International Journal of Mathematics and Mathematical Sciences

International Journal of Mathematics and Mathematical Sciences / 2010 / Article

Research Article | Open Access

Volume 2010 |Article ID 369078 |

Ma'moun Harayzeh Al-Abbadi, Maslina Darus, "Differential Subordination Defined by New Generalised Derivative Operator for Analytic Functions", International Journal of Mathematics and Mathematical Sciences, vol. 2010, Article ID 369078, 15 pages, 2010.

Differential Subordination Defined by New Generalised Derivative Operator for Analytic Functions

Academic Editor: Dorothy Wallace
Received09 Jun 2009
Revised20 Dec 2009
Accepted09 Mar 2010
Published07 Apr 2010


A new generalised derivative operator πœ‡πœ†π‘›,π‘š1,πœ†2 is introduced. This operator generalised many well-known operators studied earlier by many authors. Using the technique of differential subordination, we will study some of the properties of differential subordination. In addition we investigate several interesting properties of the new generalised derivative operator.

1. Introduction and Preliminaries

Let π’œ denote the class of functions of the form

𝑓(𝑧)=𝑧+βˆžξ“π‘˜=2π‘Žπ‘˜π‘§π‘˜,whereπ‘Žπ‘˜iscomplexnumber,(1.1) which are analytic in the open unit disc π‘ˆ={π‘§βˆˆβ„‚βˆΆ|𝑧|<1} on the complex plane β„‚. Let 𝑆,π‘†βˆ—(𝛼),𝐢(𝛼)(0≀𝛼<1) denote the subclasses of π’œ consisting of functions that are univalent, starlike of order 𝛼, and convex of order 𝛼 in π‘ˆ, respectively. In particular, the classes π‘†βˆ—(0)=π‘†βˆ—and𝐢(0)=𝐢 are the familiar classes of starlike and convex functions in π‘ˆ, respectively. A function π‘“βˆˆπΆ(𝛼) if Re(1+π‘§π‘“ξ…žξ…ž/π‘“ξ…ž)>𝛼. Furthermore a function 𝑓 analytic in π‘ˆ is said to be convex if it is univalent and 𝑓(π‘ˆ) is convex.

Let β„‹(π‘ˆ) be the class of holomorphic function in unit disc π‘ˆ={π‘§βˆˆβ„‚βˆΆ|𝑧|<1}.


π’œπ‘›=ξ€½π‘“βˆˆβ„‹(π‘ˆ)βˆΆπ‘“(𝑧)=𝑧+π‘Žπ‘›+1𝑧𝑛+1ξ€Ύ,+β‹―,(π‘§βˆˆπ‘ˆ)(1.2) with π’œ1=π’œ.

For π‘Žβˆˆβ„‚ and π‘›βˆˆβ„•={1,2,3,…} we let

β„‹[]=ξ€½π‘Ž,π‘›π‘“βˆˆβ„‹(π‘ˆ)βˆΆπ‘“(𝑧)=𝑧+π‘Žπ‘›π‘§π‘›+π‘Žπ‘›+1𝑧𝑛+1ξ€Ύ.+β‹―,(π‘§βˆˆπ‘ˆ)(1.3) Let βˆ‘π‘“(𝑧)=𝑧+βˆžπ‘˜=2π‘Žπ‘˜π‘§π‘˜βˆ‘and𝑔(𝑧)=𝑧+βˆžπ‘˜=2π‘π‘˜π‘§π‘˜be analytic in the open unit disc π‘ˆ={π‘§βˆˆβ„‚βˆΆ|𝑧|<1}. Then the Hadamard product (or convolution) π‘“βˆ—π‘” of the two functions 𝑓, 𝑔 is defined by

𝑓(𝑧)βˆ—π‘”(𝑧)=(π‘“βˆ—π‘”)(𝑧)=𝑧+βˆžξ“π‘˜=2π‘Žπ‘˜π‘π‘˜π‘§π‘˜.(1.4) Next, we state basic ideas on subordination. If 𝑓 and 𝑔 are analytic in π‘ˆ, then the function 𝑓 is said to be subordinate to 𝑔, and can be written as

𝑓≺𝑔,𝑓(𝑧)≺𝑔(𝑧),(π‘§βˆˆπ‘ˆ),(1.5) if and only if there exists the Schwarz function 𝑀, analytic in π‘ˆ, with 𝑀(0)=0and|𝑀(𝑧)|<1such that𝑓(𝑧)=𝑔(𝑀(𝑧)), (π‘§βˆˆπ‘ˆ).

Furthermore if 𝑔 is univalent in π‘ˆ, then 𝑓≺𝑔 if and only if 𝑓(0)=𝑔(0) and 𝑓(π‘ˆ)βŠ‚π‘”(π‘ˆ) (see [1, page 36]).

Let πœ“βˆΆβ„‚3Γ—π‘ˆβ†’β„‚ and let β„Ž be univalent in π‘ˆ. If 𝑝 is analytic in π‘ˆ and satisfies the (second-order) differential subordination

πœ“ξ€·π‘(𝑧),π‘§π‘ξ…ž(𝑧),𝑧2π‘ξ…žξ…žξ€Έ(𝑧);π‘§β‰Ίβ„Ž(𝑧),(π‘§βˆˆπ‘ˆ),(1.6) then 𝑝 is called a solution of the differential subordination.

