Skip to main content

Theory and Modern Applications

Differential subordinations using the Ruscheweyh derivative and the generalized Sălăgean operator

Abstract

In the present paper, we study the operator, using the Ruscheweyh derivative R m f(z) and the generalized Sălăgean operator D λ m f(z), denote by R D λ , α m : A n A n , R D λ , α m f(z)=(1α) R m f(z)+α D λ m f(z), zU, where A n ={fH(U):f(z)=z+ a n + 1 z n + 1 +,zU} is the class of normalized analytic functions. We obtain several differential subordinations regarding the operator R D λ , α m .

MSC:30C45, 30A20, 34A40.

1 Introduction

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

Let A n ={fH(U):f(z)=z+ a n + 1 z n + 1 +,zU} and H[a,n]={fH(U):f(z)=a+ a n z n + a n + 1 z n + 1 +,zU} for aC and nN.

Denote by K={f A n :Re z f ( z ) f ( z ) +1>0,zU} the class of normalized convex functions in U.

If f and g are analytic functions in U, we say that f is subordinate to g, written fg, if there is a function w analytic in U, with w(0)=0, |w(z)|<1, for all zU such that f(z)=g(w(z)) for all zU. If g is univalent, then fg if and only if f(0)=g(0) and f(U)g(U).

Let ψ: C 3 ×UC, and let h be an univalent function in U. If p is analytic in U and satisfies the (second-order) differential subordination

ψ ( p ( z ) , z p ( z ) , z 2 p ( z ) ; z ) h(z),zU,
(1.1)

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 pq for all p satisfying (1.1).

A dominant q ˜ that satisfies q ˜ q for all dominants q of (1.1) is said to be the best dominant of (1.1). The best dominant is unique up to a rotation of U.

Definition 1.1 (Al-Oboudi [1])

For f A n , λ0 and n,mN, the operator D λ m is defined by D λ m : A n A n ,

D λ 0 f ( z ) = f ( z ) , D λ 1 f ( z ) = ( 1 λ ) f ( z ) + λ z f ( z ) = D λ f ( z ) , , D λ m + 1 f ( z ) = ( 1 λ ) D λ m f ( z ) + λ z ( D λ m f ( z ) ) = D λ ( D λ m f ( z ) ) , z U .

Remark 1.1 If f A n and f(z)=z+ j = n + 1 a j z j , then D λ m f(z)=z+ j = n + 1 [ 1 + ( j 1 ) λ ] m a j z j , zU.

Remark 1.2 For λ=1, in the definition above, we obtain the Sălăgean differential operator [2].

Definition 1.2 (Ruscheweyh [3])

For f A n , n,mN, the operator R m is defined by R m : A n A n ,

R 0 f ( z ) = f ( z ) , R 1 f ( z ) = z f ( z ) , , ( m + 1 ) R m + 1 f ( z ) = z ( R m f ( z ) ) + m R m f ( z ) , z U .

Remark 1.3 If f A n , f(z)=z+ j = n + 1 a j z j , then R m f(z)=z+ j = n + 1 C m + j 1 m a j z j , zU.

Definition 1.3 [4]

Let α,λ0, n,mN. Denote by R D λ , α m the operator given by R D λ , α m : A n A n ,

R D λ , α m f(z)=(1α) R m f(z)+α D λ m f(z),zU.

Remark 1.4 If f A n , f(z)=z+ j = n + 1 a j z j , then R D λ , α m f(z)=z+ j = n + 1 {α [ 1 + ( j 1 ) λ ] m +(1α) C m + j 1 m } a j z j , zU.

This operator was studied also in [46] and [7].

Remark 1.5 For α=0, R D λ , 0 m f(z)= R m f(z), where zU and for α=1, R D λ , 1 m f(z)= D λ m f(z), where zU.

For λ=1, we obtain R D 1 , α m f(z)= L α m f(z), which was studied in [811].

For m=0, R D λ , α 0 f(z)=(1α) R 0 f(z)+α D λ 0 f(z)=f(z)= R 0 f(z)= D λ 0 f(z), where zU.

