Directory UMM :Journals:Journal_of_mathematics:OTHER:
ACTA UNIVERSITATIS APULENSIS
No 16/2008
A NEW INTEGRAL UNIVALENT OPERATOR
Daniel Breaz1 , Shigeyoshi Owa and Nicoleta Breaz2
Abstract. In this paper the authors introduced an integral operator
and proved its properties.
2000 Mathematics Subject Classification: 30C45
Keywords 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 }.
Consider S ∗ , the class of starlike functions in unit disk, defined by
∗
S =
zf ′ (z)
> 0, z ∈ U
f ∈ H (U ) : f (0) = f (0) − 1 = 0, Re
f (z)
′
.
A function f ∈ S is the starlike function of order α, 0 ≤ α < 1 and denote
this class by S ∗ (α) if f verify the inequality
zf ′ (z)
> α, z ∈ U.
Re
f (z)
Denote with K the class of convex functions in U , defined by
′′
zf (z)
′
+ 1 > 0, z ∈ U .
K = f ∈ H (U ) : f (0) = f (0) − 1 = 0, Re
f ′ (z)
1
2
Supported by GAR 19/2008
Supported by GAR 19/2008
11
D. Breaz, S. Owa, N. Breaz - A new integral univalent operator
A function f ∈ S is convex function of order α, 0 ≤ α < 1 and denote
this class by K (α) if f verify the inequality
′′
zf (z)
+ 1 > α, z ∈ U.
Re
f ′ (z)
A function f ∈ U CV if and only if
′′
zf (z)
zf ′′ (z)
, z ∈ U.
≥ ′
Re 1 + ′
f (z)
f (z)
S. N. Kudryashov in 1973 investigated the maximum value of M such
that the inequality
′′
f (z)
f ′ (z) ≤ M
implies that f is univalent in U . He showed that if M = 3, 05... and
′′
f (z)
f ′ (z) ≤ 3, 05...
1/2
where M is the solution of the equation 8 M (M − 2)3
−3 (3 − M )2 = 12,
then f is univalent in U .
Also, if
′′
f (z)
f ′ (z) ≤ 2, 8329...,
then the function f is starlike in unit disk. This result is obtained by Miller
and Mocanu.
Consider the new general integral operator defined by the formula:
Z z
α
α
(1)
(f1′ (t)) 1 ... (fn′ (t)) n dt.
Fα1 ,...,αn (z) =
0
2.Main results
In the next theorem we proved the univalence for this operator.
Theorem 1. Let αi ∈ R, i ∈ {1, ..., n} and αi > 0. Consider fi ∈ S
(univalent functions) and suppose that
12
D. Breaz, S. Owa, N. Breaz - A new integral univalent operator
′′
fi (z)
f ′ (z) ≤ M,
i
(2)
where M = 3, 05..., for all z ∈ U .
n
P
If
αi ≤ 1 the integral operator Fα1 ,...,αn is univalent.
i=1
Proof. We have
Fα′′1 ,...,αn (z)
fn′′ (z)
f1′′ (z)
+ ... + αn ′
.
= α1 ′
Fα′ 1 ,...,αn (z)
f1 (z)
fn (z)
(3)
that is equivalent with,
′′
′′
′′
Fα1 ,...,αn (z)
≤ α1 f1 (z) + ... + αn fn (z) .
′
F ′
′
f1 (z)
fn (z)
α1 ,...,αn (z)
Now, applying the inequality (2), we obtain that
′′
n
X
Fα1 ,...,αn (z)
≤ α1 M + ... + αn M = M
αi ≤ M.
F ′
α1 ,...,αn (z)
i=1
The last inequality implies that the integral operator Fα1 ,...,αn is univalent.
Theorem 2. Let αi ∈ R, i ∈ {1, ..., n} and αi > 0. Consider fi ∈ S
(univalent functions) and suppose that
′′
fi (z)
(4)
f ′ (z) ≤ M1 ,
i
where M1 = 2, 8329..., is the smallest root of equation xsinx + cosx = 1/e.
n
P
If
αi ≤ 1, the integral operator Fα1 ,...,αn is starlike.
i=1
Proof. We have
Fα′′1 ,...,αn (z)
f1′′ (z)
fn′′ (z)
.
=
α
+
...
+
α
1
n
Fα′ 1 ,...,αn (z)
f1′ (z)
fn′ (z)
that is equivalent with,
13
(5)
D. Breaz, S. Owa, N. Breaz - A new integral univalent operator
′′
′′
′′
Fα1 ,...,αn (z)
fn (z)
f1 (z)
≤ α1
F ′
f ′ (z) + ... + αn f ′ (z) .