The univalent function π‘ž is called a dominant of the solutions of the differential subordination, or more simply a dominant, if π‘β‰Ίπ‘ž for all 𝑝 satisfying (1.6). A dominant Μƒπ‘ž that satisfies Μƒπ‘žβ‰Ίπ‘ž for all dominants π‘ž of (1.6) is said to be the best dominant of (1.6). (Note that the best dominant is unique up to a rotation of π‘ˆ.)

Now, (π‘₯)π‘˜ denotes the Pochhammer symbol (or the shifted factorial) defined by

(π‘₯)π‘˜=ξƒ―1forπ‘˜=0,π‘₯βˆˆβ„‚β§΅{0},π‘₯(π‘₯+1)(π‘₯+2)β‹―(π‘₯+π‘˜βˆ’1)forπ‘˜βˆˆβ„•={1,2,3,…},π‘₯βˆˆβ„‚.(1.7) To prove our results, we need the following equation throughout the paper:

πœ‡πœ†π‘›,π‘š+11,πœ†2𝑓(𝑧)=1βˆ’πœ†1ξ€Έξ‚ƒπœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)βˆ—πœ™πœ†2ξ‚„(𝑧)+πœ†1π‘§ξ‚ƒπœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)βˆ—πœ™πœ†2ξ‚„(𝑧)ξ…ž,(π‘§βˆˆπ‘ˆ),(1.8) where 𝑛,π‘šβˆˆβ„•0={0,1,2,…}, πœ†2β‰₯πœ†1β‰₯0, and πœ™πœ†2(𝑧) is analytic function given by

πœ™πœ†2(𝑧)=𝑧+βˆžξ“π‘˜=2π‘§π‘˜1+πœ†2.(π‘˜βˆ’1)(1.9) Here πœ‡πœ†π‘›,π‘š1,πœ†2 is the generalized derivative operator which we shall introduce later in the paper. Moreover, we need the following lemmas in proving our main results.

Lemma 1.1 (see [2, page 71]). Let β„Ž be analytic, univalent, and convex in π‘ˆ, with β„Ž(0)=π‘Ž, 𝛾≠0 and, Re𝛾β‰₯0. If π‘βˆˆβ„‹[π‘Ž,𝑛]and 𝑝(𝑧)+π‘§π‘ξ…ž(𝑧)π›Ύβ‰Ίβ„Ž(𝑧),(π‘§βˆˆπ‘ˆ),(1.10) then 𝑝(𝑧)β‰Ίπ‘ž(𝑧)β‰Ίβ„Ž(𝑧),(π‘§βˆˆπ‘ˆ),(1.11) where π‘ž(𝑧)=(𝛾/𝑛𝑧𝛾/𝑛)βˆ«π‘§0β„Ž(𝑑)𝑑(𝛾/𝑛)βˆ’1𝑑𝑑,(π‘§βˆˆπ‘ˆ).

The function π‘ž is convex and is the best (π‘Ž,𝑛)-dominant.

Lemma 1.2 (see [3]). Let 𝑔 be a convex function in π‘ˆ and let β„Ž(𝑧)=𝑔(𝑧)+𝑛𝛼𝑧𝑔′(𝑧),(1.12) where 𝛼>0 and 𝑛 is a positive integer.
𝑝(𝑧)=𝑔(0)+𝑝𝑛𝑧𝑛+𝑝𝑛+1𝑧𝑛+1+β‹―,(π‘§βˆˆπ‘ˆ),(1.13) is holomorphic in π‘ˆ and 𝑝(𝑧)+π›Όπ‘§π‘ξ…ž(𝑧)β‰Ίβ„Ž(𝑧),(π‘§βˆˆπ‘ˆ),(1.14) then 𝑝(𝑧)≺𝑔(𝑧),(1.15) and this result is sharp.

Lemma 1.3 (see [4]). Let π‘“βˆˆπ’œ, if ξ‚΅Re1+π‘§π‘“ξ…žξ…ž(𝑧)π‘“ξ…žξ‚Ά1(𝑧)>βˆ’2,(1.16) then 2π‘§ξ€œπ‘§0𝑓(𝑑)𝑑𝑑,(π‘§βˆˆπ‘ˆ,𝑧≠0),(1.17) belongs to the class of convex functions.

2. Main Results

In the present paper, we will use the method of differential subordination to derive certain properties of generalised derivative operator πœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧). Note that differential subordination has been studied by various authors, and here we follow similar works done by Oros [5] and G. Oros and G. I. Oros [6].

In order to derive our new generalised derivative operator, we define the analytic function

πΉπ‘šπœ†1,πœ†2(𝑧)=𝑧+βˆžξ“π‘˜=2ξ€·1+πœ†1ξ€Έ(π‘˜βˆ’1)π‘šξ€·1+πœ†2ξ€Έ(π‘˜βˆ’1)π‘šβˆ’1π‘§π‘˜,(2.1) where π‘šβˆˆβ„•0={0,1,2,…} and πœ†2β‰₯πœ†1β‰₯0. Now, we introduce the new generalised derivative operator πœ‡πœ†π‘›,π‘š1,πœ†2 as follows.