Lemma 1.1 (Hallenbeck and Ruscheweyh [[12], Th. 3.1.6, p.71])

Let h be a convex function with h(0)=a, and let γC{0} be a complex number with Reγ0. If pH[a,n] and

p(z)+ 1 γ z p (z)h(z),zU,

then

p(z)g(z)h(z),zU,

where g(z)= γ n z γ / n 0 z h(t) t γ / n 1 dt, zU.

Lemma 1.2 (Miller and Mocanu [12])

Let g be a convex function in U, and let h(z)=g(z)+nαz g (z), for zU, where α>0 and n is a positive integer.

If p(z)=g(0)+ p n z n + p n + 1 z n + 1 +, zU, is holomorphic in U and

p(z)+αz p (z)h(z),zU,

then

p(z)g(z),zU,

and this result is sharp.

2 Main results

Theorem 2.1 Let g be a convex function, g(0)=1, and let h be the function h(z)=g(z)+ n z δ g (z), for zU.

If α,λ,δ0, n,mN, f A n and satisfies the differential subordination

( R D λ , α m f ( z ) z ) δ 1 ( R D λ , α m f ( z ) ) h(z),zU,
(2.1)

then

( R D λ , α m f ( z ) z ) δ g(z),zU,

and this result is sharp.

Proof By using the properties of operator R D λ , α m , we have

R D λ , α m f(z)=z+ j = n + 1 { α [ 1 + ( j 1 ) λ ] m + ( 1 α ) C m + j 1 m } a j z j ,zU.

Consider p(z)= ( R D λ , α m f ( z ) z ) δ = ( z + j = n + 1 { α [ 1 + ( j 1 ) λ ] m + ( 1 α ) C m + j 1 m } a j z j z ) δ =1+ p n δ z n δ + p n δ + 1 z n δ + 1 +, zU.

We deduce that pH[1,nδ].

Differentiating we obtain ( R D λ , α m f ( z ) z ) δ 1 ( R D λ , α m f ( z ) ) =p(z)+ 1 δ z p (z), zU.

Then (2.1) becomes

p(z)+ 1 δ z p (z)h(z)=g(z)+ n z δ g (z)for zU.

By using Lemma 1.2, we have

p(z)g(z),zU,i.e., ( R D λ , α m f ( z ) z ) δ g(z),zU.

 □

Theorem 2.2 Let h be a holomorphic function, which satisfies the inequality Re(1+ z h ( z ) h ( z ) )> 1 2 , zU, and h(0)=1.

If α,λ,δ0, n,mN, f A n and satisfies the differential subordination

( R D λ , α m f ( z ) z ) δ 1 ( R D λ , α m f ( z ) ) h(z),zU,
(2.2)

then

( R D λ , α m f ( z ) z ) δ q(z),zU,

where q(z)= δ n z δ n 0 z h(t) t δ n 1 dt. The function q is convex, and it is the best dominant.

Proof Let

p ( z ) = ( R D λ , α m f ( z ) z ) δ = ( z + j = n + 1 { α [ 1 + ( j 1 ) λ ] m + ( 1 α ) C m + j 1 m } a j z j z ) δ = ( 1 + j = n + 1 { α [ 1 + ( j 1 ) λ ] m + ( 1 α ) C m + j 1 m } a j z j 1 ) δ = 1 + j = n δ p j z j

for zU, pH[1,nδ].

Differentiating, we obtain ( R D λ , α m f ( z ) z ) δ 1 ( R D λ , α m f ( z ) ) =p(z)+ 1 δ z p (z), zU, and (2.2) becomes

p(z)+ 1 δ z p (z)h(z),zU.

Using Lemma 1.1, we have

p(z)q(z),zU,i.e., ( R D λ , α m f ( z ) z ) δ q(z)= δ n z δ n 0 z h(t) t δ n 1 dt,zU,

and q is the best dominant. □

Corollary 2.3 Let h(z)= 1 + ( 2 β 1 ) z 1 + z be a convex function in U, where 0β<1.

