On Certain p-Valent Analytic Functions Involving a Linear Operator and Majorization Problems

Vietnam J MathDOI 10.1007/s10013-013-0029-z

On Certain p-Valent Analytic Functions Involvinga Linear Operator and Majorization Problems

Jagannath Patel · Ashis Kumar Palit

Received: 27 December 2012 / Accepted: 11 July 2013© Vietnam Academy of Science and Technology (VAST) and Springer Science+Business Media Singapore2013

Abstract The purpose of the present paper is to derive some sufficient conditions for p-valent starlikeness and p-valent close-to-convexity for certain subclasses of p-valent ana-lytic functions defined here by means of a linear operator. We also obtain results involvingmajorization problems by applying the linear operator to p-valent analytic functions. Rele-vant connections of the results presented here with those given earlier on the subject are alsoindicated.

Keywords Analytic function · p-Valent · Subordination · Hadamard product(or convolution) · Hypergeometric function · Linear operator

Mathematics Subject Classification (2010) 30C45

1 Introduction and Definitions

Let Ap denote the class of functions f of the form

f (z) = zp +∞∑

k=1

ap+kzp+k

(p ∈N= {1,2, . . . }), (1)

which are analytic and p-valent in the open unit disk U = {z ∈ C : |z| < 1}. We writeA1 = A.

A function f ∈ Ap is said to be p-valently starlike of order α in U, if it satisfies theinequality

�{

zf ′(z)f (z)

}> α (0 � α < p; z ∈ U).

J. Patel (B)Department of Mathematics, Utkal University, Vani Vihar, Bhubaneswar 751004, Indiae-mail: [email protected]

A.K. PalitDepartment of Mathematics, Bhadrak Institute of Engineering and Technology, Bhadrak 756 113, India

mailto:[email protected]

J. Patel, A.K. Palit

Analogously, a function f ∈ Ap is said to be p-valently convex of order α in U, if it satisfiesthe inequality

�{

1 + zf ′′(z)f ′(z)

}> α (0 � α < p; z ∈U).

Further, a function f ∈ Ap is said to be p-valently close-to-convex of order α in U, if itsatisfies the inequality

�{

f ′(z)zp−1

}> α (0 � α < p; z ∈U).

We denote by S∗p(α), Kp(α) and Cp(α) the subclasses of Ap consisting of functions which

are p-valently starlike of order α, p-valently convex of order α and p-valently close-to-convex of order α in U, respectively. We note that

S∗p(α) ⊆ S∗

p(0) = S∗p and Kp(α) ⊆ Kp(0) = Kp (0 � α < p),

where S∗p , Kp denote the classes of p-valently starlike and p-valently convex functions in U.

We also observe that S∗1 (α) and K1(α) are the classes of univalent starlike functions of order

α and univalent convex functions of order α (0 � α < p) in U and will be denoted by S∗(α)

and K(α), respectively. We shall use S∗ and K to denote S∗(0) and K(0), respectively whichare the familiar classes of univalent starlike and univalent convex functions in U.

Suppose that f and g are analytic in U. We say that the function f is subordinate to g

in U, and we write f ≺ g or f (z) ≺ g(z), if there exists a Schwarz function ω such thatf (z) = g(ω(z)), z ∈ U. If g is univalent in U, then the following equivalence relationshipholds true:

f (z) ≺ g(z) ⇐⇒ f (0) = g(0) and f (U) = g(U).

Following MacGregor [8], we say that an analytic function f is majorized by an analyticfunction g in U, written f � g or f (z) � g(z) (z ∈U), if there exists a function ϕ, analyticin U such that

|ϕ(z)| � 1 and f (z) = g(z)ϕ(z) (z ∈U). (2)

In our present investigation, we shall make use of the Gauss hypergeometric function 2F1

defined in U by

2F1(a, b; c; z) =∞∑

k=0

(a)k(b)k

(c)k

zk

k!(a, b, c ∈C; c /∈ Z

−0 = {0,−1,−2, . . . }), (3)

where (x)k denotes the Pochhammer symbol (or shifted factorial) defined by

(x)k ={

x(x + 1)(x + 2) · · · (x + k − 1) (k ∈N),

1 (k = 0).