Definition 2.1. For π‘“βˆˆπ’œ the operator πœ‡πœ†π‘›,π‘š1,πœ†2 is defined by πœ‡πœ†π‘›,π‘š1,πœ†2βˆΆπ’œβ†’π’œπœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)=πΉπ‘šπœ†1,πœ†2(𝑧)βˆ—π‘…π‘›π‘“(𝑧),(π‘§βˆˆπ‘ˆ),(2.2) where 𝑛,π‘šβˆˆβ„•0=β„•βˆͺ{0},πœ†2β‰₯πœ†1β‰₯0,and𝑅𝑛𝑓(𝑧) denotes the Ruscheweyh derivative operator [7], given by 𝑅𝑛𝑓(𝑧)=𝑧+βˆžξ“π‘˜=2𝑐(𝑛,π‘˜)π‘Žπ‘˜π‘§π‘˜,ξ€·π‘›βˆˆβ„•0ξ€Έ,,π‘§βˆˆπ‘ˆ(2.3) where 𝑐(𝑛,π‘˜)=(𝑛+1)π‘˜βˆ’1/(1)π‘˜βˆ’1.
If 𝑓 is given by (1.1), then we easily find from equality (2.2) that
πœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)=𝑧+βˆžξ“π‘˜=2ξ€·1+πœ†1ξ€Έ(π‘˜βˆ’1)π‘šξ€·1+πœ†2ξ€Έ(π‘˜βˆ’1)π‘šβˆ’1𝑐(𝑛,π‘˜)π‘Žπ‘˜π‘§π‘˜,(π‘§βˆˆπ‘ˆ),(2.4) where 𝑛,π‘šβˆˆβ„•0={0,1,2…},πœ†2β‰₯πœ†1β‰₯0,and𝑐(𝑛,π‘˜)=(𝑛+π‘˜βˆ’1𝑛)=(𝑛+1)π‘˜βˆ’1/(1)π‘˜βˆ’1.

Remark 2.2. Special cases of this operator include the Ruscheweyh derivative operator in two cases πœ‡π‘›,10,πœ†2≑𝑅𝑛 and πœ‡πœ†π‘›,01,0≑𝑅𝑛 [7], the Salagean derivative operator πœ‡0,π‘š1,0≑𝑆𝑛 [8], the generalised Ruscheweyh derivative operator in two cases πœ‡πœ†π‘›,11,πœ†2β‰‘π‘…π‘›πœ† and πœ‡πœ†π‘›,01,πœ†2β‰‘π‘…π‘›πœ† [9], the generalised Salagean derivative operator πœ‡πœ†0,π‘š1,0≑𝑆𝑛𝛽 introduced by Al-Oboudi [10], and the generalised Al-Shaqsi and Darus derivative operator πœ‡πœ†π‘›,π‘š1,0β‰‘π·π‘›πœ†,𝛽 that can be found in [11].

Now, we remind the well-known Carlson-Shaffer operator 𝐿(π‘Ž,𝑐) [12] associated with the incomplete beta function πœ™(π‘Ž,𝑐;𝑧), defined by

𝐿𝐿(π‘Ž,𝑐)βˆΆπ’œβ†’π’œ,(π‘Ž,𝑐)𝑓(𝑧)∢=πœ™(π‘Ž,𝑐;𝑧)βˆ—π‘“(𝑧),(π‘§βˆˆπ‘ˆ),(2.5) where

πœ™(π‘Ž,𝑐;𝑧)=𝑧+βˆžξ“π‘˜=2(π‘Ž)π‘˜βˆ’1(𝑐)π‘˜βˆ’1π‘§π‘˜,(2.6)π‘Ž is any real number, and π‘βˆ‰π‘§βˆ’0;π‘§βˆ’0={0,βˆ’1,βˆ’2,…}.

It is easily seen that

πœ‡πœ†0,01,0𝑓(𝑧)=πœ‡0,π‘š0,0𝑓(𝑧)=πœ‡0,10,πœ†2𝑓(𝑧)=πœ‡1,20,1πœ‡π‘“(z)=𝐿(π‘Ž,π‘Ž)𝑓(𝑧)=𝑓(𝑧),πœ†1,01,0𝑓(𝑧)=πœ‡1,π‘š0,0𝑓(𝑧)=πœ‡1,10,πœ†2𝑓(𝑧)=πœ‡πœ†0,01,1𝑓(𝑧)=𝐿(2,1)𝑓(𝑧)=π‘§π‘“ξ…ž(𝑧),(2.7) and also

πœ‡πœ†π‘Žβˆ’1,01,0𝑓(𝑧)=πœ‡π‘Žβˆ’1,10,πœ†2𝑓(𝑧)=πœ‡π‘Žβˆ’1,π‘š0,0𝑓(𝑧)=𝐿(π‘Ž,1)𝑓(𝑧),(2.8) where π‘Ž=1,2,3,….

Next, we give the following.

