be a convex function with ℎ(0, 𝜁) = 𝑎, for
every 𝜁 ∈ 𝑈̅ and let 𝛾 ∈ ℂ∗ = ℂ ∖ {0} with 𝑅𝑒 (𝛾) ≥ 0. If 𝑝 ∈ ℋ∗[𝑎, 𝑛, 𝜁] and
𝑝(𝑧, 𝜁) +1
𝛾𝑧𝑝𝑧′(𝑧, 𝜁) ≺≺ ℎ(𝑧, 𝜁), (𝑧 ∈ 𝑈, 𝜁 ∈ 𝑈̅), (1.4)
then
𝑝(𝑧, 𝜁) ≺≺ 𝑞(𝑧, 𝜁) ≺≺ ℎ(𝑧, 𝜁), (𝑧 ∈ 𝑈, 𝜁 ∈ 𝑈̅),
where 𝑞(𝑧, 𝜁) = 𝛾
𝑛𝑧𝛾𝑛∫ 𝑡 𝛾
𝑛−1ℎ(𝑡, 𝜁)𝑑𝑡
𝑧
0 is convex and it is the best dominant of (1.4).
Lemma 1.2 (Miller and Mocanu, [7]) Let 𝑞(𝑧, 𝜁) be a convex function in 𝑈 × 𝑈̅ for all 𝜁 ∈
𝑈̅ and let ℎ(𝑧, 𝜁) = 𝑞(𝑧, 𝜁) + 𝑛𝛿𝑧𝑞𝑧′(𝑧, 𝜁), 𝑧 ∈ 𝑈, 𝜁 ∈ 𝑈̅, where 𝛿 > 0 and 𝑛 is a positive
integer. If
𝑝(𝑧, 𝜁) = 𝑞(0, 𝜁) + 𝑝𝑛(𝜁)𝑧𝑛 + 𝑝𝑛+1(𝜁)𝑧𝑛+1+ ⋯,
is analytic in 𝑈 × 𝑈̅ and
𝑝(𝑧, 𝜁) + 𝛿𝑧𝑝𝑧′(𝑧, 𝜁) ≺≺ ℎ(𝑧, 𝜁), (𝑧 ∈ 𝑈, 𝜁 ∈ 𝑈̅),
then
𝑝(𝑧, 𝜁) ≺≺ 𝑞(𝑧, 𝜁), (𝑧 ∈ 𝑈, 𝜁 ∈ 𝑈̅),
and this result is sharp.
2.
M
AIN RESULTTheorem 2.1. Let ℎ(𝑧, 𝜁) be a convex function such that ℎ(0, 𝜁) = 1. If 𝑓 ∈ 𝒜 𝜁∗ satisfies the
strong differential subordination:
(𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁))
𝑧 ′
then
𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁)
𝑧 ≺≺ 𝑞(𝑧, 𝜁) ≺≺ ℎ(𝑧, 𝜁),
where 𝑞(𝑧, 𝜁) =1𝑧∫ ℎ(𝑡, 𝜁)𝑑𝑡0𝑧 is convex and it is the best dominant
Proof. Suppose that
𝑝(𝑧, 𝜁) = 𝐴𝜇,𝜆,𝜏
𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁)
𝑧 , 𝑧 ∈ 𝑈, 𝜁 ∈ 𝑈̅ . (2.2)
Then the function 𝑝(𝑧, 𝜁) is analytic in 𝑈 × 𝑈̅ and 𝑝(0, 𝜁) = 1.
Simple computations from (2.2), we get
𝑝(𝑧, 𝜁) + 𝑧𝑝𝑧′(𝑧, 𝜁) = (𝐴
𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁))𝑧 ′
. (2.3)
Using (2.3), (2.1) becomes
𝑝(𝑧, 𝜁) + 𝑧𝑝𝑧′(𝑧, 𝜁) ≺≺ ℎ(𝑧, 𝜁).
An application of Lemma 1.1 with 𝑛 = 1, 𝛾 = 1 yields
𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁)
𝑧 ≺≺ 𝑞(𝑧, 𝜁) = 1
𝑧∫ ℎ(𝑡, 𝜁)𝑑𝑡
𝑧
0
≺≺ ℎ(𝑧, 𝜁).
By taking ℎ(𝑧, 𝜁) =𝜁+(2𝜌−𝜁)𝑧1+𝑧 , 0 ≤ 𝜌 < 1 in Theorem 2.1, we obtain the following corollary:
Corollary 2.1. If 𝑓 ∈ 𝒜 𝜁∗ satisfies the strong differential subordination:
(𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁))
𝑧 ′
≺≺𝜁 + (2𝜌 − 𝜁)𝑧 1 + 𝑧 ,
then
𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁)
𝑧 ≺≺
1 𝑧∫
𝜁 + (2𝜌 − 𝜁)𝑡 1 + 𝑡 𝑑𝑡
𝑧
0
= 2𝜌 − 𝜁 +2(𝜁 − 𝜌)
𝑧 ln(1 + 𝑧).
