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Kyungpook Mathematical Journal 2023; 63(2): 225-234

Published online June 30, 2023 https://doi.org/10.5666/KMJ.2023.63.2.225

Copyright © Kyungpook Mathematical Journal.

Subordination Properties for Classes of Analytic Univalent Involving Linear Operator

Amal Madhi Rashid, Abdul Rahman S. Juma, Sibel Yalçın∗

Department of Mathematics, College of Education for Pure Sciences, University of Anbar, Ramadi, Iraq
e-mail : ama19u2001@uoanbar.edu.iq and eps.abdulrahman.juma@uoanbar.edu.iq

Department of Mathematics, Faculty of Arts and Sciences, Bursa Uludag University, 16059, Görükle, Bursa, Turkey
e-mail : syalcin@uludag.edu.tr

Received: July 25, 2022; Revised: March 23, 2023; Accepted: March 29, 2023

In this paper, we use the use the linear operator τ,σx(u,v,y)f(z) and the concept of the subordination to analyse the general class of all analytic univalent functions. Our main results are implication properties between the classes of such functions and the application of these properties to special cases.

Keywords: Analytic function, Univalent function, Differential subordination, Starlike function

Let ¥ be the class of all analytic univalent functions having power series extensions of the form

f(z)=z+ m=2emzm

in the open unit disk U={z: |z|<1} when normalized with the condition f(0)=f(z)1=0.

The Hadamard product (or convolution) (f1f2)(z) of the functions

fl(z)=z+ m=2em,lzm¥,  (l=1,2)

is given by

(f1f2)(z)=z+ m=2em,1em,2zm=(f2f1)(z)(zU).

For two functions f1 and f2 which are analytic in U, we say that the function f1 is subordinate to f2, and denote this by

f1(z)f2(z),(zU),

if there exists a Schwarz function ζ(z) analytic in U with ζ(0)=0 and |ζ(z)|<1(zU) such that f1(z)=f2(ζ(z)), for zU. See [3]. In particular, if f2(z) is univalent in U, we get the following equivalence

f1(z)f2(z)f1(0)=f2(0)andf1(U)f2(U).

Now, for function f(z) of the form (1.1) we define the series expansion of the linear operator τ,σx(u,v,y)f(z) of the function as

τ,σx(u,v,y)f(z)=z+ m=2Γ(v+y)Γ(u+my)Γ(u+y)Γ(v+my)1+τ(m1)σ+1xemzm,

for x,σ>1,τ>0,y>0 and Re(u)>Re(v)>y.

The operator τ,σx(u,v,y)f(z) was introduced in [9] generalizes some known operators as follows

(i) 1,0x(u,u,y)=Dx(x0={0,1,...})(Salagean [10]),

(ii) τ,0x(u,u,y)=Dτx(x0)(Al-Oboudi [1]),

(iii) τ,σ0(u,v+t,1)=Gut(t>0,u>1)(Gao et al. [4], Jung et al.[5]),

(iv) τ,σx(u,0,1)=Jx(τ,u,σ)(x0)(Catas [2]),

(v) 1,ux(u,u,y)=Lu+1x(x0,u0)(Komatu [6]),

(vi) τ,0x(u,v,1)=Dτx(u+1,v+1)(x0)(Selvaraj and Karthikeyan [11]).

From (1.5), it is easy to show that

zτ,σx(u,v,y)f(z)=uy+1τ,σx(u+1,v,y)f(z)uyτ,σx(u,v,y)f(z),
zτ,σx(u,v,y)f(z)= σ+1ττ,σx+1(u+1,v,y)f(z) σ+1τ1τ,σx(u,v,y)f(z),
zτ,σx(u,v+1,y)f(z)=vy+1τ,σx(u,v,y)f(z)vyτ,σx(u,v+1,y)f(z).

Using the concept of subordination in (1.4) and the operator τ,σx(u,v,y)f(z), we investigate the subclass of ¥ defined as follows.

Definition 1.1. A function f¥ is in the class Tτ,σx(u,v,y;γ;V,W) if it satisfies

zτ,σx(u,v,y)f(z)(1γ)τ,σx(u,v,y)f(z)γ1γ1+Vz1+Wz,(zU),

for V,W with 1V<W1 and 0γ<1.

The following lemmas will be useful in deriving our results.

Lemma 1.2. ([8]) If 1V<W1, λ>0 and η is restricted by

Reη1V1Wλ,

then the differential equation

g(z)+zg(z)λg(z)+η=1+Vz1+Wz,(zU)

has a univalent solution given by

g(z)= zλ+η(1+Wz) λ(VW)W λ0 z tλ+η1 (1+Wt) (VW)W dtηλ, W0 z λ+ηe λVz λ0z tλ+η1 e λVzdtηλ,W=0.

If ψ is regular in U and satisfies the differential subordination

ψ(z)+zψ(z)λψ(z)+η1+Vz1+Wz,

then ψ(z)g(z)1+Vz1+Wz and g is the best dominant of the above subordination.

