MULTIVALENT FUNCTIONS INVOLVING THE GENERALIZED HYPERGEOMETRIC FUNCTIONS

MULTIVALENT FUNCTIONS INVOLVING THE GENERALIZED HYPERGEOMETRIC FUNCTIONS

ALI MUHAMMAD

In this paper, we introduce new subclasses of meromorphically multivalent func- tions. We investigate a number of inclusion relationships, radius problem and other interesting properties of meromorphically multivalent functions which are defined here by means of a linear operator involving the generalized hypergeome- tric functions.

AMS 2010 Subject Classification: Primary 30C45, Secondary 30C50.

Key words: multivalent functions, analytic functions, meromorphic functions, li- near operator, functions with positive real part, Hadamard product (or convolution).

1. INTRODUCTION For any integerm >1−p,let P

p,m

denote the class of functionsf :

(1.1) f/(z) =z^−p+

∞

X

k=m

a_kz^k, p∈N={1,2, . . .},

which are analytic and p-valent in the punctured unit disc E^∗ = {z :z ∈ C and 0<|z|<1}=E\{0}.

LetPk(ρ) be the class of functionsh(z) analytic in E withp(0) = 1 and (1.2)

Z 2π 0

Reh(z)−ρ 1−ρ

dθ≤kπ, z=re^iθ,

where k >2 and 0 ≤ρ < 1. This class was introduced by Padmanabhan et al. in [9]. We note that P_k(0) =P_k, see Pinchuk [11], P₂(ρ) =P(ρ),the class of analytic functions with positive real part greater than ρ and P2(0) = P, the class of functions with positive real part. From (1.2) we can easily deduce that h(z) ∈ Pk(ρ) if, and only if, there exists h1(z), h2(z) ∈ P(ρ) such that

REV. ROUMAINE MATH. PURES APPL.,57(2012),2, 133-144

for z∈E,

(1.3) h(z) =

k 4 +1

2

h₁(z)− k

4 −1 2

h₂(z).

For function f, g∈ P

p,m

,wheref is given by (1.1) and g is defined by g(z) =z^−p+

∞

X

k=m

b_kz^k, p∈N={1,2, . . .},

then the Hadamard product (or convolution) f∗g of the function f and g is defined by

(f ∗g)(z) =z^−p+

∞

X

k=m

a_kb_kz^k = (g∗f)(z).

For real or complex parameters α₁, . . . , α_q and β₁, . . . , β_s (β_j ∈/ Z₀⁻={0,−1,

−2,−3, . . . ,};j = 1, . . . , s),the generalized hypergeometric function qFs, see [14] is given by

(1.4) _qF_s(α₁, . . . , α_q;β₁, . . . , β_s;z) =

∞

X

k=0

(α₁)_k. . .(α_q)_k (β1)k, . . . ,(βs)k k!z^k,

q ≤ s+ 1; q, s ∈ N₀ = N∪ {0}; N = {1,2, . . . ,}; z ∈ E, where (v)_k is the Pochhammer symbol (or the shifted factorial) defined in (terms of the Gamma function) by

(v)k= Γ(v+k) Γ(v) =

( 1 ifk= 0, v∈C\{0}

v(v+ 1), . . . ,(v+k−1) ifk∈N, v∈C )

. Corresponding to the functionφ_p(α₁, . . . , α_q;β₁, . . . , β_s;z) defined by (1.5) φp(α1, . . . , αq;β1, . . . , βs;z) =z^−p qFs(α1, . . . , αq;β1, . . . , βs;z), we introduce a function φp,µ(α1, . . . , αq;β1, . . . , βs;z) defined by

φp(α1, . . . , αq;β1, . . . , βs;z)∗φp,µ(α1, . . . , αq;β1, . . . , βs;z) = (1.6)

= 1

z^p(1−z)^µ+p (µ >−p; z∈E^∗).

We now define a linear operator H^m,µ_p,q,s(α1, . . . , αq;β1, . . . , βs)f(z) : P

p,m

→ P

p,m

by

H^m,µ_p,q,s(α₁, . . . , α_q;β₁, . . . , β_s)f(z) =φ_p,µ(α₁, . . . , α_q;β₁, . . . , β_s;z)∗f(z) (1.7)

αi, βj ∈/ Z₀⁻; i= 1,2, . . . , q; j= 1, . . . , s; µ >−p; f ∈X

p,m

; z∈E^∗

.

