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Geometric properties of certain analytic functions associated with generalized fractional integral operators

Journal of Inequalities and Applications20142014:177

https://doi.org/10.1186/1029-242X-2014-177

  • Received: 22 February 2014
  • Accepted: 16 April 2014
  • Published:

Abstract

Let be the class of normalized analytic functions in the unit disk and define the class P ( β ) = { f A : φ R  such that  Re [ e i φ ( f ( z ) β ) ] > 0 , z U } . In this paper we find conditions on the number β and the non-negative weight function λ ( t ) such that the integral transform V λ ( f ) ( z ) = 0 1 λ ( t ) f ( t z ) t d t is convex of order γ ( 0 γ 1 / 2 ) when f P ( β ) . Some interesting further consequences are also considered.

MSC:30C45, 33C50.

Keywords

  • Gaussian hypergeometric function
  • integral transform
  • convex function
  • starlike function
  • fractional integral operator

1 Introduction and definitions

Let denote the class of functions of the form
f ( z ) = z + n = 2 a n z n
(1.1)

which are analytic in the open unit disk U = { z C : | z | < 1 } . Also let , S ( γ ) and K ( γ ) denote the subclasses of consisting of functions which are univalent, starlike of order γ and convex of order γ in , respectively. In particular, the classes S ( 0 ) = S and K ( 0 ) = K are the familiar ones of starlike and convex functions in , respectively.

We note that
f ( z ) K ( γ ) z f ( z ) S ( γ )
(1.2)

for 0 γ < 1 .

Let a, b, and c be complex numbers with c 0 , 1 , 2 ,  . Then the Gaussian hypergeometric function F 1 2 is defined by
F 1 2 ( a , b ; c ; z ) = n = 0 ( a ) n ( b ) n ( c ) n z n n ! ,
(1.3)
where ( λ ) n is the Pochhammer symbol defined, in terms of the Gamma function, by
( λ ) n = Γ ( λ + n ) Γ ( λ ) = { 1 ( n = 0 ) , λ ( λ + 1 ) ( λ + n 1 ) ( n N ) .
For functions f j ( z ) ( j = 1 , 2 ) of the forms
f j ( z ) : = n = 0 a j , n + 1 z n + 1 ( a j , 1 : = 1 ; j = 1 , 2 ) ,
let ( f 1 f 2 ) ( z ) denote the Hadamard product or convolution of f 1 ( z ) and f 2 ( z ) , defined by
( f 1 f 2 ) ( z ) : = n = 0 a 1 , n + 1 a 2 , n + 1 z n + 1 ( a j , 1 : = 1 ; j = 1 , 2 ) .
By using (1.3), Hohlov [1] introduced the convolution operator H a , b , c by
H a , b , c ( f ) ( z ) : = z 2 F 1 ( a , b ; c ; z ) f ( z )
(1.4)
for f A . The three-parameter family of operators given by (1.4) contains as special cases several of the known linear integral or differential operators studied by a number of authors. This operator has been studied extensively by Ponnusamy [2], Kim and Rønning [3] and many others [4, 5]. In particular, if a = 1 in (1.4), then H 1 , b , c is the operator L ( b , c ) due to Carlson and Shaffer [6] which was defined by
L ( b , c ) f ( z ) = z 2 F 1 ( 1 , b ; c ; z ) f ( z ) .
Clearly, L ( b , c ) maps onto itself, and L ( c , b ) is the inverse of L ( b , c ) , provided that b 0 , 1 , 2 ,  . Furthermore, L ( b , b ) is the unit operator and
L ( b , c ) = L ( b , e ) L ( e , c ) = L ( e , c ) L ( b , e ) ( c , e 0 , 1 , 2 , ) .
(1.5)
Also, we note that
K ( γ ) = L ( 1 , 2 ) S ( γ ) ( 0 γ < 1 )
and
S ( γ ) = L ( 2 , 1 ) K ( γ ) ( 0 γ < 1 ) .
(1.6)

Various definitions of fractional calculus operators are given by many authors. We use here the following definition due to Saigo [7] (see also [5, 8]).

