# On Some Relationships of CertainK−Uniformly Analytic Functions Associated with Mittag-Leffler Function

Authors: Adel A. Attiya; Ekram E. Ali; Taher S. Hassan; Abeer M. Albalahi
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we introduce and investigate several inclusion relationships of newk-uniformly classes of analytic functions defined by the Mittag-Leffler function. Also, integral-preserving properties of these classes associated with the certain integral operator are also obtained.


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## Body

## 1. Introduction

LetA be the class of analytic functions in the open unit disc U=z:z<1 which in the form
(1)fz=z+∑n=2∞anzn.Forfz and gz∈A, we say that the function fz is subordinate to gz, written symbolically as follows:
(2)f≺gorfz≺gz,if there exists a Schwarz functionwz, which (by definition) is analytic in U with w0=0 and wz<1, z∈U, such that fz=gwz for all z∈U. In particular, if the function gz is univalent in U, then we have the following equivalence relation (cf., e.g., [1, 2]; see also [3]):
(3)fz≺gz⇔f0≺g0andfU⊂gU.Letf be as in (1) and hz=z+∑n=2∞bnzn, then Hadamard product (or convolution) of fz and hz is given by
(4)f∗hz=z+∑k=2∞akbkzkz∈U.Forζ,η∈0,1, we denote by S∗ζ,Cζ,Kζ,η, and K∗ζ,η the subclasses of A consisting of all analytic functions which are, respectively, starlike of order ζ, convex of order ζ, close-to-convex of order ζ and type η, and quasiconvex of order ζ and type η in U.Also, let the subclassesUSμ,ζ, UCμ,ζ, USKμ,ζ,η, and UCKμ,ζ,η of Aη∈0,1<1;μ≥0 be defined as follows:
(5)USμ,ζ=f∈A:Rzf′zfz−ζ>μzf′zfz−1,UCμ,ζ=f∈A:R1+zf″zf′z−ζ>μzf″zf′z,USKμ,ζ,η=f∈A:∃h∈USμ,ζs.t.Rzf′zhz−ζ>μzf′zhz−1UCKμ,ζ,η=f∈A:∃h∈UCμ,ζs.t.Rzf′z′h′z−ζ>μzf′z′h′z−1.We note that(6)US0,ζ=S∗ζ,UC0,ζ=Cζ,USK0,ζ,η=Kζ,ηandUCK0,ζ,η=K∗ζ,η0≤ζ;η<1.Moreover, letqμ,ζz be an analytic function which maps U onto the conic domain Φμ,ζ=u+iv:u>ku−12+v2+ζ such that 1∈Φμ,ζ defined as follows:
(7)qμ,ζz=1+1−2ζz1−zμ=0,1−ζ1−μ2cos2πcos−1μilog1+z1−z−μ2−ζ1−μ20<μ<1,1+21−ζπ2log1+z1−z2μ=1,1−ζμ2−1sinπ2ςμ∫0uzμdt1−t21−μ2t2+μ2−ζμ2−1μ>1,where uz=z−μ/1−μz and ςμ is such that μ=coshπς′z/4ςz. By virtue of properties of the conic domain Φμ,ζ (cf., e.g., [4, 5]), we have
(8)Rqμ,ζz>μ+ζμ+1.Making use of the principal of subordination and the definition ofqμ,ζz, we may rewrite the subclasses USμ,ζ, UCμ,ζ, USKμ,ζ,η, and UCKμ,ζ,η as follows:
(9)USμ,ζ=f∈A:zf′zfz≺qμ,ζz,UCμ,ζ=f∈A:1+zf′′zf′z≺qμ,ζz,USKμ,ζ,η=f∈A:∃h∈USμ,ηs.t.zf′zhz≺qμ,ζzand(10)UCKμ,ζ,η=f∈A:∃h∈UCμ,ζs.t.zf′z′h′z≺qμ,ζz.Attiya [6] introduced the operator Hα,βγ,kf, where Hα,βγ,kf:A⟶A is defined by
(11)Hα,βγ,kf=μα,βγ,k∗fzz∈U,with β,γ∈ℂ,Reα>max0,Rek−1 and Rek>0. Also, Reα=0 when Rek=1; β≠0. Here, μα,βγ,k is the generalized Mittag–Leffler function defined by [7], see also [6], and the symbol (∗) denotes the Hadamard product.Due to the importance of the Mittag–Leffler function, it is involved in many problems in natural and applied science. A detailed investigation of the Mittag–Leffler function has been studied by many authors (see, e.g., [7–12]).Attiya [6] noted that
(12)Hα,βγ,kfz=z+∑n=2∞Γγ+nkΓα+βΓγ+kΓβ+αnn!anzn.Also, Attiya [6] showed that
(13)zHα,βγ,kfz′=γ+kkHα,βγ+1,kfz−γkHα,βγ,kfz,and(14)zHα,β+1γ,kfz′=α+βαHα,βγ,kfz−βαHα,β+1γ,kfz.Next, by using the operatorHα,βγ,kf, we introduce the following subclasses of analytic functions in U(15)USβγμ,ζ=f∈A:Hα,βγ,kfz∈USμ,ζ,UCβγμ,ζ=f∈A:Hα,βγ,kfz∈UCμ,ζ,USKβγμ,ζ,η=f∈A:Hα,βγ,kfz∈USKμ,ζ,η,UCKβγμ,ζ,η=f∈A:Hα,βγ,kfz∈UCKμ,ζ,η,where β,γ∈ℂ,Rα>max0,Rk−1and Rk>0. Also, Rα=0 when Rk=1; β≠0.Also, we note that(16)fz∈UCβγμ,ζ⇔zf′z∈USβγμ,ζ,(17)fz∈UCKβγμ,ζ,η⇔zf′z∈USKβγμ,ζ,η.In this paper, we introduce several inclusion properties of the classesUSβγμ,ζ, UCβγμ,ζ, USKβγμ,ζ,η, and UCKβγμ,ζ,η. Also, integral-preserving properties of these classes associated with generalized Libera integral operator are also obtained.