If α,δ,λ0, n,mN, f A n and satisfies the differential subordination

( R D λ , α m f ( z ) z ) δ 1 ( R D λ , α m f ( z ) ) h(z),zU,
(2.3)

then

( R D λ , α m f ( z ) z ) δ q(z),zU,

where q is given by q(z)=(2β1)+ 2 ( 1 β ) δ n z δ n 0 z t δ n 1 1 + t dt, zU. The function q is convex, and it is the best dominant.

Proof Following the same steps as in the proof of Theorem 2.2 and considering p(z)= ( R D λ , α m f ( z ) z ) δ , the differential subordination (2.3) becomes

p(z)+ z δ p (z)h(z)= 1 + ( 2 β 1 ) z 1 + z ,zU.

By using Lemma 1.1, for γ=δ, we have p(z)q(z), i.e.,

( R D λ , α m f ( z ) z ) δ q ( z ) = δ n z δ n 0 z h ( t ) t δ n 1 d t = δ n z δ n 0 z t δ n 1 1 + ( 2 β 1 ) t 1 + t d t = δ n z δ n 0 z [ ( 2 β 1 ) t δ n 1 + 2 ( 1 β ) t δ n 1 1 + t ] d t = ( 2 β 1 ) + 2 ( 1 β ) δ n z δ n 0 z t δ n 1 1 + t d t , z U .

 □

Remark 2.1 For n=1, λ= 1 2 , α=2, δ=1, we obtain the same example as in [[13], Example 4.2.1, p.125].

Theorem 2.4 Let g be a convex function such that g(0)=1, and let h be the function h(z)=g(z)+ n z δ g (z), zU.

If α,λ,δ0, n,mN, f A n and the differential subordination

z δ + 1 δ R D λ , α n f ( z ) ( R D λ , α n + 1 f ( z ) ) 2 + z 2 δ R D λ , α n f ( z ) ( R D λ , α n + 1 f ( z ) ) 2 [ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) 2 ( R D λ , α n + 1 f ( z ) ) R D λ , α n + 1 f ( z ) ] h ( z ) , z U
(2.4)

holds, then

z R D λ , α m f ( z ) ( R D λ , α m + 1 f ( z ) ) 2 g(z),zU,

and this result is sharp.

Proof For f A n , f(z)=z+ j = n + 1 a j z j , we have

R D λ , α m f(z)=z+ j = n + 1 { α [ 1 + ( j 1 ) λ ] m + ( 1 α ) C m + j 1 m } a j z j ,zU.

Consider p(z)=z R D λ , α m f ( z ) ( R D λ , α m + 1 f ( z ) ) 2 , and we obtain

p(z)+ z δ p (z)=z δ + 1 δ R D λ , α n f ( z ) ( R D λ , α n + 1 f ( z ) ) 2 + z 2 δ R D λ , α n f ( z ) ( R D λ , α n + 1 f ( z ) ) 2 [ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) 2 ( R D λ , α n + 1 f ( z ) ) R D λ , α n + 1 f ( z ) ] .

Relation (2.4) becomes

p(z)+ z δ p (z)h(z)=g(z)+ n z δ g (z),zU.

By using Lemma 1.2, we have

p(z)g(z),zU,i.e.,z R D λ , α m f ( z ) ( R D λ , α m + 1 f ( z ) ) 2 g(z),zU.

 □

Theorem 2.5 Let h be a holomorphic function, which satisfies the inequality Re(1+ z h ( z ) h ( z ) )> 1 2 , zU, and h(0)=1.

If α,λ,δ0, n,mN, f A n and satisfies the differential subordination

z δ + 1 δ R D λ , α n f ( z ) ( R D λ , α n + 1 f ( z ) ) 2 + z 2 δ R D λ , α n f ( z ) ( R D λ , α n + 1 f ( z ) ) 2 [ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) 2 ( R D λ , α n + 1 f ( z ) ) R D λ , α n + 1 f ( z ) ] h ( z ) , z U ,
(2.5)

then

z R D λ , α m f ( z ) ( R D λ , α m + 1 f ( z ) ) 2 q(z),zU,

where q(z)= δ n z δ n 0 z h(t) t δ n 1 dt. The function q is convex, and it is the best dominant.

