Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 5, Article 177, 2006
SOME CONVEXITY PROPERTIES FOR A GENERAL INTEGRAL OPERATOR
DANIEL BREAZ AND NICOLETA BREAZ
D EPARTMENT OF M ATHEMATICS
" 1 D ECEMBRIE 1918 " U NIVERSITY
A LBA I ULIA
ROMANIA .
dbreaz@uab.ro
nbreaz@uab.ro
Received 12 September, 2006; accepted 18 October, 2006
Communicated by G. Kohr
A BSTRACT. In this paper we consider the classes of starlike functions, starlike functions of
order α, convex functions, convex functions of order α and the classes of the univalent functions
denoted by SH (β), SP and SP (α, β). On these classes we study the convexity and α- order
convexity for a general integral operator.
Key words and phrases: Univalent function, Integral operator, Convex function, Analytic function, Starlike function.
2000 Mathematics Subject Classification. 30C45.
1. I NTRODUCTION
Let U = {z ∈ C, |z| < 1} be the unit disc of the complex plane and denote by H (U ), the
class of the holomorphic functions in U. Consider
A = f ∈ H (U ) , f (z) = z + a2 z 2 + a3 z 3 + · · · , z ∈ U
the class of analytic functions in U and S = {f ∈ A : f is univalent in U }. We denote by S ∗
the class of starlike functions that are defined as holomorphic functions in the unit disc with the
properties f (0) = f 0 (0) − 1 = 0 and
Re
zf 0 (z)
> 0,
f (z)
z ∈ U.
A function f ∈ A is a starlike function by the order α, 0 ≤ α < 1 if f satisfies the inequality
Re
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
250-06
zf 0 (z)
> α,
f (z)
z ∈ U.
2
DANIEL B REAZ AND N ICOLETA B REAZ
We denote this class by S ∗ (α). Also, we denote by K the class of convex functions that are
defined as holomorphic functions in the unit disc with the properties f (0) = f 0 (0) − 1 = 0 and
00
zf (z)
Re
+ 1 > 0,
z ∈ U.
f 0 (z)
A function f ∈ A is a convex function by the order α, 0 ≤ α < 1 if f verifies the inequality
00
zf (z)
+ 1 > α,
z ∈ U.
Re
f 0 (z)
We denote this class by K (α).
In the paper [5] J. Stankiewicz and A. Wisniowska introduced the class of univalent functions,
SH (β), β > 0 defined by:
0
√
√
√ zf 0 (z)
zf (z)
(1.1)
2 − 1 < Re
2
+ 2β
2−1 ,
f ∈ S,
f (z) − 2β
f (z)
for all z ∈ U .
Also, in the paper [3] F. Ronning introduced the class of univalent functions, SP , defined by
zf 0 (z) zf 0 (z)
(1.2)
Re
>
− 1 ,
f ∈ S,
f (z)
f (z)
for all z ∈ U . The geometric interpretation of the relation (1.2) is that the class SP is the class
of all functions f ∈ S for which the expression zf 0 (z) /f (z) , z ∈ U takes all values in the
parabolic region
Ω = {ω : |ω − 1| ≤ Re ω} = ω = u + iv : v 2 ≤ 2u − 1 .
In the paper [3] F. Ronning introduced the class of univalent functions SP (α, β) , α > 0,
β ∈ [0, 1), as the class of all functions f ∈ S which have the property:
0
0
zf (z)
≤ Re zf (z) + α − β,
(1.3)
−
(α
+
β)
f (z)
f (z)
for all z ∈ U . Geometric interpretation: f ∈ SP (α, β) if and only if zf 0 (z) /f (z) , z ∈ U
takes all values in the parabolic region
Ωα,β = {ω : |ω − (α + β)| ≤ Re ω + α − β}
= ω = u + iv : v 2 ≤ 4α (u − β) .
We consider the integral operator Fn , defined by:
α
α
Z z
f1 (t) 1
fn (t) n
(1.4)
Fn (z) =
· ··· ·
dt
t
t
0
and we study its properties.
