DIFFERENTIAL SUBORDINATIONS USING RUSCHEWEYH DERIVATIVE AND SALAGEAN OPERATOR

DIFFERENTIAL SUBORDINATIONS USING RUSCHEWEYH DERIVATIVE AND S ˘AL ˘AGEAN OPERATOR

ALB LUPAS¸ ALINA1 §

Abstract. In the present paper we study the operator defined by using the Ruscheweyh derivativeRm_{f(z) and the S˘al˘agean operatorS}m_{f(z), denotedL}m

α :An→ An, Lmαf(z) =

(1−α)Rmf(z)+αSmf(z), z∈U,whereAn={f∈ H(U) :f(z) =z+an+1zn+1+. . . , z∈ U}is the class of normalized analytic functions. We obtain several differential subordi-nations regarding the operatorLm

α.

Keywords: differential subordination, convex function, best dominant, differential oper-ator, S˘al˘agean operator, Ruscheweyh derivative.

AMS Subject Classification: 30C45, 30A20, 34A40.

1. Introduction

Denote by U the unit disc of the complex plane,U ={z∈C:|z|<1} and H(U) the space of holomorphic functions inU. Let

An={f ∈ H(U) :f(z) =z+an+1zn+1+. . . , z ∈U}

and

H[a, n] ={f ∈ H(U) :f(z) =a+anzn+an+1zn+1+. . . , z ∈U}

fora∈C and n∈N. Denote by K =

{

f ∈ An: Re zf_f′′′₍(_zz₎)+ 1>0, z∈U

}

, the class of normalized convex functions in U.

Iff andg are analytic functions inU, we say thatf is subordinate tog, written f ≺g, if there is a functionw analytic in U, with w(0) = 0, |w(z)|<1, for all z ∈U,such that f(z) =g(w(z)) for allz ∈U. Ifg is univalent, then f ≺g if and only if f(0) =g(0) and f(U)⊆g(U). Let ψ:C3×U →C andh be a univalent function in U. Ifp is analytic in U and satisfies the (second-order) differential subordination

ψ(p(z), zp′(z), z2p′′(z);z)≺h(z), z∈U, (1) then p is called a solution of the differential subordination. The univalent function q is called a dominant of the solutions of the differential subordination, or more simply a dominant, if p ≺ q for all p satisfying (1). A dominant qethat satisfies qe ≺ q for all dominants q of (1) is said to be the best dominant of (1). The best dominant is unique up to a rotation ofU.

1

Department of Mathematics and Computer Science University of Oradea str. Universitatii nr. 1, 410087 Oradea, Romania,

e-mail: [email protected]

§ Submitted for GFTA’13, held in I¸sık University on October 12, 2013.

TWMS Journal of Applied and Engineering Mathematics, Vol.4, No.1; c⃝I¸sık University, Department of Mathematics 2014; all rights reserved.

Definition 1.1. (S˘al˘agean [6]) For f ∈ An, n, m ∈ N, the operator Sm is defined by Sm_{:An→ An,}

S0f(z) =f(z), S1f(z) =zf′(z), ... Sm+1f(z) =z(Smf(z))′, z∈U.

Remark 1.1. If f ∈ An, f(z) =z+∑∞_j=n+1ajzj, then

Smf(z) =z+

∞ ∑

j=n+1

jmajzj, z∈U

.

Definition 1.2. ([5]) Forf ∈ An,n, m∈N, the operatorRmis defined byRm :An→ An, R0f(z) = f(z), R1f(z) =zf′(z), ...

(m+ 1)Rm+1f(z) = z(Rmf(z))′+mRmf(z), z∈U.

Remark 1.2.Iff ∈ An,f(z) =z+∑∞_j=n+1ajzj, thenRmf(z) =z+

∑_∞

j=n+1Cmm+j−1ajzj,

z∈U.

Definition 1.3. ([1]) Let α ≥ 0, n, m ∈ N. Denote by Lm_α the operator given by Lm_α :

An→ An,

Lm_αf(z) = (1−α)Rmf(z) +αSmf(z), z∈U.

Remark 1.3. If f ∈ An, f(z) =z+∑∞_j=n+1ajzj, then

Lm_αf(z) =z+∑∞_j=n+1

(

αjm+ (1−α)C_mm_+j−1

)

ajzj, z ∈U.

This operator was studied also in[1], [2].