For convenience, we write

H_p,q,s^m,µ(α₁) =H^m,µ_p,q,s(α₁, . . . , α_q;β₁, . . . , β_s).

If f ∈ P

p,m

is given by (1.1),then from (1.7),we deduce that H^m,µ_p,q,s(α₁)f(z) =z^−p+ X

k=m

(µ+p)_p+k(β1)_p+k. . .(βs)_p+k (α1)p+k. . .(αq)p+k

a_kz^k, (1.8)

(µ >−p; z∈E^∗), and it is easily verified from (1.8)

(1.9) z(H^m,µ_p,q,s(α₁)f(z))⁰= (µ+p)H_p,q,s^m,µ+1(α₁)f(z)−(µ+ 2p)H^m,µ_p,q,s(α₁)f(z).

and

(1.10) z(H^m,µ_p,q,s(α₁+ 1)f(z))⁰=α₁H^m,µ_p,q,s(α₁)f(z)−(p+α₁)H^m,µ_p,q,s(α₁+ 1)f(z).

We note that the linear operatorH_p,q,s^m,µ(α₁) is closely related to the Choi-Saigo- Srivastava operator [4] for analytic functions and is essentially motivated by the operators defined and studied in [2]. The linear operator H^0,µ_1,q,s(α1) was investigated recently by Cho and Kim [1],whereas H^1−p_p,2,1(c,1;a;z) =L_p(a, c) (c ∈ R, a ∈ Z⁻₀) is the operator studied in [6]. In particular, we have the following observations:

(i)H^m,0_p,s+1,s(p+ 1, β₁, . . . , β_s;β₁, . . . , β_s)f(z) = _z^p2p

Rz

0 t^2p−1f(t)dt;

(ii)H^m,0_p,s+1,s(p, β₁, . . . , β_s;β₁, . . . , β_s)f(z) =H^m,1_p,s+1,s(p+ 1, β₁, . . . , β_s;β₁, . . . , β_s)f(z) =f(z);

(iii) H_p,s+1,s^m,1 (p, β₁, . . . , β_s;β₁, . . . , β_s)f(z) = ^zf0(z)+2pf_p ^(z);

(iv) H_p,s+1,s^m,2 (p+ 1, β1, . . . , βs;β1, . . . , βs)f(z) = ^zf0(z)+(2p+1)f(z)

p+1 ;

(v)H^1−p,n_p,s+1,s(β1, . . . , βs; 1;β1, . . . , βs)f(z) = _zp(1−z)^{1 n+p} =D^n+p−1f(z) (nis an integer >−p) the operator studied in [5],and

(vi)H^m,1−p_p,s+1,s(δ+ 1, β2, . . . , βs; 1;δ;β2, . . . , βs)f(z) = _zδ+p^δ

R_z

0 t^δ+p−1f(t)dt (δ >0;z∈E^∗).

Using the operatorH^m,µ_p,q,s(α₁) we now define a new classes of meromorphic multivalent functions.

Definition 1.1. Letf ∈ P

p,m

.Then f ∈ P

p,m,k

(λ, µ, p, q, s, ρ),if and only if

−(1−λ)z^p+1(H^m,µ_p,q,s(α₁)f(z))⁰−λz^p+1(H^m,µ+1_p,q,s (α₁)f(z))⁰

p ∈P_k(ρ), z∈E,

where λ∈C, k≥2 and 0≤ρ < p.

Definition 1.2. Letf ∈ P

p,m

. Thenf ∈Mp,m,k(λ, µ, p, q, s, ρ),if and only if (1−λ)z^p(H^m,µ_p,q,s(α1)f(z)) +λz^p(H_p,q,s^m,µ+1(α1)f(z))∈Pk(ρ), z∈E, where λ∈C, k≥2 and 0≤ρ < p.