## 2. Inclusion Properties Associated withHα,βγ,kfz

Lemma 1 (see [13]).
Ifhz is convex univalent in U with h0=1 and Rξhz+ζ>0ζ∈ℂ. Let pz be analytic in U with p0=1 which satisfy the following subordination relation
(18)pz+zp′zξpz+ζ≺hz,then
(19)pz≺hz.Lemma 2 (see [2]).
Ifhz is convex univalent in U and let w be analytic in U with Rwz≥0. Let pz be analytic in U and p0=h0 which satisfy the following subordination relation
(20)pz+wzzp′z≺hz,then
(21)pz≺hz.Theorem 3.
IfRγ/k>−μ+ζ/μ+1, then USβγ+1μ,ζ⊂USβγμ,ζ.Proof.
Letfz∈USβγ+1μ,ζ, put
(22)pz=zHα,βγ,kfz′Hα,βγ,kfzz∈U,we note that pz is analytic in U and p0=1. From (13) and (22), we have
(23)Hα,βγ+1,kfzHα,βγ,kfz=kγ+kpz+γk.
Differentiating (23) with respect to z, we obtain
(24)zHα,βγ+1,kfz′Hα,βγ+1,kfz=pz+zp′zpz+γ/k.
From the above relation and using (7), we may write
(25)pz+zp′zpz+γ/k≺qμ,ζzz∈U.
SinceRqμ,ζz>μ+ζ/μ+1, we see that
(26)Rqμ,ζz+γk>0z∈U.Applying Lemma1, it follows that pz≺qμ,ζz, that is, fz∈USβγμ,ζ.Using the same technique in Theorem3 with relation (14), we have the following theorem.Theorem 4.
IfRα/β>−μ+ζ/μ+1, then USβγμ,ζ⊂USβ+1γμ,ζ.Theorem 5.
IfRγ/k>−μ+ζ/μ+1, then UCβγ+1μ,ζ⊂UCβγμ,ζ.Proof.
Applying Theorem3 and relation (16), we observe that
(27)fz∈UCβγ+1μ,ζ⇔zf′z∈USβγ+1μ,ζ⇒zf′z∈USβγμ,ζ⇔fz∈UCβγμ,ζ,which evidently proves Theorem 5.Similarly, we can prove the following theorem.Theorem 6.
IfRα/β>−μ+ζ/μ+1, then UCβγμ,ζ⊂UCβ+1γμ,ζ.Theorem 7.
IfRγ/k>−μ+ζ/μ+1, then USKβγ+1μ,ζ,η⊂USKβγμ,ζ,η.Proof.
Letfz∈USKβγ+1μ,ζ,η. Then, there exists a function rz∈USμ,ζ such that
(28)zHα,βγ+1,kfz′rz≺qμ,ζz.
We can choose the functionhz such that Hα,βγ+1,khz=rz. Then, hz∈USβγ+1μ,ζ and
(29)zHα,βγ+1,kfz′Hα,βγ+1,khz≺qμ,ζz.
Now, let(30)pz=zHα,βγ,kfz′Hα,βγ,khz,where pz is analytic in U with p0=1. Since hz∈USβγ+1μ,ζ, by Theorem 3, we know that hz∈USβγμ,ζ. Let