Lemma 1.1. (Hallenbeck and Ruscheweyh [4, Th. 3.1.6, p. 71]) Let h be a convex function with h(0) = a, and let γ ∈ C\{0} be a complex number with Re γ ≥ 0. If p ∈ H[a, n] and p(z) + _γ1zp′(z) ≺ h(z), z ∈ U, then p(z) ≺ g(z) ≺ h(z), z ∈ U, where g(z) = _nzγγ/n

∫z

0 h(t)t

γ/n−1_{dt, z ∈U.}

Lemma 1.2. (Miller and Mocanu [4]) Let g be a convex function in U and let h(z) = g(z) +nαzg′(z), for z∈U, where α >0 and n is a positive integer.

If p(z) =g(0) +pnzn+pn+1zn+1+. . . , z∈U,is holomorphic in U andp(z) +αzp′(z)≺

h(z), z∈U,then p(z)≺g(z), z∈U,and this result is sharp.

2. Main results

Theorem 2.1.Letgbe a convex function,g(0) = 1and lethbe the functionh(z) =g(z)+

nz

δ g′(z), z∈U.If α, δ ≥0, n, m∈N, f ∈ An and satisfies the differential subordination

(

Lm_αf(z) z

)δ−1

(Lm_αf(z))′ ≺h(z), z∈U, (2)

then

(

Lm αf(z)

z

)δ

≺g(z), z∈U,and this result is sharp.

Theorem 2.2. Let h be an holomorphic function which satisfies the inequality Re

(

1 +zh_h′′′₍(_zz₎)

)

>−1₂, z ∈ U,and h(0) = 1. If α, δ ≥0, n, m∈N, f ∈ An and satisfies the differential subordination

(

Lm_αf(z) z

)δ−1

then

(

Lm αf(z)

z

)δ

≺q(z), z ∈U, where q(z) = δ

nznδ ∫z

0 h(t)t δ

n−1dt. The function q is convex

and it is the best dominant.

Corollary 2.1. Let h(z) = 1+(2₁₊β−1)_z z be a convex function in U, where 0 ≤ β < 1. If α, δ≥0, n, m∈N, f ∈ An and satisfies the differential subordination

(

Lm_αf(z) z

)δ−1

(Lm_αf(z))′ ≺h(z), z∈U, (4)

then

(

Lm αf(z)

z

)δ

≺q(z), z ∈ U, where q is given by q(z) = (2β−1) + 2(1−β)δ

nzδn ∫z

0

tnδ−1 1+t dt,

z∈U.The function q is convex and it is the best dominant.

Remark 2.1. For n = 1, m = 1, α = 2, δ = 1 we obtain the same example as in [3, Example 2.2.1, p. 26].

Theorem 2.3. Let g be a convex function such that g(0) = 1 and let h be the function h(z) = g(z) + nz_γ g′(z), z ∈ U, where γ > 0. If α ≥ 0, n, m ∈ N, f ∈ An and the differential subordination

(γ+ 1)z γ

Lm αf(z)

(

Lmα+1f(z)

)2 +z

2

γ Lm

αf(z)

(

Lmα+1f(z)

)2 [

(Lm αf(z))′

Lm αf(z)

−2

(

Lm+1

α f(z)

)_′

Lmα+1f(z)

]

≺h(z), z∈U

(5) holds, then z Lmαf(z)

(Lm+1α f(z))

2 ≺g(z), z∈U,and this result is sharp.

Theorem 2.4. Let h be an holomorphic function which satisfies the inequality Re

(

1 +zh_h′′′₍(_zz₎)

)

>−1₂, z ∈U, and h(0) = 1. If α ≥0, γ ∈ C\{0} be a complex number withRe γ ≥0, n, m∈N, f ∈ An and satisfies the differential subordination

(γ+ 1)z γ

Lm_αf(z)

(

Lmα+1f(z)

)2+z

2

γ

Lm_αf(z)

(

Lmα+1f(z)

)2 [

(Lm_αf(z))′ Lm

αf(z) −2

(

Lm_α+1f(z))′ Lmα+1f(z)

]

≺h(z), z∈U,

(6) then z Lmαf(z)

(Lm+1α f(z))

2 ≺ q(z), z ∈ U, where q(z) =

γ nzγ/n

∫z

0 h(t)tγ/n−1dt. The function q is

convex and it is the best dominant.

Theorem 2.5. Let g be a convex function such that g(0) = 1 and let h be the function h(z) =g(z) +nz_γg′(z), z∈U, whereγ >0.Ifα≥0,n, m∈N, f ∈ An and the differential subordination

(γ+ 2)z2 γ

(Lm_αf(z))′ Lm

αf(z)

+z

3

γ

[

(Lm_αf(z))′′ Lm

αf(z) −

(

(Lm_αf(z))′ Lm

αf(z)

)2]

≺h(z), z∈U (7)

holds, then z2 (Lmαf(z))′

Lm

αf(z) ≺g(z), z∈U.This result is sharp.