Lemma 1.3.([7]) Let g be univalent in U and Ψ, ϑ be analytic functions in a domain D containing g(D) with Ψ(ρ)0 for ρg(D). Set T(z)=zg(z)Ψ(g(z)) and g(z)=ϑ(g(z))+T(z). Suppose that

(i) T is starlike univalent in U,

(ii) Rezg(z)T(z)>0, zU.

If h is analytic in U for h(0)=g(0), h(U)D and

ϑ(h(z))+zh(z)Ψ(h(z))ϑ(g(z))+zg(z)Ψ(g(z)),

then

h(z)g(z),

and g is the best dominant.

For x, σ>1, τ>0, y>0, 1V<W1 and 0γ<1, the following theorems are obtained.

Theorem 2.1. Let fTτ,σx(u+1,v,y;γ;V,W) such that τ,σx(u,v,y)f(z)0, zU=U\{0} and (1γ)(V1)(1W)γ+uy, then

z(τ,σx(u,v,y)f(z))(1γ)τ,σx(u,v,y)f(z)γ1γg1(z)1+Vz1+Wz, (zU)

where

g1(z)=11γ1(z)γuy,

and

(z)= 01 tuy 1+Wzt1+Wz (1γ) VW1dt, W0 01 t uy e (1γ)(t1)Vzdt,W=0.%

Further, g1(z) is the best dominant of (2.1).

Proof. Since fTτ,σx(u+1,v,y;γ;V,W), if we consider the function

ψ(z)=z(τ,σx(u,v,y)f(z))(1γ)τ,σx(u,v,y)f(z)γ1γ, (zU)

then ψ(z) is analytic in U and ψ(0)=1. Using identity (1.6),yields

(1γ)ψ(z)+γ+uy=uy+1τ,σx(u+1,v,y)f(z)τ,σx(u,v,y)f(z)

then differentiating (2.5) with respect to z and multiplying by z, we have

ψ(z)+zψ(z)(1γ)ψ(z)+γ+uy          =z(τ,σx(u+1,v,y)f(z))(1γ)τ,σx(u+1,v,y)f(z)γ1γ1+Vz1+Wz.

Applying Lemma 1.2 for λ=1γ and η=γ+uy, we get

ψ(z)g1(z)1+Vz1+Wz,

where g1(z) defined in (2.2) and it is the best dominant. Hence, the proof is completed.

Theorem 2.2. If fTτ,σx+1(u,v,y;γ;V,W) is such that τ,σx(u,v,y)f(z)0, zU=U\{0} and (1γ)(V1)(1W)σ+1τ1+γ, then

z(τ,σx(u,v,y)f(z))(1γ)τ,σx(u,v,y)f(z)γ1γg2(z)1+Vz1+Wz, (zU)

such that

g2(z)=11γ1χ(z)σ+1τ+1γ,

and

χ(z)= 01 tσ+1τ1 1+Wzt1+Wz (1γ) VW1dt, W0 01 t σ+1τ1 e (1γ)(t1)Vzdt,W=0.

Further, g2(z) is the best dominant of (2.6).

Proof. Suppose that fTτ,σx+1(u,v,y;γ;V,W), define

ψ(z)=z(τ,σx(u,v,y)f(z))(1γ)τ,σx(u,v,y)f(z)γ1γ, (zU).

Thus, the function ψ(z) is analytic in U and ψ(0)=1. From (1.7), the equation (2.9) becomes

(1γ)ψ(z)+σ+1τ1+γ=σ+1ττ,σx+1(u,v,y)f(z)τ,σx(u,v,y)f(z),

then differentiating (2.10) with respect to z and multiplying by z, we obtain

ψ(z)+zψ(z)(1γ)ψ(z)+σ+1τ1+γ          =z(τ,σx+1(u,v,y)f(z))(1γ)τ,σx+1(u,v,y)f(z)γ1γ1+Vz1+Wz.

By make use of Lemma 1.2 for λ=1γ and η=σ+1τ1+γ, we get

ψ(z)g2(z)1+Vz1+Wz,

where g2(z) defined in (2.7) and it is the best dominant. Thus, the proof is completed.

Theorem 2.3. If fTτ,σx(u,v,y;γ;V,W) such that τ,σx(u,v+1,y)f(z)0, zU=U\{0} and (1γ)(V1)(1W)vy+γ, then

z(τ,σx(u,v+1,y)f(z))(1γ)τ,σx(u,v+1,y)f(z)γ1γg3(z)1+Vz1+Wz, (zU)

where

g3(z)=11γ1ϕ(z)vyγ,

and

ϕ(z)= 01 tvy 1+Wzt1+Wz (1γ) VW1dt, W0 01 t vy e (1γ)(t1)Vzdt,W=0.

Further, g3(z) is the best dominant of (2.11).

Proof. Suppose that fTτ,σx(u,v,y;γ;V,W) and consider that the function

ψ(z)=z(τ,σx(u,v+1,y)f(z))(1γ)τ,σx(u,v+1,y)f(z)γ1γ, (zU).

Subsequently, ψ(z) is analytic in U together ψ(0)=1. Applying (1.8), yields

(1γ)ψ(z)+vy+γ=vy+1τ,σx(u,v,y)f(z)τ,σx(u,v+1,y)f(z).

