# Majorization for Certain Classes of Analytic Functions Defined by Fournier–Ruscheweyh Integral Operator

Authors: Murli Manohar Gour; Pranay Goswami; Basem Aref Frasin; Saad Althobaiti
Journal: Journal of Mathematics
 (2022)

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## Abstract

In this paper, we introduce three new classesSkaM,N;μ,Rkaμ, and Tkaθ of analytic functions defined by Fournier–Ruscheweyh integral operator. For these classes, we investigate the majorization problem. Furthermore, a number of new results are shown to follow upon specializing the parameters involved in our main results.


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## Body

## 1. Introduction and Definitions

For the two functionsu and v which are analytic in the open unit disk D=z∈ℂ:z<1, we can define the majorization for these two functions as follows (see [1]):(1)uz≪vzz∈D.If there exists a functionψz that is analytic in D, then(2)ψz≤1anduz=ψzvzz∈D.For the two functionsu and v, we say that the function u is subordinate to the function v defined as uz≺vz, if there is a Schwarz function w, that is analytic in D with w0=0 and |wz|<1, z∈D, such that uz=vwz, z∈D.Now, on combining subordination and majorization, we define quasi-subordination as follows. For two functionsu and v, we say that u is quasi-subordinate to v (see [2]) and it is defined as(3)uz≺qvzz∈D,If there are two functionsψz and wz that are analytic in D, then(uz/ψz) is analytic in D and(4)|ψz≤1andw0=0,wz|≤z≤1z∈D,satisfying(5)uz=ψzvwzz∈D.Remark 1.
(i)
If we putψz=1 in (5), we have the usual definition of subordination(ii)
If we putwz=z in (6), we have the usual definition of majorization
LetA denote the class of all functions of the form(6)fz=z+∑m=0∞amzm,z∈D,which are analytic in open unit disk D.
The function classΦ has been introduced and studied by Li and Srivastava [3] and is defined as(7)Φ=kt:kt≥0,0≤t≤1,∫01ktdt=1.
Fournier and Ruscheweyh [3, 4] considered an integral operator with a nonnegative function:(8)ka:0,1⟶ℝsuch that∫01katdt=1.
By substituting suitable values of parametera, there are lots of special cases of function kat. We therefore consider the Fournier–Ruscheweyh integral operator to be in the following modified form [3] (see [5]):(9)ℐkafz=∫01katftztdt,f∈A.where the real-valued functions ka and ka−1 fulfill the requirements:(1)
For an acceptable parametera,(10)ka−1t∈Φ,kat∈Φandka1=0.(2)
There exists a constantλ−1<λ≤2 such that(11)λkat−tka′t=λ+1ka−1t,wheret∈0,1 and −1<λ≤2.
Forℐka operator, we have(12)z(ℐkauz)′=−λℐkauz+λ+1ℐka−1uz.Remark 2.
(i)
If we take(13)kat=2aμatlog1ta−1=k1a>0,in (9), we get the integral operator Pa as(14)Pat=2azμa∫0zlogzta−1ftdtf∈A,a>0.The integral operatorPa is exactly the same as the transformation Ik given by Flett [6] and studied subsequently by Li [7], Li and Srivastava [8], and many others. In the case when a>1, then we have λ=1.(ii)
If we take(15)kat=a+baa1−ta−1tb=k2,a>0,b>−1,in (9), we get the Jung–Kim–Srivastava integral operator Qba [9] (see [10–12]) as(16)Qbafz=a+baazb∫0z1−tza−1tb−1ftdtf∈A,a>0,b>−1,where(17)ab=μa+1μb+1μa−b+1=ab,a,b∈ℂ.
In terms of known Gamma functions, the integral operatorQbauz is analogous to the convolution operator La,d by Carlson and Shaffer [13]. In the case when a>1,b>−1, and 0<a+b≤3, we have λ=a+b−1.
Now, we describe the following classes of analytical functions using integral operator (9).Definition 1.
The functionf∈A is said to be in the class SkaM,N;μ if and only if(18)1+1μzℐkafz′ℐkafz−1≺1+Mz1+Nz,with −1≤N<M≤1,k∈Φ,μ∈ℂ∗=ℂ∖0.
If we take the value ofk as defined in (13) and (15), this class becomes Sk1aM,N;μ and Sk2aM,N;μ, respectively.Definition 2.
The functionf∈A is said to be in the class Rkaμ if and only if(19)z(ℐkafz)′ℐkafz−μz(ℐkafz)′ℐkafz−1≺ez,z∈D,where a≥0,k∈Φ, and μ≥0.
If we take the value ofk as defined in (13) and (15), this class becomes Rk1aμ and Rk2aμ, respectively.Definition 3.
The functionf∈A is said to be in the class Tkaθ if and only if(20)eiθzℐkafz′ℐkafz≺ezcosθ+isinθ,z∈D,where a≥0,k∈Φ, and −Π/2<θ<Π/2.
If we take the value ofk as defined in (13) and (15), this class becomes Tk1aθ and Tk2aθ, respectively.
A majorization problem for the normalized class of starlike functions has been investigated by MacGregor [1] and further studied by Altintas et al. [14]. Recently, a number of researchers have studied several majorization problems for univalent and multivalent functions or meromorphic and multivalent meromorphic functions involving different operators and different classes [14–20, 22–24]. By motivating the above work, the majorization problems of the classes SkaM,N,μ,Rakμ, and Takθ are investigated as follows.