We note that the series defined by (3) converges absolutely for z ∈ U and hence representsan analytic function in the open unit disk U [19, Chapter 14].

Motivated by the multiplier transformation introduced by Cho and Srivastava [3] on A,we define a function φp(n,λ) on U by

φp(n,λ)(z) = zp +∞∑

k=1

(p + k + λ

p + λ

)n

zp+k(λ > −p;n ∈ Z = {0,±1,±2, . . . }).

On Certain p-Valent Analytic Functions Involving a Linear Operator

The function φp(n,λ) is closely related to the multiplier transformation studied by Flett [4].Corresponding to the function φp(n,λ), we define a new function φ(†)

p (n,λ) in terms ofthe Hadamard product by

φp(n,λ)(z) ∗ φ(†)p (n,λ)(z) = zp

(1 − z)p+μ(μ > −p; z ∈ U). (4)

We now introduce the operator I np (λ,μ) : Ap −→ Ap by

I np (λ,μ)f (z) = φ(†)

p (n,λ)(z) ∗ f (z) (n ∈ Z;λ,μ > −p). (5)

If the function f is given by (1), then from (4) and (5), we deduce that

I np (λ,μ)f (z) = zp +

∞∑

k=1

(p + μ)k

(1)k

(p + λ

p + k + λ

)n

ap+kzp+k (z ∈U).

In view of (5), it follows that

z(I np (λ,μ)f

)′(z) = (p + λ)In−1

p (λ,μ)f (z) − λInp (λ,μ)f (z)

(f ∈ Ap, n ∈ Z; z ∈U). (6)

In particular, we note that

I−2p (0,1 − p)f (z) = z2f ′′(z) + zf ′(z)

p2, I−1

p (0,1 − p)f (z) = zf ′(z)p

,

I 0p(0,1 − p)f (z) = f (z), I−1

p (λ,1 − p)f (z) = zf ′(z) + λf (z)

p + λ, and

I 1p(λ,1 − p)f (z) = p + λ

zλ

∫ z

0tλ−1f (t) dt.

The operator I np (0,1 − p) is closely related to the Salagean [18] derivative operator. The

operator I nλ = I n

1 (λ,0) was studied recently by Cho and Srivastava [3], Cho and Kim [1]and Cho and Kim [2]. For any n ∈ Z, the operator In = I n

1 (1,0) was studied by Uralegaddiand Somanatha [18]. I−n

1 (0,0) is the well-known Salagean [17] derivative operator Dn,defined as:

Dnf (z) = z +∞∑

k=2

knakzk

(n ∈N0 = N∪ {0};f ∈ A

).

In the present paper, we derive some sufficient conditions for p-valent starlikeness andp-valent close-to-convexity for certain subclasses of Ap defined here by means of the linearoperator I n

p (λ,μ). We also obtain results involving majorization problems by applying thelinear operator I n

p (λ,μ) to functions in Ap . Our main tool is the following well-known JackLemma [6] (see also [9]).

Lemma Let ω be a non-constant analytic function in U with ω(0) = 0. If |ω| attains itsmaximum value at a point z0 ∈ U on the circle |z| = r < 1, then

z0ω′(z0) = cω(z0) (c is real and c � 1).


2 Main Results

Unless otherwise mentioned, we assume throughout this paper that

n ∈ Z, λ,μ > −p, δ > 0 and − 1 � B < A� 1.