(31)tz=zHα,βγ,khz′Hα,βγ,khzz∈U,where tz is analytic in U with Rtz>μ+ζ/μ+1. Also, from(30), we note that
(32)zHα,βγ,kfz′=Hα,βγ,kzf′z=Hα,βγ,khzpz.
Differentiating both sides of (32) with respect to z, we obtain
(33)zHα,βγ,kzf′z′Hα,βγ,khz=zHα,βγ,khz′Hα,βγ,khzpz+zp′z=tzpz+zp′z.
Now, using (13) and (33), we obtain
(34)zHα,βγ,kfz′Hα,βγ,khz=Hα,βγ+1,kzf′zHα,βγ+1,khz=zHα,βγ,kzf′z′+γ/kHα,βγ,kzf′zzHα,βγ,khz′+γ/kHα,βγ,khz=zHα,βγ,kzf′z′/Hα,βγ,khz+γ/kzHα,βγ,kfz′/Hα,βγ,khzzHα,βγ,khz′/Hα,βγ,khz+γ/k=tzpz+zp′z+γ/kpztz+γ/k=pz+zp′ztz+γ/k.
SinceRγ/k>−μ+ζ/μ+1, we see that
(34)Rtz+γk>0z∈U.Hence, applying Lemma2, we can show that pz≺qμ,ζz, so that fz∈USKβγμ,ζ,η. This completes the proof of Theorem 7.Similarly, we can prove the following theorem.Theorem 8.
IfRα/β>−μ+ζ/μ+1, then USKβγμ,ζ,η⊂USKβ+1γμ,ζ,η.We can also prove Theorem9 by using Theorem 7 and relation (17).Theorem 9.
IfRγ/k>−μ+ζ/μ+1, then UCKβγ+1μ,ζ,η⊂UCKβγμ,ζ,η.Also, we obtain the following theorem.Theorem 10.
IfRα/β>−μ+ζ/μ+1, then UCKβγμ,ζ,η⊂UCKβ+1γμ,ζ,η.Now, we obtain squeeze theorems for inclusion by combining the above theorems as follows:Combining both theorems3 and 4, we have the following corollary.Corollary 11.
Ifμ+ζ/μ+1>−minRγ/k,Rα/β, then
(36)USβγ+1μ,ζ⊂USβγμ,ζ⊂USβ+1γμ,ζ.Combining both theorems5 and 6, we have the following corollary.Corollary 12.
Ifμ+ζ/μ+1>−minRγ/k,Rα/β, then
(37)UCβγ+1μ,ζ⊂UCβγμ,ζ⊂UCβ+1γμ,ζ.Combining both theorems7 and 8, we have the following corollary.Corollary 13.
Ifμ+ζ/μ+1>−minRγ/k,Rα/β, then
(38)USKβγ+1μ,ζ,η⊂USKβγμ,ζ,η⊂USKβ+1γμ,ζ,η.Combining both theorems9 and 10, we have the following corollary.Corollary 14.
Ifμ+ζ/μ+1>−minRγ/k,Rα/β, then
(39)UCKβγ+1μ,ζ,η⊂UCKβγμ,ζ,η⊂UCKβ+1γμ,ζ,η.


