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Properties of multivalent functions associated with the integral operator defined by the hypergeometric function

Abstract

In this paper, we introduce a new class of multivalent functions by using a generalized integral operator defined by the hypergeometric function. Some properties such as inclusion, radius problem and integral preserving are considered.

MSC:30C45, 30C50.

1 Introduction and preliminaries

Let A p denote the class of functions f(z) of the form

f(z)= z p + n = p + 1 a n z n ( p N = { 1 , 2 , 3 , } ) ,
(1.1)

which are analytic in the open unit disc E. Also A 1 =A, the usual class of analytic functions defined in the open unit disc E={z:|z|<1}. A function f A p is a p-valent starlike function of order ρ if and only if

Re z f ( z ) f ( z ) >ρ,0ρ<p,zE.

This class of functions is denoted by S p (ρ). It is noted that S p (0)= S p . Let f(z) and g(z) be analytic in E, we say f(z) is subordinate to g(z), written fg or f(z)g(z) if there exists a Schwarz function w(z), w(0)=0 and |w(z)|<1 in E, then f(z)=g(w(z)). In particular, if g is univalent in E, then we have the following equivalence

f(z)g(z)f(0)=g(0)andf(E)g(E).

For any two analytic functions f(z) and g(z) with

f(z)= n = 0 b n z n + 1 andg(z)= n = 0 c n z n + 1 ,zE,

the convolution (the Hadamard product) is given by

(fg)(z)= n = 0 b n c n z n + 1 ,zE.

A function fA is said to be in the class, denoted by SD(k,δ) (0δ<1), if and only if

Re { z f ( z ) f ( z ) } >k| z f ( z ) f ( z ) 1|+δ,k0,zE.
(1.2)

Similarly, a function fA is said to be in the class, denoted by CD(k,δ) of k-uniformly convex of order δ (0δ<1), if

Re { 1 + z f ( z ) f ( z ) } >k| z f ( z ) f ( z ) |+δ,k0,zE.
(1.3)

Geometric interpretation The functions fSD(k,δ) and fCD(k,δ) if and only if z f ( z ) f ( z ) and z f ( z ) f ( z ) +1, respectively, take all the values in the conic domain Ω k , δ defined by

Ω k , δ = { u + i v : u > k ( u 1 ) 2 + v 2 + δ }

with p(z)= z f ( z ) f ( z ) or p(z)= z f ( z ) f ( z ) +1 and considering the functions which map E onto the conic domain Ω k , δ such that 1 Ω k , δ . One may rewrite the conditions (1.2) or (1.3) in the form

p(z) q k , δ (z).

The function q k , δ (z) plays the role of extremal for these classes and is given by

q k , δ (z)={ 1 + ( 1 2 δ ) z 1 z , k = 0 , 1 + 2 δ γ π 2 ( log 1 + z 1 z ) 2 , k = 1 , 1 + 2 δ 1 k 2 sinh 2 [ ( 2 π arccos k ) arctan h z ] , 0 < k < 1 , 1 + δ k 2 1 sin ( π 2 R ( t ) 0 u ( z ) t 1 1 x 2 1 ( t x ) 2 d x ) + δ k 2 1 , k > 1 .
(1.4)

For f(z) in A p , the operator D μ + p 1 : A p A p is defined by

D μ + p 1 f(z)= z p ( 1 z ) μ + p f(z)(μ>p),

or equivalently

D μ + p 1 f(z)= z p ( z μ 1 f ( z ) ) μ + p 1 ( μ + p 1 ) ! ,
(1.5)

where μ is any integer greater than −p. If f(z) is given by (1.1), then it follows that

D μ + p 1 f(z)= z p + n = p + 1 ( μ + n 1 ) ! ( n p ) ! ( μ + p 1 ) ! a n z n .