Where and through this paper unless otherwise mentioned, the parame- ters α, µ, p, k andρ are constrained as follows:

λ∈C; µ >−p; k≥2; 0≤ρ < p, mis any integer andp∈N. 2. PRELIMINARY RESULTS

Lemma 2.1 [12]. If p(z) is analytic in E with p(0) = 1, and if λ1 is a complex number satisfying Re (λ₁)≥0, λ₁ 6= 0, then

Re

p(z) +λ₁zp⁰(z) > β, 0≤β <1.

Implies

Rep(z)> β+ (1−β)(2γ−1), where γ is given by

γ =γ(Reλ1) = Z 1

0

1 +t^Reλ1

dt,

which is an increasing function of Reλ₁ and ¹₂ ≤γ <1. The estimate is sharp in the sense that the bound cannot be improved.

Lemma 2.2 [13]. If p(z) is analytic in E, p(0) = 1 and Rep(z) > ¹₂, z∈E,then for any function F analytic in E, the functionp∗F takes values in the convex hull of the image of E under F.

Lemma 2.3(cf., e.g., Pashkouleva [10]).Let p(z) = 1 +b1z+b2z²+· · · ∈ P(ρ). Then

Rep(z)≥2ρ−1 +2(1−ρ) 1 +|z| . 3. MAIN RESULTS Theorem 3.1. Let f ∈ P

p,m,k

(λ, µ, p, q, s, ρ) withReλ >0.Then

−z^p+1(H^m,µ_p,q,s(α₁)f(z))⁰

p ∈Pk(η),

where η is given by

(3.1) η=ρ+ (1−ρ)(2γ−1),

and

γ= Z 1

0

1 +t^Reµ+pλ −1

dt.

Proof. Let

(3.2) −z^p+1(H^m,µ_p,q,s(α₁)f(z))⁰

p =h(z) =

k 4 +1

2

h₁(z)− k

4 +1 2

h₂(z).

Then h(z) is analytic in E with h(0) = 1.Making use of the identity (1.9) in (3.2) and differentiating the resulting equation with respect toz,we have

−(1−λ)z^p+1(H^m,µ_p,q,s(α1)f(z))⁰−λz^p+1(H^m,µ+1_p,q,s (α1)f(z))⁰

p =

h(z)+λzh⁰(z) (µ+p)

. Since f ∈ P

p,m,k

(λ, µ, p, q, s, ρ), so n

h(z) +^λzh_(µ+p)^0(z) o

∈ Pk(ρ) for z ∈ E. This implies that

h_i(z) +λzh⁰_i(z) (µ+p)

> ρ, i= 1,2.

Now, by applying Lemma 2.1 we see that Re{h_i(z)}> η,whereη is given by (3.1).Consequently, h∈Pk(η) for z∈E,and the proof is complete.

Now, we take the converse case of Theorem 3.1.

Theorem 3.2. Let f ∈ P

p,m,k

(0, µ, p, q, s, ρ) for z ∈ E. Then f ∈ P

p,m,k

(λ, µ, p, q, s, ρ) for |z|< R(λ, µ, p, m), where

(3.3) R(λ, µ, p, m) =

"

(µ+p)

|λ|(m+p) + q

|λ|²(m+p)²+ (µ+p)²

#_p+m¹ .

Proof. Setting

−z^p+1(H^m,µ_p,q,s(α₁)f(z))⁰

p = (p−ρ)h(z) +ρ, h∈P_k. Now, proceeding as in Theorem 3.1, we have

−(1−λ)z^p+1(H^m,µ_p,q,s(α1)f(z))⁰−λz^p+1(H_p,q,s^m,µ+1(α1)f(z))⁰−ρ

p(p−ρ) =

(3.4)

=

h(z) +λzh⁰(z) (µ+p)

=

= k

4 +1

2 h₁(z) +λzh⁰₁(z) (µ+p)

− k

4 −1

2 h₂(z) +λzh⁰₂(z) (µ+p)

,

where we have used (1.3) and h1, h2 ∈ P, z ∈ E. Using the following well known estimate, see [7],

zh⁰_i(z)

≤ 2(p+m)r^p+m

1−r^2(p+m) Re{h_i(z)}, |z|=r <1, i= 1,2, we have

Re

hi(z) + λzh⁰_i(z) (µ+p)

≥Re

hi(z)− |λ| |zh⁰_i(z)|

(µ+p)

≥Reh_i(z) (

1− 2|λ|(m+p)r^m+p (µ+p) 1−r^2(p+m)

) . The right hand side of this inequality is positive if r < R(λ, µ, p, m), where R(λ, µ, p, m) is given by (3.3). Consequently, it follows from (3.4) that f ∈

P

p,m,k

(λ, µ, p, q, s, ρ) for |z|< R(λ, µ, p, m).

