Classes of meromorphic multivalent functions with Montel’s normalization

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A N N A L E S

U N I V E R S I T A T I S M A R I A E C U R I E - S K Ł O D O W S K A L U B L I N – P O L O N I A

VOL. LXVI, NO. 1, 2012 SECTIO A 13–23

B. A. FRASIN

On certain general integral operators of analytic functions

Abstract. In this paper, we obtain new sufficient conditions for the operators Fα₁,α₂,...,α_n,β(z) and Gα₁,α₂,...,α_n,β(z) to be univalent in the open unit disc U , where the functions f1, f2, . . . , fn belong to the classes S^?(a, b) and K(a, b). The order of convexity for the operators Fα₁,α₂,...,αn,β(z) and Gα₁,α₂,...,α_n,β(z) is also determined. Furthermore, and for β = 1, we obtain sufficient conditions for the operators Fn(z) and Gn(z) to be in the class K(a, b). Several corollaries and consequences of the main results are also con- sidered.

1. Introduction and definitions. Let A denote the class of functions of the form

f (z) = z +

∞

X

n=2

anzⁿ

which are analytic in the open unit disc U = {z : |z| < 1}. Further, by S we shall denote the class of all functions in A which are univalent in U . A function f (z) ∈ A is said to be starlike of order γ (0 ≤ γ < 1) if it satisfies

(1.1) Re zf⁰(z)

f (z)

> γ (z ∈ U ).

1991 Mathematics Subject Classification. 30C45.

Key words and phrases. Analytic functions, starlike and convex functions, integral operator.

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Also, we say that a function f (z) ∈ A is said to be convex of order γ (0 ≤ γ < 1) if it satisfies

(1.2) Re

1 +zf⁰⁰(z) f⁰(z)

> γ (z ∈ U ).

We denote by S^?(γ) and K(γ) , respectively, the usual classes of starlike and convex functions of order γ (0 ≤ γ < 1) in U .

A function f ∈ A is said to be in the class S^?(a, b) if

(1.3)

zf⁰(z) f (z) − a

< b (z ∈ U ; |a − 1| < b ≤ a) and a function f ∈ A is said to be in the class K(a, b) if (1.4)

1 +zf⁰⁰(z) f⁰(z) − a

< b (z ∈ U ; |a − 1| < b ≤ a).

From (1.3) and (1.4), we have Re zf⁰(z)

f (z)

> a − b (z ∈ U ; |a − 1| < b ≤ a) and

Re

1 +zf⁰⁰(z) f⁰(z)

> a − b (z ∈ U ; |a − 1| < b ≤ a).

The class S^?(a, b) was introduced by Jakubowski [12]. It is clear that a > ¹₂, S^?(a, b) ⊂ S^?(a − b) ⊂ S^?(0) ≡ S^? and K(a, b) ⊂ K(a − b) ⊂ K(0) ≡ K.

Further, applying the Briot-Bouquet differential subordination [9], we can easily see that K(a, b) ⊂ S^?(a, b).

Several authors (e.g., see [4, 5, 6, 8, 10, 11, 15, 16]), obtained many sufficient conditions for the univalency of the integral operators

(1.5) F_α₁_,α₂_,...,α_n_,β(z) =





 β

z

Z

0

t^β−1

n

Y

i=1

f_i(t) t

αi

dt







1 β

and

(1.6) G_α₁_,α₂_,...,α_n_,β(z) =





 β

Zz

0

t^β−1

n

Y

i=1

f_i⁰(t)αi

dt







1 β

,

where the functions f₁, f2, . . . , fn belong to the class A and the parame- ters α₁, α₂, . . . , α_n, and β are complex numbers such that the integrals in (1.5) and (1.6) exist. Here and throughout in the sequel every many-valued function is taken with the principal branch.