1
α1 ,...,αn (z)
n
Now, applying the inequality (4), we obtain that
′′
n
X
Fα1 ,...,αn (z)
≤ α1 M1 + ... + αn M1 = M1
αi ≤ M1 .
F ′
α1 ,...,αn (z)
i=1
The last inequality implies that the integral operator Fα1 ,...,αn is starlike.
Theorem 3. Let αi ∈ R, i ∈ {1, ..., n} and αi > 0. Consider fi ∈ K
(convex functions), for all i ∈ {1, ..., n}. Then the integral operator Fα1 ,...,αn
is convex.
Proof.
Re
X
′′
n
zFα′′1 ,...,αn (z)
zfi (z)
+ 1 ≥ 0.
+1 =
αi Re
Fα′ 1 ,...,αn (z)
fi′ (z)
i=1
Since
Re
zfi′′ (z)
+1 ≥0
fi′ (z)
for all i ∈ {1, ..., n}, we have the integral operator Fα1 ,...,αn is convex.
Theorem 4. Let αi ∈ R, i ∈ {1, ..., n} and αi > 0. Consider fi ∈ K (βi ),
0 ≤ βi < 1, for all i ∈ {1, ..., n}. In these conditions, the integral operator
n
n
P
P
Fα1 ,...,αn is convex of order
αi (βi − 1)+1, where 0 ≤
αi (βi − 1)+1 < 1.
i=1
i=1
Proof. We have
Re
X
′′
n
zFα′′1 ,...,αn (z)
zfi (z)
+1 =
+1 .
αi Re
′
Fα′ 1 ,...,αn (z)
f
(z)
i
i=1
Since fi ∈ K (βi ), we have
′′
zfi (z)
Re
+ 1 ≥ βi ,
fi′ (z)
14
(6)
D. Breaz, S. Owa, N. Breaz - A new integral univalent operator
for all i ∈ {1, ..., n}.
We apply this property in (6) and we obtain that:
Re
X
n
n
n
X
X
zFα′′1 ,...,αn (z)
+1 =
αi βi + 1 −
αi =
αi (βi − 1) + 1. (7)
Fα′ 1 ,...,αn (z)
i=1
i=1
i=1
The relation (7) implies that the integral operator Fα1 ,...,αn is convex of
n
P
order
αi (βi − 1) + 1.
i=1
Theorem 5. Let αi ∈ R, i ∈ {1, ..., n} and αi > 0. Consider fi ∈ U CV ,
for all i ∈ {1, ..., n}. In these conditions, the integral operator Fα1 ,...,αn is
n
n
P
P
convex of order 1 −
αi , where 1 −
αi ≥ 0.
i=1
i=1
Proof. We have
Re
X
′′
n
zFα′′1 ,...,αn (z)
zfi (z)
+1 =
+1 .
αi Re
′
Fα′ 1 ,...,αn (z)
f
(z)
i
i=1
(8)
Because fi ∈ U CV , we have
′′
zfi (z)
zfi′′ (z)
, z ∈ U.
≥ ′
Re 1 + ′
fi (z)
fi (z)
We apply this inequality in relation (8) and obtain that:
Re
′′
X
n
n
n
X
X
zfi (z)
zFα′′1 ,...,αn (z)
+1 ≥
αi ′
+1−
αi ≥ 1 −
αi . (9)
Fα′ 1 ,...,αn (z)
fi (z)
i=1
i=1
i=1
The relation (9) implies that the integral operator Fα1 ,...,αn is convex of
n
P
order 1 −
αi .
i=1
References
[1] S.S. Miller and P.T. Mocanu, Differential Subordinations. Theory and
Applications, Marcel Dekker, INC., New York, Basel, 2000.
15
D. Breaz, S. Owa, N. Breaz - A new integral univalent operator
[2] F. Ronning, A survey on uniformly convex and uniformly starlike functions, Annales Universitatis Mariae Curie-Sklodowska, vol. XLVII, 13, 123134, Lublin-Polonia.
Authors:
Daniel Breaz
Department of Mathematics
”1 Decembrie 1918” University
Alba Iulia
Romania
e-mail:[email protected]
Shigeyoshi Owa
Department of Mathematics
Kinki University
Higashi-Osaka, Osaka 577-8502
Japan
e-mail:[email protected]
Nicoleta Breaz
Department of Mathematics
”1 Decembrie 1918” University
Alba Iulia
Romania
e-mail:[email protected]
16
No 16/2008
A NEW INTEGRAL UNIVALENT OPERATOR
Daniel Breaz1 , Shigeyoshi Owa and Nicoleta Breaz2
Abstract. In this paper the authors introduced an integral operator
and proved its properties.