Sharpness of this result follows by taking hi(z) = ^1+z_1−z^p+mp+m in (3.4), i= 1,2.

For a functionf∈P

p,m

,we consider the integral operator F_δ,p defined by F_δ,p(z) =F_δ,p(f)(z) = δ

z^δ+p Z _z

0

t^δ+p−1f(t)dt (3.5)

=

z^−p+X

k=m

δ δ+p+kz^k

∗f(z), δ >0, z∈E^∗. It follows from (3.5) thatF_δ,p(f)∈ P

p,m

and

(3.6) z(H^m,µ_p,q,s(α1)F_δ,pf)⁰(z) =δH^m,µ_p,q,s(α1)f(z)−(δ+p)H^m,µ_p,q,s(α1)F_δ,p(f)(z).

Theorem 3.3. Let f ∈ P

p,m

and the function F_δ,p(f), defined by (3.5) satisfies

−(1−λ)z^p+1(H_p,q,s^m,µ(α₁)F_δ,p(f)(z))⁰−λz^p+1(H^m,µ+1_p,q,s (α₁)f(z))⁰

p ∈P_k(ρ), z∈E,

then

−z^p+1(H^m,µ_p,q,s(α₁)F_δ,p(f)(z))⁰

p ∈P_k(σ), z∈E,

where σ is given by

(3.7) σ=ρ+ (1−ρ)(2γ1−1), γ1 = Z 1

0

1 +t^Reλδ −1

dt.

Proof. If we let

−z^p+1(H^m,µ_p,q,s(α₁)F_δ,p(f)(z))⁰

p =h(z) =

k 4+1

2

h1(z)− k

4+1 2

h2(z).

(3.8)

Then h(z) is analytic in E with h(0) = 1. Using the identity (3.6) in (3.8) followed by the differentiation of the resulting equation, we obtain

−(1−λ)z^p+1(H^m,µ_p,q,s(α1)F_δ,p(f)(z))⁰−λz^p+1(H^m,µ+1_p,q,s (α1)f(z))⁰

p =

=h(z) +λ

δzh⁰(z)∈P_k(ρ) forz∈E.

Using Lemma 2.1, we see that −^{zp+1(Hm,µp,q,s(α}_p^{1)Fδ,p(f)(z))0} ∈ P_k(σ), for z ∈ E, where σ is given by (3.7),and the proof is complete.

Remark 1. Under the hypothesis of Theorem 3.3 and using the fact that z^p+1(H_p,q,s^m,µ(α₁)F_δ,p(f)(z))⁰= δ

z^δ Z z

0

t^δ+p((H^m,µ_p,q,s(α₁)f(z)⁰dt δ >0, z∈E, we obtain

− δ

z^δ Z _z

0

t^δ+p((H^m,µ_p,q,s(α₁)f(z)⁰dt

∈P_k(σ), z∈E, where σ is given as in Theorem 3.3.

Corollary 3.4. Let f ∈ P

p,m,k

(0, µ, p, q, s, ρ).Then

−z^p+1(H^m,µ_p,q,s(α1)F_δ,p(f)(z))⁰

p ∈P_k(σ),

where σ is given as in Theorem 3.3 with γ₁=

Z 1 0

1 +t^Re1δ−1

dt.

Proof. Setting

(3.9) −z^p+1(H^m,µ_p,q,s(α₁)F_δ,p(f)(z))⁰

p =h(z) =

k 4 + 1

2

h1(z)−

k 4 +1

2

h2(z).

Then h(z) is analytic in E and h(0) = 1.Using the operator identity (3.6) in (3.9) and differentiating the resulting equation with respect toz,we find that

−z^p+1(H^m,µ_p,q,s(α₁)f(z))⁰

p =h(z) +1

δzh⁰(z)∈P_k(ρ), z∈E.

Now the remaining part of Corollary 3.4 follows by employing the techniques that we used in proving Theorem 3.3 above.

