Skip to main content

Some differential subordinations using Ruscheweyh derivative and Sălăgean operator

Abstract

In the present paper we study the operator defined by using the Ruscheweyh derivative R n f(z) and the Sălăgean operator S n f(z), denoted by L α n :AA, L α n f(z)=(1α) R n f(z)+α S n 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 with A 1 =A. We obtain several differential subordinations regarding the operator L α n .

MSC:30C45, 30A20, 34A40.

1 Introduction

Denote by U the unit disc of the complex plane, U={zC:|z|<1}, and by H(U) the space of holomorphic functions in U. Let A n ={fH(U):f(z)=z+ a n + 1 z n + 1 +,zU} with A 1 =A 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={fA: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 a 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 (Sălăgean [1])

For fA, nN, the operator S n is defined by S n :AA,

S 0 f ( z ) = f ( z ) , S 1 f ( z ) = z f ( z ) , S n + 1 f ( z ) = z ( S n f ( z ) ) , z U .

Remark 1.1 If fA, f(z)=z+ j = 2 a j z j , then S n f(z)=z+ j = 2 j n a j z j , zU.

Definition 1.2 (Ruscheweyh [2])

For fA, nN, the operator R n is defined by R n :AA,

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

Remark 1.2 If fA, f(z)=z+ j = 2 a j z j , then R n f(z)=z+ j = 2 C n + j 1 n a j z j , zU.

Definition 1.3 ([3])

Let α0, nN. Denote by L α n the operator given by L α n :AA,

L α n f(z)=(1α) R n f(z)+α S n f(z),zU.

Remark 1.3 If fA, f(z)=z+ j = 2 a j z j , then L α n f(z)=z+ j = 2 (α j n +(1α) C n + j 1 n ) a j z j , zU.

This operator was studied also in [35].

Lemma 1.1 (Hallenbeck and Ruscheweyh [[6], 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 [6])

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)+ z δ g (z), zU.

If α,δ0, nN, fA and satisfies the differential subordination

( L α n f ( z ) z ) δ 1 ( L α n f ( z ) ) h(z),zU,
(2.1)

then

( L α n f ( z ) z ) δ g(z),zU,

and this result is sharp.

Proof By using the properties of the operator L α n , we have

L α n f(z)=z+ j = 2 ( α j n + ( 1 α ) C n + j 1 n ) a j z j ,zU.

Consider p(z)= ( L α n f ( z ) z ) δ = ( z + j = 2 ( α j n + ( 1 α ) C n + j 1 n ) a j z j z ) δ =1+ p δ z δ + p δ + 1 z δ + 1 + , zU.

We deduce that pH[1,δ].

Differentiating, we obtain ( L α n f ( z ) z ) δ 1 ( L α n 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)+ z δ g (z),zU.

By using Lemma 1.2, we have

p(z)g(z),zU, i.e. , ( L α n 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,δ0, nN, fA and satisfies the differential subordination

( L α n f ( z ) z ) δ 1 ( L α n f ( z ) ) h(z),zU,
(2.2)

then

( L α n f ( z ) z ) δ q(z),zU,

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

Proof Let

p ( z ) = ( L α n f ( z ) z ) δ = ( z + j = 2 ( α j n + ( 1 α ) C n + j 1 n ) a j z j z ) δ = ( 1 + j = 2 ( α j n + ( 1 α ) C n + j 1 n ) a j z j 1 ) δ = 1 + j = δ + 1 p j z j 1

for zU, pH[1,δ].

Differentiating, we obtain ( L α n f ( z ) z ) δ 1 ( L α n 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. , ( L α n f ( z ) z ) δ q(z)= δ z δ 0 z h(t) t δ 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,δ0, nN, fA and satisfies the differential subordination

( L α n f ( z ) z ) δ 1 ( L α n f ( z ) ) h(z),zU,
(2.3)

then

( L α n f ( z ) z ) δ q(z),zU,

where q is given by q(z)=(2β1)+ 2 ( 1 β ) δ z δ 0 z t δ 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)= ( L α n 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.,

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

 □

Remark 2.1 For n=1, α=2, δ=1, we obtain the same example as in [[7], Example 2.2.1, p.26].

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

If α0, nN, fA and the differential subordination

( γ + 1 ) z γ L α n f ( z ) ( L α n + 1 f ( z ) ) 2 + z 2 γ L α n f ( z ) ( L α n + 1 f ( z ) ) 2 [ ( L α n f ( z ) ) L α n f ( z ) 2 ( L α n + 1 f ( z ) ) L α n + 1 f ( z ) ] h(z),zU,
(2.4)

holds, then

z L α n f ( z ) ( L α n + 1 f ( z ) ) 2 g(z),zU,

and this result is sharp.