Definition 2.3. For 𝑛,π‘šβˆˆβ„•0,πœ†2β‰₯πœ†1β‰₯0, and 0≀𝛼<1, let π‘…πœ†π‘›,π‘š1,πœ†2(𝛼) denote the class of functions π‘“βˆˆπ’œ which satisfy the condition ξ‚€πœ‡Reπœ†π‘›,π‘š1,πœ†2𝑓(𝑧)ξ…ž>𝛼,(π‘§βˆˆπ‘ˆ).(2.9) Also let πΎπœ†π‘›,π‘š1,πœ†2(𝛿) denote the class of functions π‘“βˆˆπ’œ which satisfy the condition ξ‚€πœ‡Reπœ†π‘›,π‘š1,πœ†2𝑓(𝑧)βˆ—πœ™πœ†2(𝑧)ξ…ž>𝛿,(π‘§βˆˆπ‘ˆ).(2.10)

Remark 2.4. It is clear that π‘…πœ†0,11,0(𝛼)≑𝑅(πœ†1,𝛼), and the class of functions π‘“βˆˆπ’œ satisfying ξ€·πœ†Re1π‘§π‘“ξ…žξ…ž(𝑧)+π‘“ξ…žξ€Έ(𝑧)>𝛼,(π‘§βˆˆπ‘ˆ)(2.11) is studied by Ponnusamy [13] and others.

Now we begin with the first result as follows.

Theorem 2.5. Let β„Ž(𝑧)=1+(2π›Όβˆ’1)𝑧1+𝑧,(π‘§βˆˆπ‘ˆ),(2.12) be convex in π‘ˆ, with h(0)=1 and 0≀𝛼<1. If 𝑛,π‘šβˆˆβ„•0,πœ†2β‰₯πœ†1β‰₯0, and the differential subordination ξ‚€πœ‡π‘›,πœ†π‘š+11,πœ†2𝑓(𝑧)ξ…žβ‰Ίβ„Ž(𝑧),(π‘§βˆˆπ‘ˆ),(2.13) holds, then ξ‚€πœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)βˆ—πœ™πœ†2(𝑧)ξ…žβ‰Ίπ‘ž(𝑧)=2π›Όβˆ’1+2(1βˆ’π›Ό)πœ†1𝑧1/πœ†1πœŽξ‚΅1πœ†1ξ‚Ά,(2.14) where 𝜎 is given by ξ€œπœŽ(π‘₯)=𝑧0𝑑π‘₯βˆ’11+𝑑𝑑𝑑,(π‘§βˆˆπ‘ˆ).(2.15) The function π‘ž is convex and is the best dominant.

Proof. By differentiating (1.8), with respect to 𝑧, we obtain ξ‚€πœ‡πœ†π‘›,π‘š+11,πœ†2𝑓(𝑧)ξ…ž=ξ‚ƒπœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)βˆ—πœ™πœ†2ξ‚„(𝑧)ξ…ž+πœ†1π‘§ξ‚ƒπœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)βˆ—πœ™πœ†2ξ‚„(𝑧)ξ…žξ…ž.(2.16) Using (2.16) in (2.13), differential subordination (2.13) becomes ξ‚ƒπœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)βˆ—πœ™πœ†2ξ‚„(𝑧)ξ…ž+πœ†1π‘§ξ‚ƒπœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)βˆ—πœ™πœ†2ξ‚„(𝑧)ξ…žξ…žβ‰Ίβ„Ž(𝑧)=1+(2π›Όβˆ’1)𝑧1+𝑧.(2.17) Let ξ‚ƒπœ‡π‘(𝑧)=πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)βˆ—πœ™πœ†2ξ‚„(𝑧)ξ…ž=𝑧+βˆžξ“π‘˜=2ξ€·1+πœ†1ξ€Έ(π‘˜βˆ’1)π‘šξ€·1+πœ†2ξ€Έ(π‘˜βˆ’1)π‘šπ‘(𝑛,π‘˜)π‘Žπ‘˜π‘§π‘˜ξƒ­ξ…ž=1+𝑝1𝑧+𝑝2𝑧2[]+β‹―,(π‘βˆˆβ„‹1,1,π‘§βˆˆπ‘ˆ).(2.18) Using (2.18) in (2.17), the differential subordination becomes 𝑝(𝑧)+πœ†1π‘§π‘ξ…ž(𝑧)β‰Ίβ„Ž(𝑧)=1+(2π›Όβˆ’1)𝑧.1+𝑧(2.19) By using Lemma 1.1, we have 1𝑝(𝑧)β‰Ίπ‘ž(𝑧)=πœ†1𝑧1/πœ†1ξ€œπ‘§0β„Ž(𝑑)𝑑(1/πœ†1)βˆ’1=1π‘‘π‘‘πœ†1𝑧1/πœ†1ξ€œπ‘§0ξ‚΅1+(2π›Όβˆ’1)𝑑𝑑1+𝑑(1/πœ†1)βˆ’1𝑑𝑑=2π›Όβˆ’1+2(1βˆ’π›Ό)πœ†1𝑧1/πœ†1πœŽξ‚΅1πœ†1ξ‚Ά,(2.20) where 𝜎 is given by (2.15), so we get ξ‚ƒπœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)βˆ—πœ™πœ†2ξ‚„(𝑧)ξ…žβ‰Ίπ‘ž(𝑧)=2π›Όβˆ’1+2(1βˆ’π›Ό)πœ†1𝑧1/πœ†1πœŽξ‚΅1πœ†1ξ‚Ά.(2.21) The function π‘ž is convex and is the best dominant. The proof is complete.

