General Mathematics Vol. 15, Nr. 2–3(2007), 177-183
A convexity property for an integral
operator on the class Sp (β)
Daniel Breaz
Abstract
In this paper we consider an integral operator Fn (z) for analytic
functions fi (z) in the open unit disk U . The object of this paper is
to prove the convexity properties for the integral operator Fn (z) on
the class Sp (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 denote by H (U ), the class of the olomorphic functions in U. Consider A =
{f ∈ H (U ) , f (z) = z + a2 z 2 + a3 z 3 + ..., z ∈ U } be the class of analytic functions in U and S = {f ∈ A : f is univalent in U }.
177
178
Daniel Breaz
Denote with K the class of convex functions in U , defined by
½
½ 00
¾
¾
zf (z)
0
K = f ∈ H (U ) : f (0) = f (0) − 1 = 0, Re
+ 1 > 0, z ∈ U .
f 0 (z)
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
½
Re
zf 00 (z)
+1
f 0 (z)
¾
> α, z ∈ U.
Consider the class Sp (β), was is introduced by F. Ronning in the paper
[3] and is defined by:
(1)
¯ 0
¯
¾
½ 0
¯ zf (z)
¯
zf (z)
¯
¯
f ∈ Sp (β) ⇔ ¯
− 1¯ ≤ Re
−β
f (z)
f (z)
where β is the real number with the property −1 ≤ β < 1.
For fi (z) ∈ A and αi > 0, i ∈ {1, ..., n}, we define the integral operator
Fn (z) given by
Z
(2)
z
Fn (z) =
0
µ
f1 (t)
t
¶α1
µ
· ... ·
fn (t)
t
¶αn
dt.
This integral operator was first defined by Breaz and Breaz in [1]. It is
easy to see that Fn (z) ∈ A.
2
Main results
Theorem 1. Let αi > 0, for i ∈ {1, ..., n}, βi is the real numbers with the
property −1 ≤ βi < 1 and fi ∈ Sp (βi ) for i ∈ {1, ..., n}.
A convexity property for an integral operator on the class Sp (β)
179
If
(3)
0<
n
X
αi (1 − βi ) ≤ 1,
i=1
the integral operator Fn is convex by the order 1 +
n
P
αi (βi − 1).
i=1
Proof. We calculate for Fn the derivatives of the first and second order.
From (2) we obtain:
µ
Fn0
(z) =
f1 (z)
z
¶α1
µ
· ... ·
fn (z)
z
¶ αn
and
Fn00
(z) =
n
X
µ
αi
i=1
fi (z)
z
¶αi −1 µ
zfi0 (z) − fi (z)
zfi (z)
¶Y
n µ
j=1
fj (z)
z
¶αj
.
j6=i
After the calculus we obtain that:
Fn00 (z)
= α1
Fn0 (z)
µ
zf10 (z) − f1 (z)
zf1 (z)
¶
µ
+ ... + αn
zfn0 (z) − fn (z)
zfn (z)
¶
These relation is equivalent with:
(4)
Fn00 (z)
= α1
Fn0 (z)
µ
f10 (z) 1
−
f1 (z) z
¶
µ
+ ... + αn
fn0 (z) 1
−
fn (z) z
¶
.
Multiply the relation (4) with z we obtain:
n
(5)
zFn00 (z) X
αi
=
Fn0 (z)
i=1
µ
¶ X
n
n
zfi0 (z)
zfi0 (z) X
αi
αi .
−1 =
−
fi (z)
f
(z)
i
i=1
i=1
.
180
Daniel Breaz
The relation (5) is equivalent with
n
n
X zf 0 (z) X
zFn00 (z)
+1=
αi i
−
αi + 1.