Remark 1.1.RWe observe that for n = 1 and α1 = 1 we obtain the integral operator of Alexanz
der, F (z) = 0 f (t)
dt.
t
J. Inequal. Pure and Appl. Math., 7(5) Art. 177, 2006
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C ONVEXITY P ROPERTIES FOR A G ENERAL I NTEGRAL O PERATOR
3
2. M AIN R ESULTS
Theorem 2.1. Let αi , i ∈ {1, . . . , n} be real numbers with the properties αi > 0 for i ∈
{1, . . . , n} and
n
X
αi ≤ n + 1.
i=1
We suppose that the functions fi , i = {1, . . . , n} are the starlike functions by order α1i , i ∈
{1, . . . , n}, that is fi ∈ S ∗ α1i for all i ∈ {1, . . . , n} . In these conditions the integral operator
defined in (1.4) is convex.
Proof. We calculate for Fn the derivatives of the first and second order. From (1.4) we obtain:
α
α
f1 (z) 1
fn (z) n
0
Fn (z) =
· ··· ·
z
z
and
Fn00
(z) =
n
X
αi
i=1
fi (z)
z
αi −1 zfi0 (z) − fi (z)
zfi (z)
Y
n j=1
fj (z)
z
α j
.
j6=i
Fn00 (z)
Fn0 (z)
(2.1)
zf10
zfn0
(z) − f1 (z)
(z) − fn (z)
= α1
+ · · · + αn
zf1 (z)
zfn (z)
0
0
Fn00 (z)
f1 (z) 1
fn (z) 1
= α1
−
+ · · · + αn
−
.
Fn0 (z)
f1 (z) z
fn (z) z
,
By multiplying the relation (2.1) with z we obtain:
0
X
n
n
zFn00 (z) X
zfi (z)
zfi0 (z)
=
α
−
1
=
α
− α1 − · · · − αn .
(2.2)
i
i
Fn0 (z)
f
(z)
f
(z)
i
i
i=1
i=1
The relation (2.2) is equivalent with
(2.3)
zf10 (z)
zfn0 (z)
zFn00 (z)
+ 1 = α1
+ · · · + αn
− α1 − · · · − αn + 1.
Fn0 (z)
f1 (z)
fn (z)
From (2.3) we obtain that:
00
zFn (z)
zf10 (z)
zfn0 (z)
(2.4)
Re
+
1
=
α
Re
+
·
·
·
+
α
Re
− α1 − · · · − αn + 1.
1
n
Fn0 (z)
f1 (z)
fn (z)
zf 0 (z)
∗
But fi ∈ S α1i , for all i ∈ {1, . . . , n} , so Re fii(z) > α1i , for all i ∈ {1, . . . , n} . We apply
this affirmation in the equality (2.4) and obtain:
00
zFn (z)
1
1
(2.5)
Re
+ 1 > α1 + · · · + αn
− α1 − · · · − αn + 1
0
Fn (z)
α1
αn
n
X
=n+1−
αi .
i=1
But, in accordance with the hypothesis, we obtain:
00
zFn (z)
Re
+1 >0
Fn0 (z)
so, Fn is a convex function.
J. Inequal. Pure and Appl. Math., 7(5) Art. 177, 2006
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4
DANIEL B REAZ AND N ICOLETA B REAZ
Theorem 2.2. Let αi , i ∈ {1, . . . , n}, be real numbers with the properties αi > 0 for i ∈
{1, . . . , n} and
n
X
αi ≤ 1.
i=1
We suppose that the functions fi , i = {1, . . . , n},
Pare the starlike functions. Then the integral
operator defined in (1.4) is convex by order, 1 − ni=1 αi .
Proof. Following the same steps as in Theorem 2.1, we obtain:
0
X
n
n
zFn00 (z) X
zfi (z)
zfi0 (z)
(2.6)
=
α
−
1
=
α
− α1 − · · · − αn .
i
i
Fn0 (z)
f
(z)
f
(z)
i
i
i=1
i=1
The relation (2.6) is equivalent with
(2.7)
zFn00 (z)
zf10 (z)
zfn0 (z)
+
1
=
α
+
·
·
·
+
α
− α1 − · · · − αn + 1.