Theorem 2.6. If 𝑛,π‘šβˆˆβ„•0,πœ†2β‰₯πœ†1β‰₯0, and 0≀𝛼<1, then one has π‘…πœ†π‘›,π‘š+11,πœ†2(𝛼)βŠ‚πΎπœ†π‘›,π‘š1,πœ†2(𝛿),(2.22) where 𝛿=2π›Όβˆ’1+2(1βˆ’π›Ό)πœ†1πœŽξ‚΅1πœ†1ξ‚Ά,(2.23) and 𝜎 is given by (2.15).

Proof. Let π‘“βˆˆπ‘…πœ†π‘›,π‘š+11,πœ†2(𝛼), then from (2.9) we have ξ‚€πœ‡Reπœ†π‘›,π‘š+11,πœ†2𝑓(𝑧)ξ…ž>𝛼,(π‘§βˆˆπ‘ˆ),(2.24) which is equivalent to ξ‚€πœ‡πœ†π‘›,π‘š+11,πœ†2𝑓(𝑧)ξ…žβ‰Ίβ„Ž(𝑧)=1+(2π›Όβˆ’1)𝑧.1+𝑧(2.25) Using Theorem 2.5, we have ξ‚ƒπœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)βˆ—πœ™πœ†2ξ‚„(𝑧)ξ…žβ‰Ίπ‘ž(𝑧)=2π›Όβˆ’1+2(1βˆ’π›Ό)πœ†1𝑧1/πœ†1πœŽξ‚΅1πœ†1ξ‚Ά.(2.26) Since π‘ž is convex and π‘ž(π‘ˆ) is symmetric with respect to the real axis, we deduce that ξ‚ƒπœ‡Reπœ†π‘›,π‘š1,πœ†2𝑓(𝑧)βˆ—πœ™πœ†2ξ‚„(𝑧)ξ…žξ€·>Reπ‘ž(1)=𝛿=𝛿𝛼,πœ†1ξ€Έ=2π›Όβˆ’1+2(1βˆ’π›Ό)πœ†1πœŽξ‚΅1πœ†1ξ‚Ά,(2.27) from which we deduce π‘…πœ†π‘›,π‘š+11,πœ†2(𝛼)βŠ‚πΎπœ†π‘›,π‘š1,πœ†2(𝛿). This completes the proof of Theorem 2.6.

Remark 2.7. Special case of Theorem 2.6 with πœ†2=0 was given earlier in [11].

Theorem 2.8. Let π‘ž be a convex function in π‘ˆ, with π‘ž(0)=1, and let β„Ž(𝑧)=π‘ž(𝑧)+πœ†1π‘§π‘žξ…ž(𝑧),(π‘§βˆˆπ‘ˆ).(2.28) If 𝑛,π‘šβˆˆβ„•0,πœ†2β‰₯πœ†1β‰₯0,and π‘“βˆˆπ’œ and satisfies the differential subordination ξ‚€πœ‡πœ†π‘›,π‘š+11,πœ†2𝑓(𝑧)ξ…žβ‰Ίβ„Ž(𝑧),(π‘§βˆˆπ‘ˆ),(2.29) then ξ‚ƒπœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)βˆ—πœ™πœ†2ξ‚„(𝑧)ξ…žβ‰Ίπ‘ž(𝑧),(π‘§βˆˆπ‘ˆ),(2.30) and this result is sharp.

Proof. Using (2.18) in (2.16), differential subordination (2.29) becomes 𝑝(𝑧)+πœ†1π‘§π‘ξ…ž(𝑧)β‰Ίβ„Ž(𝑧)=π‘ž(𝑧)+πœ†1π‘§π‘žξ…ž(𝑧),(π‘§βˆˆπ‘ˆ).(2.31) Using Lemma 1.2, we obtain 𝑝(𝑧)β‰Ίπ‘ž(𝑧),(π‘§βˆˆπ‘ˆ).(2.32) Hence ξ‚ƒπœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)βˆ—πœ™πœ†2ξ‚„(𝑧)ξ…žβ‰Ίπ‘ž(𝑧),(π‘§βˆˆπ‘ˆ).(2.33) And the result is sharp. This completes the proof of the theorem.

We give a simple application for Theorem 2.8.