## 3. Integral Preserving Properties Associated withFδ

The generalized Libera integral operatorFδ (see [14–16], also, see related topics [17–19]) is defined by
(40)Fδfz=δ+1zδ∫0ztδ−1ftdt,where fz∈A and δ>−1.Theorem 15.
Letδ>−μ+ζ/μ+1. If f∈USβγμ,ζ, then Fδf∈USβγμ,ζ.Proof.
Letf∈USβγμ,ζ and set
(41)pz=zHα,βγ,kFδfz′Hα,βγ,kFδfzz∈U,where pz is analytic in U with p0=1. From definition of Hα,βγ,kf and (40), we have
(42)zHα,βγ,kFδfz′=δ+1Hα,βγ,kfz−δHα,βγ,kFδfz.
Then, by using (41) and (42), we obtain
(43)δ+1Hα,βγ,kfzHα,βγ,kFδfz=pz+δ.
Taking the logarithmic differentiation on both sides of (43) and simple calculations, we have
(44)pz+zp′zpz+δ=zHα,βγ,kfz′Hα,βγ,kfz≺qμ,ζz.
SinceRqμ,ζ+δ>μ+ζ/μ+1+δ>0, by virtue of Lemma 1, we conclude that pz≺qμ,ζz in U, which implies that Fδf∈USβγμ,ζ.Theorem 16.
Letδ>−μ+ζ/μ+1. If f∈UCβγμ,ζ, then Fδf∈UCβγμ,ζ.Proof.
By applying Theorem15, it follows that
(45)fz∈UCβγμ,ζ⇔zf′z∈USβγμ,ζ⇒Fδzf′z∈USβγμ,ζ⇔zFδfz′∈USβγμ,ζ⇔Fδfz∈UCβγμ,ζ,which proves Theorem 16.Theorem 17.
Letδ>−μ+ζ/μ+1. If f∈USKβγμ,ζ,η, then Fδf∈USKβγμ,ζ,η.Proof.
Letfz∈USKβγμ,ζ,η. Then, there exists a function hz∈USβγμ,ζ such that
(46)zHα,βγ,kfz′Hα,βγ,khz≺qμ,ζz.
Thus, we set(47)pz=zHα,βγ,kFδfz′Hα,βγ,kFδhzz∈U,where pz is analytic in U with p0=1. Since hz∈USβγμ,ζ, we see from Theorem 15 that Fδh∈USβγμ,ζ. Let
(48)tz=zHα,βγ,kFδhz′Hα,βγ,kFδhz,where tz is analytic in U with Rtz>μ+ζ/μ+1. Using (47), we have
(49)Hα,βγ,kzFδ′fz=Hα,βγ,kFδhzpz.
Differentiating both sides of (49) with respect to z and simple calculations, we obtain
(50)zHα,βγ,kzFδ′fz′Hα,βγ,kFδhz=zHα,βγ,kFδhz′Hα,βγ,kFδhzpz+zp′z=tzpz+zp′z.
Now, using the identity (42) and (50), we obtain
(51)zHα,βγ,kfz′Hα,βγ,khz=Hα,βγ,kzf′zHα,βγ,khz=zHα,βγ,kzFδ′fz′+δHα,βγ,kzFδ′fzzHα,βγ,kFδhz′+δHα,βγ,kFδhz=zHα,βγ,kzFδ′fz′/Hα,βγ,kFδhz+δzHα,βγ,kFδfz′/Hα,βγ,kFδhzzHα,βγ,kFδhz′/Hα,βγ,kFδhz+δ=tzpz+zp′z+δpztz+δ=pz+zp′ztz+δ.
Sinceδ>−μ+ζ/μ+1 and Rtz>μ+ζ/μ+1, we see that
(52)Rtz+δ>0z∈U.
Applying Lemma2 into relation (51), it follows that pz≺qμ,ζz, which is Fδf∈USKβγμ,ζ,η.We can deduce the integral-preserving property asserted by18 by using Theorem 17 and relation (17).Theorem 18.
Letδ>−μ+ζ/μ+1. If f∈UCKβγμ,ζ,η, then Fδf∈UCKβγμ,ζ,η.



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