0
Fn (z)
fi (z)
i=1
i=1
This relation is equivalent with:
n
X
zFn00 (z)
+
1
=
αi
Fn0 (z)
i=1
µ
zfi0 (z)
− βi
fi (z)
¶
+
n
X
αi βi −
i=1
n
X
αi + 1.
i=1
We calculate the real part from both terms of the above equality and
obtain:
µ
Re
zFn00 (z)
+1
Fn0 (z)
¶
=
n
X
µ
αi Re
i=1
zfi0 (z)
− βi
fi (z)
¶
+
n
X
αi βi −
i=1
n
X
αi + 1.
i=1
Because fi ∈ Sp (βi ) for i = {1, ..., n}, we apply in the above relation the
inequality (1) and obtain:
µ
Re
zFn00 (z)
+1
Fn0 (z)
¶
>
n
X
i=1
¯
¯ 0
n
¯ X
¯ zfi (z)
¯
¯
− 1¯ +
αi ¯
αi (βi − 1) + 1.
fi (z)
i=1
¯ 0
¯
¯ zfi (z)
¯
Because αi ¯¯
− 1¯¯ > 0 for all i ∈ {1, ..., n}, obtain that
fi (z)
µ
Re
zFn00 (z)
+1
Fn0 (z)
So, Fn is convex by the order
¶
n
P
>
n
X
αi (βi − 1) + 1.
i=1
αi (βi − 1) + 1.
i=1
Theorem 2. Let αi , i ∈ {1, ..., n} the real positive numbers and fi ∈ Sp (β)
for i ∈ {1, ..., n}.
A convexity property for an integral operator on the class Sp (β)
If
0<
n
X
αi ≤
i=1
181
1
,
1−β
the integral operator Fn is convex by the order (β − 1)
n
P
αi + 1.
i=1
Proof. Since
µ
Fn0
¶α1
(z) =
f1 (z)
z
µ
¶αi −1 µ
µ
· ... ·
fn (z)
z
¶ αn
and
Fn00
(z) =
n
X
αi
i=1
fi (z)
z
zfi0 (z) − fi (z)
zfi (z)
¶Y
n µ
j=1
fj (z)
z
¶ αj
j6=i
we obtain
n
n
X zf 0 (z) X
zFn00 (z)
−
αi + 1.
+
1
=
αi i
Fn0 (z)
f
(z)
i
i=1
i=1
Thus we see, for fi (z) ∈ Sp (β), for all i ∈ {1, ..., n}
¯
¯ 0
¶ X
n
n
X
¯
¯ zfi (z)
zFn00 (z)
¯
¯
Re
−
1
+
1
>
α
−
(β
−
1)
αi + 1.
i¯
¯
Fn0 (z)
f
(z)
i
i=1
i=1
¯ 0
¯
¯ zf (z)
¯
Because αi ¯ fii(z) − 1¯ > 0 for all i ∈ {1, ..., n}, obtain that
µ
µ
Re
zFn00 (z)
+1
Fn0 (z)
¶
> (β − 1)
n
X
αi + 1.
i=1
n
P
1
obtain that 0 ≤ (β − 1) αi
1−β
i=1
i=1
n
P
+1 < 1. So Fn is convex by the order (β − 1) αi + 1.
Because −1 ≤ β < 1 and 0 <
n
P
αi ≤
i=1
182
Daniel Breaz
Remark 1.If β = 0 and
n
P
αi = 1 then
i=1
µ
Re
zFn00 (z)
+1
Fn0 (z)
¶
>0
so, Fn is the convex function.
Corollary 1. Let γ the real number, γ > 0. We suppose that the functions
1
f ∈ Sp (β) and 0 < γ ≤
. In this conditions the integral operator
1−β
¶γ
Zz µ
f (t)
dt is convex by the order (β − 1) γ + 1.
F1 (z) =
t
0
Proof. In the Theorem 2, we consider n = 1.
Corollary 2. Let f ∈ Sp (β) and consider the integral operator of AlexanZz
f (t)
der, F (z) =
dt. In this condition F is convex by the order β.
t
0
Proof. We have:
zF 00 (z)
zf 0 (z)
=
− 1.
F 0 (z)
f (z)
(6)
From (6) we have:
µ
(7) Re
zF 00 (z)
+1
F 0 (z)
¶
µ
= Re
¯ 0
¯
¶
¯ zf (z)
¯
zf 0 (z)
¯
− β +β > k ¯
− 1¯¯ +β > β.
f (z)
f (z)
So, the relation (7) imply that the Alexander operator is convex.
A convexity property for an integral operator on the class Sp (β)
183
References
[1] D. Breaz, N. Breaz, Two integral operators, Studia Universitatis Babeş
-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: dbreaz@uab.ro