Definition 1 For λ > 0 , μ , ν R , the fractional integral operator I 0 , z λ , μ , ν is defined by
I 0 , z λ , μ , ν f ( z ) = z λ μ Γ ( λ ) 0 z ( z ζ ) λ 1 2 F 1 ( λ + μ , ν ; λ ; 1 ζ z ) f ( ζ ) d ζ ,
where f ( z ) is taken to be an analytic function in a simply connected region of the z-plane containing the origin with the order
f ( z ) = O ( | z | ϵ ) ( z 0 )
for ϵ > max { 0 , μ ν } 1 , and the multiplicity of ( z ζ ) λ 1 is removed by requiring log ( z ζ ) to be real when z ζ > 0 . With the aid of the above definition, Owa et al. [9] defined a modification of the fractional integral operator J 0 , z λ , μ , ν by
J 0 , z λ , μ , ν f ( z ) = Γ ( 2 μ ) Γ ( 2 + λ + ν ) Γ ( 2 μ + ν ) z μ I 0 , z λ , μ , ν f ( z )
for f ( z ) A and min { λ + ν , μ + ν , μ } > 2 . Then it is observed that J 0 , z λ , μ , ν also maps onto itself and
J 0 , z λ , μ , ν f ( z ) = L ( 2 , 2 μ ) L ( 2 μ + ν , 2 + λ + ν ) f ( z ) .
(1.7)
The function
s α ( z ) = z ( 1 z ) 2 ( 1 α ) ( 0 α < 1 )
is the well-known extremal function for the class S ( α ) . A function f ( z ) A is said to be in the class R ( α , γ ) if
( f s α ) ( z ) S ( γ ) ( 0 α < 1 ; 0 γ < 1 ) .
Note that
R ( α , γ ) = L ( 1 , 2 2 α ) S ( γ )
(1.8)
and R ( α , α ) R ( α ) is the subclass of consisting of prestarlike functions of order α which was introduced by Suffridge [10]. In [11], it is shown that R ( α ) S if and only if α 1 / 2 . For β < 1 we denote the class
P ( β ) = { f A : φ R  such that  Re [ e i φ ( f ( z ) β ) ] > 0 , z U } .
Throughout this paper we let λ : [ 0 , 1 ] R be a non-negative function with
0 1 λ ( t ) d t = 1 .
(1.9)
For certain specific subclasses of f A , many authors considered the geometric properties of the integral transform of the form
V λ ( f ) ( z ) = 0 1 λ ( t ) f ( t z ) t d t .
(1.10)

More recently, starlikeness of this general operator V λ ( f ) was discussed by Fournier and Ruscheweyh [12] by assuming that f P ( β ) . The method of proof is the duality principle developed mainly by Ruscheweyh [13]. This result was later extended by Ponnusamy and Rønning [14] by means of finding conditions such that V λ ( f ) carries P ( β ) into starlike functions of order γ, 0 γ 1 / 2 .

In this paper, we find conditions on β and the function λ ( t ) such that V λ ( f ) carries P ( β ) into K ( γ ) . As a consequence of this investigation, a number of new results are established.

2 Preliminaries

We begin by recalling the following results.

Lemma 1 ([15]; see also [5])