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For β = 1, we obtain the integral operators

(1.7) F_n(z) =

z

Z

0

f₁(t) t

α1

. . . f_n(t) t

αn

dt

and

(1.8) G_n(z) =

z

Z

0

f₁⁰(t)α1

. . . f_n⁰(t)αn

dt

introduced and studied by Breaz and Breaz [5] and Breaz et al. [7], respectively.

In this paper, we obtain new sufficient conditions for the operators Fα1,α2,...,αn,β(z) and Gα1,α2,...,αn,β(z) defined by (1.5) and (1.6) to be univalent in the open unit disc U , where the functions f₁, f₂, . . . , f_nbelong to the above classes S^?(a, b) and K(a, b). The order of convexity for the operators F_α₁_,α₂_,...,α_n_,β(z) and G_α₁_,α₂_,...,α_n_,β(z) is also determined. Furthermore, we obtain sufficient conditions for the operators F_n(z) and G_n(z) defined by (1.5) and (1.6) to be in the class K(a, b).

In the proofs of our main results we need the following univalence cri- teria. The first result, i.e. Lemma 1.1 is a generalization of the well- known univalence criterion of Becker [2] (which in fact corresponds to the case β = δ = 1), while the second, i.e. Lemma 1.2 is a generalization of Ahlfors’ and Becker’s univalence criterion [1, 3] (which corresponds to the case β = 1).

Lemma 1.1 ([13]). Let δ be a complex number with Re(δ) > 0. If f ∈ A satisfies

1 − |z|^{2 Re(δ)} Re(δ)

zf⁰⁰(z) f⁰(z)

≤ 1,

for all z ∈ U , then, for any complex number β with Re(β) ≥ Re(δ), the integral operator

F_β(z) =





 β

z

Z

0

t^β−1f⁰(t)dt







1 β

is in the class S.

Lemma 1.2 ([14]). Let β be a complex number with Re(β) > 0 and c be a complex number with |c| ≤ 1, c 6= −1. If f ∈ A satisfies

c |z|^2β +

1 − |z|^2βzf⁰⁰(z) βf⁰(z)

≤ 1

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for all z ∈ U , then the integral operator

F_β(z) =





 β

z

Z

0

t^β−1f⁰(t)dt







1 β

is in the class S.

2. Univalence conditions for F_α₁_,α₂_,...,α_n_,β(z). We first prove

Theorem 2.1. Let f_i(z) ∈ S^?(a_i, b_i); |a_i− 1| < b_i ≤ a_i, α_i ∈ C for all i = 1, . . . , n, and δ ∈ C with

(2.1) Re(δ) ≥ 2

n

X

i=1

|α_i| b_i.

Then for any β ∈ C with Re(β) ≥ Re(δ), the integral operator Fα1,α2,...,αn,β(z) defined by (1.5) is analytic and univalent in U .

Proof. Defining

h(z) =

z

Z

0 n

Y

i=1

f_i(t) t

αi

dt,

we observe that h(0) = h⁰(0) − 1 = 0, where

(2.2) h⁰(z) =

n

Y

i=1

fi(z) z

αi

.

Differentiating both sides of (2.2) logarithmically, we obtain zh⁰⁰(z)

h⁰(z) =

n

X

i=1

α_i zf_i⁰(z) f_i(z) − 1

which is equivalent to (2.3) zh⁰⁰(z)

h⁰(z) =

n

X

i=1

α_i zf_i⁰(z) fi(z) − a_i

+

n

X

i=1

α_ia_i−

n

X

i=1

α_i.

Since f_i(z) ∈ S^?(a_i, b_i); |a_i− 1| < b_i ≤ a_i for all i = 1, 2, . . . , n, it follows from (2.3) that

(2.4)

zh⁰⁰(z) h⁰(z)

≤

n

X

i=1

|α_i|

zf_i⁰(z) f_i(z) − a_i

+

n

X

i=1

|α_i| |a_i− 1|

< 2

n

X

i=1

|α_i| b_i.