## 2. Problem of Majorization for the ClassesSkaM,N,μ, Rkaμ, and Tkaθ

Theorem 1.
Let the functionf∈A, and assume that g∈SkaM,N,μ. If ℐkafz is majorized by ℐkagz in D, then(21)ℐka−1fz≤ℐka−1gzforz≤ρ0,where ρ0 is the smallest positive root of the equation(22)μM−N+1+λNρ3−2N+λ+1ρ2−2+μM−N+λ+1Nρ+λ+1=0,where −1≤N<M≤1,k∈Φ,μ∈ℂ∗,−1<λ≤2, and λ+1≥μM−N+λ+1N.Proof.
Sinceg∈SkaM,N,μ, then, from (18),(23)1+1μzℐkagz′ℐkagz−1=1+Mwz1+Nwz,where w is the analytic function in D, with w0=0 and |wz|≤z<1∀z∈D.
Now, from the previous equality,(24)z(ℐkagz)′ℐkagz=1+μM−N+Nwz1+Nwz.
Now, we make use of relation (12), that is,(25)z(ℐkagz)′=−λℐkagz+λ+1ℐka−1gz,
For−1<λ≤2, then, from (24), we have(26)ℐka−1gzℐkagz=λ+1+μM−N+λ+1Nwzλ+11+Nwz,which implies that(27)|ℐkagz|≤λ+11+Nzℐka−1gzλ+1−μM−N+λ+1N‖z.
Now, sinceℐkafz is majorized by ℐkagz in open unit disk D, then(28)ℐkafz=ψzℐkagz.
Differentiating the previous equality with respect toz and then multiplying by z, we get(29)z(ℐkafz)′=zψz(ℐkagz)′+zψ′zℐkagz.
On using relation (12), we have(30)λ+1ℐka−1fz=zψ′zℐkagz+λ+1ψzℐka−1gz.
This implies(31)λ+1ℐka−1fz≤zψ′zℐkagz+λ+1ψzℐka−1gz.
Note, therefore, that the Schwarz functionψ satisfies the inequality (see [21])(32)ψ′z≤1−ψz21−z2;z∈D.
On using (27) and (32) in (31), we have(33)ℐka−1fz≤z1−ψz21+Nz1−z2λ+1−μM−N+λ+1Nz+ψzℐka−1gz.
Settingz=ρandψz=c, then inequality (33) leads to(34)ℐka−1fz≤ζρ,cℐka−1gz|1−ρ2λ+1−μM−N+λ+1Nρ,where(35)ζρ,c=ρ1−c21+Nρ+c1−ρ2λ+1−μM−N+λ+1Nρ.
Then, from (34),(36)ℐka−1fz≤ηρ,cℐka−1gz,where(37)ηρ,c=ζρ,c1−ρ2λ+1−μM−N+λ+1Nρ.
From relation (36), in order to prove our result, we need to determine(38)ρ0=maxρ∈[0,1);η(ρ,c)≤1∀c∈[0,1]=maxρ∈[0,1);G(ρ,c)≥0∀c∈[0,1],where(39)Gρ,c=1−ρ21−cλ+1−μM−N+λ+1Nρ−ρ1−c21+Nρ.
A simple calculation shows that theGρ,c≥0 inequality is equivalent to(40)uρ,c=λ+1−μM−N+λ+1Nρ1−ρ2−ρ1+c1+Nρ≥0.
However, the functionuρ,c has a minimum value at c=1, that is,(41)minuρ,c:c∈0,1=uρ,1=vρ,where(42)uρ=μM−N+λ+1Nρ3−2N+λ+1ρ2−2+μM−N+λ+1Nρ+λ+1=0.