If f A and c a + 1 > b > 0 , then
H a , b , c ( f ) ( z ) = Γ ( c ) Γ ( a ) Γ ( b ) 0 1 ( 1 t ) c a b Γ ( c a b + 1 ) t b 2 2 F 1 ( c a , 1 a ; c a b + 1 ; 1 t ) f ( t z ) d t .
Remark 1 In view of Lemma 1, we see that the convolution operator (1.4) is an integral operator of the form (1.10) with
λ ( t ) = Γ ( c ) t b 1 ( 1 t ) c a b Γ ( a ) Γ ( b ) Γ ( c a b + 1 ) 2 F 1 ( c a , 1 a ; c a b + 1 ; 1 t ) .
For Λ : [ 0 , 1 ] R being integrable and positive on ( 0 , 1 ) , we define
L Λ ( h γ ) = inf z U 0 1 Λ ( t ) [ Re h γ ( z t ) z t 1 γ ( 1 + t ) ( 1 γ ) ( 1 + t ) 2 ] d t
and
M Λ ( h γ ) = inf z U 0 1 Λ ( t ) [ Re h γ ( z t ) 1 t γ ( 1 + t ) ( 1 γ ) ( 1 + t ) 3 ] d t ,
where 0 γ < 1 and
h γ ( z ) = z ( 1 + ϵ + 2 γ 1 2 2 γ z ) ( 1 z ) 2 , | ϵ | = 1 .
(2.1)

In [16], Ponnusamy and Rønning proved the following lemmas.

Lemma 2 Let Λ ( t ) be integrable on [ 0 , 1 ] and positive on ( 0 , 1 ) . If Λ ( t ) / ( 1 + t ) ( 1 t ) 1 + 2 γ is decreasing on ( 0 , 1 ) , then for 0 γ 1 / 2 we have L Λ ( h γ ) 0 .

Lemma 3 Let 0 γ < 1 and let λ ( t ) be given by (1.9). Define β < 1 by
β 1 β = 0 1 λ ( t ) [ 1 + γ ( 1 γ ) t ( 1 γ ) ( 1 + t ) 2 γ 1 γ log ( 1 + t ) t ] d t .
Assume that lim t 0 + t Λ ( t ) = 0 , where
Λ ( t ) = t 1 λ ( s ) d s / s .

Then V λ ( P ( β ) ) S ( γ ) if and only if L Λ ( h γ ) 0 .

We now find conditions on β and the non-negative weight function λ ( t ) such that V λ ( P ( β ) ) K ( γ ) .

Lemma 4 (i) Let Λ ( t ) be monotone decreasing on [ 0 , 1 ] satisfying Λ ( 1 ) = 0 and lim t 0 + t Λ ( t ) = 0 . For 0 γ 1 / 2 if t Λ ( t ) / ( 1 + t ) ( 1 t ) 1 + 2 γ is increasing on ( 0 , 1 ) , then M Λ ( h γ ) 0 .
  1. (ii)
    Let 0 γ 1 / 2 and let λ ( t ) and Λ ( t ) be as in Lemma 3. Define β < 1 by
    β 1 2 1 β = 0 1 λ ( t ) 1 γ ( 1 + t ) ( 1 γ ) ( 1 + t ) 2 d t .
     

Then V λ ( P ( β ) ) K ( γ ) if and only if M Λ ( h γ ) 0 .

Proof (i) Let M Λ ( h γ ) = inf z U I γ . Then, by using the conditions Λ ( 1 ) = 0 and lim t 0 + t Λ ( t ) = 0 , an integration by parts yields
I γ = 0 1 Λ ( t ) [ Re h γ ( z t ) 1 t γ ( 1 + t ) ( 1 γ ) ( 1 + t ) 3 ] d t = 0 1 Λ ( t ) d d t [ Re h γ ( z t ) z t ( 1 γ ( 1 + t ) ) ( 1 γ ) ( 1 + t ) 2 ] d t = 0 1 t Λ ( t ) [ Re h γ ( z t ) z t 1 γ ( 1 + t ) ( 1 γ ) ( 1 + t ) 2 ] d t .
Since t Λ ( t ) / ( 1 + t ) ( 1 t ) 1 + 2 γ is increasing on ( 0 , 1 ) , by Lemma 2, inf z U I γ 0 , which evidently completes the proof of (i).
  1. (ii)
    We state this proof only in outline here because the proof is similar to that of [[3], Theorem 2.1]. Let F ( z ) = V λ ( f ) ( z ) . Then, by convolution theory [[13], p.94] and (1.2), we have
    F ( z ) K ( γ ) 1 z ( z F ( z ) h γ ( z ) ) 0 ,
    (2.2)
     