Proof For fA, f(z)=z+ j = 2 a j z j , we have L α n f(z)=z+ j = 2 (α j n +(1α) C n + j 1 n ) a j z j , zU.

Consider p(z)=z L α n f ( z ) ( L α n + 1 f ( z ) ) 2 and we obtain p(z)+ z γ p (z)= ( γ + 1 ) z γ L α n f ( z ) ( L α n + 1 f ( z ) ) 2 + z 2 γ L α n f ( z ) ( L α n + 1 f ( z ) ) 2 [ ( L α n f ( z ) ) L α n f ( z ) 2 ( L α n + 1 f ( z ) ) L α n + 1 f ( z ) ].

Relation (2.4) becomes

p(z)+ z γ p (z)h(z)=g(z)+γ g (z),zU.

By using Lemma 1.2, we have

p(z)g(z),zU, i.e. , z L α n f ( z ) ( L α n + 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 α0, γC{0} is a complex number with Reγ0, nN, fA and satisfies the differential subordination

( γ + 1 ) z γ L α n f ( z ) ( L α n + 1 f ( z ) ) 2 + z 2 γ L α n f ( z ) ( L α n + 1 f ( z ) ) 2 [ ( L α n f ( z ) ) L α n f ( z ) 2 ( L α n + 1 f ( z ) ) L α n + 1 f ( z ) ] h(z),zU,
(2.5)

then

z L α n f ( z ) ( L α n + 1 f ( z ) ) 2 q(z),zU,

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

Proof Let p(z)=z L α n f ( z ) ( L α n + 1 f ( z ) ) 2 , zU, pH[1,1]. Differentiating, we obtain p(z)+ z γ p (z)= ( γ + 1 ) z γ L α n f ( z ) ( L α n + 1 f ( z ) ) 2 + z 2 γ L α n f ( z ) ( L α n + 1 f ( z ) ) 2 [ ( L α n f ( z ) ) L α n f ( z ) 2 ( L α n + 1 f ( z ) ) L α n + 1 f ( z ) ], zU, and (2.5) becomes

p(z)+ z γ p (z)h(z),zU.

Using Lemma 1.1, we have

p(z)q(z),zU, i.e. , z L α n f ( z ) ( L α n + 1 f ( z ) ) 2 q(z)= γ z γ 0 z h(t) t γ 1 dt,zU,

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)+ z γ g (z), zU, where γ>0.

If α0, nN, fA and the differential subordination

( γ + 2 ) z 2 γ ( L α n f ( z ) ) L α n f ( z ) + z 3 γ [ ( L α n f ( z ) ) L α n f ( z ) ( ( L α n f ( z ) ) L α n f ( z ) ) 2 ] h(z),zU,
(2.6)

holds, then

z 2 ( L α n f ( z ) ) L α n f ( z ) g(z),zU.

This result is sharp.

Proof Let p(z)= z 2 ( L α n f ( z ) ) L α n f ( z ) . We deduce that pH[0,1].

Differentiating, we obtain p(z)+ z γ p (z)= ( γ + 2 ) z 2 γ ( L α n f ( z ) ) L α n f ( z ) + z 3 γ [ ( L α n f ( z ) ) L α n f ( z ) ( ( L α n f ( z ) ) L α n f ( z ) ) 2 ], zU.

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

p(z)+ 1 γ z p (z)h(z)=g(z)+ z γ g (z).

By using Lemma 1.2, we have

p(z)g(z),zU, i.e. , z 2 ( L α n f ( z ) ) L α n f ( z ) g(z),zU,

and this result is sharp. □

Theorem 2.7 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 α0, γC{0} is a complex number with Reγ0, nN, fA and satisfies the differential subordination

( γ + 2 ) z 2 γ ( L α n f ( z ) ) L α n f ( z ) + z 3 γ [ ( L α n f ( z ) ) L α n f ( z ) ( ( L α n f ( z ) ) L α n f ( z ) ) 2 ] h(z),zU,
(2.7)

then

z 2 ( L α n f ( z ) ) L α n f ( z ) q(z),zU,

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

Proof Let p(z)= z 2 ( L α n f ( z ) ) L α n f ( z ) , zU, pH[0,1].

Differentiating, we obtain p(z)+ z γ p (z)= ( γ + 2 ) z 2 γ ( L α n f ( z ) ) L α n f ( z ) + z 3 γ [ ( L α n f ( z ) ) L α n f ( z ) ( ( L α n f ( z ) ) L α n f ( z ) ) 2 ], zU, and (2.7) becomes

p(z)+ 1 γ z p (z)h(z),zU.