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Multiplying both sides of (2.4) by ^1−|z|_Re(δ)^{2 Re(δ)} and making use of (2.1), we obtain

1 − |z|^{2 Re(δ)} Re(δ)

zh⁰⁰(z) h⁰(z)

≤ 2 1 − |z|^{2 Re(δ)} Re(δ)

! _n X

i=1

|α_i| b_i

< 2 Re(δ)

n

X

i=1

|α_i| b_i ≤ 1.

Applying Lemma 1.1 for the function h(z), we prove that F_α₁_,α₂_,...,α_n_,β(z) ∈

S.

Letting n = 1, α₁ = α, a₁ = a, b₁ = b and f₁ = f in Theorem 2.1, we have

Corollary 2.2. Let f (z) ∈ S^?(a, b); |a − 1| < b ≤ a, α ∈ C and δ ∈ C with Re(δ) > 2 |α| b. Then for any β ∈ C with Re(β) ≥ Re(δ), the integral operator

(2.5) F_α,β(z) =





 β

Zz

0

t^β−1 f (t) t

α

dt







1 β

is analytic and univalent in U .

Making use of Lemma 1.2, we prove the following theorem:

Theorem 2.3. Let f_i(z) ∈ S^?(a_i, b_i); |a_i− 1| < b_i ≤ a_i, α_i ∈ C for all i = 1, 2, . . . , n, and β ∈ C with

Re(β) ≥ 2

n

X

i=1

|α_i| b_i

and

(2.6) |c| ≤ 1 − 2

Re(β)

n

X

i=1

|α_i| b_i (c ∈ C).

Then the integral operator F_α₁_,α₂_,...,α_n_,β(z) defined by (1.5) is analytic and univalent in U .

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Proof. Let f_i(z) ∈ S^?(a_i, b_i); |a_i− 1| < b_i ≤ a_i for all i = 1, 2, . . . , n, it follows from (2.4) that

c |z|^2β+ (1 − |z|^2β)zh⁰⁰(z) βh⁰(z)

≤ |c| +

1 − |z|^2β β

zh⁰⁰(z) h⁰(z)

≤ |c| + 2

1 − |z|^2β β

n

X

i=1

|α_i| b_i

< |c| + 2

|β|

n

X

i=1

|α_i| b_i

< |c| + 2 Re(β)

n

X

i=1

|α_i| b_i which, in the light of the hypothesis (2.6), yields

c |z|^2β + (1 − |z|^2β)zh⁰⁰(z) βh⁰(z)

≤ 1.

Finally, by applying Lemma 1.2, we conclude that F_α₁_,α₂_,...,α_n_,β(z) ∈ S.

Corollary 2.4. Let f (z) ∈ S^?(a, b); |a − 1| < b ≤ a, α ∈ C, and β ∈ C with

Re(β) ≥ 2 |α| b and

|c| ≤ 1 − 2

Re(β)|α| b (c ∈ C).

Then the integral operator F_α,β(z) defined by (2.5) is analytic and univalent in U .

3. Univalence conditions for G_α₁_,α₂_,...,α_n_,β(z). Now, we prove

Theorem 3.1. Let f_i(z) ∈ K(a_i, b_i); |a_i− 1| < b_i ≤ a_i, α_i ∈ C for all i = 1, . . . , n, and δ ∈ C with

Re(δ) ≥ 2

n

X

i=1

|α_i| b_i.

Then for any β ∈ C with Re(β) ≥ Re(δ), the integral operator Gα1,α2,...,αn,β(z) defined by (1.6) is analytic and univalent in U .

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Proof. Defining

h(z) =

z

Z

0 n

Y

i=1

f_i⁰(t)αi

dt,

we observe that h(0) = h⁰(0) − 1 = 0. On the other hand, it is easy to see that

(3.1) h⁰(z) =

n

Y

i=1

f_i⁰(z)αi

.