Proof Let p(z)=z R D λ , α m f ( z ) ( R D λ , α m + 1 f ( z ) ) 2 , zU, pH[1,n].

Differentiating, we obtain p(z)+ z δ p (z)=z δ + 1 δ R D λ , α n f ( z ) ( R D λ , α n + 1 f ( z ) ) 2 + z 2 δ R D λ , α n f ( z ) ( R D λ , α n + 1 f ( z ) ) 2 [ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) 2 ( R D λ , α n + 1 f ( z ) ) R D λ , α n + 1 f ( z ) ], zU, and (2.5) becomes

p(z)+ z δ p (z)h(z),zU.

Using Lemma 1.1, we have

p ( z ) q ( z ) , z U , i.e. , z R D λ , α m f ( z ) ( R D λ , α m + 1 f ( z ) ) 2 q ( z ) = δ n z δ n 0 z h ( t ) t δ n 1 d t , z U ,

and q is the best dominant. □

Theorem 2.6 Let g be a convex function such that g(0)=1, and let h be the function h(z)=g(z)+ n z δ g (z), zU.

If α,λ,δ0, n,mN, f A n and the differential subordination

z 2 δ + 2 δ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) + z 3 δ [ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) ( ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) ) 2 ] h(z),zU
(2.6)

holds, then

z 2 ( R D λ , α m f ( z ) ) R D λ , α m f ( z ) g(z),zU.

This result is sharp.

Proof Let p(z)= z 2 ( R D λ , α m f ( z ) ) R D λ , α m f ( z ) . We deduce that pH[0,n].

Differentiating, we obtain p(z)+ z δ p (z)= z 2 δ + 2 δ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) + z 3 δ [ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) ( ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) ) 2 ], zU.

Using the notation in (2.6), the differential subordination becomes

p(z)+ 1 δ z p (z)h(z)=g(z)+ n z δ g (z).

By using Lemma 1.2, we have

p(z)g(z),zU,i.e., z 2 ( R D λ , α m f ( z ) ) R D λ , α m f ( z ) g(z),zU,

and this result is sharp. □

Theorem 2.7 Let h be an holomorphic function, which satisfies the inequality Re(1+ z h ( z ) h ( z ) )> 1 2 , zU, and h(0)=1.

If α,λ,δ0, n,mN, f A n and satisfies the differential subordination

z 2 δ + 2 δ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) + z 3 δ [ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) ( ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) ) 2 ] h(z),zU,
(2.7)

then

z 2 ( R D λ , α m f ( z ) ) R D λ , α m f ( z ) q(z),zU,

where q(z)= δ n z δ n 0 z h(t) t δ n 1 dt. The function q is convex, and it is the best dominant.

Proof Let p(z)= z 2 ( R D λ , α m f ( z ) ) R D λ , α m f ( z ) , zU, pH[0,n].

Differentiating, we obtain p(z)+ z δ p (z)= z 2 δ + 2 δ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) + z 3 δ [ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) ( ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) ) 2 ], zU, and (2.7) becomes

p(z)+ 1 δ z p (z)h(z),zU.

Using Lemma 1.1, we have

p(z)q(z),zU,i.e., z 2 ( R D λ , α m f ( z ) ) R D λ , α m f ( z ) q(z)= δ n z δ n 0 z h(t) t δ n 1 dt,zU,

and q is the best dominant. □

Theorem 2.8 Let g be a convex function such that g(0)=1, and let h be the function h(z)=g(z)+nz g (z), zU.

If α,λ0, n,mN, f A n and the differential subordination

1 R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) [ ( R D λ , α m f ( z ) ) ] 2 h(z),zU
(2.8)

holds, then

R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) g(z),zU.

This result is sharp.

Proof Let p(z)= R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) . We deduce that pH[1,n].

Differentiating, we obtain 1 R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) [ ( R D λ , α m f ( z ) ) ] 2 =p(z)+z p (z), zU.