Using Lemma 1.1, we have

p(z)q(z),zU, i.e. , z 2 ( L α n f ( z ) ) L α n f ( z ) q(z)= γ z γ 0 z h(t) t γ 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)+z g (z), zU.

If α0, nN, fA and the differential subordination

1 L α n f ( z ) ( L α n f ( z ) ) [ ( L α n f ( z ) ) ] 2 h(z),zU,
(2.8)

holds, then

L α n f ( z ) z ( L α n f ( z ) ) g(z),zU.

This result is sharp.

Proof Let p(z)= L α n f ( z ) z ( L α n f ( z ) ) . We deduce that pH[1,1].

Differentiating, we obtain 1 L α n f ( z ) ( L α n f ( z ) ) [ ( L α n 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)+z g (z).

By using Lemma 1.2, we have

p(z)g(z),zU, i.e. , L α n f ( z ) z ( L α n 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 α0, nN, fA and satisfies the differential subordination

1 L α n f ( z ) ( L α n f ( z ) ) [ ( L α n f ( z ) ) ] 2 h(z),zU,
(2.9)

then

L α n f ( z ) z ( L α n f ( z ) ) q(z),zU,

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

Proof Let p(z)= L α n f ( z ) z ( L α n f ( z ) ) , zU, pH[0,1].

Differentiating, we obtain 1 L α n f ( z ) ( L α n f ( z ) ) [ ( L α n 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. , L α n f ( z ) z ( L α n f ( z ) ) q(z)= 1 z 0 z h(t)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 α0, nN, fA and satisfies the differential subordination

1 L α n f ( z ) ( L α n f ( z ) ) [ ( L α n f ( z ) ) ] 2 h(z),zU,
(2.10)

then

L α n f ( z ) z ( L α n f ( z ) ) q(z),zU,

where q is given by q(z)=(2β1)+2(1β) ln ( 1 + z ) z , 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)= L α n f ( z ) z ( L α n 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.,

L α n f ( z ) z ( L α n f ( z ) ) q ( z ) = 1 z 0 z h ( t ) d t = 1 z 0 z 1 + ( 2 β 1 ) t 1 + t d t = 1 z 0 z [ ( 2 β 1 ) + 2 ( 1 β ) 1 + t ] d t = ( 2 β 1 ) + 2 ( 1 β ) ln ( 1 + z ) z , 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, α=2, we obtain L 2 1 f(z)= R 1 f(z)+2 S 1 f(z)=z f (z)+2z f (z)=z f (z)=z+2 z 2 .

Then ( L 2 1 f ( z ) ) =1+4z,

L 2 1 f ( z ) z ( L 2 1 f ( z ) ) = z + 2 z 2 z ( 1 + 4 z ) = 1 + 2 z 1 + 4 z , 1 L 2 1 f ( z ) ( L 2 1 f ( z ) ) [ ( L 2 1 f ( z ) ) ] 2 = 1 ( z + 2 z 2 ) 4 ( 1 + 4 z ) 2 = 8 z 2 + 4 z + 1 ( 1 + 4 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

8 z 2 + 4 z + 1 ( 1 + 4 z ) 2 1 z 1 + z ,zU,

induce

1 + 2 z 1 + 4 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)+z g (z), zU.

If α0, nN, fA and the differential subordination

[ ( L α n f ( z ) ) ] 2 + L α n f(z) ( L α n f ( z ) ) h(z),zU,
(2.11)

holds, then

L α n f ( z ) ( L α n f ( z ) ) z g(z),zU.

This result is sharp.

Proof Let p(z)= L α n f ( z ) ( L α n f ( z ) ) z . We deduce that pH[0,1].

Differentiating, we obtain [ ( L α n f ( z ) ) ] 2 + L α n f(z) ( L α n 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)+z g (z).

By using Lemma 1.2, we have

p(z)g(z),zU, i.e. , L α n f ( z ) ( L α n 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 α0, nN, fA and satisfies the differential subordination

[ ( L α n f ( z ) ) ] 2 + L α n f(z) ( L α n f ( z ) ) h(z),zU,
(2.12)

then

L α n f ( z ) ( L α n f ( z ) ) z q(z),zU,

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

Proof Let p(z)= L α n f ( z ) ( L α n f ( z ) ) z , zU, pH[0,1].