Theorem 2.6. Let h be an holomorphic function which satisfies the inequality Re

(

1 +zh_h′′′₍(_zz₎)

)

>−1₂, z ∈U, and h(0) = 1. If α ≥0, γ ∈ C\{0} be a complex number withRe γ ≥0, n, m∈N, f ∈ An and satisfies the differential subordination

(γ+ 2)z2 γ

(Lm_αf(z))′ Lm

αf(z)

+z

3

γ

[

(Lm_αf(z))′′ Lm

αf(z) −

(

(Lm_αf(z))′ Lm

αf(z)

)2]

≺h(z), z∈U, (8)

then z2 (Lmαf(z))′

Lm

αf(z) ≺ q(z), z ∈ U, where q(z) =

γ nzγ/n

∫z

0 h(t)t

γ/n−1_{dt. The function q is}

Theorem 2.7. Let g be a convex function such that g(0) = 1 and let h be the function h(z) =g(z)+nzg′(z), z∈U.Ifα≥0,n, m∈N,f ∈ Anand the differential subordination

1−L

m

αf(z)·(Lmαf(z))′′

[

(Lm αf(z))′

]2 ≺h(z), z∈U (9)

holds, then Lmαf(z)

z(Lm

αf(z))′ ≺g(z), z∈U.This result is sharp.

Theorem 2.8. Let h be an holomorphic function which satisfies the inequality Re

(

1 +zh_h′′′((zz)) )

>−1₂, z∈U, and h(0) = 1.If α≥0, n, m∈N, f ∈ An and satisfies the differential subordination

1−L

m

αf_[(z)·(Lmαf(z))′′

(Lm αf(z))′

]2 ≺h(z), z∈U, (10)

then Lmαf(z)

z(Lm

αf(z))′ ≺ q(z), z ∈ U,where q(z) = 1

nz1n ∫z

0 h(t)t 1

n−1dt. The function q is convex

and it is the best dominant.

Corollary 2.2. Let h(z) = 1+(2₁₊β−_z1)z be a convex function in U, where 0 ≤ β < 1. If α≥0, n, m∈N, f ∈ An and satisfies the differential subordination

1−L

m

αf(z)·(Lmαf(z))′′

[

(Lm αf(z))′

]2 ≺h(z), z∈U, (11)

then Lmαf(z)

z(Lm

αf(z))′ ≺ q(z), z ∈ U, where q is given by q(z) = (2β−1) + 2(1−β)

nz1n ∫z

0

tn1−1 1+t dt,

z∈U.The function q is convex and it is the best dominant.

Example 2.1.Leth(z) = ₁₊1−z_z a convex function inU withh(0) = 1andRe

(

zh′′(z)

h′(z) + 1 )

>

−1 2.

Let f(z) =z+z2_{, z∈U. For n= 1, m= 1, α= 2, we obtain}

L1₂f(z) =−R1f(z) + 2S1f(z) =−zf′(z) + 2zf′(z) =zf′(z) =z+ 2z2.

Then (L1₂f(z))′ = 1 + 4z,

L1₂f(z) z(L1

2f(z)

)_′ = z+ 2z

2

z(1 + 4z) =

1 + 2z 1 + 4z,

1−L

1 2f(z)·

(

L1₂f(z))′′

[(

L1₂f(z))′

]2 = 1− (

z+ 2z2)·4 (1 + 4z)2 =

8z2+ 4z+ 1 (1 + 4z)2 .

We have

q(z) = 1 z

∫ z

0

1−t

1 +tdt=−1 +

2 ln (1 +z)

z .

Using Theorem 2.8 we obtain 8z2+4z+1

(1+4z)2 ≺ 1−z

1+z, z ∈ U, induce

1+2z

1+4z ≺ −1 +

2 ln(1+z)

z ,

z∈U.

Theorem 2.9. Let g be a convex function such that g(0) = 0 and let h be the function h(z) =g(z)+nzg′(z), z∈U.Ifα≥0,n, m∈N,f ∈ Anand the differential subordination

[

(Lm_αf(z))′]2+Lm_αf(z)·(Lm_αf(z))′′≺h(z), z∈U (12)

holds, then Lmαf(z)·(Lmαf(z))′

Theorem 2.10. Let h be an holomorphic function which satisfies the inequality Re

(

1 +zh_h′′′((zz)) )

>−1₂, z∈U, and h(0) = 0.If α≥0, n, m∈N, f ∈ An and satisfies the differential subordination

[

(Lm_αf(z))′]2+Lm_αf(z)·(Lm_αf(z))′′≺h(z), z∈U, (13)

then Lmαf(z)·(Lmαf(z))′

z ≺ q(z), z ∈ U,where q(z) =

1

nzn1 ∫z

0 h(t)t 1

n−1dt. The function q is

convex and it is the best dominant.