1
n
Fn0 (z)
f1 (z)
fn (z)
From (2.7) we obtain that:
00
zFn (z)
zf10 (z)
zfn0 (z)
(2.8)
Re
+
1
=
α
Re
+
·
·
·
+
α
Re
− α1 − · · · − αn + 1.
1
n
Fn0 (z)
f1 (z)
fn (z)
zf 0 (z)
But fi ∈ S ∗ for all i ∈ {1, . . . , n} , so Re fii(z) > 0 for all i ∈ {1, . . . , n} . We apply this
affirmation in the equality (2.8) and obtain that:
00
n
X
zFn (z)
(2.9)
Re
+ 1 > α1 · 0 + · · · + αn · 0 − α1 − · · · − αn + 1 = 1 −
αi .
Fn0 (z)
i=1
But in accordance with the inequality (2.9), obtain that
00
n
X
zFn (z)
Re
+1 >1−
αi
Fn0 (z)
i=1
P
so, Fn is a convex function by order 1 − ni=1 αi .
Theorem 2.3. Let αi , i ∈ {1, . . . , n}, be real numbers with the properties αi > 0, for i ∈
{1, . . . , n} and
√
n
X
2
√
√ .
(2.10)
αi ≤
2β
2−1 + 2
i=1
We suppose that fi ∈ SH (β), for i = {1, . . . , n} and β > 0. In these conditions, the integral
operator defined in (1.4) is convex.
Proof. Following the same steps as in Theorem 2.1, we obtain that:
n
n
X
zFn00 (z)
zfi0 (z) X
(2.11)
+1=
αi
−
αi + 1.
Fn0 (z)
fi (z)
i=1
i=1
√
We multiply the relation (2.11) with 2 and obtain:
X
n
n
√ zFn00 (z)
√ zfi0 (z) √ X
√
(2.12)
2
+
1
=
2α
−
2
α
+
2.
i
i
Fn0 (z)
fi (z)
i=1
i=1
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C ONVEXITY P ROPERTIES FOR A G ENERAL I NTEGRAL O PERATOR
5
The equality (2.12) is equivalent with:
√
2
X
n √
√ zfi0 (z)
zFn00 (z)
2−1
+1 =
αi 2
+ 2αi β
Fn0 (z)
fi (z)
i=1
n
n
√
√ X
X
√
−
2αi β
2−1 − 2
αi + 2.
i=1
i=1
We calculate the real part from both terms of the above equality and obtain:
√
2 Re
X
n √
√ zfi0 (z)
zFn00 (z)
+1 =
αi Re
2
+ 2β
2−1
Fn0 (z)
f
(z)
i
i=1
n
n
√
√ X
X
√
−
2αi β
2−1 − 2
αi + 2.
i=1
i=1
Because fi ∈ SH (β) for i = {1, . . . , n}, we apply in the above relation the inequality (1.1)
and obtain:
0
00
X
n
√
√
zfi (z)
zFn (z)
2 Re
+1 >
αi − 2β
2 − 1 Fn0 (z)
f
(z)
i
i=1
n
n
√ X
√
X
√
−
2αi β
2−1 − 2
αi + 2.
i=1
i=1
0
√
zf (z)
Because αi fii(z) − 2β
2 − 1 > 0, for all i ∈ {1, . . . , n}, we obtain that
(2.13)
√
2 Re
n
n
√
√ X
X
√
zFn00 (z)
+
1
>
−
2α
β
2
−
1
−
2
α
+
2.
i
i
Fn0 (z)
i=1
i=1
Using the hypothesis (2.10), we have:
(2.14)
Re
zFn00 (z)
+ 1 > 0,
Fn0 (z)
so, Fn is a convex function.