The symbol D μ + p 1 when p=1, was introduced by Ruscheweyh [1] and D μ + p 1 is called the (μ+p1)th order Ruscheweyh derivative. We now introduce a function ( z 2 p F 1 ( a , b , c ; z ) ) 1 given by

( z p 2 F 1 ( a , b , c ; z ) ) ( z p 2 F 1 ( a , b , c ; z ) ) 1 = z p ( 1 z ) μ + p (μ>p),

and the following linear operator

I μ , p (a,b,c)f(z)= ( z p 2 F 1 ( a , b , c ; z ) ) 1 f(z),
(1.6)

where a, b, c are real or complex numbers other than 0,1,2, , μ>p, zE and f(z) A p . This operator was recently introduced in [2]. In particular, for p=1, this operator is studied by Noor [3]. For b=1, this operator reduces to the well-known Cho-Kwon-Srivastava operator I μ , p (a,c), which was studied by Cho et al. [4], and for μ=1, b=c, a=n+p, see [5]. For a=n+p, b=c=1, this operator was investigated by Liu [6] and Liu and Noor [7].

Simple computations yield

I μ , p (a,b,c)f(z)= z p + n = p + 1 ( c ) n ( μ + p ) n ( a ) n ( b ) n a n z n .

From (1.6), we note that

I μ , 1 ( a , b , c ) f ( z ) = I μ ( a , b , c ) f ( z ) (see [3]) , I 0 , p ( a , p , a ) f ( z ) = f ( z ) , I 1 , p ( a , p , a ) f ( z ) = z f ( z ) p .

Also, it can be easily seen that

z ( I μ , p ( a , b , c ) f ( z ) ) =(μ+p) I μ + 1 , p (a,b,c)f(z)μ I μ , p (a,b,c)f(z),
(1.7)

and

z ( I μ , p ( a + 1 , b , c ) f ( z ) ) =a I μ , p (a,b,c)f(z)(ap) I μ , p (a,b,c)f(z).

We define the following class of multivalent analytic functions by using the operator I μ , p (a,b,c)f(z) above.

Definition 1.1 Let f A p for pN. Then fU B μ , p α (a,b,c;γ,k,δ) for a,b,cR Z 0 , μ>p, α>0, k0, 0δ<1 and γ>0 if and only if

( 1 γ ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α + γ I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α 1 q k , δ ( z ) ,
(1.8)

where g A p is such that

q(z)= I μ + 1 , p ( a , b , c ) g ( z ) I μ , p ( a , b , c ) g ( z ) P(ρ),ρ= k + δ k + 1 ,zE.
(1.9)

Furthermore, for different choices of parameters being involved, we obtain many other well-known subclasses of the class A p and A as special cases.

  1. (i)

    a=c, b=1, k=0, μ=m N 0 , we have B m , p α (γ,δ) studied in [8].

  2. (ii)

    a=c=b=p=γ=1, k=μ=0, g(z)=z, the class U B μ , p α (a,b,c;γ,k,δ) reduces to the class

    B α (δ)= { f A ( 1 ) : z f ( z ) f ( z ) ( f ( z ) z ) α P ( δ ) }

    studied in [9].

  3. (iii)

    a=c=b=p=γ=1, k=μ=0, g(z)=z, the U B μ , p α (a,b,c;γ,k,δ) reduces to B(α) is the class of Bazilevich functions investigated by Singh [10].

  4. (iv)

    a=c=b=p=α=1, γ=0, k=μ=0, g(z)=z, the class U B μ , p α (a,b,c;γ,k,δ) reduces to the class

    B δ = { f A ( 1 ) : f ( z ) z P ( δ ) } ,

the class studied by Chen [11].

Let f A p and F η , p : A p A p be defined by

F η , p (z)= ( η + p ) z η 0 z t η 1 f(t)dt,η>p.
(1.10)

We need the following lemmas which will be used in our main results.

Lemma 1.2 [12]

Let u= u 1 +i u 2 and v= v 1 +i v 2 , and let ψ:D C 2 C be a complex-valued function satisfying the conditions:

  1. (i)

    ψ(u,v) is continuous in a domain D C 2 ,

  2. (ii)

    (1,0)D and ψ(1,0)>0,

  3. (iii)

    Reψ(i u 2 , v 1 )0, whenever (i u 2 , v 1 )D and v 1 1 2 (1+ u 2 2 ).

If h(z)=1+ c 1 z+ c 2 z 2 + is analytic in E such that (h,z h )D and Reψ(h(z),z h (z))>0 for zE, then Reh(z)>0.