Theorem 1 Let 0 < δ < 1 and (1 − A) − δ(1 − B)� 0. If f ∈ Ap satisfies

(1 − δ)I np (λ,μ)f (z)

zp+ δ

I n−1p (λ,μ)f (z)

I np (λ,μ)f (z)

≺ 1 + Az

1 + Bz(z ∈ U),

then

I np (λ,μ)f (z)

zp≺ δ

(p + λ)(1 − δ)Q(z)= q(z) ≺ 1 + (A−δB

1−δ)z

1 + Bz(z ∈U), (7)

where Q is given by

Q(z) =⎧⎨

⎩

∫ 10 t

(p+λ)(1−δ)δ

−1(

1+Btz1+Bz

) (p+λ)(A−B)δ dt (B �= 0),

∫ 10 t

(p+λ)(1−δ)δ

−1 exp(

(p+λ)

δA(t − 1)z

)dt (B = 0)

(8)

and q is the best dominant of (7). Furthermore, if

A � min

{1 − δ(1 − B),−δB(1 − p − λ)

p + λ

}with − 1 � B < 0,

then

�{

I np (λ,μ)f (z)

zp

}>

[2F1

(1,

p + λ

δ

(B − A

B

); (p + λ)(1 − δ)

δ+ 1; B

B − 1

)]−1

(z ∈ U). (9)

The result in (9) is the best possible.

Proof Letting

ϕ(z) = I np (λ,μ)f (z)

zp(z ∈U), (10)

we see that ϕ(z) = 1 + c1z + c2z2 + · · · is analytic in U. Taking logarithmic differentiation

in (10) and using the identity (6), we deduce that

(1 − δ)I np (λ,μ)f (z)

zp+ δ


I np (λ,μ)f (z)

= P (z) + zP ′(z)p+λ

δP (z) − (p + λ)

≺ 1 + Az

1 + Bz(z ∈ U), (11)

where P (z) = (1 − δ)ϕ(z) + δ. Thus, by using a result [10, Corollary 3.2] in (11), we get

ϕ(z) ≺ q(z) = δ

(p + λ)(1 − δ)Q(z)≺ 1 + (A−δB

1−δ)z

1 + Bz(z ∈U),


where q is the best dominant of (7) and Q is given by (8). The proof of the remaining partof the theorem can be deduced on the same lines as in [15, Theorem 1]. This evidentlycompletes the proof of Theorem 1. �

Taking n = 0, λ = 0, μ = 1 − p, A = 1 − (2α/p) and B = −1 in Theorem 1, we obtain

Corollary 1 Let 0 < δ < 1 and max{δp,p−δ(1−p)

2 } � α < p. If f ∈ Ap satisfies

�{(1 − δ)

f (z)

zp+ δ

p

zf ′(z)f (z)

}>

α

p(0 � α < p; z ∈U),

then

�{

f (z)

zp

}>

[2F1

(1,

2(p − α)

δ; p(1 − δ)

δ+ 1; 1

2

)]−1

(z ∈ U).

The result is the best possible.

For n = −m (m ∈ N0), p = 1, λ = 0 and μ = 0 in Theorem 1, we have

Corollary 2 Let 0 < δ < 1,−1 � B < 0 and A � min{0,1 − δ(1 − B)}. If f ∈ A satisfies

(1 − δ)Dmf (z)

z+ δ

Dm+1f (z)

Dmf (z)≺ 1 + Az

1 + Bz(m ∈N0; z ∈U),

then

�{

Dmf (z)

z

}>

[2F1

(1,

B − A

δB; 1

δ; B

B − 1

)]−1

(z ∈ U).


Remark 1 For n = −1, λ = 0, μ = 1 − p, A = 1 − (2α/p) (0 � α < p) and B = −1 inTheorem 1, we get the result due to Patel and Cho [15, Theorem 1].

Theorem 2 Let 0 < δ � 1. If f ∈ Ap satisfies

�{(1 − δ)

I np (λ,μ)f (z)

zp+ δ


I np (λ,μ)f (z)

}<

[2(p + λ) + 1]δ2(p + λ)

(z ∈ U), (12)

for some δ(δ > (2p + λ)/{2(p + λ) + 1}), then∣∣∣∣I np (λ,μ)f (z)

zp− 1

∣∣∣∣ < 1 (z ∈ U).