Using the notation in (2.8), the differential subordination becomes

p(z)+z p (z)h(z)=g(z)+nz g (z).

By using Lemma 1.2, we have

p(z)g(z),zU,i.e., R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) g(z),zU,

and this result is sharp. □

Theorem 2.9 Let h be a holomorphic function, which satisfies the inequality Re(1+ z h ( z ) h ( z ) )> 1 2 , zU, and h(0)=1.

If α,λ0, n,mN, f A n and satisfies the differential subordination

1 R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) [ ( R D λ , α m f ( z ) ) ] 2 h(z),zU,
(2.9)

then

R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) q(z),zU,

where q(z)= 1 n z 1 n 0 z h(t) t 1 n 1 dt. The function q is convex, and it is the best dominant.

Proof Let p(z)= R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) , zU, pH[0,n].

Differentiating, we obtain 1 R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) [ ( R D λ , α m f ( z ) ) ] 2 =p(z)+z p (z), zU, and (2.9) becomes

p(z)+z p (z)h(z),zU.

Using Lemma 1.1, we have

p(z)q(z),zU,i.e., R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) q(z)= 1 n z 1 n 0 z h(t) t 1 n 1 dt,zU,

and q is the best dominant. □

Corollary 2.10 Let h(z)= 1 + ( 2 β 1 ) z 1 + z be a convex function in U, where 0β<1.

If α,λ0, n,mN, f A n and satisfies the differential subordination

1 R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) [ ( R D λ , α m f ( z ) ) ] 2 h(z),zU,
(2.10)

then

R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) q(z),zU,

where q is given by q(z)=(2β1)+ 2 ( 1 β ) n z 1 n 0 z t 1 n 1 1 + t dt, zU. The function q is convex, and it is the best dominant.

Proof Following the same steps as in the proof of Theorem 2.9 and considering p(z)= R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) , the differential subordination (2.10) becomes

p(z)+z p (z)h(z)= 1 + ( 2 β 1 ) z 1 + z ,zU.

By using Lemma 1.1 for γ=1, we have p(z)q(z), i.e.,

R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) q ( z ) = 1 n z 1 n 0 z h ( t ) t 1 n 1 d t = 1 n z 1 n 0 z 1 + ( 2 β 1 ) t 1 + t t 1 n 1 d t = 1 n z 1 n 0 z t 1 n 1 [ ( 2 β 1 ) + 2 ( 1 β ) 1 + t ] d t = ( 2 β 1 ) + 2 ( 1 β ) n z 1 n 0 z t 1 n 1 1 + t d t , z U .

 □

Example 2.1 Let h(z)= 1 z 1 + z be a convex function in U with h(0)=1 and Re( z h ( z ) h ( z ) +1)> 1 2 .

Let f(z)=z+ z 2 , zU. For n=1, m=1, λ= 1 2 , α=2, we obtain R D 1 2 , 2 1 f(z)= R 1 f(z)+2 D 1 2 1 f(z)=z f (z)+2( 1 2 f(z)+ 1 2 z f (z))=f(z)=z+ z 2 , zU.

Then ( R D 1 2 , 2 1 f ( z ) ) = f (z)=1+2z,

R D 1 2 , 2 1 f ( z ) z ( R D 1 2 , 2 1 f ( z ) ) = z + z 2 z ( 1 + 2 z ) = 1 + z 1 + 2 z , 1 R D 1 2 , 2 1 f ( z ) ( R D 1 2 , 2 1 f ( z ) ) [ ( R D 1 2 , 2 1 f ( z ) ) ] 2 = 1 ( z + z 2 ) 2 ( 1 + 2 z ) 2 = 2 z 2 + 2 z + 1 ( 1 + 2 z ) 2 .

We have q(z)= 1 z 0 z 1 t 1 + t dt=1+ 2 ln ( 1 + z ) z .

Using Theorem 2.9, we obtain

2 z 2 + 2 z + 1 ( 1 + 2 z ) 2 1 z 1 + z ,zU,

induce

1 + z 1 + 2 z 1+ 2 ln ( 1 + z ) z ,zU.