Lemma 1.3 [13]

Let h be convex in the unit disc E, and let A0. Suppose that B(z) is analytic in E with ReB(z)A. If g is analytic in E and g(0)=h(0). Then

A z 2 g (z)+B(z)z g (z)+g(z)h(z)implies thatg(z)h(z).

Lemma 1.4 [14]

Let F be analytic and convex in E. If f,g A p and f,gF. Then

σf+(1σ)gF,0σ1.

Lemma 1.5 [15]

Let h be convex in E with h(0)=a and βC such that Reβ0. If pH[a,n] and

p(z)+ z p ( z ) β h(z),

then p(z)q(z)h(z), where

q(z)= β n z β / n 0 z h(t) t β / n 1 dt

and q(z) is the best dominant.

2 Main results

Theorem 2.1 Let fU B μ , p α (a,b,c;γ,k,δ) for a,b,cR Z 0 , μ>p, pN, α>0, k0, 0δ<1 and γ>0. Then fU B μ , p α (a,b,c;0,k,δ).

Proof Consider

h(z)= ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α ,
(2.1)

where h is analytic in E with h(0)=1, and g A p satisfies condition (1.9). Differentiating (2.1) logarithmically and using (1.7), we have

z h ( z ) h ( z ) =α(μ+p) { ( I μ + 1 , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) I μ , p ( a , b , c ) g ( z ) ) } .

Using (2.1) and simplifying, we obtain

h ( z ) + γ z h ( z ) α ( μ + p ) q ( z ) = ( 1 γ ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α + γ I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α 1 .

Since fU B μ , p α (a,b,c;γ,k,δ), therefore, we can write

h(z)+ γ z h ( z ) α ( μ + p ) q ( z ) q k , δ (z),zE.

Now using Lemma 1.3 for A=0 and B(z)= γ α ( μ + p ) q ( z ) with Req(z)>0, we have ReB(z)0, therefore, h(z) q k , δ (z). Hence fU B μ , p α (a,b,c;0,k,δ). □

Theorem 2.2 Let fU B μ , p α (a,b,c;γ,0,δ) for a,b,cR Z 0 , μ>p, pN. Then fU B μ , p α (a,b,c;0,0, δ 1 ), where

δ 1 = 2 α δ ( p + μ ) | q ( z ) | 2 + γ ρ 2 α ( p + μ ) | q ( z ) | 2 + γ ρ .

Proof Consider

h(z)= 1 ( 1 δ 1 ) { ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α δ 1 } ,
(2.2)

where h is analytic in E with h(0)=1, and g A p satisfies condition (1.9). Differentiating (2.2), we have

( 1 δ 1 ) α h ( z ) = ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α 1 × { ( I μ , p ( a , b , c ) f ( z ) ) I μ , p ( a , b , c ) g ( z ) I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ( I μ , p ( a , b , c ) g ( z ) ) I μ , p ( a , b , c ) g ( z ) } .

Using (1.7) and simplifying, we obtain

( 1 γ ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α + γ I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α 1 = ( 1 δ 1 ) h ( z ) + δ 1 + γ ( 1 δ 1 ) z h ( z ) α ( p + μ ) q ( z ) .

Since fU B μ , p α (a,b,c;γ,0,δ), therefore we have

(1 δ 1 )h(z)+ δ 1 + γ ( 1 δ 1 ) z h ( z ) α ( p + μ ) q ( z ) 1 + ( 1 2 δ ) z 1 z ,0δ<1,zE.

This implies that

1 1 δ { ( 1 δ 1 ) h ( z ) + δ 1 δ + γ ( 1 δ 1 ) z h ( z ) α ( p + μ ) q ( z ) } q 0 , 0 (E)=P.

To obtain our desired result, we show that hP, for zE. Let u= u 1 +i u 2 , v= v 1 +i v 2 , and let Ψ:D C 2 C be a complex-valued function such that u=h(z), v=z h (z). Then

Ψ(u,v)=(1 δ 1 )u+ δ 1 δ+ γ ( 1 δ 1 ) v α ( p + μ ) q ( z ) .