where h γ ( z ) is given by (2.1). Since f P ( β ) , by the duality principle [[13], p.23], it is enough to verify this with f given by
f ( z ) = ( 1 β ) 1 x z 1 y z + β ( | x | = | y | = 1 ) .
In the same way as in [[3], Theorem 2.1], we conclude that (2.2) holds if and only if
Re 0 1 λ ( t ) [ h γ ( z t ) z t 1 γ ( 1 + t ) ( 1 γ ) ( 1 + t ) 2 ] d t > 0 .
(2.3)
Integrating by parts, we find that the inequality (2.3) is equivalent to
Re 0 1 Λ ( t ) [ h γ ( z t ) 1 t γ ( 1 + t ) ( 1 γ ) ( 1 + t ) 3 ] d t 0 ,

which again is equivalent to M Λ ( h γ ) 0 . □

Remark 2 In particular, taking γ = 0 in Lemma 4, we obtain the result due to Ali and Singh [[17], Theorem 1].

3 Main results

We define
φ ( 1 t ) = 1 + n = 1 b n ( 1 t ) n ( b n 0 )
(3.1)
and
λ ( t ) = C t b 1 ( 1 t ) c a b φ ( 1 t ) ,
(3.2)
where C is a constant satisfying the condition (1.9). For f A Balasubramanian et al. [4] defined the operator P a , b , c by
P a , b , c ( f ) ( z ) = 0 1 λ ( t ) f ( t z ) t d t ,
where λ ( t ) is given by (3.2). Special choices of φ ( 1 t ) and C led to various interesting geometric properties concerning certain linear operators. For example, if we take φ ( 1 t ) = 2 F 1 ( c a , 1 a ; c a b + 1 ; 1 t ) and
C = Γ ( c ) Γ ( a ) Γ ( b ) Γ ( c a b + 1 ) ,
by virtue of Remark 1,
P a , b , c ( f ) ( z ) = H a , b , c ( f ) ( z ) .
(3.3)

First, by applying Lemma 4, we prove the following.

Theorem 1 Let 0 γ 1 / 2 , a > 0 , 0 < b 1 , and c a + b + 2 γ + 1 , and let λ ( t ) be given by (3.2). Define β = β ( a , b , c , γ ) by
β 1 2 1 β = 0 1 λ ( t ) 1 γ ( 1 + t ) ( 1 γ ) ( 1 + t ) 2 d t .

If f ( z ) P ( β ) , then P a , b , c ( f ) ( z ) K ( γ ) . The value of β is sharp.

Proof Let C > 0 and
Λ ( t ) = t 1 λ ( s ) s d s ,
where λ ( t ) is given by (3.2). Then it is easily seen that Λ ( t ) is monotone decreasing on [ 0 , 1 ] and lim t 0 + t Λ ( t ) = 0 . In order to apply Lemma 4, we want to prove that the function
u ( t ) = λ ( t ) ( 1 + t ) ( 1 t ) 1 + 2 γ
(3.4)
is decreasing on ( 0 , 1 ) , where λ ( t ) is given by (3.2). Making use of the logarithmic differentiation of both sides in (3.4), we have
u ( t ) u ( t ) = λ ( t ) λ ( t ) + 2 ( γ + ( 1 + γ ) t ) 1 t 2 .
(3.5)
Since
λ ( t ) = C t b 2 ( 1 t ) c a b 1 [ φ ( 1 t ) { ( b 1 ) ( 1 t ) t ( c a b ) } t ( 1 t ) φ ( 1 t ) ] ,
from (3.4) and (3.5) we find that u ( t ) 0 on ( 0 , 1 ) is equivalent to
( c a 3 2 γ ) t 2 + ( c a b 2 γ ) t + 1 b t ( 1 t 2 ) φ ( 1 t ) φ ( 1 t ) ( 0 < t < 1 ) .
(3.6)