Example 2.9. For 𝑛=0,π‘š=1,πœ†2β‰₯πœ†1β‰₯0,π‘ž(𝑧)=(1+𝑧)/(1βˆ’π‘§),π‘“βˆˆπ’œ, and π‘§βˆˆπ‘ˆ and applying Theorem 2.8, we have β„Ž(𝑧)=1+𝑧1βˆ’π‘§+πœ†1𝑧1+𝑧1βˆ’π‘§ξ…ž=1+2πœ†1π‘§βˆ’π‘§2(1βˆ’π‘§)2.(2.34) By using (1.8) we find that πœ‡πœ†0,11,πœ†2𝑓(𝑧)=1βˆ’πœ†1𝑓(𝑧)+πœ†1π‘§π‘“ξ…ž(𝑧).(2.35) Now, πœ‡πœ†0,11,πœ†2𝑓(𝑧)βˆ—πœ™πœ†2ξ€·(𝑧)=1βˆ’πœ†1𝑧+βˆžξ“π‘˜=2π‘Žπ‘˜π‘§π‘˜1+πœ†2ξƒ­(π‘˜βˆ’1)+πœ†1𝑧+βˆžξ“π‘˜=2π‘Žπ‘˜π‘˜π‘§π‘˜1+πœ†2ξƒ­.(π‘˜βˆ’1)(2.36) A straightforward calculation gives the following: ξ‚ƒπœ‡πœ†0,11,πœ†2𝑓(𝑧)βˆ—πœ™πœ†2ξ‚„(𝑧)ξ…ž=1+βˆžξ“π‘˜=2ξ€·1+πœ†1ξ€Έ(π‘˜βˆ’1)π‘˜π‘Žπ‘˜1+πœ†2𝑧(π‘˜βˆ’1)π‘˜βˆ’1=βˆ‘π‘§+βˆžπ‘˜=2ξ€·π‘˜π‘Žπ‘˜ξ€·1+πœ†1ξ€Έ/ξ€·(π‘˜βˆ’1)1+πœ†2𝑧(π‘˜βˆ’1)ξ€Έξ€Έπ‘˜π‘§=𝑓(𝑧)βˆ—πœ™πœ†2ξ€»βˆ—ξ€Ίβˆ‘(𝑧)𝑧+βˆžπ‘˜=2π‘˜ξ€·1+πœ†1𝑧(π‘˜βˆ’1)π‘˜ξ€»π‘§.(2.37) Similarly, using (1.8), we find that πœ‡πœ†0,21,πœ†2𝑓(𝑧)=1βˆ’πœ†1ξ€Έξ‚ƒπœ‡πœ†0,11,πœ†2𝑓(𝑧)βˆ—πœ™πœ†2ξ‚„(𝑧)+πœ†1π‘§ξ‚ƒπœ‡πœ†0,11,πœ†2𝑓(𝑧)βˆ—πœ™πœ†2ξ‚„(𝑧)ξ…ž,(2.38) then ξ‚€πœ‡πœ†0,21,πœ†2𝑓(𝑧)ξ…ž=ξ‚€πœ‡πœ†0,11,πœ†2𝑓(𝑧)βˆ—πœ™πœ†2(𝑧)ξ…ž+πœ†1π‘§ξ‚€πœ‡πœ†0,11,πœ†2𝑓(𝑧)βˆ—πœ™πœ†2(𝑧)ξ…žξ…ž.(2.39) By using (2.37) we obtain ξ‚€πœ‡πœ†0,11,πœ†2𝑓(𝑧)βˆ—πœ™πœ†2(𝑧)ξ…žξ…ž=βˆžξ“π‘˜=2π‘˜(π‘˜βˆ’1)π‘Žπ‘˜ξ€·1+πœ†1ξ€Έ(π‘˜βˆ’1)1+πœ†2𝑧(π‘˜βˆ’1)π‘˜βˆ’2.(2.40) We get ξ‚€πœ‡πœ†0,21,πœ†2𝑓(𝑧)ξ…ž=1+βˆžξ“π‘˜=2π‘˜π‘Žπ‘˜ξ€·1+πœ†1ξ€Έ(π‘˜βˆ’1)1+πœ†2𝑧(π‘˜βˆ’1)π‘˜βˆ’1+πœ†1βˆžξ“π‘˜=2π‘˜(π‘˜βˆ’1)π‘Žπ‘˜ξ€·1+πœ†1ξ€Έ(π‘˜βˆ’1)1+πœ†2𝑧(π‘˜βˆ’1)π‘˜βˆ’1=1+βˆžξ“π‘˜=2π‘˜π‘Žπ‘˜ξ€·1+πœ†1ξ€Έ(π‘˜βˆ’1)21+πœ†2𝑧(π‘˜βˆ’1)π‘˜βˆ’1=𝑓(𝑧)βˆ—πœ™πœ†2ξ€»βˆ—ξ‚ƒβˆ‘(𝑧)𝑧+βˆžπ‘˜=2π‘˜ξ€·1+πœ†1ξ€Έ(π‘˜βˆ’1)2π‘§π‘˜ξ‚„π‘§.(2.41) From Theorem 2.8 we deduce that 𝑓(𝑧)βˆ—πœ™πœ†2ξ€»βˆ—ξ‚ƒβˆ‘(𝑧)𝑧+βˆžπ‘˜=2π‘˜ξ€·1+πœ†1ξ€Έ(π‘˜βˆ’1)2π‘§π‘˜ξ‚„π‘§β‰Ί1+2πœ†1π‘§βˆ’π‘§2(1βˆ’π‘§)2,(2.42) implies that 𝑓(𝑧)βˆ—πœ™πœ†2ξ€»βˆ—ξ€Ίβˆ‘(𝑧)𝑧+βˆžπ‘˜=2π‘˜ξ€·1+πœ†1𝑧(π‘˜βˆ’1)π‘˜ξ€»π‘§β‰Ί1+𝑧,1βˆ’π‘§(π‘§βˆˆπ‘ˆ).(2.43)

Theorem 2.10. Let π‘ž be a convex function in π‘ˆ, with π‘ž(0)=1 and let β„Ž(𝑧)=π‘ž(𝑧)+π‘§π‘žξ…ž(𝑧),(π‘§βˆˆπ‘ˆ).(2.44) If 𝑛,π‘šβˆˆβ„•0,πœ†2β‰₯πœ†1β‰₯0, and π‘“βˆˆπ’œ and satisfies the differential subordination ξ‚€πœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)ξ…žβ‰Ίβ„Ž(𝑧),(2.45) then πœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)π‘§β‰Ίπ‘ž(𝑧),(π‘§βˆˆπ‘ˆ).(2.46) And the result is sharp.