The first two conditions of Lemma 1.2 are easily verified. To verify the third condition, we consider

Re Ψ ( i u 2 , v 1 ) = Re { ( 1 δ 1 ) i u 2 + δ 1 δ + γ ( 1 δ 1 ) v 1 α ( p + μ ) q ( z ) } = δ 1 δ + Re γ ( 1 δ 1 ) v 1 α ( p + μ ) q ( z ) δ 1 δ Re γ ( 1 δ 1 ) ( 1 + u 2 2 ) q ( z ) ¯ 2 α ( p + μ ) | q ( z ) | 2 δ 1 δ γ ( 1 δ 1 ) ( 1 + u 2 2 ) ρ 2 α ( p + μ ) | q ( z ) | 2 = A + B u 2 C ,

where A=2α(p+μ)( δ 1 δ)|q(z) | 2 γρ(1 δ 1 ), B=γρ(1 δ 1 )0 if 0 δ 1 <1 and C=2α(p+μ)|q(z) | 2 >0. From the relation δ 1 = 2 α δ ( p + μ ) | q ( z ) | 2 + γ ρ 2 α ( p + μ ) | q ( z ) | 2 + γ ρ , we have A0. This implies that ReΨ(i u 2 , v 1 )0. Using Lemma 1.2, we have hP for zE. This completes the proof. □

Theorem 2.3 Let a,b,cR Z 0 , μ>p, pN, α>0, k0 and 0δ<1. Then

U B μ , p α (a,b,c; γ 2 ,k,δ)U B μ , p α (a,b,c; γ 1 ,k,δ),0 γ 1 < γ 2 ,zE.

Proof Since fU B μ , p α (a,b,c; γ 2 ,k,δ), therefore, we have

( 1 γ 2 ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α + γ 2 I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α 1 = h 1 ( z ) q k , δ ( z ) ,
(2.3)

where g A p satisfies condition (1.9). From Theorem 2.1, we write

( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α = h 2 (z) q k , δ (z),zE.
(2.4)

Now, for γ 1 0, we obtain

( 1 γ 1 ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α + γ 1 I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α 1 = ( 1 γ 1 γ 2 ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α + γ 1 γ 2 { ( 1 γ 2 ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α + γ 2 I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α 1 } = γ 1 γ 2 h 1 ( z ) + ( 1 γ 1 γ 2 ) h 2 ( z ) .

Using the convexity of the class of the function q k , δ (z) and Lemma 1.4, we write

γ 1 γ 2 h 1 (z)+ ( 1 γ 1 γ 2 ) h 2 (z) q k , δ (z),zE,

where h 1 and h 2 are given by (2.3) and (2.4), respectively. This implies that fU B μ , p α (a,b,c; γ 1 ,k,δ). Hence the proof of the theorem is completed. □

Theorem 2.4 Let fU B μ , p 1 (a,b,c;γ,k,δ), a,b,cR Z 0 , μ>p, pN, α>0, k0, γ1 and 0δ<1. Then fU B μ + 1 , p 1 (a,b,c;1,k,δ).

Proof Since fU B μ , p 1 (a,b,c;γ,k,δ), therefore, we have

(1γ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) +γ I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) q k , δ (z).

Now, consider

γ I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) = ( 1 γ ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) + γ I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) + ( γ 1 ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) .

This implies that

I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) = 1 γ { ( 1 γ ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) + γ I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) } + ( 1 1 γ ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) .

Using Theorem 2.1, Lemma 1.4 and the convexity of q k , δ (z), we have the required result. □

Now, using the operators I μ , p (a,b,c) and F η , p defined by (1.7) and (1.10), respectively, we have

z ( I p μ ( a , b , c ) F η , p ( f ) ( z ) ) =(p+η) I p μ (a,b,c)f(z)η I p μ (a,b,c) F η , p (f)(z),η>p.
(2.5)

Theorem 2.5 Let f A p and F η , p be given by (1.10). If

(1γ) I μ , p ( a , b , c ) F η , p ( f ) ( z ) z p +γ I μ , p ( a , b , c ) ( f ( z ) ) z p q k , δ (z),zE,
(2.6)

with a,b,cR Z 0 , μ,η>p, pN, γ>0, then

I μ , p ( a , b , c ) F η , p ( f ( z ) ) z p h(z) q k , δ (z),zE,

where

h(z)= p + η γ z ( p + η ) / γ 0 z q k , δ (z) t ( p + η ) / γ 1 dt.