Theorem 2.11 Let g be a convex function such that g(0)=0, and let h be the function h(z)=g(z)+nz g (z), zU.

If α,λ0, n,mN, f A n and the differential subordination

[ ( R D λ , α m f ( z ) ) ] 2 +R D λ , α m f(z) ( R D λ , α m f ( z ) ) h(z),zU
(2.11)

holds, then

R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) z g(z),zU.

This result is sharp.

Proof Let p(z)= R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) z . We deduce that pH[0,n].

Differentiating, we obtain [ ( R D λ , α m f ( z ) ) ] 2 +R D λ , α m f(z) ( R D λ , α m f ( z ) ) =p(z)+z p (z), zU.

Using the notation in (2.11), the differential subordination becomes

p(z)+z p (z)h(z)=g(z)+nz g (z).

By using Lemma 1.2, we have

p(z)g(z),zU,i.e., R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) z g(z),zU,

and this result is sharp. □

Theorem 2.12 Let h be a holomorphic function, which satisfies the inequality Re(1+ z h ( z ) h ( z ) )> 1 2 , zU, and h(0)=0.

If α,λ0, n,mN, f A n and satisfies the differential subordination

[ ( R D λ , α m f ( z ) ) ] 2 +R D λ , α m f(z) ( R D λ , α m f ( z ) ) h(z),zU,
(2.12)

then

R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) z q(z),zU,

where q(z)= 1 n z 1 n 0 z h(t) t 1 n 1 dt. The function q is convex, and it is the best dominant.

Proof Let p(z)= R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) z , zU, pH[0,n].

Differentiating, we obtain [ ( R D λ , α m f ( z ) ) ] 2 +R D λ , α m f(z) ( R D λ , α m f ( z ) ) =p(z)+z p (z), zU, and (2.12) becomes

p(z)+z p (z)h(z),zU.

Using Lemma 1.1, we have

p ( z ) q ( z ) , z U , i.e. , R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) z q ( z ) = 1 n z 1 n 0 z h ( t ) t 1 n 1 d t , z U ,

and q is the best dominant. □

Corollary 2.13 Let h(z)= 1 + ( 2 β 1 ) z 1 + z be a convex function in U, where 0β<1.

If α,λ0, n,mN, f A n and satisfies the differential subordination

[ ( R D λ , α m f ( z ) ) ] 2 +R D λ , α m f(z) ( R D λ , α m f ( z ) ) h(z),zU,
(2.13)

then

R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) z q(z),zU,

where q is given by q(z)=(2β1)+ 2 ( 1 β ) n z 1 n 0 z t 1 n 1 1 + t dt, zU. The function q is convex, and it is the best dominant.

Proof Following the same steps as in the proof of Theorem 2.12 and considering p(z)= R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) z , the differential subordination (2.13) becomes

p(z)+z p (z)h(z)= 1 + ( 2 β 1 ) z 1 + z ,zU.

By using Lemma 1.1 for γ=1, we have p(z)q(z), i.e.,

R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) z q ( z ) = 1 n z 1 n 0 z h ( t ) t 1 n 1 d t = 1 n z 1 n 0 z 1 + ( 2 β 1 ) t 1 + t t 1 n 1 d t = 1 n z 1 n 0 z t 1 n 1 [ ( 2 β 1 ) + 2 ( 1 β ) 1 + t ] d t = ( 2 β 1 ) + 2 ( 1 β ) n z 1 n 0 z t 1 n 1 1 + t d t , z U .

 □

Example 2.2 Let h(z)= 1 z 1 + z be a convex function in U with h(0)=1 and Re( z h ( z ) h ( z ) +1)> 1 2 .

Let f(z)=z+ z 2 , zU. For n=1, m=1, λ= 1 2 , α=2, we obtain R D 1 2 , 2 1 f(z)= R 1 f(z)+2 D 1 2 1 f(z)=z f (z)+2( 1 2 f(z)+ 1 2 z f (z))=f(z)=z+ z 2 , zU.