√
Corollary 2.4. Let α be real numbers with the properties 0 < α ≤
√ 2
√ ,
2β ( 2−1)+ 2
β > 0. We
suppose that the functions f ∈ SH (β). In these conditions the integral operator, F (z) =
R z f (t) α
dt is convex.
t
0
Proof. In Theorem 2.3, we consider n = 1, α1 = α and f1 = f .
Theorem 2.5. Let αi , i ∈ {1, . . . , n} be real numbers with the properties αi > 0 for i ∈
{1, . . . , n},
(2.15)
n
X
αi < 1
i=1
P
and 1 − ni=1 αi ∈ [0, 1). We consider the functions fi , fi ∈ SP P
for i = {1, . . . , n}. In these
conditions, the integral operator defined in (1.4) is convex by 1 − ni=1 αi order.
J. Inequal. Pure and Appl. Math., 7(5) Art. 177, 2006
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6
DANIEL B REAZ AND N ICOLETA B REAZ
Proof. Following the same steps as in Theorem 2.1, we have:
n
n
X zf 0 (z) X
zFn00 (z)
+1=
αi i
−
αi + 1.
0
Fn (z)
fi (z)
i=1
i=1
(2.16)
We calculate the real part from both terms of the above equality and obtain:
X
0
X
00
n
n
zFn (z)
zfi (z)
(2.17)
Re
+1 =
αi Re
−
αi + 1.
Fn0 (z)
fi (z)
i=1
i=1
Because fi ∈ SP for i = {1, . . . , n} we apply in the above relation the inequality (1.2) and
obtain:
0
X
X
00
n
n
zfi (z)
zFn (z)
(2.18)
Re
+1 >
αi − 1 −
αi + 1.
Fn0 (z)
fi (z)
i=1
i=1
0
zfi (z)
Because αi fi (z) − 1 > 0, for all i ∈ {1, . . . , n}, we get
Re
(2.19)
n
X
zFn00 (z)
+1 >1−
αi .
Fn0 (z)
i=1
Using the hypothesis, we obtain that Fn is a convex function by 1 −
P
Remark 2.6. If ni=1 αi = 1 then
00
zFn (z)
+ 1 > 0,
(2.20)
Re
Fn0 (z)
Pn
i=1
αi order.
so, Fn is a convex function.
Corollary 2.7. Let γ be a real number with the property
0 < γ < 1. We suppose that f ∈ SP .
R z f (t) γ
In these conditions the integral operator F (z) = 0
dt is convex of 1 − γ order.
t
Proof. In Theorem 2.5, we consider n = 1, α1 = γ and f1 = f .
Theorem 2.8. We suppose that f ∈ SP . In this condition, the integral operator of Alexander,
defined by
Z z
f (t)
(2.21)
F1 (z) =
dt,
t
0
is convex.
Proof. We have:
(2.22)
Re
zF100 (z)
zf 0 (z) zf 0 (z)
+ 1 = Re
>
− 1 > 0.
0
F1 (z)
f (z)
f (z)
So, the relation (2.22) implies that the Alexander operator is convex.
Remark 2.9. Theorem 2.8 can be obtained from Corollary 2.7, for γ = 1.
Theorem 2.10. Let αi , i ∈ {1, . . . , n} be real numbers with the properties αi > 0 for i ∈
{1, . . . , n},
(2.23)
n
X
i=1
αi <
1
,
α−β+1
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α > 0, β ∈ [0, 1)
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C ONVEXITY P ROPERTIES FOR A G ENERAL I NTEGRAL O PERATOR
7
P
and (β − α − 1) ni=1 αi + 1 ∈ (0, 1). We suppose that fi ∈ SP (α, β), for i =P
{1, . . . , n}.
In these conditions, the integral operator defined in (1.4) is convex by (β − α − 1) ni=1 αi + 1
order.