Theorem 3.5. If f ∈ P

p,m

and the function F_δ,p(f) defined by (3.5) satisfies

(1−λ)z^p(H^m,µ_p,q,s(α1)F_δ,p(f)(z)) +λz^p(H_p,q,s^m,µ+1(α1)f(z))∈Pk(ρ), z∈E, then

z^p(H^m,µ_p,q,s(α₁)F_δ,p(f)(z))∈P_k(σ), z∈E, where σ is given as in Theorem 3.3.

Theorem 3.6. Let f ∈ M_p,m,k(λ, µ, p, q, s, ρ) and let φ(z) satisfy the following inequality

(3.10) Re{z^pφ(z)}> 1

2, z∈E.

Then φ∗f ∈M_p,m,k(λ, µ, p, q, s, ρ).

Proof. LetG=φ∗f. Then we have

(1−λ)z^p(H^m,µ_p,q,s(α1)G(z)) +λz^p(H^m,µ+1_p,q,s (α1)G(z)) =

=

(1−λ)z^p(H^m,µ_p,q,s(α₁)(φ∗f)(z) +λz^p(H^m,µ+1_p,q,s (α₁)(φ∗f)(z)) =

=z^p(φ(z))∗H(z), H ∈P_k(ρ), where

H(z) =

(1−λ)z^p(H_p,q,s^m,µ(α1)f(z)) + (H^m,µ+1_p,q,s (α1)f(z)) ∈Pk(ρ).

Therefore, we have z^p(φ(z))∗H(z) =

k 4 +1

2

{(p−ρ) (z^p(φ(z))∗h1(z)) +ρ} −

− k

4 −1 2

{(p−ρ) (z^p(φ(z))∗h2(z)) +ρ}, h1, h2∈P.

Since Re{z^p(φ(z))} > ¹₂, z ∈ E and so, using Lemma 2.2, we conclude that G=φ∗f ∈M_p,m,k(λ, µ, p, q, s, ρ).

Theorem 3.7.Letf ∈ P

p,m,k

(0, µ, p, q, s, ρ)and letφ(z)satisfy the inequa- lity (3.10).Thenφ∗f ∈ P

p,m,k

(0, µ, p, q, s, ρ).

Proof. We have

−z^p+1(H^m,µ_p,q,s(α₁)(φ∗f)(z))⁰

p =−z^p+1(H_p,q,s^m,µ(α₁)f(z))⁰

p ∗z^pφ(z), z∈E.

Now, the remaining part of Theorem 3.7 follows by employing the tech- niques that we used in proving Theorem 3.6 above.

Theorem 3.8. Let 0≤λ2 < λ1.Then

M_p,m,k(λ₁, µ, p, q, s, ρ)⊂M_p,m,k(λ₂, µ, p, q, s, ρ).

Proof. For λ₂ = 0, the proof is immediate. Let λ₂ > 0 and let f ∈ Mp,m,k(λ1, µ, p, q, s, ρ).Then

(1−λ₂)z^p(H^m,µ_p,q,s(α₁)f(z)) +λ₂z^p(H^m,µ+1_p,q,s (α₁)f(z)) =

= λ₂ λ1

"

(^λ_λ¹

2 −1)z^p(H^m,µ_p,q,s(α1)f(z)) + (1−λ1)z^p(H^m,µ_p,q,s(α1)f(z)) +λ₁z^p(H^m,µ+1_p,q,s (α₁)f(z))

#

=

=

1−λ₂ λ₁

H₁(z) + λ₂

λ₁H₂(z), H₁(z), H₂(z)∈P_k(ρ).

SincePk(ρ) is a convex set, see [8],we conclude thatf ∈Mp,m,k(λ2, µ, p, q, s, ρ) for z∈E.

Now, by using Theorem 3.1 and the lines of proof of Theorem 3.8, we have the following theorem.

Theorem 3.9. Let 0≤λ₂ < λ₁.Then X

p,m,k(λ1, µ, p, q, s, ρ)⊂X

p,m,k(λ2, µ, p, q, s, ρ).