Proof We set

I np (λ,μ)f (z)

zp= 1 + ω(z) (z ∈U). (13)

Clearly, the function ω is analytic in U with ω(0) = 1. Differentiating (13) logarithmicallyand using the identity (6) in the resulting equation, we get


I np (λ,μ)f (z)

= 1 + zω′(z)(p + λ)(1 + ω(z))

.


Thus, by making use of (13) and the above equality, we obtain

(1 − δ)I np (λ,μ)f (z)

zp+ δ


I np (λ,μ)f (z)

= 1 + (1 − δ)ω(z) + δ

p + λ

zω′(z)1 + ω(z)

.

We claim that |ω(z)| < 1 in U. For otherwise, there exists a point z0 ∈ U such thatmax|z|�|z0| |ω(z)| = |ω(z0)| = 1, z ∈U. Then by applying the Lemma, we can write ω(z0) =eiθ and z0ω(z0) = ceiθ (c � 1). Therefore,

�{(1 − δ)

I np (λ,μ)f (z0)

zp

0

+ δI n−1p (λ,μ)f (z0)

I np (λ,μ)f (z0)

}

= 1 + �((1 − δ)eiθ

) + cδ

p + λ�

(eiθ

1 + eiθ

)

= 1 + (1 − δ) cos θ + cδ

2(p + λ)

� δ + δ

2(p + λ),

which contradicts our assumption (12). Hence, |ω(z)| < 1 holds true for all z ∈ U and weconclude that

∣∣∣∣I np (λ,μ)f (z)

zp− 1

∣∣∣∣ < 1 (z ∈U),

which completes the proof of Theorem 2. �

Remark 2

1. For n = −1, λ = 0, μ = 1 − p and δ = 1, Theorem 2 yields

�{

1 + zf ′′(z)f ′(z)

}<

2p + 1

2(z ∈U) ⇒

∣∣∣∣f ′(z)zp−1

− p

∣∣∣∣ < p (z ∈U).

This improves a result of Ozaki [14] for p = 1 and also gives a sufficient condition forfunctions in Ap to be p-valent close-to-convex in U.

2. For n = 0, λ = 0 and μ = 1 − p in Theorem 2, we get the result contained in [15].

Theorem 3 If f ∈ Ap satisfies

�{


I np (λ,μ)f (z)

}>

α

p(0 � α < p; z ∈ U), (14)

then

�{

I np (λ,μ)f (z)

zp

}> 2−2(p+λ)(p−α)/p (z ∈U). (15)



Proof Define the function ω by

I np (λ,μ)f (z)

zp= 1

(1 − ω(z))2(p+λ)(1−α/p)(z ∈U), (16)

where ω(z) �= 1. Clearly, ω is analytic in U. Making use of the identity (6) in the logarithmicdifferentiation of (16), we have


I np (λ,μ)f (z)

= 1 + 2

(1 − α

p

)zω′(z)

1 − ω(z).

Suppose that |ω(z)| ≮ 1 for some z ∈ U. Then by using the Lemma, there exists a pointz0 ∈ U such that z0ω

′(z0) = cω(z0) (c � 1) with |ω(z0)| = 1. This implies that

�{

I n−1p (λ,μ)f (z0)

I np (λ,μ)f (z0)

}= 1 + 2

(1 − α

p

)�

{z0ω

′(z0)

1 − ω(z0)

}� α

p,

which contradicts (14). Thus, we have |ω(z)| < 1 for all z ∈U and the proof of the assertion(15) is complete.

To show that the bound in (15) is the best possible, we consider the function

f0(z) = zp +∞∑

k=1

(2(p + λ)(1 − αp))k

(p + μ)k

(p + k + λ

p + λ

)n

zp+k (z ∈ U).