Theorem 2.2. Let 𝑞(𝑧, 𝜁) be a convex function such that 𝑞(0, 𝜁) = 1 and let ℎ be the
function ℎ(𝑧, 𝜁) = 𝑞(𝑧, 𝜁) + 𝑧𝑞𝑧′(𝑧, 𝜁). If 𝑓 ∈ 𝒜
𝜁
∗ satisfies the strong differential subordination:
(𝑧 𝐴𝜇,𝜆,𝜏
𝜂+1(𝛼, 𝛽)𝑓(𝑧, 𝜁)
𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁) )
𝑧 ′
≺≺ ℎ(𝑧, 𝜁), (2.4)
𝐴𝜇,𝜆,𝜏𝜂+1(𝛼, 𝛽)𝑓(𝑧, 𝜁)
𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁)≺≺ 𝑞(𝑧, 𝜁).
Proof. Suppose that
𝑝(𝑧, 𝜁) =𝐴𝜇,𝜆,𝜏
𝜂+1(𝛼, 𝛽)𝑓(𝑧, 𝜁)
𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁) , 𝑧 ∈ 𝑈, 𝜁 ∈ 𝑈̅ . (2.5)
Then the function 𝑝(𝑧, 𝜁) is analytic in 𝑈 × 𝑈̅ and 𝑝(0, 𝜁) = 1.
Differentiating both sides of (2.5) with respect to 𝑧 and using (2.4), we have
𝑝(𝑧, 𝜁) + 𝑧𝑝𝑧′(𝑧, 𝜁)
=𝐴𝜇,𝜆,𝜏
𝜂+1(𝛼, 𝛽)𝑓(𝑧, 𝜁)
𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁)
+𝐴𝜇,𝜆,𝜏
𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁) (𝐴
𝜇,𝜆,𝜏𝜂+1(𝛼, 𝛽)𝑓(𝑧, 𝜁))𝑧 ′
− 𝐴𝜇,𝜆,𝜏𝜂+1(𝛼, 𝛽)𝑓(𝑧, 𝜁) (𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁))
𝑧 ′
[𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁)]2
=𝐴𝜇,𝜆,𝜏
𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁) (𝑧 𝐴 𝜇,𝜆,𝜏
𝜂+1(𝛼, 𝛽)𝑓(𝑧, 𝜁)) 𝑧 ′
− 𝑧𝐴𝜇,𝜆,𝜏𝜂+1(𝛼, 𝛽)𝑓(𝑧, 𝜁) (𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁))
𝑧 ′
[𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁)]2
= (𝑧 𝐴𝜇,𝜆,𝜏
𝜂+1(𝛼, 𝛽)𝑓(𝑧, 𝜁)
𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁) )
𝑧 ′
≺≺ ℎ(𝑧, 𝜁).
An application of Lemma 1.2, we obtain
𝐴𝜇,𝜆,𝜏𝜂+1(𝛼, 𝛽)𝑓(𝑧, 𝜁)
𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁)≺≺ 𝑞(𝑧, 𝜁).
Theorem 2.3. Let ℎ(𝑧, 𝜁) be a convex function such that ℎ(0, 𝜁) = 1. If 0 ≤ 𝜎 < 𝑝, 𝜃 ∈ ℂ
and 𝑓 ∈ 𝒜 𝜁∗ satisfies the strong differential subordination:
1 − 𝜃 1 − 𝜎 (
𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁)
𝑧 − 𝜎) +
𝜃
1 − 𝜎 ((𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁))𝑧 ′
− 𝜎) ≺≺ ℎ(𝑧, 𝜁), (2.6)
then
1 1 − 𝜎 (
𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁)
where 𝑞(𝑧, 𝜁) =1𝜃𝑧− 1𝜃∫ 𝑡 1
𝜃−1ℎ(𝑡, 𝜁)𝑑𝑡
𝑧
0 is convex and it is the best dominant.
Proof. Suppose that
𝑝(𝑧, 𝜁) = 1 1 − 𝜎 (
𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁)
𝑧 − 𝜎) , 𝑧 ∈ 𝑈, 𝜁 ∈ 𝑈̅ . (2.7)
Then the function 𝑝(𝑧, 𝜁) is analytic in 𝑈 × 𝑈̅ and 𝑝(0, 𝜁) = 1.
Differentiating both sides of (2.7) with respect to 𝑧, we have
𝑝(𝑧, 𝜁) + 𝜃𝑧𝑝𝑧′(𝑧, 𝜁)
=1 − 𝜃 1 − 𝜎 (
𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁)
𝑧 − 𝜎) +
𝜃
1 − 𝜎 ((𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁))𝑧 ′
− 𝜎). (2.8)
From (2.6) and (2.8), we get
𝑝(𝑧, 𝜁) + 𝜃𝑧𝑝𝑧′(𝑧, 𝜁) ≺≺ ℎ(𝑧, 𝜁).