Proof. Following the same steps as in Theorem 2.1, we obtain that:
0
n
n
X
zFn00 (z) X
zfi (z)
(2.24)
=
α
+
α
−
β
+
(β
−
α
−
1)
αi .
i
Fn0 (z)
f
(z)
i
i=1
i=1
and
n
(2.25)
X
zFn00 (z)
+
1
=
αi
Fn0 (z)
i=1
zfi0 (z)
+α−β
fi (z)
+ (β − α − 1)
n
X
αi + 1.
i=1
We calculate the real part from both terms of the above equality and get:
( n
)
00
X zf 0 (z)
zFn (z)
i
(2.26)
Re
+ 1 = Re
αi
+α−β
Fn0 (z)
fi (z)
i=1
+ (β − α − 1)
n
X
αi + 1.
i=1
Because fi ∈ SP (α, β) for i = {1, . . . , n} we apply in the above relation the inequality (1.3)
and obtain:
0
00
X
n
n
X
zfi (z)
zFn (z)
+1 ≥
− (α + β) + (β − α − 1)
αi + 1.
(2.27)
Re
αi Fn0 (z)
fi (z)
i=1
i=1
0
zfi (z)
Since αi fi (z) − (α + β) > 0, for all i ∈ {1, . . . , n}, using the inequality (1.3), we have
00
n
X
zFn (z)
(2.28)
Re
+ 1 ≥ (β − α − 1)
αi + 1 > 0.
Fn0 (z)
i=1
Pn
From (2.28), since (β − α − 1) P
i=1 αi +1 ∈ (0, 1), we obtain that the integral operator defined
in (1.4) is convex by (β − α − 1) ni=1 αi + 1 order.
1
,
α−β+1
α > 0, β ∈ [0, 1). We
R z f (t) γ
suppose that f ∈ SP (α, β). In these conditions, the integral operator F (z) = 0
dt is
t
convex.
Corollary 2.11. Let γ be a real number with the property 0 < γ <
Proof. In Theorem 2.10, we consider n = 1, α1 = γ and f1 = f .
For α = β ∈ (0, 1) we obtain the class S (α, α) that is characterized by the property
0
zf (z)
zf 0 (z)
(2.29)
f (z) − 2α ≤ Re f (z) .
Corollary 2.12. Let αi , i ∈ {1, . . . , n} be real numbers with the properties αi > 0 for i ∈
{1, . . . , n} and
n
X
(2.30)
1−
αi ∈ [0, 1) .
i=1
We consider the functions fi , fi ∈ SP (α, α), i = {1,P
. . . , n}, α ∈ (0, 1). In these conditions,
the integral operator defined in (1.4) is convex by 1 − ni=1 αi order.
J. Inequal. Pure and Appl. Math., 7(5) Art. 177, 2006
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8
DANIEL B REAZ AND N ICOLETA B REAZ
Proof. From (1.4) we obtain
n
(2.31)
n
zFn00 (z) X zfi0 (z) X
=
αi
−
αi ,
Fn0 (z)
f
(z)
i
i=1
i=1
which is equivalent with
00
X
n
n
zFn (z)
zfi0 (z) X
(2.32)
Re
+
1
=
α
Re
−
αi + 1.
i
Fn0 (z)
f
(z)
i
i=1
i=1
From (2.31) and (2.32), we have:
0
X
00
n
n
X
zfi (z)
zFn (z)
(2.33)
Re
+1 >
αi − 2α + 1 −
αi .
Fn0 (z)
fi (z)
i=1
i=1
0
P
zf (z)
Since ni=1 αi fii(z) − 2α > 0, for all i ∈ {1, . . . , n}, from (2.33), we get:
00
n
X
zFn (z)
+
1
>
1
−
(2.34)
Re
αi .
Fn0 (z)
i=1
P
Now, from (2.34) we obtain that the operator defined in (1.4) is convex by 1− ni=1 αi order.
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[3] F. RONNING, Integral reprezentations of bounded starlike functions, Ann. Polon. Math., LX(3)
(1995), 289–297.
[4] F. RONNING, Uniformly convex functions and a corresponding class of starlike functions, Proc.
Amer. Math. Soc., 118(1) (1993), 190–196.
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J. Inequal. Pure and Appl. Math., 7(5) Art. 177, 2006
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