Then

I np (λ,μ)f0(z) = zp +

∞∑

k=1

(2(p + λ)(1 − αp))k

(1)k

zp+k = zp

(1 − z)2(p+λ)(1−α/p)

and

I n−1p (λ,μ)f0(z)

I np (λ,μ)f0(z)

= 1 + 2

(1 − α

p

)z

1 − z(z ∈ U).

Thus

infz∈U

{I np (λ,μ)f0(z)

zp

}= 2−2(p+λ)(p−α)/p and inf

z∈U

{I n−1p (λ,μ)f0(z)

I np (λ,μ)f0(z)

}= α

p.

This completes the proof of Theorem 3. �

Letting n = 0, λ = 0 and μ = 1 − p (resp. n = −1, λ = 0 and μ = 1 − p) in Theorem 3,we get

Corollary 3 For 0 � α < p, we have

f ∈ S∗p(α) =⇒ �

{f (z)

zp

}>

1

22(p−α)(z ∈U)

and

f ∈ Kp(α) =⇒ �{

f ′(z)zp−1

}>

p

22(p−α)(z ∈U).


The estimates are the best possible.


∣∣∣∣I n−1p (λ,μ)f (z)

I np (λ,μ)f (z)

− I np (λ,μ)f (z)

I n+1p (λ,μ)f (z)

∣∣∣∣ <A − B

(p + λ)(1 + |A|)(1 + |B|) (z ∈U), (17)

then

I np (λ,μ)f (z)

I n+1p (λ,μ)f (z)

≺ 1 + Az

1 + Bz(z ∈U). (18)

Proof We consider the function ω defined by

I np (λ,μ)f (z)

I n+1p (λ,μ)f (z)

= 1 + Aω(z)

1 + Bω(z)(z ∈U). (19)

Then the function ω is analytic in U with ω(0) = 0. Logarithmic differentiation of (19)implies that


I np (λ,μ)f (z)

− I np (λ,μ)f (z)

I n+1p (λ,μ)f (z)

= (A − B)zω′(z)(p + λ)(1 + Aω(z))(1 + Bω(z))

(z ∈ U). (20)

Suppose that there exists a point z0 ∈ U such that max|z|�|z0| |ω(z)| = |ω(z0)| = 1 (z ∈ U).Then by the Lemma, we have ω(z0) = eiθ and z0ω

′(z0) = ceiθ (c � 1). Thus, from (20), weimmediately obtain

∣∣∣∣I n−1p (λ,μ)f (z0)

I np (λ,μ)f (z0)

− I np (λ,μ)f (z0)

I n+1p (λ,μ)f (z0)

∣∣∣∣ =∣∣∣∣

c(A − B)eiθ

(p + λ)(1 + Aeiθ )(1 + Beiθ )

∣∣∣∣

� A − B

(p + λ)(1 + |A|)(1 + |B|) ,

which contradicts our hypothesis (17). Therefore, we conclude that |ω(z)| < 1 for all z ∈U.This proves our assertion (18). �

Setting n = −1, λ = 0, μ = 1 − p, A = 1 − (α/p) and B = 0 in Theorem 4, we obtainthe following result due to Irmak and Raina [5, p. 395].

Corollary 4 If f ∈ Ap satisfies

∣∣∣∣1 + zf ′′(z)f ′(z)

− zf ′(z)f (z)

∣∣∣∣ <p − α

2p − α(0 � α < p; z ∈U),

then∣∣∣∣zf ′(z)f (z)

− p

∣∣∣∣ < p − α (z ∈U)

and hence f is p-valent starlike of order α in U.



∣∣∣∣I n−1p (λ,μ)f (z)

I np (λ,μ)f (z)

− 1

∣∣∣∣β ∣∣∣∣

I np (λ,μ)f (z)

I n+1p (λ,μ)f (z)

− 1

∣∣∣∣γ

< Mp(α,β, γ,λ) (z ∈U) (21)

for some α (0 � α < p), β (β � 0) and γ (γ � 0) with β + γ > 0, then

�{

I np (λ,μ)f (z)

I n+1p (λ,μ)f (z)

}>

α

p(z ∈U), (22)

where

Mp(α,β, γ,λ) =⎧⎨

⎩(1 − α

p)γ

{1 + 1

2(p+λ)− α

p

}β (0 � α � p

2

),

(1 − αp)β+γ

(1 + 1

p+λ

)β (p

2 � α < p).

