1Introduction DanielBreaz S ( β ) Aconvexitypropertyforanintegraloperatorontheclass

A convexity property for an integral operator on the class S

(β)

Daniel Breaz

Abstract

In this paper we consider an integral operatorF_n(z) for analytic functions f_i(z) in the open unit disk U. The object of this paper is to prove the convexity properties for the integral operator F_n(z) on the class S_p(β).

2000 Mathematical Subject Classification: 30C45

Key words and phrases: analytic functions, univalent functions, integral operator, convex functions.

1 Introduction

Let U = {z ∈C,|z|<1} be the unit disc of the complex plane and de- note by H(U), the class of the olomorphic functions in U. Consider A = {f ∈H(U), f(z) = z+a₂z²+a₃z³+..., z ∈U}be the class of analytic func- tions in U and S={f ∈A :f is univalent in U}.

177

Denote with K the class of convex functions in U, defined by

K =

½

f ∈H(U) :f(0) =f⁰(0)−1 = 0,Re

½zf⁰⁰(z) f⁰(z) + 1

¾

>0, z ∈U

¾ . A function f ∈S is the convex function by the order α,0≤ α <1 and denote this class by K(α) if f verify the inequality

Re

½zf⁰⁰(z) f⁰(z) + 1

¾

> α, z∈U.

Consider the class S_p(β), was is introduced by F. Ronning in the paper [3] and is defined by:

(1) f ∈S_p(β)⇔

¯¯

¯¯zf⁰(z) f(z) −1

¯¯

¯¯≤Re

½zf⁰(z) f(z) −β

¾

where β is the real number with the property −1≤β <1.

Forf_i(z)∈A and α_i >0, i∈ {1, ..., n}, we define the integral operator F_n(z) given by

(2) F_n(z) =

Z _z

0

µf1(t) t

¶_α1

·...·

µfn(t) t

¶_αn dt.

This integral operator was first defined by Breaz and Breaz in [1]. It is easy to see that F_n(z)∈A.

2 Main results

Theorem 1. Let α_i >0, for i∈ {1, ..., n}, β_i is the real numbers with the property −1≤βi <1 and fi ∈Sp(βi) for i∈ {1, ..., n}.

If

(3) 0<

Xn

i=1

α_i(1−β_i)≤1,

the integral operator Fn is convex by the order 1 +Pⁿ

i=1

αi(βi −1).

Proof. We calculate for F_n the derivatives of the first and second order.

From (2) we obtain:

F_n⁰ (z) =

µf₁(z) z

¶_α1

·...·

µf_n(z) z

¶_αn

and

F_n⁰⁰(z) = Xn

i=1

α_i

µf_i(z) z

¶_αi−1µ

zf_i⁰(z)−f_i(z) zf_i(z)

¶Y_n

j=1

j6=i

µf_j(z) z

¶_αj .

After the calculus we obtain that:

F_n⁰⁰(z) F_n⁰ (z) =α₁

µzf₁⁰ (z)−f₁(z) zf₁(z)

¶

+...+α_n

µzf_n⁰ (z)−f_n(z) zf_n(z)

¶ . These relation is equivalent with:

(4) F_n⁰⁰(z) F_n⁰ (z) =α₁

µf₁⁰(z) f₁(z)− 1

z

¶

+...+α_n

µf_n⁰ (z) f_n(z) − 1

z

¶ . Multiply the relation (4) with z we obtain:

(5) zF_n⁰⁰(z) F_n⁰ (z) =

Xn

i=1

α_i

µzf_i⁰(z) f_i(z) −1

¶

= Xn

i=1

α_izf_i⁰(z) f_i(z) −

Xn

i=1

α_i.

The relation (5) is equivalent with zF_n⁰⁰(z)

F_n⁰ (z) + 1 = Xn

i=1

α_izf_i⁰(z) f_i(z) −

Xn

i=1

α_i+ 1.

This relation is equivalent with:

zF_n⁰⁰(z)

F_n⁰ (z) + 1 = Xn

i=1

α_i

µzf_i⁰(z) fi(z) −β_i

¶ +

Xn

i=1

α_iβ_i− Xn

i=1

α_i+ 1.

We calculate the real part from both terms of the above equality and obtain:

Re

µzF_n⁰⁰(z) F_n⁰ (z) + 1

¶

= Xn

i=1

α_iRe

µzf_i⁰(z) f_i(z) −β_i

¶ +

Xn

i=1

α_iβ_i− Xn

i=1

α_i+ 1.