Then ( R D 1 2 , 2 1 f ( z ) ) = f (z)=1+2z,

R D 1 2 , 2 1 f ( z ) ( R D 1 2 , 2 1 f ( z ) ) z = ( z + z 2 ) ( 1 + 2 z ) z = 2 z 2 + 3 z + 1 , [ ( R D 1 2 , 2 1 f ( z ) ) ] 2 + R D 1 2 , 2 1 f ( z ) ( R D 1 2 , 2 1 f ( z ) ) = ( 1 + 2 z ) 2 + ( z + z 2 ) 2 = 6 z 2 + 6 z + 1 .

We have q(z)= 1 z 0 z 1 t 1 + t dt=1+ 2 ln ( 1 + z ) z .

Using Theorem 2.12, we obtain

6 z 2 +6z+1 1 z 1 + z ,zU,

induce

2 z 2 +3z+11+ 2 ln ( 1 + z ) z ,zU.

Theorem 2.14 Let g be a convex function such that g(0)=0, and let h be the function h(z)=g(z)+ n z 1 δ g (z), zU.

If α,λ0, δ(0,1), n,mN, f A n and the differential subordination

( z R D λ , α m f ( z ) ) δ R D λ , α m + 1 f ( z ) 1 δ ( ( R D λ , α m + 1 f ( z ) ) R D λ , α m + 1 f ( z ) δ ( R D λ , α m f ( z ) ) R D λ , α m f ( z ) ) h(z),zU
(2.14)

holds, then

R D λ , α m + 1 f ( z ) z ( z R D λ , α m f ( z ) ) δ g(z),zU.

This result is sharp.

Proof Let p(z)= R D λ , α m + 1 f ( z ) z ( z R D λ , α m f ( z ) ) δ . We deduce that pH[1,n].

Differentiating, we obtain ( z R D λ , α m f ( z ) ) δ R D λ , α m + 1 f ( z ) 1 δ ( ( R D λ , α m + 1 f ( z ) ) R D λ , α m + 1 f ( z ) δ ( R D λ , α m f ( z ) ) R D λ , α m f ( z ) )=p(z)+ 1 1 δ z p (z), zU.

Using the notation in (2.14), the differential subordination becomes

p(z)+ 1 1 δ z p (z)h(z)=g(z)+ n z 1 δ g (z).

By using Lemma 1.2, we have

p(z)g(z),zU,i.e., R D λ , α m + 1 f ( z ) z ( z R D λ , α m f ( z ) ) δ g(z),zU,

and this result is sharp. □

Theorem 2.15 Let h be a holomorphic function, which satisfies the inequality Re(1+ z h ( z ) h ( z ) )> 1 2 , zU, and h(0)=1.

If α,λ0, δ(0,1), n,mN, f A n and satisfies the differential subordination

( z R D λ , α m f ( z ) ) δ R D λ , α m + 1 f ( z ) 1 δ ( ( R D λ , α m + 1 f ( z ) ) R D λ , α m + 1 f ( z ) δ ( R D λ , α m f ( z ) ) R D λ , α m f ( z ) ) h(z),zU,
(2.15)

then

R D λ , α m + 1 f ( z ) z ( z R D λ , α m f ( z ) ) δ q(z),zU,

where q(z)= 1 δ n z 1 δ n 0 z h(t) t 1 δ n 1 dt. The function q is convex, and it is the best dominant.

Proof Let p(z)= R D λ , α m + 1 f ( z ) z ( z R D λ , α m f ( z ) ) δ , zU, pH[0,n].

Differentiating, we obtain ( z R D λ , α m f ( z ) ) δ R D λ , α m + 1 f ( z ) 1 δ ( ( R D λ , α m + 1 f ( z ) ) R D λ , α m + 1 f ( z ) δ ( R D λ , α m f ( z ) ) R D λ , α m f ( z ) )=p(z)+ 1 1 δ z p (z), zU, and (2.15) becomes

p(z)+ 1 1 δ z p (z)h(z),zU.