Proof We follow the method similar to that of Lin and Owa [7]. Let 0 � α � p/2 and wedefine the function ω by

I np (λ,μ)f (z)

I n+1p (λ,μ)f (z)

= 1 + (1 − 2αp

)ω(z)

1 − ω(z)(z ∈U) (23)

with ω(z) �= 1. Then ω is analytic in U. It follows from (23) by using the identity (22) that

∣∣∣∣I n−1p (λ,μ)f (z)

I np (λ,μ)f (z)

− 1

∣∣∣∣β ∣∣∣∣

I np (λ,μ)f (z)

I n+1p (λ,μ)f (z)

− 1

∣∣∣∣γ

={

2

(1 − α

p

)}β+γ ∣∣∣∣ω(z)

1 − ω(z)

∣∣∣∣β+γ

×∣∣∣∣1 + zω′(z)

(p + λ)ω(z){1 + (1 − 2αp

)ω(z)}∣∣∣∣β

(z ∈U).

Suppose that there exists a point z0 ∈ U such that max|z|�|z0| |ω(z)| = ω(z0)| = 1 (z ∈ U).Then by the Lemma, we have ω(z0) = eiθ and z0ω

′(z0) = ceiθ (c � 1). Thus,

∣∣∣∣I n−1p (λ,μ)f (z0)

I np (λ,μ)f (z0)

− 1

∣∣∣∣β ∣∣∣∣

I np (λ,μ)f (z0)

I n+1p (λ,μ)f (z0)

− 1

∣∣∣∣γ

={

2

(1 − α

p

)}β+γ ∣∣∣∣eiθ

1 − eiθ

∣∣∣∣β+γ ∣∣∣∣1 + c

(p + λ){1 + (1 − 2αp

)eiθ }∣∣∣∣β

�(

1 − α

p

)β+γ ∣∣∣∣1 + c

(p + λ){1 + (1 − 2αp

)}∣∣∣∣β

�(

1 − α

p

)γ {1 + 1

2(p + λ)− α

p

}β

(z ∈ U),

which contradicts the condition (21) for 0 � α � p/2. This yields |ω(z)| < 1 for all z ∈ U

and the assertion (22) for 0 � α � p/2 is proved.


For the case p/2 � α < p, we consider the function ω defined by

I np (λ,μ)f (z)

I n+1p (λ,μ)f (z)

= α

α − (p − α)ω(z)(z ∈U).

Then by the preceding argument, we deduce that

∣∣∣∣I n−1p (λ,μ)f (z0)

I np (λ,μ)f (z0)

− 1

∣∣∣∣β ∣∣∣∣

I np (λ,μ)f (z0)

I n+1p (λ,μ)f (z0)

− 1

∣∣∣∣γ

= (p − α)β+γ

(p + λ)β

|ceiθ + (p + λ)eiθ |β |eiθ |γ|α − (p − α)eiθ |β+γ

�(

1 − α

p

)β+γ (1 + 1

p + λ

)β

,

which contradicts (21) for the case p/2 � α < p. Thus, we conclude that the assertion (22)holds true and the proof of the theorem is complete. �

Letting n = −1, λ = 0 and μ = 1−p in Theorem 5, we obtain the following result whichin turn yields the corresponding work of Lin and Owa [7, Theorem 1] for p = 1.

Corollary 5 If f ∈ Ap satisfies

∣∣∣∣

(1 + zf ′′(z)

f ′(z)

)− p

∣∣∣∣β ∣∣∣∣

zf ′(z)f (z)

− p

∣∣∣∣γ

<

⎧⎨

⎩(p − α)γ

(p + 1

2 − α)β (

0 � α � p

2

),

(1 + 1

p

)β(p − α)β+γ

(p

2 � α < p)

for some α (0 � α < p), β (β � 0) and γ (γ � 0) with β + γ > 0, then f ∈ S∗p(α).