Becausef_i ∈S_p(β_i) fori={1, ..., n}, we apply in the above relation the inequality (1) and obtain:

Re

µzF_n⁰⁰(z) F_n⁰ (z) + 1

¶

>

Xn

i=1

α_i

¯¯

¯¯zf_i⁰(z) f_i(z) −1

¯¯

¯¯+ Xn

i=1

α_i(β_i−1) + 1.

Because α_i

¯¯

¯¯zf_i⁰(z) f_i(z) −1

¯¯

¯¯>0 for all i∈ {1, ..., n}, obtain that

Re

µzF_n⁰⁰(z) F_n⁰ (z) + 1

¶

>

Xn

i=1

α_i(β_i−1) + 1.

So, F_n is convex by the order Pⁿ

i=1

α_i(β_i−1) + 1.

Theorem 2. Let α_i, i∈ {1, ..., n} the real positive numbers and f_i ∈S_p(β) for i∈ {1, ..., n}.

If

0<

Xn

i=1

α_i ≤ 1 1−β,

the integral operator F_n is convex by the order (β−1)Pⁿ

i=1

α_i+ 1.

Proof. Since

F_n⁰ (z) =

µf1(z) z

¶_α1

·...·

µfn(z) z

¶_αn

and

F_n⁰⁰(z) = Xn

i=1

α_i

µf_i(z) z

¶_αi−1µ

zf_i⁰(z)−f_i(z) zf_i(z)

¶Y_n

j=1

j6=i

µf_j(z) z

¶_αj

we obtain

zF_n⁰⁰(z)

F_n⁰ (z) + 1 = Xn

i=1

α_izf_i⁰(z) fi(z) −

Xn

i=1

α_i+ 1.

Thus we see, forf_i(z)∈S_p(β), for all i∈ {1, ..., n}

Re

µzF_n⁰⁰(z) F_n⁰ (z) + 1

¶

>

Xn

i=1

αi

¯¯

¯¯zf_i⁰(z) f_i(z) −1

¯¯

¯¯−(β−1) Xn

i=1

αi+ 1.

Because α_i

¯¯

¯^zf_fiⁱ⁰_(z)^(z) −1

¯¯

¯>0 for all i∈ {1, ..., n}, obtain that

Re

µzF_n⁰⁰(z) F_n⁰ (z) + 1

¶

>(β−1) Xn

i=1

α_i+ 1.

Because−1≤β <1 and 0<Pⁿ

i=1

α_i ≤ 1

1−β obtain that 0≤(β−1)Pⁿ

i=1

α_i +1 <1. So F_n is convex by the order (β−1)Pⁿ

i=1

α_i + 1.

Remark 1.If β = 0 and Pⁿ

i=1

α_i = 1 then

Re

µzF_n⁰⁰(z) F_n⁰ (z) + 1

¶

>0 so, F_n is the convex function.

Corollary 1. Let γ the real number, γ >0. We suppose that the functions f ∈ S_p(β) and 0 < γ ≤ 1

1−β. In this conditions the integral operator F1(z) =

Zz

0

µf(t) t

¶_γ

dt is convex by the order (β−1)γ+ 1.

Proof. In the Theorem 2, we consider n= 1.

Corollary 2. Let f ∈ S_p(β) and consider the integral operator of Alexan- der, F(z) =

Zz

0

f(t)

t dt. In this condition F is convex by the order β.

Proof. We have:

(6) zF⁰⁰(z)

F⁰(z) = zf⁰(z) f(z) −1.

From (6) we have:

(7) Re

µzF⁰⁰(z) F⁰(z) + 1

¶

=Re

µzf⁰(z) f(z) −β

¶

+β > k

¯¯

¯¯zf⁰(z) f(z) −1

¯¯

¯¯+β > β.

So, the relation (7) imply that the Alexander operator is convex.

References

[1] D. Breaz, N. Breaz,Two integral operators, Studia Universitatis Babe¸s -Bolyai, Mathematica, Cluj Napoca, No. 3-2002, pp. 13-21.

[2] G. Murugusundaramoorthy, N. Maghesh, A new subclass of uniforlmly convex functions and a corresponding subclass of starlike functions with fixed second coefficient, Journal of Inequalities in Pure and Applied Mathematics, Vol. 5, Issue 4, 2005, pp. 1-10.

[3] F. Ronning,Uniformly convex functions, Ann. Polon. Math., 57(1992), 165-175.

Department of Mathematics Faculty of Science

University ”1 Decembrie 1918” of Alba Iulia Romania

Email addresses: [email protected]