AN INTEGRAL UNIVALENT OPERATOR
D. BREAZ and N. BREAZ
˛ 2 0
˛
˛ z f (z)
˛
Abstract. In this paper we consider the class of univalent functions defined by the condition ˛ f 2 (z) − 1˛ < 1, z ∈ U,
where f (z) = z + a2 z 2 + . . . , is an analytic function in the unit disc U = {z ∈ C : |z| < 1} . We present univalence
conditions for the operator
1
1
0
n(α−1)+1
Zz
α−1
α−1
A
@
.
(t) . . . gn (t) dt
Gα,n (z) = (n (α − 1) + 1) g1
0
1.
Introduction
Let U be the unit disc, U = {z ∈ C : |z| < 1}. Denote by H (U ) the class of holomorphic functions on U and
consider the set of analytic functions
(1)
An = f ∈ H (U ) : f (z) = z + an+1 z n+1 + . . .
If n = 1 then A1 = A. Let Hu (U ) be the class of univalent functions on the unit disc U and S the class of
regular and univalent functions f (z) = z + a2 z 2 + . . . in U, which satisfy the condition f (0) = f 0 (0) − 1 = 0.
In their paper [3], Ozaki and Nunokawa proved the following results:
Received October 18, 2004.
2000 Mathematics Subject Classification. Primary 30C45.
Key words and phrases. univalent operator, unit disc, univalent functions.
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Theorem 1. If we assumethat g ∈ A satisfies the condition
2 0
z f (z)
< 1, z ∈ U
(2)
−
1
f 2 (z)
then f is univalent in U .
Lemma 1 (The Schwartz Lemma). Let the analytic function g be regular in the unit disc U and g (0) = 0. If
|g (z)| ≤ 1, ∀z ∈ U, then
(3)
|g (z)| ≤ |z| ,
∀z ∈ U
and equality holds only if g (z) = εz, where |ε| = 1.
Theorem 2. Let α be a complex number with Re α > 0, and let f = z + a2 z 2 + . . . be a regular function on
U . If
2Re α 1 − |z|
zf 00 (z) (4)
∀z ∈ U
f 0 (z) ≤ 1,
Re α
then for any complex number β with Re β ≥ Re α the function
 z
 β1
Z
(5)
Fβ (z) = β tβ−1 f 0 (t) dt = z + . . .
0
is regular and univalent in U .
Theorem 3. Assume that g ∈ A satisfies condition (2), and let α be a complex number with
(6)
|α − 1| ≤
Re α
.
3
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If
(7)
|g (z)| ≤ 1,
∀z ∈ U
then the function

(8)
Zz
Gα (z) = α
 α1
g (α−1) (t) dt
0
is of class S.
2.
Main results
Theorem 4. Let gi ∈ A, ∀i = 1, . . . , n, n ∈ N∗ , satisfy the properties
z 2 g 0 (z)
i
− 1 < 1, ∀z ∈ U, ∀i = 1, . . . , n
(9)
2
gi (z)
and α ∈ C, with
(10)
|α − 1| ≤
Re α
.
3n
If |gi (z)| ≤ 1, ∀z ∈ U , ∀i = 1, . . . , n, then the function
1

 n(α−1)+1
Zz
Gα,n (z) = (n (α − 1) + 1) g1α−1 (t) . . . gnα−1 (t) dt
(11)
0
is univalent.
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Proof. From (11) we have:

(12)
Zz
Gα,n (z) = (n(α − 1) + 1)
tn(α−1)
g1 (t)
t
α−1
· ... ·
gn (t)
t
α−1
1
 n(α−1)+1
dt
0
We consider the function
Zz f (z) =
(13)
g1 (t)
t
α−1
...
gn (t)
t
α−1
dt.
0
The function f is regular in U , and from (13) we obtain
0
(14)
f (z) =
g1 (z)
z
α−1
...
gn (z)
z
α−1
and
00
f (z) = (α − 1)
(15)
+
g1 (z)
z
g1 (z)
z
α−1
α−2
α−1
α−1
zg10 (z) − g1 (z) g2 (z)
gn (z)
...
+ ...
z2
z
z
α−1
α−2 0
gn−1 (z)
gn (z)
zgn (z) − gn (z)
...
(α − 1)
.
z
z
z2
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Next we calculate the expresion
(16)
zf 00
f0 .
α−2 0
α−1
α−1 zg1 (z)−g1 (z) g2 (z)
g1 (z)
gn (z)
z
(α
−
1)
.
.
.
z
z2
z
z
zf 00 (z)
=
+ ...
α−1
α−1
f 0 (z)
g1 (z)
. . . gnz(z)
z
α−1
α−1
α−2 0
zgn (z)−gn (z)
gn−1 (z)
g1 (z)
gn (z)
z
...
(α − 1)
z
z
z
z2
+
α−1
α−1
g1 (z)
. . . gnz(z)
z
= (α − 1)
zg10 (z) − 1
zg 0 (z) − 1
+ · · · + (α − 1) n
.
g1 (z)
gn (z)
The modulus
(17)
00 zf f0 can then be evaluated as
00
0
0
zf (z) = (α − 1) zg1 (z) − 1 + · · · + (α − 1) zgn (z) − 1 f 0 (z) g1 (z)
gn (z) 0
zg1 (z) − 1 zgn0 (z) − 1 (18)
≤ (α − 1)
+ · · · + (α − 1)
g1 (z) gn (z) 0
0
zg1 (z) − 1 zgn (z) − 1 .
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By multiplying the first and the last terms of (18) with
2Re α
1 − |z|
Re α
(19)
1−|z|2Re α
Re α
> 0, we obtain
00
0
2Re α
zg1 (z) zf (z) ≤ 1 − |z|
+ 1 + ...
|α
−
1|
g1 (z) f 0 (z) Re α
0
2Re α
zgn (z) 1 − |z|
+
|α − 1| +1
Re α
gn (z) 2Re α 2 0
1 − |z|
z g1 (z) |g1 (z)| + 1 + . . .
≤ |α − 1|
g 2 (z) |z|
Re α
1
2Re α
z 2 gn0 (z) |gn (z)|
1 − |z|
+ |α − 1|
g 2 (z) |z| + 1 .
Re α
n
By applying the Schwartz Lemma and using (19), we obtain
2Re α
(20)
1 − |z|
Re α
00
2Re α 2 0
zf (z) ≤ |α − 1| 1 − |z|
z g1 (z) − 1 + 2 + . . .
f 0 (z) g 2 (z)
Re α
1
2Re α 2 0
1 − |z|
z gn (z)
+2 .
−
1
+ |α − 1|
g 2 (z)
Reα
n
Since gi satisfies the condition (2) ∀i = 1, . . . , n, then from (20) we obtain:
2Re α
(21)
1 − |z|
Re α
00
2Re α
2Re α
zf (z) 1 − |z|
≤ 3 |α − 1| 1 − |z|
+
·
·
·
+
3
|α
−
1|
f 0 (z) Re α
Re α
3 |α − 1|
3 |α − 1|
3n |α − 1|
≤
+ ··· +
=
.
Re α
Re α
Reα
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But |α − 1| ≤
Re α
3n
so from (7) we obtain that
2Re α
1 − |z|
Re α
(22)
00
zf (z) f 0 (z) ≤ 1,
for all z ∈ U and the Theorem 2 implies that the function Gα,n is in the class S.
Corollary 1. Let g ∈ A satisfy (2) and α a complex number such that
|α − 1| ≤
(23)
Re α
,
3k
If |g (z)| ≤ 1, ∀z ∈ U, then the function

(24)
Gkα (z) = (k (α − 1) + 1)
Zz
k ∈ N∗
1
 k(α−1)+1
g k(α−1) (t) dt
0
is univalent.
Proof. We consider the functions

(25)
Gkα (z) = (k (α − 1) + 1)
Zz
tk(α−1)
g (t)
t
k(α−1)
1
 k(α−1)+1
dt
0
and
Zz (26)
f (z) =
g (t)
t
k(α−1)
dt.
0
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The function f is regular in U. From (26) we obtain
f 0 (z) =
g (z)
z
k(α−1)
and
00
f (z) = k (α − 1)
g (z)
z
k(α−1)−1
zg 0 (z) − g (z)
.
z2
Next we have
2Re α
(27)
1 − |z|
Re α
00
0
zf (z) 1 − |z|2Re α
zg (z) ≤
+1 ,
k
|α
−
1|
f 0 (z) g (z) Re α
∀z ∈ U.
Applying the Schwartz Lemma and using (27) we obtain
2Re α 2Re α 2 0
00
1 − |z|
zf (z) ≤ |α − 1| 1 − |z|
z g (z) − 1 + 2 .
(28)
f 0 (z) g 2 (z)
Re α
Re α
Since g satisfies conditions (2) then from (28) and (23) we obtain
2Re α 3k |α − 1|
1 − |z|
zf 00 (z) 3k |α − 1| 2Re α
(29)
≤
1
−
|z|
≤
≤ 1.
f 0 (z) Re α
Re α
Re α
Now Theorem 2 and (29) imply that Gkα ∈ S.
Corollary 2. Let f, g ∈ A, satisfy (1) and α the complex number with the property
(30)
|α − 1| ≤
Re α
.
6
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If |f (z)| ≤ 1, ∀z ∈ U and |g (z)| ≤ 1, ∀z ∈ U , then the function
1

 2α−1
Zz
Gα (z) = (2α − 1) f (α−1) (t) g (α−1) (t) dt
(31)
0
is univalent.
Proof. In Theorem 1 we set n = 2, g1 = f , g2 = g.
Remark 1. Theorem 1 is a generalization of Theorem 3.
Remark 2. From Corollary 1, for k = 1, we obtain Theorem 3.
1.
2.
3.
4.
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Ozaki S. and Nunokawa M., The Schwartzian derivative and univalent functions, Proc. Amer. Math. Soc. 33(2) (1972), 392–394.
Pascu N. N., On univalent criterion II, Itinerant seminar on functional equations approximation and convexity, Cluj-Napoca,
Preprint nr. 6, (1985), 153–154.
5. Pascu N. N., An improvement of Beker’s univelence criterion, Proceedings of the Commemorative Session Cimion Stoilow, Brasov
(1987), 43–48.
6. Pescar V., New criteria for univalence of certain integral operators, Demonstratio Mathematica, XXXIII (2000), 51–54.
D. Breaz, Department of Mathematics, “1 Decembrie 1918” University, Alba Iulia, Romania, e-mail: dbreaz@uab.ro
N. Breaz, Department of Mathematics, “1 Decembrie 1918” University, Alba Iulia, Romania, e-mail: nbreaz@uab.ro
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