In view of (3.1), φ ( 1 t ) > 0 and φ ( 1 t ) 0 on ( 0 , 1 ) , so that the right hand side of the inequality (3.6) is non-positive for all t ( 0 , 1 ) . If we assume that 0 γ 1 / 2 , a > 0 , 0 < b 1 , and c a + b + 2 γ + 1 , then ( c a 3 2 γ ) t 2 + ( c a b 2 γ ) t + 1 b 0 for t ( 0 , 1 ) . Thus, the inequality (3.6) holds for all t ( 0 , 1 ) . Hence, from Lemma 4 we obtain P a , b , c ( f ) ( z ) K ( γ ) . □

The same techniques as in the proof of [[5], Theorem 1] show that the value β is sharp.

By using (3.3) and Theorem 1, we have the following.

Corollary 1 Let 0 γ 1 / 2 , 0 < a 1 , 0 < b 1 , and c a + b + 2 γ + 1 . Define β = β ( a , b , c , γ ) by
β 1 2 1 β = Γ ( c ) Γ ( a ) Γ ( b ) 0 1 ( 1 t ) c a b t b 1 Γ ( c a b + 1 ) 1 γ ( 1 + t ) ( 1 γ ) ( 1 + t ) 2 × 2 F 1 ( c a , 1 a ; c a b + 1 ; 1 t ) d t .

If f ( z ) P ( β ) , then H a , b , c ( f ) ( z ) K ( γ ) . The value of β is sharp.

Proof If we put
λ ( t ) = Γ ( c ) t b 1 ( 1 t ) c a b Γ ( a ) Γ ( b ) Γ ( c a b + 1 ) 2 F 1 ( c a , 1 a ; c a b + 1 ; 1 t ) ,

then, by applying (3.3) and Theorem 1, we obtain the desired result. □

Setting a = 1 in Corollary 1, we obtain the following.

Corollary 2 Let 0 γ 1 / 2 , 0 < b 1 , and c b + 2 γ + 2 . Also let
β ( 1 , b , c , γ ) = 1 1 γ 2 [ 1 2 F 1 ( 2 , b ; c ; 1 ) γ ( 1 2 F 1 ( 1 , b ; c ; 1 ) ) ] .

If β ( 1 , b , c , γ ) < β < 1 and f ( z ) P ( β ) , then L ( b , c ) f ( z ) K ( γ ) .

Next we find a univalence criterion for the operator J 0 , z λ , μ , ν .

Theorem 2 Let 0 γ 1 / 2 , 0 μ < 2 , λ 2 ( 1 + γ ) μ , and μ 2 < ν μ 1 . Define β = β ( λ , μ , ν , γ ) by
β = 1 1 γ 2 [ 1 2 F 1 ( 2 , 2 μ + ν ; 2 + λ + ν ; 1 ) γ ( 1 2 F 1 ( 1 , 2 μ + ν ; 2 + λ + ν ; 1 ) ) ] .

If f ( z ) P ( β ) , then J 0 , z λ , μ , ν f ( z ) R ( μ / 2 , γ ) .

Proof Making use of (1.5) and (1.7), we note that
J 0 , z λ , μ , ν f ( z ) = L ( 2 , 2 μ ) L ( 2 μ + ν , 2 + λ + ν ) f ( z ) = L ( 1 , 2 μ ) L ( 2 , 1 ) L ( 2 μ + ν , 2 + λ + ν ) f ( z ) .
(3.7)
By using Corollary 2, we obtain
L ( 2 μ + ν , 2 + λ + ν ) f ( z ) K ( γ ) .

Since 0 μ < 2 , from (1.6), (1.8) and (3.7) we have J 0 , z λ , μ , ν f ( z ) R ( μ / 2 , γ ) , which completes the proof of Theorem 2. □

Taking μ = 2 γ in Theorem 2, we get the following.