Using Lemma 1.1, we have

p ( z ) q ( z ) , z U , i.e. , R D λ , α m + 1 f ( z ) z ( z R D λ , α m f ( z ) ) δ q ( z ) = 1 δ n z 1 δ n 0 z h ( t ) t 1 δ n 1 d t , z U ,

and q is the best dominant. □

Author’s contributions

The author drafted the manuscript, read and approved the final manuscript.

References

  1. Al-Oboudi FM: On univalent functions defined by a generalized Sălăgean operator. Int. J. Math. Math. Sci. 2004, 27: 1429–1436.

    Article  MathSciNet  MATH  Google Scholar 

  2. Sălăgean GS Lecture Notes in Math. 1013. In Subclasses of Univalent Functions. Springer, Berlin; 1983:362–372.

    Google Scholar 

  3. Ruscheweyh S: New criteria for univalent functions. Proc. Am. Math. Soc. 1975, 49: 109–115. 10.1090/S0002-9939-1975-0367176-1

    Article  MathSciNet  MATH  Google Scholar 

  4. Lupaş AA: On special differential subordinations using a generalized Sălăgean operator and Ruscheweyh derivative. J. Comput. Anal. Appl. 2011, 13(1):98–107.

    MathSciNet  MATH  Google Scholar 

  5. Lupaş AA: On a certain subclass of analytic functions defined by a generalized Sălăgean operator and Ruscheweyh derivative. Carpath. J. Math. 2012, 28(2):183–190.

    MATH  Google Scholar 

  6. Lupaş AA: On special differential superordinations using a generalized Sălăgean operator and Ruscheweyh derivative. Comput. Math. Appl. 2011, 61: 1048–1058. 10.1016/j.camwa.2010.12.055

    Article  MathSciNet  MATH  Google Scholar 

  7. Lupaş AA: Certain special differential superordinations using a generalized Sălăgean operator and Ruscheweyh derivative. An. Univ. Oradea, Fasc. Mat. 2011, XVIII: 167–178.

    MATH  Google Scholar 

  8. Lupaş AA: On special differential subordinations using Sălăgean and Ruscheweyh operators. Math. Inequal. Appl. 2009, 12(4):781–790.

    MathSciNet  MATH  Google Scholar 

  9. Lupaş AA: On a certain subclass of analytic functions defined by Sălăgean and Ruscheweyh operators. J. Math. Appl. 2009, 31: 67–76.

    MathSciNet  Google Scholar 

  10. Lupaş AA, Breaz D: On special differential superordinations using Sălăgean and Ruscheweyh operators. Geometric Function Theory and Applications 2010, 98–103. (Proc. of International Symposium, Sofia, 27–31 August 2010)

    Google Scholar 

  11. Lupaş AA: Some differential subordinations using Ruscheweyh derivative and Sălăgean operator. Adv. Differ. Equ. 2013., 2013: Article ID 150 10.1186/1687-1847-2013-150

    Google Scholar 

  12. Miller SS, Mocanu PT: Differential Subordinations: Theory and Applications. Dekker, New York; 2000.

    MATH  Google Scholar 

  13. Lupaş DAA: Subordinations and Superordinations. Lap Lambert Academic Publishing, Saarbrücken; 2011.

    Google Scholar 

Download references

Acknowledgements

The author thanks the referee for his/her valuable suggestions to improve the present article.

Author information

Authors and Affiliations

Authors

Corresponding author

Correspondence to Loriana Andrei.

Additional information

Competing interests

The author declares that she has no competing interests.

Rights and permissions

Open Access This article is distributed under the terms of the Creative Commons Attribution 2.0 International License (https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Reprints and permissions

About this article

Cite this article

Andrei, L. Differential subordinations using the Ruscheweyh derivative and the generalized Sălăgean operator. Adv Differ Equ 2013, 252 (2013). https://doi.org/10.1186/1687-1847-2013-252

Download citation

  • Received:

  • Accepted:

  • Published:

  • DOI: https://doi.org/10.1186/1687-1847-2013-252

Keywords