Differentiating, we obtain [ ( L α n f ( z ) ) ] 2 + L α n f(z) ( L α n 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),zU, i.e. , L α n f ( z ) ( L α n f ( z ) ) z q(z)= 1 z 0 z h(t)dt,zU,

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 α0, nN, fA and satisfies the differential subordination

[ ( L α n f ( z ) ) ] 2 + L α n f(z) ( L α n f ( z ) ) h(z),zU,
(2.13)

then

L α n f ( z ) ( L α n f ( z ) ) z q(z),zU,

where q is given by q(z)=(2β1)+2(1β) ln ( 1 + z ) z , 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)= L α n f ( z ) ( L α n 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.,

L α n f ( z ) ( L α n f ( z ) ) z q ( z ) = 1 z 0 z h ( t ) d t = 1 z 0 z 1 + ( 2 β 1 ) t 1 + t d t = 1 z 0 z [ ( 2 β 1 ) + 2 ( 1 β ) 1 + t ] d t = ( 2 β 1 ) + 2 ( 1 β ) ln ( 1 + z ) z , 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, α=2, we obtain L 2 1 f(z)= R 1 f(z)+2 S 1 f(z)=z f (z)+2z f (z)=z f (z)=z+2 z 2 , zU.

Then ( L 2 1 f ( z ) ) =1+4z,

L 2 1 f ( z ) ( L 2 1 f ( z ) ) z = ( z + 2 z 2 ) ( 1 + 4 z ) z = 8 z 2 + 6 z + 1 , [ ( L 2 1 f ( z ) ) ] 2 + L 2 1 f ( z ) ( L 2 1 f ( z ) ) = ( 1 + 4 z ) 2 + ( z + 2 z 2 ) 4 = 24 z 2 + 12 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

24 z 2 +12z+1 1 z 1 + z ,zU,

induce

8 z 2 +6z+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)+ z 1 δ g (z), zU.

If α0, δ(0,1), nN, fA and the differential subordination

( z L α n f ( z ) ) δ L α n + 1 f ( z ) 1 δ ( ( L α n + 1 f ( z ) ) L α n + 1 f ( z ) δ ( L α n f ( z ) ) L α n f ( z ) ) h(z),zU,
(2.14)

holds, then

L α n + 1 f ( z ) z ( z L α n f ( z ) ) δ g(z),zU.

This result is sharp.

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

Differentiating, we obtain ( z L α n f ( z ) ) δ L α n + 1 f ( z ) 1 δ ( ( L α n + 1 f ( z ) ) L α n + 1 f ( z ) δ ( L α n f ( z ) ) L α n 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)+ z 1 δ g (z).

By using Lemma 1.2, we have

p(z)g(z),zU, i.e. , L α n + 1 f ( z ) z ( z L α n 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 α0, δ(0,1), nN, fA and satisfies the differential subordination

( z L α n f ( z ) ) δ L α n + 1 f ( z ) 1 δ ( ( L α n + 1 f ( z ) ) L α n + 1 f ( z ) δ ( L α n f ( z ) ) L α n f ( z ) ) h(z),zU,
(2.15)

then

L α n + 1 f ( z ) z ( z L α n f ( z ) ) δ q(z),zU,

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

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

Differentiating, we obtain ( z L α n f ( z ) ) δ L α n + 1 f ( z ) 1 δ ( ( L α n + 1 f ( z ) ) L α n + 1 f ( z ) δ ( L α n f ( z ) ) L α n 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. , L α n + 1 f ( z ) z ( z L α n f ( z ) ) δ q ( z ) = 1 δ z 1 δ 0 z h ( t ) t δ d t , z U ,

and q is the best dominant. □

References

  1. 1.

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

    Google Scholar 

  2. 2.

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

    MathSciNet  Article  Google Scholar 

  3. 3.

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

    MathSciNet  Google Scholar 

  4. 4.

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

    MathSciNet  Google Scholar 

  5. 5.

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

    Google Scholar 

  6. 6.

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

    Google Scholar 

  7. 7.

    Alb Lupaş, DA: Subordinations and Superordinations. Lambert Academic Publishing (2011)

    Google Scholar 

Download references

Acknowledgements

Dedicated to Professor Hari M Srivastava.

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

Author information

Affiliations

Authors

Corresponding author

Correspondence to Alina Alb Lupaş.

Additional information

Competing interests

The author declares that she has no competing interests.

Authors’ contributions

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

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

Alb Lupaş, A. Some differential subordinations using Ruscheweyh derivative and Sălăgean operator. Adv Differ Equ 2013, 150 (2013). https://doi.org/10.1186/1687-1847-2013-150

Download citation

Keywords

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