An application of Lemma 1.1 with 𝑛 = 1, 𝛾 =𝜃1 yields
1 1 − 𝜎 (
𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁)
𝑧 − 𝜎) ≺≺ 𝑞(𝑧, 𝜁) = 1 𝜃𝑧
− 𝜃1∫ 𝑡𝑧 1𝜃−1ℎ(𝑡, 𝜁)𝑑𝑡
0
≺≺ ℎ(𝑧, 𝜁).
Theorem 2.4. Let 𝑞(𝑧, 𝜁) be a convex function such that 𝑞(0, 𝜁) = 1 and let ℎ be the
function ℎ(𝑧, 𝜁) = 𝑞(𝑧, 𝜁) + 1
1+(𝜆−𝛼)𝛽+𝑛𝜏𝜇+𝜆 𝑧𝑞𝑧
′(𝑧, 𝜁), where 𝜇+𝜆
(𝜆−𝛼)𝛽+𝑛𝜏> 0. Suppose that
𝐹(𝑧, 𝜁) =
1 +(𝜆 − 𝛼)𝛽 + 𝑛𝜏𝜇 + 𝜆
𝑧(𝜆−𝛼)𝛽+𝑛𝜏𝜇+𝜆
∫ 𝑡𝑧 (𝜆−𝛼)𝛽+𝑛𝜏−1𝜇+𝜆 𝑓(𝑡, 𝜁)𝑑𝑡 0
, 𝑧 ∈ 𝑈, 𝜁 ∈ 𝑈̅. (2.9)
If 𝑓 ∈ 𝒜 𝜁∗(𝑝) satisfies the strong differential subordination
(𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁))
𝑧 ′
≺≺ ℎ(𝑧, 𝜁), (2.10)
then
(𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝐹(𝑧, 𝜁))
𝑧 ′
≺≺ 𝑞(𝑧, 𝜁).
Proof. Suppose that
𝑝(𝑧, 𝜁) = (𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝐹(𝑧, 𝜁))
𝑧 ′
Then the function 𝑝(𝑧, 𝜁) is analytic in 𝑈 × 𝑈̅ and 𝑝(0, 𝜁) = 1.
From (2.9), we have
𝑧(𝜆−𝛼)𝛽+𝑛𝜏𝜇+𝜆 𝐹(𝑧, 𝜁) = (1 + 𝜇 + 𝜆
(𝜆 − 𝛼)𝛽 + 𝑛𝜏) ∫ 𝑡
𝜇+𝜆
(𝜆−𝛼)𝛽+𝑛𝜏−1𝑓(𝑡, 𝜁)𝑑𝑡 𝑧
0
. (2.12)
Differentiating both sides of (2.12) with respect to 𝑧, we get
(1 + 𝜇 + 𝜆
(𝜆 − 𝛼)𝛽 + 𝑛𝜏) 𝑓(𝑧, 𝜁) =
𝜇 + 𝜆
(𝜆 − 𝛼)𝛽 + 𝑛𝜏𝐹(𝑧, 𝜁) + 𝑧𝐹𝑧′(𝑧, 𝜁)
and
(1 + 𝜇 + 𝜆
(𝜆 − 𝛼)𝛽 + 𝑛𝜏) 𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁)
= 𝜇 + 𝜆
(𝜆 − 𝛼)𝛽 + 𝑛𝜏𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝐹(𝑧, 𝜁) + 𝑧 (𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝐹(𝑧, 𝜁))𝑧 ′
.
So
(𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁))
𝑧 ′
= (𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝐹(𝑧, 𝜁))
𝑧 ′
+𝑧 (𝐴𝜇,𝜆,𝜏
𝜂 (𝛼, 𝛽)𝐹(𝑧, 𝜁)) 𝑧2
′′
1 +(𝜆 − 𝛼)𝛽 + 𝑛𝜏𝜇 + 𝜆 . (2.13)
From (2.11) and (2.13), we obtain
𝑝(𝑧, 𝜁) + 1
1 +(𝜆 − 𝛼)𝛽 + 𝑛𝜏𝜇 + 𝜆 𝑧𝑝𝑧
′(𝑧, 𝜁) = (𝐴 𝜇,𝜆,𝜏
𝜂 (𝛼, 𝛽)𝑓(𝑧, 𝜁)) 𝑧 ′
. (2.14)
Using (2.14), (2.10) becomes
𝑝(𝑧, 𝜁) + 1
1 +(𝜆 − 𝛼)𝛽 + 𝑛𝜏𝜇 + 𝜆 𝑧𝑝𝑧
′(𝑧, 𝜁) ≺≺ 𝑞(𝑧, 𝜁) + 1
1 +(𝜆 − 𝛼)𝛽 + 𝑛𝜏𝜇 + 𝜆 𝑧𝑞𝑧
′(𝑧, 𝜁).
An application of Lemma 1.2 yields 𝑝(𝑧, 𝜁) ≺≺ 𝑞(𝑧, 𝜁). By using (2.10), we obtain
(𝐴𝜇,𝜆,𝜏𝜂 (𝛼, 𝛽)𝐹(𝑧, 𝜁))
𝑧 ′
≺≺ 𝑞(𝑧, 𝜁).
Conflict of Interests
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