Proof Let

I μ , p ( a , b , c ) F η , p ( f ( z ) ) z p = h 1 (z),zE,

where h 1 is analytic in E with h 1 (0)=1. Then

z ( I μ , p ( a , b , c ) F η , p ( f ( z ) ) ) =p z p h 1 (z)+ z p + 1 h 1 (z).

Using (2.5), we have

γ(p+η) I μ , p ( a , b , c ) f ( z ) z p γη I μ , p ( a , b , c ) F η , p ( f ) ( z ) z p =pγ h 1 (z)+γz h 1 (z).

Thus,

(1γ) I μ , p ( a , b , c ) F η , p ( f ( z ) ) z p +γ I μ , p ( a , b , c ) ( f ( z ) ) z p = h 1 (z)+γ z h 1 ( z ) p + η .
(2.7)

From (2.7), it follows that

h 1 (z)+γ z h 1 ( z ) p + η q k , δ (z),zE.

Using Lemma 1.5, for β 1 = p + η γ , n=1 and a=1, we obtain h 1 (z)h(z) q k , δ (z). That is, I μ , p ( a , b , c ) F η , p ( f ( z ) ) z p q k , δ (z). □

Theorem 2.6 Let fU B μ , p α (a,b,c;0,0,δ) for a,b,cR Z 0 , μ>p, pN, α,γ>0, 0δ<1. Then fU B μ , p α (a,b,c;γ,0,δ), for |z|< r 0 , where

r 0 = α ( p + μ)+γ γ 2 + 2 α γ ( p + μ ) α(p+μ).
(2.8)

Proof Let fU B μ , p α (a,b,c;0,0,δ). Then we have

( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α =(1δ)h(z)+δ,
(2.9)

where g A p satisfies the condition

q(z)= I μ + 1 , p ( a , b , c ) g ( z ) I μ , p ( a , b , c ) g ( z ) P,zE

and hP. Differentiating (2.9) and then using (1.7), we obtain

γ I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α 1 γ ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α = γ ( p δ ) z h ( z ) α ( p + μ ) q ( z ) .

This implies that

1 1 δ { ( 1 γ ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α + γ I μ + 1 , p ( a , b , c ) f ( z ) I μ + 1 , p ( a , b , c ) g ( z ) ( I μ , p ( a , b , c ) f ( z ) I μ , p ( a , b , c ) g ( z ) ) α 1 δ } = h ( z ) + γ z h ( z ) α ( p + μ ) q ( z ) , z E .
(2.10)

Now, using the well-known distortion result for class P, we have

|z h (z)| 2 r Re h ( z ) 1 r 2 andReh(z) 1 r 1 + r ,|z|<r<1,zE.

Thus, due to the applications of these inequalities, we have

Re ( h ( z ) + γ z h ( z ) α ( p + μ ) q ( z ) ) Re h ( z ) γ | z h ( z ) | α ( p + μ ) | q ( z ) | Re h ( z ) ( 1 2 γ r α ( p + μ ) ( 1 r ) 2 ) = Re h ( z ) ( α ( p + μ ) ( 1 r ) 2 2 γ r α ( p + μ ) ( 1 r ) 2 ) .

For |z|< r 0 , where r 0 is given in (2.8), the inequality above is positive. Sharpness of the result follows by taking h(z)= 1 + z 1 z . Hence from (2.10), we have the required result. □

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Correspondence to Mohsan Raza.

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MR and SNM jointly discussed and presented the ideas of this article. MR made the text file and corresponded it to the journal. Both authors read and approved the final manuscript.

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Raza, M., Malik, S.N. Properties of multivalent functions associated with the integral operator defined by the hypergeometric function. J Inequal Appl 2013, 458 (2013). https://doi.org/10.1186/1029-242X-2013-458

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Keywords

  • Bazilevic functions
  • multivalent functions
  • integral operator
  • hypergeometric functions