2000 Mathematics Subject Classification: 30C45
Keywords 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 }.
Consider S ∗ , the class of starlike functions in unit disk, defined by
∗
S =
zf ′ (z)
> 0, z ∈ U
f ∈ H (U ) : f (0) = f (0) − 1 = 0, Re
f (z)
′
.
A function f ∈ S is the starlike function of order α, 0 ≤ α < 1 and denote
this class by S ∗ (α) if f verify the inequality
zf ′ (z)
> α, z ∈ U.
Re
f (z)
Denote with K the class of convex functions in U , defined by
′′
zf (z)
′
+ 1 > 0, z ∈ U .
K = f ∈ H (U ) : f (0) = f (0) − 1 = 0, Re
f ′ (z)
1
2
Supported by GAR 19/2008
Supported by GAR 19/2008
11
D. Breaz, S. Owa, N. Breaz - A new integral univalent operator
A function f ∈ S is convex function of order α, 0 ≤ α < 1 and denote
this class by K (α) if f verify the inequality
′′
zf (z)
+ 1 > α, z ∈ U.
Re
f ′ (z)
A function f ∈ U CV if and only if
′′
zf (z)
zf ′′ (z)
, z ∈ U.
≥ ′
Re 1 + ′
f (z)
f (z)
S. N. Kudryashov in 1973 investigated the maximum value of M such
that the inequality
′′
f (z)
f ′ (z) ≤ M
implies that f is univalent in U . He showed that if M = 3, 05... and
′′
f (z)
f ′ (z) ≤ 3, 05...
1/2
where M is the solution of the equation 8 M (M − 2)3
−3 (3 − M )2 = 12,
then f is univalent in U .
Also, if
′′
f (z)
f ′ (z) ≤ 2, 8329...,
then the function f is starlike in unit disk. This result is obtained by Miller
and Mocanu.
Consider the new general integral operator defined by the formula:
Z z
α
α
(1)
(f1′ (t)) 1 ... (fn′ (t)) n dt.
Fα1 ,...,αn (z) =
0
2.Main results
In the next theorem we proved the univalence for this operator.
Theorem 1. Let αi ∈ R, i ∈ {1, ..., n} and αi > 0. Consider fi ∈ S
(univalent functions) and suppose that
12
D. Breaz, S. Owa, N. Breaz - A new integral univalent operator
′′
fi (z)
f ′ (z) ≤ M,
i
(2)
where M = 3, 05..., for all z ∈ U .
n
P
If
αi ≤ 1 the integral operator Fα1 ,...,αn is univalent.
i=1
Proof. We have
Fα′′1 ,...,αn (z)
fn′′ (z)
f1′′ (z)
+ ... + αn ′
.
= α1 ′
Fα′ 1 ,...,αn (z)
f1 (z)
fn (z)
(3)
that is equivalent with,
′′
′′
′′
Fα1 ,...,αn (z)
≤ α1 f1 (z) + ... + αn fn (z) .
′
F ′
′
f1 (z)
fn (z)
α1 ,...,αn (z)
Now, applying the inequality (2), we obtain that
′′
n
X
Fα1 ,...,αn (z)
≤ α1 M + ... + αn M = M
αi ≤ M.
F ′
α1 ,...,αn (z)
i=1
The last inequality implies that the integral operator Fα1 ,...,αn is univalent.
Theorem 2. Let αi ∈ R, i ∈ {1, ..., n} and αi > 0. Consider fi ∈ S
(univalent functions) and suppose that
′′
fi (z)
(4)
f ′ (z) ≤ M1 ,
i
where M1 = 2, 8329..., is the smallest root of equation xsinx + cosx = 1/e.
n
P
If
αi ≤ 1, the integral operator Fα1 ,...,αn is starlike.
i=1
Proof. We have
Fα′′1 ,...,αn (z)
f1′′ (z)
fn′′ (z)
.
=
α
+
...
+
α
1
n
Fα′ 1 ,...,αn (z)
f1′ (z)
fn′ (z)
that is equivalent with,
13
(5)
D. Breaz, S. Owa, N. Breaz - A new integral univalent operator
′′
′′
′′
Fα1 ,...,αn (z)
fn (z)
f1 (z)
≤ α1
F ′
f ′ (z) + ... + αn f ′ (z) .