It follows thatvρ≥0∀ρ∈0,ρ0, where ρ0=ρ0μ,λ,M,N is the smallest positive root of equation (22), which proves conclusion (21).Theorem 2.
Let the functionf∈A, and assume that g∈Rkaμ. If ℐkafz is majorized by ℐkagz in D, then(43)ℐka−1fz|≤|ℐka−1gzforz≤ρ1,where ρ1 is the smallest positive root of the equation(44)eρ+μλ+1−λρ2−21+μρ+λ−μλ+1−eρ=0,where a≥0,k∈Φ,μ≥0,−1<λ≤2, and λ>μλ+1+e.Proof.
Sinceg∈Rkaμ, then, from (19) and the subordination relation,(45)z(ℐkagz)′ℐkagz−μz(ℐkagz)′ℐkagz−1=ewzz∈D,where w is the analytic function in D, with w0=0 and wz≤z≤1,∀z∈D. Now, let(46)W=z(ℐkagz)′ℐkagz.
In (45), we have(47)W−μW−1=ewz,which implies that(48)W−μW−1eiϕ=ewz.
Then, we have(49)W=ewz−μeiϕ1−μeiϕ.
From (46 and 49), we have(50)z(ℐkagz)′ℐkagz=ewz−μeiϕ1−μeiϕ.
Now, on using (12) in (50), for −1<λ≤2, we have the following:(51)ℐkagzℐka−1gz=ewz+λ−λ+1μeiϕλ+11−μeiϕ,which implies that(52)ℐkagz≤λ+11+μλ−μλ+1−ezℐka−1gz.
Now, sinceℐkafz is majorized by ℐkagz in D, then we have(53)ℐkafz=ψzℐkagz.
Differentiating the previous equality with respect toz and then multiplying by z, we get(54)z(ℐkafz)′=zψz(ℐkagz)′+zψ′zℐkagz.
On using relation (12), we have(55)λ+1ℐka−1fz=zψ′zℐkagz+λ+1ψzℐka−1gz.
This implies(56)λ+1ℐka−1fz≤zψ′zℐkagz+λ+1ψzℐka−1gz.
Thus, note that the Schwarz functionψ satisfies the inequality (see [21])(57)ψ′z≤1−ψz21−z2;z∈D.
On using (52) and (57) in (56), we have(58)|ℐka−1fz≤z1+μ1−ψz21−z2λ−μλ+1−ez+ψzℐka−1gz|.
Settingz=ρandψz=c0≤c≤1, then inequality (58) leads to(59)ℐka−1fz≤ζ1ρ,c1−ρ2λ−μλ+1−eρℐka−1gz,where(60)ζ1ρ,c=ρ1+μ1−c2+c1−ρ2λ−μλ+1−eρ.
Then, from (59),(61)ℐka−1fz≤η1ρ,cℐka−1gz.Here,(62)η1ρ,c=ζ1ρ,c1−ρ2λ−μλ+1−eρ.
From relation (61), in order to prove our result, we need to determine(63)ρ1=maxρ∈0,1;η1ρ,c≤1∀c∈0,1,=maxρ∈0,1;G1ρ,c≥0∀c∈0,1,where(64)G1ρ,c=1−ρ21−cλ−μλ+1−eρ−ρ1+μ1−c2.
A simple calculus shows that the inequalityG1ρ,c≥0 is equivalent to(65)u1ρ,c=1−ρ2λ−μλ+1−eρ−ρ1+μ1+c≥0.
However, the functionu1ρ,c takes its minimum value at c=1, that is,(66)minu1ρ,c:c∈0,1=u1ρ,1=v1ρ,where(67)v1ρ=1−ρ2λ−μλ+1−eρ−2ρ1+μ=0.