Remark 3

1. If f ∈ A satisfies (letting p = 1 and β = γ = 1 in Corollary 5)

∣∣∣∣zf ′′(z)f ′(z)

(zf ′(z)f (z)

− 1

)∣∣∣∣ <

{(1 − α)

(32 − α

) (0 � α � 1

2

),

2(1 − α)2(

12 � α < 1

),

then f ∈ S∗(α), which improves a result of Obradovic [12, p. 229].2. We note that Corollary 4 is an improvement and extension of a result due to Owa and

Srivastava [13, Lemma 3].

We now obtain certain results involving majorization problems by applying the linear oper-ator I n

p (λ,μ) to the functions in Ap .

Theorem 6 Let 0 � α < p,p � |λ+ (p +λ){1 − (2α/p)}| and κ = 2 +p +|λ+ (p +λ) ×{1 − (2α/p)}| (0 � α < p). If f ∈ Ap , g ∈ Ap satisfies

�{

I n−1p (λ,μ)g(z)

I np (λ,μ)g(z)

}>

α

p(z ∈U) (24)


and I np (λ,μ)f � I n

p (λ,μ)g in U, then

∣∣(I np (λ,μ)f

)′(z)

∣∣ �∣∣(I n

p (λ,μ)g)′(z)

∣∣ (|z| � R1(p,α,λ)), (25)

where

R1(p,α,λ) = κ − √κ2 − 4p|λ + (p + λ){1 − (2α/p)}|2|λ + (p + λ){1 − (2α/p)}| . (26)

Proof From (24), we get


I np (λ,μ)g(z)

= 1 + (1 − 2αp

)ω(z)

1 − ω(z)(z ∈U),

where ω is an analytic function in U with

ω(0) = 0 and∣∣ω(z)

∣∣ � |z| (z ∈U), (27)

which in view of (6) immediately yields

z(I np (λ,μ)g)′(z)

I np (λ,μ)g(z)

= p + {λ + (p + λ)(1 − 2αp

)}ω(z)

1 − ω(z)(z ∈ U). (28)

Now, by using (27) in (28), we get

∣∣I np (λ,μ)g(z)

∣∣ � (1 + |z|)|z|p − |λ + (p + λ)(1 − 2α

p)||z|

∣∣(I np (λ,μ)g

)′(z)

∣∣ (z ∈U). (29)

Since I np (λ,μ)f � I n

p (λ,μ)g in U, from (2), we obtain

(I np (λ,μ)f

)′(z) = (

I np (λ,μ)g

)′(z)ϕ(z) + I n

p (λ,μ)g(z)ϕ′(z), (30)

where ϕ satisfies the inequality (cf. [11, p. 168])

|ϕ′(z)| � 1 − |ϕ(z)|21 − |z|2 (z ∈U). (31)

Applying (29) and (31) in (30), we get∣∣∣∣(I n

p (λ,μ)f )′(z)(I n

p (λ,μ)g)′(z)

∣∣∣∣� |ϕ(z)| + 1 − |ϕ(z)|21 − |z|

|z|p − |λ + (p + λ)(1 − 2α

p)||z| (z ∈U),

which upon setting |z| = r and |ϕ(z)| = x (0 � x � 1) leads us to the inequality


)′(z)

∣∣ � Θ(x)

(1 − r){p − |λ + (p + λ)(1 − 2αp

)|r}∣∣(I n

p (λ,μ)g)′(z)

∣∣

(z ∈U), (32)

where

Θ(x) = −rx2 + (1 − r)

{p −

∣∣∣∣λ + (p + λ)

(1 − 2α

p

)∣∣∣∣r}x + r (0 � x � 1).