By differentiation (2.15) with respect to z and multiplying by z, we have

ψ(z)+zψ(z)(1γ)ψ(z)+vy+γ          =z(τ,σx(u,v,y)f(z))(1γ)τ,σx+1(u,v,y)f(z)γ1γ1+Vz1+Wz.

Thus, from Lemma 1.2 for λ=1γ and η=vy+γ, we obtain

ψ(z)g3(z)1+Vz1+Wz,

where g3(z) defined in (2.12) and it is the best dominant.

Theorem 2.4. Suppose that g(z) is a univalent function in U with g(0)=1 and

Re1+zg(z)g(z)zg(z)g(z)>0, (zU).

Let f¥ satisfy the condition

θτ,σx(u+1,v,y)f(z)+κτ,σx(u,v,y)f(z)z(θ+κ)0,

where θ, κ and θ+κ0. If

δθzτ,σx(u+1,v,y)f(z)+κzτ,σx(u,v,y)f(z)θτ,σx(u+1,v,y)f(z)+κτ,σx(u,v,y)f(z)1zg(z)g(z),

then

θτ,σx(u+1,v,y)f(z)+κτ,σx(u,v,y)f(z)z(θ+κ)δg(z),

for δ\{0}, and g(z) is the best dominant of (2.17).

Proof. From Lemma 1.3, we can prove the results above for

Ψ(ρ)=1ρ, ϑ(ρ)=0, T(z)=zg(z)Ψ(g(z))=zg(z)g(z), g(z)=T(z), ρ.

Since T(0)=g(0)0, from (2.16) the function T is a starlike univalent in U and

Rezg(z)T(z)=Re1+zg(z)g(z)zg(z)g(z)>0, (zU).

Hence, both (i) and (ii) of Lemma 1.3 are satisfied. Consider the function h be defined by

h(z)=θτ,σx(u+1,v,y)f(z)+κτ,σx(u,v,y)f(z)z(θ+κ)δ, (zU)

subsequently, the function h is analytic in U, h(0)=g(0)=1, and

zh(z)h(z)=δθzτ,σx(u+1,v,y)f(z)+κzτ,σx(u,v,y)f(z)θτ,σx(u+1,v,y)f(z)+κτ,σx(u,v,y)f(z)1,

from (2.20),(2.17) becomes

zh(z)h(z)zg(z)g(z),

equivalent to

ϑ(h(z))+zh(z)Ψ(h(z))ϑ(g(z))+zg(z)Ψ(g(z)).

Thus, applying Lemma 1.3 it follows that h(z)g(z) and that g is the best dominant of (2.17). Taking θ=0, κ=1 and g(z)=1+Vz1+Wz(1V<W1), it is easy to show

Re1+zg(z)g(z)zg(z)g(z)=1ReVz1+Vz+Wz1+Wz    >1V1+V+W1+W=1VW1+V1+W>0, (zU).

From Theorem 2.4, we obtain the following result.

Corollary 2.5. Let f¥ satisfy the condition

τ,σx(u,v,y)f(z)z0, (zU).

If

δzτ,σx(u,v,y)f(z)τ,σx(u,v,y)f(z)1(VW)z(1+Vz)(1+Wz),

then

τ,σx(u,v,y)f(z)zδ1+Vz1+Wz,

with δ\{0} and 1+Vz1+Wz is the best dominant of (2.21).

We took V=1 and W=1 in Corollary 2.5 as a special case and we got:

Corollary 2.6. Let f¥ satisfies the condition

τ,σx(u,v,y)f(z)z0, (zU).

If

δzτ,σx(u,v,y)f(z)τ,σx(u,v,y)f(z)12z1z2,

then

τ,σx(u,v,y)f(z)zδ1+z1z.

For δ\{0} and 1+z1z is the best dominant of (2.22).

Taking θ=1 and κ=0 and g(z)=1+Vz1+Wz (1V<W1) in Theorem 2.4, it follows that.

Corollary 2.7. Suppose that f¥ satisfies the condition

τ,σx(u+1,v,y)f(z)z0, (zU).

If

δzτ,σx(u+1,v,y)f(z)τ,σx(u+1,v,y)f(z)1(VW)z(1+Vz)(1+Wz),

then

τ,σx(u+1,v,y)f(z)zδ1+Vz1+Wz.

For δ\{0} and 1+Vz1+Wz is the best dominant of (2.23).

We took V=1 and W=1 in Corollary 2.7 as a special case and we got:

Corollary 2.8. Let f¥ satisfies the condition

τ,σx(u+1,v,y)f(z)z0, (zU).

If

δzτ,σx(u+1,v,y)f(z)τ,σx(u+1,v,y)f(z)12z1z2,

then

τ,σx(u+1,v,y)f(z)zδ1+z1z

For δ\{0} and 1+z1z is the best dominant of (2.24).

In the present work, we were able to obtain the best results, or best dominants of the subordination. Our main results give an interesting process for the study of many analytic univalent classes earlier defined by several authors. These classes expand and generalize many of those defined by many specialists in this field. Furthermore, the general subordination theorems lead us to some special cases that were used to determine new results connected with the classes we investigated.

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