Differentiating both sides of (3.1) logarithmically, we obtain zh⁰⁰(z)

h⁰(z) =

n

X

i=1

α_i zf_i⁰⁰(z) f_i⁰(z)

. Thus, we have

(3.2) zh⁰⁰(z) h⁰(z) =

n

X

i=1

αi

1 +zf_i⁰⁰(z) f_i⁰(z) − a_i

+

n

X

i=1

αi(ai− 1).

Let f_i(z) ∈ K(ai, bi); |ai− 1| < b_i ≤ a_i, for all i = 1, 2, . . . , n, and following the same steps in the proof of Theorem 2.1, we get the required result. Letting n = 1, α₁ = α, a₁ = a, b₁ = b and f₁ = f in Theorem 3.1, we have

Corollary 3.2. Let f (z) ∈ K(a, b); |a − 1| < b ≤ a, α and δ ∈ C with Re(δ) ≥ 2 |α| b. Then for any β ∈ C with Re(β) ≥ Re(δ), the integral operator

(3.3) Gα,β(z) =





 β

z

Z

0

t^β−1 f⁰(t)α

dt







1 β

is analytic and univalent in U .

Using (3.1), (1.4) and applying Lemma 1.2, we prove the following theorem:

Theorem 3.3. Let f_i(z) ∈ K(a_i, b_i); |a_i− 1| < b_i ≤ a_i, α_i ∈ C for all i = 1, . . . , n and β ∈ C with

Re(β) ≥ 2

n

X

i=1

|α_i| b_i and

|c| ≤ 1 − 2 Re(β)

n

X

i=1

|α_i| b_i (c ∈ C).

Then the integral operator G_α₁_,α₂_,...,α_n_,β(z) defined by (1.6) is analytic and univalent in U .

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Corollary 3.4. Let f (z) ∈ K(a, b); |a − 1| < b ≤ a, α and β ∈ C with Re(β) ≥ 2 |α| b

and

|c| ≤ 1 − 2

Re(β)|α| b.

Then the integral operator G_α,β(z) defined by (3.3) is analytic and univalent in U .

4. Order of convexity. Now, we prove

Theorem 4.1. Let f_i(z) ∈ S^?(a_i, b_i); |a_i− 1| < b_i≤ a_i, and α_i > 0 for all i = 1, . . . , n, with

0 ≤ 1 −

n

X

i=1

α_i

b_i+1

2

< 1 and

n

X

i=1

α_i

b_i+1

2

≤ 1.

Then the integral operator F_n(z) defined by (1.7) is in the class K 1 −

n

X

i=1

αi

bi+1

2

! . Proof. From (1.7), it follows that

(4.1) F_n⁰(z) =

n

Y

i=1

f_i(z) z

αi

.

Differentiating both sides of (4.1) logarithmically, we obtain 1 +zF_n⁰⁰(z)

F_n⁰(z) =

n

X

i=1

α_i zf_i⁰(z) fi(z)

−

n

X

i=1

α_i+ 1.

Since f_i(z) ∈ S^?(a_i, b_i); |a_i− 1| < b_i ≤ a_i and a_i > ¹₂ for all i = 1, 2, . . . , n, we have

(4.2)

Re

1 +zF_n⁰⁰(z) F_n⁰(z)

=

n

X

i=1

α_iRe zf_i⁰(z) fi(z)

−

n

X

i=1

α_i+ 1

≥

n

X

i=1

α_i(a_i− b_i− 1) + 1

> 1 −

n

X

i=1

α_i

b_i+1

2

. Therefore, F_n(z) is convex of order 1 −P_n

i=1α_i b_i+¹₂ in U.

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Corollary 4.2. Let f (z) ∈ S^?(a, b); |a − 1| < b ≤ a, and α > 0 with 0 ≤ 1 − α b + ¹₂ < 1 and α b +¹₂ ≤ 1. Then

Z z 0

_{f (t)}

t

α

dt ∈ K(1 − α(b +¹₂)).