1
α1 ,...,αn (z)
n
Now, applying the inequality (4), we obtain that
′′
n
X
Fα1 ,...,αn (z)
≤ α1 M1 + ... + αn M1 = M1
αi ≤ M1 .
F ′
α1 ,...,αn (z)
i=1
The last inequality implies that the integral operator Fα1 ,...,αn is starlike.
Theorem 3. Let αi ∈ R, i ∈ {1, ..., n} and αi > 0. Consider fi ∈ K
(convex functions), for all i ∈ {1, ..., n}. Then the integral operator Fα1 ,...,αn
is convex.
Proof.
Re
X
′′
n
zFα′′1 ,...,αn (z)
zfi (z)
+ 1 ≥ 0.
+1 =
αi Re
Fα′ 1 ,...,αn (z)
fi′ (z)
i=1
Since
Re
zfi′′ (z)
+1 ≥0
fi′ (z)
for all i ∈ {1, ..., n}, we have the integral operator Fα1 ,...,αn is convex.
Theorem 4. Let αi ∈ R, i ∈ {1, ..., n} and αi > 0. Consider fi ∈ K (βi ),
0 ≤ βi < 1, for all i ∈ {1, ..., n}. In these conditions, the integral operator
n
n
P
P
Fα1 ,...,αn is convex of order
αi (βi − 1)+1, where 0 ≤
αi (βi − 1)+1 < 1.
i=1
i=1
Proof. We have
Re
X
′′
n
zFα′′1 ,...,αn (z)
zfi (z)
+1 =
+1 .
αi Re
′
Fα′ 1 ,...,αn (z)
f
(z)
i
i=1
Since fi ∈ K (βi ), we have
′′
zfi (z)
Re
+ 1 ≥ βi ,
fi′ (z)
14
(6)
D. Breaz, S. Owa, N. Breaz - A new integral univalent operator
for all i ∈ {1, ..., n}.
We apply this property in (6) and we obtain that:
Re
X
n
n
n
X
X
zFα′′1 ,...,αn (z)
+1 =
αi βi + 1 −
αi =
αi (βi − 1) + 1. (7)
Fα′ 1 ,...,αn (z)
i=1
i=1
i=1
The relation (7) implies that the integral operator Fα1 ,...,αn is convex of
n
P
order
αi (βi − 1) + 1.
i=1
Theorem 5. Let αi ∈ R, i ∈ {1, ..., n} and αi > 0. Consider fi ∈ U CV ,
for all i ∈ {1, ..., n}. In these conditions, the integral operator Fα1 ,...,αn is
n
n
P
P
convex of order 1 −
αi , where 1 −
αi ≥ 0.
i=1
i=1
Proof. We have
Re
X
′′
n
zFα′′1 ,...,αn (z)
zfi (z)
+1 =
+1 .
αi Re
′
Fα′ 1 ,...,αn (z)
f
(z)
i
i=1
(8)
Because fi ∈ U CV , we have
′′
zfi (z)
zfi′′ (z)
, z ∈ U.
≥ ′
Re 1 + ′
fi (z)
fi (z)
We apply this inequality in relation (8) and obtain that:
Re
′′
X
n
n
n
X
X
zfi (z)
zFα′′1 ,...,αn (z)
+1 ≥
αi ′
+1−
αi ≥ 1 −
αi . (9)
Fα′ 1 ,...,αn (z)
fi (z)
i=1
i=1
i=1
The relation (9) implies that the integral operator Fα1 ,...,αn is convex of
n
P
order 1 −
αi .
i=1
References
[1] S.S. Miller and P.T. Mocanu, Differential Subordinations. Theory and
Applications, Marcel Dekker, INC., New York, Basel, 2000.
15
D. Breaz, S. Owa, N. Breaz - A new integral univalent operator
[2] F. Ronning, A survey on uniformly convex and uniformly starlike functions, Annales Universitatis Mariae Curie-Sklodowska, vol. XLVII, 13, 123134, Lublin-Polonia.
Authors:
Daniel Breaz
Department of Mathematics
”1 Decembrie 1918” University
Alba Iulia
Romania
e-mail:[email protected]
Shigeyoshi Owa
Department of Mathematics
Kinki University
Higashi-Osaka, Osaka 577-8502
Japan
e-mail:[email protected]
Nicoleta Breaz
Department of Mathematics
”1 Decembrie 1918” University
Alba Iulia
Romania
e-mail:[email protected]
16