Proof. Let πœ‡π‘(𝑧)=πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)𝑧=βˆ‘π‘§+βˆžπ‘˜=2ξ‚€ξ€·1+πœ†1ξ€Έ(π‘˜βˆ’1)π‘š/ξ€·1+πœ†2ξ€Έ(π‘˜βˆ’1)π‘šβˆ’1𝑐(𝑛,π‘˜)π‘Žπ‘˜π‘§π‘˜π‘§=1+𝑝1𝑧+𝑝2𝑧2[]+β‹―,(π‘βˆˆβ„‹1,1,π‘§βˆˆπ‘ˆ).(2.47) Differentiating (2.47), with respect to 𝑧, we obtain ξ‚€πœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)ξ…ž=𝑝(𝑧)+π‘§π‘ξ…ž(𝑧),(π‘§βˆˆπ‘ˆ).(2.48) Using (2.48), (2.45) becomes 𝑝(𝑧)+π‘§π‘ξ…ž(𝑧)β‰Ίβ„Ž(𝑧)=π‘ž(𝑧)+π‘§π‘žξ…ž(𝑧).(2.49) Using Lemma 1.2, we deduce that 𝑝(𝑧)β‰Ίπ‘ž(𝑧),(π‘§βˆˆπ‘ˆ),(2.50) and using (2.47), we have πœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)π‘§β‰Ίπ‘ž(𝑧),(π‘§βˆˆπ‘ˆ).(2.51) This proves Theorem 2.10.

We give a simple application for Theorem 2.10.

Example 2.11. For 𝑛=0,π‘š=1,πœ†2β‰₯πœ†1β‰₯0,π‘ž(𝑧)=1/(1βˆ’π‘§),π‘“βˆˆπ’œ, and π‘§βˆˆπ‘ˆ and applying Theorem 2.10, we have 1β„Ž(𝑧)=ξ‚€11βˆ’π‘§+𝑧1βˆ’π‘§ξ…ž=1(1βˆ’π‘§)2.(2.52) From Example 2.9, we have πœ‡πœ†0,11,πœ†2𝑓(𝑧)=1βˆ’πœ†1𝑓(𝑧)+πœ†1𝑧𝑓′(𝑧),(2.53) so ξ‚€πœ‡πœ†0,11,πœ†2𝑓(𝑧)ξ…ž=𝑓′(𝑧)+πœ†1π‘§π‘“ξ…žξ…ž(𝑧).(2.54) Now, from Theorem 2.10 we deduce that 𝑓′(𝑧)+πœ†1π‘§π‘“ξ…žξ…ž1(𝑧)β‰Ί(1βˆ’π‘§)2(2.55) implies that ξ€·1βˆ’πœ†1𝑓(𝑧)+πœ†1𝑧𝑓′(𝑧)𝑧≺1.1βˆ’π‘§(2.56)

Theorem 2.12. Let β„Ž(𝑧)=1+(2π›Όβˆ’1)𝑧1+𝑧,(π‘§βˆˆπ‘ˆ),(2.57) be convex in π‘ˆ, with β„Ž(0)=1 and 0≀𝛼<1. If 𝑛,π‘šβˆˆβ„•0,πœ†2β‰₯πœ†1β‰₯0,π‘“βˆˆπ’œ, and the differential subordination holds as ξ‚€πœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)ξ…žβ‰Ίβ„Ž(𝑧),(2.58) then πœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)π‘§β‰Ίπ‘ž(𝑧)=2π›Όβˆ’1+2(1βˆ’π›Ό)ln(1+𝑧)𝑧.(2.59) The function π‘ž is convex and is the best dominant.

Proof. Let πœ‡π‘(𝑧)=πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)𝑧=βˆ‘π‘§+βˆžπ‘˜=2ξ‚€ξ€·1+πœ†1ξ€Έ(π‘˜βˆ’1)π‘š/ξ€·1+πœ†2ξ€Έ(π‘˜βˆ’1)π‘šβˆ’1𝑐(𝑛,π‘˜)π‘Žπ‘˜π‘§π‘˜π‘§=1+𝑝1𝑧+𝑝2𝑧2[]+β‹―,(π‘βˆˆβ„‹1,1,π‘§βˆˆπ‘ˆ).(2.60) Differentiating (2.60), with respect to 𝑧, we obtain ξ‚€πœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)ξ…ž=𝑝(𝑧)+π‘§π‘ξ…ž(𝑧),(π‘§βˆˆπ‘ˆ).(2.61) Using (2.61), the differential subordination (2.58) becomes 𝑝(𝑧)+π‘§π‘ξ…ž(𝑧)β‰Ίβ„Ž(𝑧)=1+(2π›Όβˆ’1)𝑧1+𝑧,(π‘§βˆˆπ‘ˆ).(2.62) From Lemma 1.1, we deduce that 1𝑝(𝑧)β‰Ίπ‘ž(𝑧)=π‘§ξ€œπ‘§0=1β„Ž(𝑑)π‘‘π‘‘π‘§ξ€œπ‘§0ξ‚΅1+(2π›Όβˆ’1)𝑑=11+π‘‘π‘‘π‘‘π‘§ξ‚Έξ€œπ‘§01ξ€œ1+𝑑𝑑𝑑+(2π›Όβˆ’1)𝑧0𝑑1+𝑑𝑑𝑑=2π›Όβˆ’1+2(1βˆ’π›Ό)ln(1+𝑧)𝑧.(2.63) Using (2.60), we have πœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)π‘§β‰Ίπ‘ž(𝑧)=2π›Όβˆ’1+2(1βˆ’π›Ό)ln(1+𝑧)𝑧.(2.64) The proof is complete.