Next, we prove

Theorem 4.3. Let f_i(z) ∈ K(a_i, b_i); |a_i− 1| < b_i ≤ a_i, and α_i > 0 for all i = 1, . . . , n, with

0 ≤ 1 −

n

X

i=1

α_i

b_i+1

2

< 1 and

n

X

i=1

α_i

b_i+1

2

≤ 1.

Then the integral operator G_n(z) defined by (1.8) is in the class K 1 −

n

X

i=1

α_i

b_i+1

2

! . Proof. From (1.8), we have

(4.3) 1 +zG⁰⁰_n(z) G⁰_n(z) =

n

X

i=1

αi

1 +zf_i⁰⁰(z) f_i⁰(z)

−

n

X

i=1

αi+ 1.

Let f_i(z) ∈ K(ai, bi); |ai− 1| < b_i ≤ a_i; a_i> ¹₂ for all i = 1, 2, . . . , n, and following the same steps in the proof of Theorem 4.1, we get the required

result.

Letting n = 1, α₁ = α, a1 = a, b1 = b and f1 = f in Theorem 4.3, we have

Corollary 4.4. Let f (z) ∈ K(a, b); |a − 1| < b ≤ a, and α > 0 with 0 ≤ 1 − α b +¹₂ < 1 and α b + ¹₂ ≤ 1. Then R₀^z(f⁰(t))^αdt ∈ K(1 − α(b +¹₂)).

5. Sufficient conditions for the operators Fn(z) and Gn(z).

Theorem 5.1. Let fi(z) ∈ S^?(γi); 0 ≤ γi < 1, for all i = 1, 2, . . . , n.

Then the integral operator F_n(z) defined by (1.7) is in the class K(a_i, b_i), where a_i = Pn

i=1αiγi+ 1, bi = Pn

i=1αi and Pn

i=1αi(1 − γi) ≤ 1 for all i = 1, 2, . . . , n.

Proof. Let fi(z) ∈ S^?(γi); 0 ≤ γi< 1, for all i = 1, 2, . . . , n. Then it follows from (4.2) that

Re

1 +zF_n⁰⁰(z) F_n⁰(z)

=

n

X

i=1

αiRe zf_i⁰(z) f_i(z)

+ 1 −

n

X

i=1

αi

>

n

X

i=1

αiγi+ 1 −

n

X

i=1

αi

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which proves that F_n(z) ∈ K(a_i, b_i), where a_i = Pn

i=1α_iγ_i + 1 and b_i = Pn

i=1αi for all i = 1, 2, . . . , n.

Letting n = 1, α₁ = α, γ1 = γ, a1 = a, b1 = b and f1 = f in Theorem 5.1, we have

Corollary 5.2. Let f (z) ∈ S^?(γ); 0 ≤ γ < 1. Then Z z

0

_{f (t)}

t

α

dt ∈ K(αγ + 1, α), where 0 < α(1 − γ) ≤ 1.

Using (4.3), we can prove the following theorem:

Theorem 5.3. Let fi(z) ∈ K(γi); 0 ≤ γi < 1, for all i = 1, 2, . . . , n.

Then the integral operator G_n(z) defined by (1.8) is in the class K(a_i, b_i), where a_i = Pn

i=1αiγi+ 1, bi = Pn

i=1αi and Pn

i=1αi(1 − γi) ≤ 1 for all i = 1, 2, . . . , n.

Letting n = 1, α₁ = α, γ1 = γ, a1 = a, b1 = b and f1 = f in Theorem 5.3, we have

Corollary 5.4. Let f (z) ∈ K(γ); 0 ≤ γ < 1. Then Rz

0 (f⁰(t))^αdt ∈ K(αγ + 1, α), where 0 < α(1 − γ) ≤ 1.

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B. A. Frasin

Department of Mathematics Al al-Bayt University P.O. Box: 130095 Mafraq Jordan

e-mail: bafrasin@yahoo.com Received April 20, 2011