Theorem 3.10. Letf ∈M_p,m,k(λ, µ, p, q, s, ρ₁)andg∈M_p,m,k(λ, µ, p, q, s, ρ2) and let F =f∗g. ThenF ∈Mp,m,k(λ, µ, p, q, s, ρ3), where

(3.11) ρ₃= 1−4(1−ρ₁)(1−ρ₂)

"

1−µ+p λ

Z 1 0

u^{µ+pλ −1} 1 +u du

# . Proof. Since f ∈Mp,m,k(λ, µ, p, q, s, ρ1),it follows that

H(z) =

(1−λ)z^p(H_p,q,s^m,µ(α₁)f(z)) +λz^p(H^m,µ+1_p,q,s (α₁)f(z)) ∈P_k(ρ₁), and so, using identity (1.9) in the above equation, we have

(3.12) H^m,µ_p,q,s(α₁)f(z) = µ+p

λ z^{−p−µ+pλ} Z z

0

t^{µ+pλ −1}H(t)dt.

Similarly,

(3.13) H^m,µ_p,q,s(α₁)g(z) = µ+p

λ z^{−p−µ+pλ} Z z

0

t^{µ+pλ −1}H^∗(t)dt, where H^∗∈P_k(ρ₃).Using (3.12) and (3.13),we have

(3.14) H^m,µ_p,q,s(α1)F(z) = µ+p

λ z^{−p−µ+pλ} Z z

0

t^{µ+pλ −1}Q(t)dt,

where

Q(z) = k

4 +1 2

q₁(z)− k

4 −1 2

q₂(z) (3.15)

= µ+p λ z^−µ+pλ

Z z

0

t^{µ+pλ −1}(H∗H^∗)dt.

Now,

H(z) = k

4 +1 2

h1(z)− k

4 −1 2

h2(z), (3.16)

H^∗(z) = k

4 +1 2

h^∗₁(z)− k

4 −1 2

h^∗₂(z), where hi ∈P(ρ1) and h^∗_i ∈P(ρ2), i= 1,2.

Since

P_i^∗= h^∗_i(z)−ρ2

2(1−ρ2) +1 2 ∈P

1 2

, i= 1,2, we obtain that (hi∗p^∗_i)∈P(ρ2), by using Herglotz formula.

Thus

(hi∗h^∗_i)∈P(ρ3), with

(3.17) ρ3= 1−2(1−ρ1)(1−ρ2).

Using (3.14),(3.15),(3.16),(3.17) and Lemma 2.3, we have Req_i(z) = µ+p

λ Z 1

0

u^{µ+pλ −1}Re{(h_i∗h^∗_i)(uz)}du

≥ µ+p λ

Z 1 0

u^{µ+pλ −1}

2ρ3−1 +2(1−ρ₃) 1 +u|z|

du

≥ µ+p λ

Z 1 0

u^{µ+pλ −1}

2ρ3−1 +2(1−ρ3) 1 +u

du

= 1−4(1−ρ₁)(1−ρ₂)

"

1−µ+p λ

Z 1 0

u^{µ+pλ −1} 1 +u du

# .

From this we conclude that F ∈ Mp,m,k(λ, µ, p, q, s, ρ3) where ρ3 is given by (3.11).

We discuss the sharpness as follows:

We take H(z) =

k 4 + 1

2

1 + (1−2ρ1)z

1−z −

k 4 −1

2

1−(1−2ρ1)z 1 +z , H^∗(z) =

k 4 + 1

2

1 + (1−2ρ₂)z

1−z −

k 4 −1

2

1−(1−2ρ₂)z 1 +z . Since

1 + (1−2ρ1)z 1−z

∗

1 + (1−2ρ2)z 1−z

= 1−4(1−ρ₁)(1−ρ₂)+4(1−ρ1)(1−ρ2)

1−z .

It follows from (3.10) that qi(z) = µ+p

λ Z 1

0

u^{µ+pλ −1}

1−4(1−ρ2)(1−ρ3) +4(1−ρ2)(1−ρ3) 1−z

du

→1−4(1−ρ₁)(1−ρ₂)

"

1−µ+p λ

Z 1 0

u^{µ+pλ −1} 1 +u du

#

asz→ −1.

This completes the proof.

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Received 9 October 2011 University of Engineering and Technology Peshawar Department of Basic Sciences

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