It is easily seen that the function Θ attains its maximum value at x = 1 with r = R1(p,α,λ)

given by (26). Furthermore, if 0 � σ �R1(p,α,λ), then the function Λ defined by

Λ(x) = −σx2 + (1 − σ)

{p −

∣∣∣∣λ + (p + λ)

(1 − 2α

p

)∣∣∣∣σ}x + σ

is seen to be increasing on the interval 0 � x � 1, so that

Λ(x)� Λ(1) = (1 − σ)

{p −

∣∣∣∣λ + (p + λ)

(1 − 2α

p

)∣∣∣∣σ}

(0 � x � 1;0 � σ � R1(p,α,λ)

).

Hence, by setting x = 1 in (32), we conclude that the inequality (25) holds true for|z| � R1(p,α,λ), where R1(p,α,λ) is given by (26). This evidently completes the proofof Theorem 3. �

Putting n = 0, λ = 0 and μ = 1 − p in Theorem 6, we get the following result.

Corollary 6 If f ∈ Ap , g ∈ S∗p(α) (0 � α < p) and f � g in U, then

∣∣f ′(z)∣∣ �

∣∣g′(z)∣∣ (|z| � R2(p,α)

),

where

R2(p,α) =

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

(p+1−α)−√

(p+1−α)2−p(p−2α)

p−2α

(0 � α <

p

2

),

p

p+2

(α = p

2

),

(1+α)−√

(1+α)2−p(2α−p)

2α−p

(p

2 < α < p).

It is known [16] that if g ∈ Ap and p(p + λ − 1)/{2(p + λ)}� α < p, then

�{



}>

α

p(z ∈ U) =⇒ �

{I n−1p (λ,μ)g(z)

I np (λ,μ)g(z)

}> ρ (z ∈U), (33)

where

ρ ={

2F1

(1,2(p − α);p + λ + 1; 1

2

)}−1

(34)

and the result is the best possible.Using (33) and following the lines of proof of Theorem 6, we can prove

Theorem 7 Let p(p +λ− 1)/{2(p +λ)}� α < p, σ = 2 +p + |λ+ (p +λ)(1 − 2ρ)| andp � |λ + (p + λ)(1 − 2ρ)|. If f ∈ Ap , g ∈ Ap satisfies

�{



}>

α

p(z ∈U)

and I np (λ,μ)f � I n

p (λ,μ)g in U, then


)′(z)

∣∣ �∣∣(I n

p (λ,μ)g)′(z)

∣∣ (|z| � R3(p,α,λ)),


where

R3(p,α,λ) = σ − √σ 2 − 4p|λ + (p + λ)(1 − 2ρ)|2|λ + (p + λ)(1 − 2ρ)|

and ρ is given by (34).

For n = 0, λ = 0 and μ = 1 − p, Theorem 7 yields

Corollary 7 Let (p − 1)/2 � α < p. If f ∈ Ap , g ∈ Kp(α) and f � g in U, then

|f ′(z)| � |g′(z)| (|z| � R4(p,α)),

where

R4(p,α) =

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

1+2p(1−�)−√

{1+2p(1−�)}2−p2(1−2�)

p(1−2�)

(0 < � < 1

2

),

p

p+2

(� = 1

2

),

(1+p�)−√

(1+p�)2−p2(2�−1)

p(2�−1)

(12 < � < 1

)

and � = {2F1(1,2(p − α);p + 1; 12 )}−1.

Remark 4 In the special case when p = 1 and α = 0 in Corollary 6 and Corollary 7, we getthe following result of MacGregor [8, p. 96, Theorem 1B and Theorem 1C].

(i) If f ∈ A, g ∈ S∗ and f � g in U, then |f ′(z)| � |g′(z)| for |z| � 2 − √3.

(ii) If f ∈ A, g ∈ K and f � g in U, then |f ′(z)| � |g′(z)| for |z| � 1/3.

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On Certain p-Valent Analytic Functions Involving a Linear Operator and Majorization Problems

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