Corollary 3 Let 0 γ 1 / 2 , λ 2 , and 2 ( γ 1 ) < ν 2 γ 1 . Define β = β ( λ , ν , γ ) by
β = 1 1 γ 2 [ 1 2 F 1 ( 2 , 2 ( 1 γ ) + ν ; 2 + λ + ν ; 1 ) γ ( 1 2 F 1 ( 1 , 2 ( 1 γ ) + ν ; 2 + λ + ν ; 1 ) ) ] .

If f ( z ) P ( β ) , then J 0 , z λ , 2 γ , ν f ( z ) R ( γ ) S .

Proof If we put μ = 2 γ in Theorem 2, then
J 0 , z λ , 2 γ , ν f ( z ) R ( γ , γ ) = R ( γ ) .

Since γ 1 / 2 , R ( γ ) S , so that the proof is completed. □

Remark 3 In [4], Balasubramanian et al. found the conditions on the number β and the function λ ( t ) such that P a , b , c ( f ) ( z ) S ( γ ) ( 0 γ 1 / 2 ). Since J 0 , z λ , μ , ν f ( z ) = P 1 ν , 2 , λ ν + 2 ( f ) ( z ) with φ ( 1 t ) = 2 F 1 ( λ + μ , ν ; λ ; 1 t ) and
C = Γ ( 2 μ ) Γ ( 2 + λ + ν ) Γ ( λ ) Γ ( 2 μ + ν ) ,

the condition on β and λ ( t ) is easily found such that J 0 , z λ , μ , ν f ( z ) S ( γ ) .

Finally, by using Lemma 4 again, we investigate convexity of the operator J 0 , z λ , μ , ν .

Theorem 3 Let 0 γ 1 / 2 , 0 < λ 1 + 2 γ , 2 < μ < 3 , 0 < ν 1 , and ν > μ 2 . Define β = β ( λ , μ , ν , γ ) by
β 1 2 1 β = Γ ( 2 μ ) Γ ( 2 + λ + ν ) Γ ( λ ) Γ ( 2 μ + ν ) 0 1 t ( 1 t ) λ 1 ( 1 γ ( 1 + t ) ) ( 1 γ ) ( 1 + t ) 2 2 F 1 ( λ + μ , ν ; λ ; 1 t ) d t .

If f ( z ) P ( β ) , then J 0 , z λ , μ , ν f ( z ) K ( γ ) . The value of β is sharp.

Proof Let 0 γ 1 / 2 , 0 < λ 1 + 2 γ , 2 < μ < 3 , and ν > μ 2 , and let
λ ( t ) = Γ ( 2 μ ) Γ ( 2 + λ + ν ) Γ ( λ ) Γ ( 2 μ + ν ) t ( 1 t ) λ 1 2 F 1 ( λ + μ , ν ; λ ; 1 t ) .
(3.8)

Then we can easily see that 0 1 λ ( t ) d t = 1 , Λ ( t ) = t 1 λ ( s ) d s / s is monotone decreasing on [ 0 , 1 ] and lim t 0 + t Λ ( t ) = 0 . Also we find that the function u ( t ) = λ ( t ) / ( 1 + t ) ( 1 t ) 1 + 2 γ is decreasing on ( 0 , 1 ) , where λ ( t ) is given by (3.8). Hence, t Λ ( t ) / ( 1 + t ) ( 1 t ) 1 + 2 γ = u ( t ) is increasing on ( 0 , 1 ) . From Lemma 4, we obtain the desired result. □

Declarations

Acknowledgements

The first author was supported by Yeungnam University (2013).

Authors’ Affiliations

(1)
Department of Mathematics Education, Yeungnam University, Gyongsan, 712-749, Korea
(2)
Department of Mathematics Education, Daegu National University of Education, Daegu, 705-715, Korea

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© Kim and Choi; licensee Springer. 2014

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