From Theorem 2.12, we deduce the following corollary.

Corollary 2.13. If π‘“βˆˆπ‘…πœ†π‘›,π‘š1,πœ†2(𝛼), then ξƒ©πœ‡Reπœ†π‘›,π‘š1,πœ†2𝑓(𝑧)𝑧ξƒͺ>(2π›Όβˆ’1)+2(1βˆ’π›Ό)ln2,(π‘§βˆˆπ‘ˆ).(2.65)

Proof. Since π‘“βˆˆπ‘…πœ†π‘›,π‘š1,πœ†2(𝛼), from Definition 2.3 we have ξ‚€πœ‡Reπœ†π‘›,π‘š1,πœ†2𝑓(𝑧)ξ…ž>𝛼,(π‘§βˆˆπ‘ˆ),(2.66) which is equivalent to ξ‚€πœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)ξ…žβ‰Ίβ„Ž(𝑧)=1+(2π›Όβˆ’1)𝑧.1+𝑧(2.67) Using Theorem 2.12, we have πœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)π‘§β‰Ίπ‘ž(𝑧)=(2π›Όβˆ’1)+2(1βˆ’π›Ό)ln(1+𝑧)𝑧.(2.68) Since π‘ž is convex and π‘ž(π‘ˆ) is symmetric with respect to the real axis, we deduce that ξƒ©πœ‡Reπœ†π‘›,π‘š1,πœ†2𝑓(𝑧)𝑧ξƒͺ>Reπ‘ž(1)=(2π›Όβˆ’1)+2(1βˆ’π›Ό)ln2,(π‘§βˆˆπ‘ˆ).(2.69)

Theorem 2.14. Let β„Žβˆˆβ„‹(π‘ˆ), with β„Ž(0)=1, β„Žβ€²(0)β‰ 0 which satisfy the inequality ξ‚΅Re1+π‘§β„Žξ…žξ…žξ…ž(𝑧)β„Žξ…ž(ξ‚Ά1𝑧)>βˆ’2,(π‘§βˆˆπ‘ˆ).(2.70) If 𝑛,π‘šβˆˆβ„•0,πœ†2β‰₯πœ†1β‰₯0,and π‘“βˆˆπ’œ and satisfies the differential subordination ξ‚€πœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)ξ…žβ‰Ίβ„Ž(𝑧),(π‘§βˆˆπ‘ˆ),(2.71) then πœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)𝑧1β‰Ίπ‘ž(𝑧)=π‘§ξ€œπ‘§0β„Ž(𝑑)𝑑𝑑.(2.72)

Proof. Let πœ‡π‘(𝑧)=πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)𝑧=βˆ‘π‘§+βˆžπ‘˜=2ξ‚€ξ€·1+πœ†1ξ€Έ(π‘˜βˆ’1)π‘š/ξ€·1+πœ†2ξ€Έ(π‘˜βˆ’1)π‘šβˆ’1𝑐(𝑛,π‘˜)π‘Žπ‘˜π‘§π‘˜π‘§=1+𝑝1𝑧+𝑝2𝑧2[]+β‹―,(π‘βˆˆβ„‹1,1,π‘§βˆˆπ‘ˆ).(2.73) Differentiating (2.73), with respect to 𝑧, we have ξ‚€πœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)ξ…ž=𝑝(𝑧)+π‘§π‘ξ…ž(𝑧),(π‘§βˆˆπ‘ˆ).(2.74) Using (2.74), the differential subordination (2.71) becomes 𝑝(𝑧)+π‘§π‘ξ…ž(𝑧)β‰Ίβ„Ž(𝑧),(π‘§βˆˆπ‘ˆ).(2.75) From Lemma 1.1, we deduce that 1𝑝(𝑧)β‰Ίπ‘ž(𝑧)=π‘§ξ€œπ‘§0β„Ž(𝑑)𝑑𝑑,(2.76) and using (2.73), we obtain πœ‡πœ†π‘›,π‘š1,πœ†2𝑓(𝑧)𝑧1β‰Ίπ‘ž(𝑧)=π‘§ξ€œπ‘§0β„Ž(𝑑)𝑑𝑑.(2.77) From Lemma 1.3, we have that the function π‘ž is convex, and from Lemma 1.1, π‘ž is the best dominant for subordination (2.71). This completes the proof of Theorem 2.14.

3. Conclusion

We remark that several subclasses of analytic univalent functions can be derived and studied using the operator πœ‡πœ†π‘›,π‘š1,πœ†2.


This work is fully supported by UKM-ST-06-FRGS0107-2009, MOHE, Malaysia.


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Copyright © 2010 Ma'moun Harayzeh Al-Abbadi and Maslina Darus. 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.

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