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Differential subordinations using the Ruscheweyh derivative and the generalized Sălăgean operator

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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.

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

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Keywords

  • differential subordination
  • convex function
  • best dominant
  • differential operator
  • generalized Sălăgean operator
  • Ruscheweyh derivative