It follows thatv1ρ≥0∀ρ∈0,ρ1, where ρ1=ρ1μ,λ is the smallest positive root of (44), which prove conclusion (43).Theorem 3.
Let the functionf∈A, and assume that g∈Tkaθ. If ℐkafz is majorized by ℐkagz in D, then(68)ℐka−1fz≤ℐka−1gzforz≤ρ∗,where ρ∗ is the smallest positive root of the equation(69)eρ+λ+1tanθ−λρ2−2secθρ+λ−λ+1tanθ−eρ=0,where a≥0,k∈Φ,−Π/2<θ<Π/2,−1<λ≤2, and λ>λ+1tanθ+e.Proof.
Sinceg∈Tkaθ, then, from (20) and the subordination relation,(70)eiθzℐkagz′ℐkagz=ewzcosθ+isinθ,where w is the analytic function in D, with w0=0 and |wz|≤z≤1,∀z∈D. From (70), we have(71)z(ℐkagz)′ℐkagz=ewz+itanθ1+itanθ.
Now, on using (12) in (71), for −1<λ≤2, we have the following:(72)ℐka−1gzℐkagz=ewz+λ+λ+1itanθλ+11+itanθ,which implies that(73)ℐkagz≤λ+1secθλ−λ+1tanθ−ezℐka−1gz.
Now, sinceℐkafz is majorized by ℐkagz in D, then we have(74)ℐkafz=ψzℐkagz.
Differentiating the previous equality with respect toz and then multiplying by z, we get(75)z(ℐkafz)′=zψz(ℐkagz)′+zψ′zℐkagz.
On using relation (12), we have(76)λ+1ℐka−1fz=zψ′zℐkagz+λ+1ψzℐka−1gz.
This implies(77)λ+1ℐka−1fz|≤|zψ′zℐkagz+λ+1ψzℐka−1gz.
Thus, note that the Schwarz functionϕ satisfies the inequality (see [21])(78)ψ′z≤1−ψz21−z2;z∈D.
On using (71) and (78) in (77), we have(79)ℐka−1fz≤z1−ψz2secθ1−z2λ−λ+1tanθ−ez+ψzℐka−1gz.
Settingz=ρandψz=c0≤c≤1, then inequality (79) leads to(80)ℐka−1fz≤ζ2ρ,c1−ρ2λ−λ+1tanθ−eρℐka−1gz,where(81)ζ2ρ,c=ρ1−c2secθ+c1−ρ2λ−λ+1tanθ−eρ.
Then, from (80),(82)ℐka−1fz≤η2ρ,cℐka−1gz,where(83)η2ρ,c=ζ2ρ,c1−ρ2λ−λ+1tanθ−eρ.
From relation (82), in order to prove our result, we need to determine(84)ρ∗=maxρ∈0,1;η2ρ,c≤1∀c∈0,1=maxρ∈0,1;G2ρ,c≥0∀c∈0,1,]where(85)G2ρ,c=1−ρ21−cλ−λ+1tanθ−eρ−ρ1−c2secθ.
A simple calculus shows that the inequalityG2r,ρ≥0 is equivalent to(86)u2ρ,c=1−ρ2λ−λ+1tanθ−eρ−ρ1+csecθ≥0.
However, the functionu2ρ,c takes its minimum value at c=1, that is,(87)minu2ρ,c:c∈0,1=u2ρ,1=v2ρ,where(88)v2ρ=1−ρ2λ−λ+1tanθ−eρ−2ρsecθ=0.
It follows thatv2ρ≥0∀ρ∈0,ρ∗, where ρ∗=ρ∗θ,λ is the smallest positive root of (69), which proves conclusion (68).