Corollary 2.3. Let h(z) = 1+(2₁₊β−_z1)z be a convex function in U, where 0 ≤ β < 1. If α≥0, n, m∈N, f ∈ An and satisfies the differential subordination

[

(Lm_αf(z))′]2+Lm_αf(z)·(Lm_αf(z))′′≺h(z), z∈U, (14)

then Lmαf(z)·(Lmαf(z))′

z ≺q(z), z ∈U,where q is given byq(z) = (2β−1) +

2(1−β)

nz1n ∫z

0

tn1−1 1+t dt,

z∈U.The function q is convex and it is the best dominant.

Example 2.2.Leth(z) = ₁₊1−z_z a convex function inU withh(0) = 1andRe

(

zh′′(z)

h′(z) + 1 )

>

−1

2. Let f(z) =z+z

2_{, z∈U. For n= 1, m= 1, α= 2, we obtain}

L1₂f(z) =−R1f(z) + 2S1f(z) =−zf′(z) + 2zf′(z) =zf′(z) =z+ 2z2, z∈U

Then ₍

L1₂f(z))′ = 1 + 4z,

L1₂f(z)·(L1₂f(z))′

z =

(

z+ 2z2)(1 + 4z)

z = 8z

2_{+ 6z+ 1,} [(

L1₂f(z))′

]2

+L1₂f(z)·(L1₂f(z))′′ = (1 + 4z)2+(z+ 2z2)·4 = 24z2+ 12z+ 1.

We have

q(z) = 1 z

∫ z

0

1−t

1 +tdt=−1 +

2 ln (1 +z)

z .

Using Theorem 2.10 we obtain 24z2+ 12z+ 1≺ ₁₊1−z_z, z∈U, induce

8z2+ 6z+ 1≺ −1 +2 ln (1 +z)

z , z∈U.

Theorem 2.11. Let g be a convex function such that g(0) = 0 and let h be the function h(z) =g(z) + ₁nz_−δg′(z), z ∈U.If α ≥0, δ ∈(0,1), n, m∈N, f ∈ An and the differential subordination

(

z Lm

αf(z)

)δ

Lm_α+1f(z) 1−δ

((

Lm_α+1f(z))′ Lmα+1f(z)

−δ(L

m αf(z))′

Lm αf(z)

)

≺h(z), z∈U (15)

holds, then Lm+1α f(z)

z ·

(

z Lm

αf(z) )δ

≺g(z), z∈U.This result is sharp.

Theorem 2.12. Let h be an holomorphic function which satisfies the inequality Re

(

1 +zh_h′′′₍(_zz₎)

)

>−1₂, z ∈U, and h(0) = 1. If α≥0, δ ∈(0,1), n, m ∈N, f ∈ An and satisfies the differential subordination

(

z Lm

αf(z)

)δ _Lm+1

α f(z)

1−δ

((

Lm+1

α f(z)

)_′

Lmα+1f(z)

−δ(L

m αf(z))′

Lm αf(z)

)

then Lm+1α f(z)

z ·

(

z Lm

αf(z) )δ

≺q(z), z∈U,whereq(z) = 1−δ

nz1−nδ ∫z

0 h(t)t 1−δ

n −1dt.The function

q is convex and it is the best dominant.

References

[1] A. Alb Lupa¸s,On special differential subordinations using S˘al˘agean and Ruscheweyh operators, Math-ematical Inequalities and Applications, Volume 12, Issue 4, 2009, 781-790.

[2] A. Alb Lupa¸s, D. Breaz,On special differential superordinations using S˘al˘agean and Ruscheweyh op-erators, Geometric Function Theory and Applications’ 2010 (Proc. of International Symposium, Sofia, 27-31 August 2010), 98-103.

[3] D.A. Alb Lupa¸s,Subordinations and Superordinations, Lap Lambert Academic Publishing, 2011. [4] S.S. Miller, P.T. Mocanu, Differential Subordinations. Theory and Applications, Marcel Dekker Inc.,

New York, Basel, 2000.

[5] St. Ruscheweyh, New criteria for univalent functions, Proc. Amet. Math. Soc., 49(1975), 109-115. [6] G. St. S˘al˘agean, Subclasses of univalent functions, Lecture Notes in Math., Springer Verlag, Berlin,