## 3. Corollaries and Consequences

If we take the values ofk defined in (13) and (15), then the above theorems give the following corollaries.Corollary 1.
Let the functionf∈A, and assume that g∈Sk1aM,N,μ. If Pafz is majorized by Pagz in D, then(89)Pa−1fz≤Pa−1gzforz≤ρ2,where ρ2 is the smallest positive root of the equation(90)μM−N+2Nρ3−2N+1ρ2−2+μM−N+2Nρ+2=0,where −1≤N<M≤1,μ∈ℂ∗, and 2≥|μM−N+2N|.Corollary 2.
Let the functionf∈A, and assume that g∈Rk1aμ. If Pafz is majorized by Pagz in D, then(91)Pa−1fz≤Pa−1gzforz≤ρ3,where ρ3 is the smallest positive root of the equation(92)eρ+2μ−1ρ2−21+μρ+1−2μ−eρ=0,where a≥0,μ≥0, and 1>2μ+e.Corollary 3.
Let the functionf∈A, and assume that g∈Tk1aθ. If Pafz is majorized by Pagz in D, then(93)Pa−1fz|≤|Pa−1gzforz≤ρ1∗,where ρ1∗ is the smallest positive root of the equation(94)eρ+2tanθ−1ρ2−2secθρ+1−2tanθ−eρ=0,where a≥0,−π/2<θ<π/2, and 1>2|tanθ|+e.Corollary 4.
Let the functionf∈A, and assume that g∈Sk2aM,N,μ. If Qbafz is majorized by Qbagz in D, then(95)Qba−1fz≤Qba−1gzforz≤ρ4,where ρ4 is the smallest positive root of the equation(96)μM−N+a+bNρ3−2N+a+bρ2−2+μM−N+a+bNρ+a+b=0,where −1≤N<M≤1,μ∈ℂ∗,a>1,b>−1, and a+b≥μM−N+a+bN.Corollary 5.
Let the functionf∈A, and assume that g∈Rk2aμ. If Qbafz is majorized by Qbagz in D, then(97)Qba−1fz|≤|Qba−1gzforz≤ρ5,where ρ5 is the smallest positive root of the equation(98)eρ+μa+b−a+b−1ρ2−21+μρ+a+b−1−μa+b−eρ=0,where a≥0,b>−1,μ≥0, and a+b−1>μa+b+e.Corollary 6.
Let the functionf∈A, and assume that g∈Tk2aθ. If Qbafz is majorized by Qbagz in D, then(99)Qba−1fz≤Qba−1gzforz≤ρ2∗,where ρ2∗ is the smallest positive root of the equation(100)eρ+a+btanθ−a+b−1ρ2−2secθρ+a+b−1−a+btanθ−eρ=0,where a≥0,b>−1,−π/2<θ<π/2, and a+b−1>a+btanθ+e.If we takeM=1 and N=−1, then Theorem 1, Corollary 1, and Corollary 4 give the following results.Corollary 7.
Let the functionf∈A, and assume that g∈Ska1,−1,μ. If ℐkafz is majorized by ℐkagz in D, then(101)ℐka−1fz≤ℐka−1gzforz≤ρ0,where ρ0 is the smallest positive root of the equation(102)2μ−1+λρ3−λ+3ρ2−2+2μ−λ+1ρ+λ+1=0,where k∈Φ,μ∈ℂ∗,−1<λ≤2, and λ+1≥2μ−λ+1.Corollary 8.
Let the functionf∈A, and assume that g∈Sk1a1,−1,μ. If Pafz is majorized by Pagz in D, then(103)Pa−1fz≤Pa−1gzforz≤ρ2,where ρ2 is the smallest positive root of the equation(104)μ−1ρ3−2ρ2−1+μ−1ρ+1=0,where μ∈ℂ∗ and 1≥μ−1.Corollary 9.
Let the functionf∈A, and assume that g∈Sk2a1,−1,μ. If Qbafz is majorized by Qbagz in D, then(105)Qba−1fz≤Qba−1gzforz≤ρ4,where ρ4 is the smallest positive root of the equation(106)2μ−a+bρ3−2+a+bρ2−2+2μ−a+bρ+a+b=0,where μ∈ℂ∗,a>1,b>−1, and a+b≥2μ−a+b.



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