Volume 9 (2008), Issue 4, Article 117, 4 pp.
AN EXTENSION OF OZAKI AND NUNOKAWA’S UNIVALENCE CRITERION
HORIANA TUDOR
FACULTY OF M ATHEMATICS AND I NFORMATICS
"T RANSILVANIA " U NIVERSITY
2200 B RA ŞOV, ROMANIA
htudor@unitbv.ro
Received 11 May, 2008; accepted 14 October, 2008
Communicated by N.E. Cho
A BSTRACT. In this paper we obtain a sufficient condition for the analyticity and the univalence
of the functions defined by an integral operator. In a particular case we find the well known
condition for univalency established by S. Ozaki and M. Nunokawa.
Key words and phrases: Univalent functions, Univalence criteria.
2000 Mathematics Subject Classification. 30C55.
1. I NTRODUCTION
We denote by Ur = {z ∈ C : |z| < r} a disk of the z-plane, where r ∈ (0, 1], U1 = U and
I = [0, ∞). Let A be the class of functions f analytic in U such that f (0) = 0, f 0 (0) = 1.
Theorem 1.1 ([1]). Let f ∈ A. If for all z ∈ U
2 0
z f (z)
(1.1)
f 2 (z) − 1 < 1,
then the function f is univalent in U .
2. P RELIMINARIES
In order to prove our main result we need the theory of Löewner chains; we recall the basic
result of this theory, from Pommerenke.
Theorem 2.1 ([2]). Let L(z, t) = a1 (t)z + a2 (t)z 2 + · · · , a1 (t) 6= 0 be analytic in Ur , for all
t ∈ I, locally absolutely continuous in I and locally uniform with respect to Ur .For almost all
t ∈ I, suppose that
∂L(z, t)
∂L(z, t)
z
= p(z, t)
, ∀z ∈ Ur ,
∂z
∂t
where p(z, t) is analytic in U and satisfies the condition Re p(z, t) > 0, for all z ∈ U , t ∈ I. If
|a1 (t)| → ∞ for t → ∞ and {L(z, t)/a1 (t)} forms a normal family in Ur , then for each t ∈ I,
the function L(z, t) has an analytic and univalent extension to the whole disk U .
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2
H ORIANA T UDOR
3. M AIN R ESULTS
Theorem 3.1. Let f ∈ A and α be a complex number, Re α > 0. If the following inequalities
2 0
z f (z)
<1
(3.1)
−
1
f 2 (z)
and
(3.2)
2 0
z f (z)
1 − |z|2α z 2 f 0 (z)
2α
−1
f 2 (z) − 1 |z| + 2
α
f 2 (z)
2 0
(1 − |z|2α )2
z f (z)
f (z)
≤1
+
−
1
+
(1
−
α)
−
1
α2 |z|2α
f 2 (z)
z
are true for all z ∈ U \ {0}, then the function Fα ,
Z z
α1
α−1 0
(3.3)
Fα (z) = α
u f (u)du
0
is analytic and univalent in U , where the principal branch is intended.
Proof. Let us consider the function g1 (z, t) given by
e2αt − 1 f (e−t z)
g1 (z, t) = 1 −
−1 .
α
e−t z
For all t ∈ I and z ∈ U we have e−t z ∈ U and because f ∈ A, the function g1 (z, t) is analytic
in U and g1 (0, t) = 1. Then there is a disk Ur1 , 0 < r1 < 1 in which g1 (z, t) 6= 0, for all t ∈ I.
For the function
Z e−t z
g2 (z, t) = α
uα−1 f 0 (u)du ,
0
g2 (z, t) = z · g3 (z, t), it can be easily shown that g3 (z, t) is analytic in Ur1 and g3 (0, t) = e−αt .
It follows that the function
−t 2
z)
(eαt − e−αt ) f (e
e−t z
g4 (z, t) = g3 (z, t) +
g1 (z, t)
α
is also analytic in a disk Ur2 , 0 < r2 ≤ r1 and g4 (0, t) = eαt . Therefore, there is a disk
Ur3 , 0 < r3 ≤ r2 in which g4 (z, t) 6= 0, for all t ∈ I and we can choose an analytic branch of
[g4 (z, t)]1/α , denoted by g(z, t). We choose the branch which is equal to et at the origin.
From these considerations it follows that the function
L(z, t) = z · g(z, t) = et z + a2 (t)z 2 + · · ·
is analytic in Ur3 , for all t ∈ I and can be written as follows

 α1
Z e−t z
2αt
(2−α)t α−2 2 −t
(e − 1)e
z f (e z) 
−t
(3.4)
L(z, t) = α
uα−1 f 0 (u)du +
.
2αt
z)
0
1 − e α−1 f (e
−
1
e−t z
From the analyticity of L(z, t) in Ur3 , it follows that there is a number r4 , 0 < r4 < r3 , and a
constant K = K(r4 ) such that
|L(z, t)/et | < K,
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 117, 4 pp.
∀z ∈ Ur4 ,
t ∈ I,
http://jipam.vu.edu.au/
O ZAKI AND N UNOKAWA’ S U NIVALENCE C RITERION
3
and then {L(z, t)/et } is a normal family in Ur4 . From the analyticity of ∂L(z, t)/∂t, for all
fixed numbers T > 0 and r5 , 0 < r5 < r4 , there exists a constant K1 > 0 (that depends on T
and r5 ) such that
∂L(z, t) ∀z ∈ Ur5 , t ∈ [0, T ].
∂t < K1 ,
It follows that the function L(z, t) is locally absolutely continuous in I, locally uniform with
respect to Ur5 . We also have that the function
∂L(z, t)
∂L(z, t)
p(z, t) = z
∂z
∂t
is analytic in Ur , 0 < r < r5 , for all t ∈ I.
In order to prove that the function p(z, t) has an analytic extension with positive real part in
U for all t ∈ I, it is sufficient to show that the function w(z, t) defined in Ur by
w(z, t) =
p(z, t) − 1
p(z, t) + 1
can be continued analytically in U and that |w(z, t)| < 1 for all z ∈ U and t ∈ I.
By simple calculations, we obtain
−2t 2 0 −t
e z f (e z)
1 − e−2αt e−2t z 2 f 0 (e−t z)
−2αt
(3.5) w(z, t) =
−1 e
+2
−1
f 2 (e−t z)
α
f 2 (e−t z)
−2t 2 0 −t
e z f (e z)
f (e−t z)
(1 − e−2αt )2
− 1 + (1 − α)
−1 .
+
α2 e−2αt
f 2 (e−t z)
e−t z
From (3.1) and (3.2) we deduce that the function w(z, t) is analytic in the unit disk and
2 0
z f (z)
(3.6)
|w(z, 0)| = 2
− 1 < 1.
f (z)
We observe that w(0, t) = 0. Let t be a fixed number, t > 0, z ∈ U, z 6= 0. Since |e−t z| ≤
e−t < 1 for all z ∈ U = {z ∈ C : |z| ≤ 1} we conclude that the function w(z, t) is analytic in
U . Using the maximum modulus principle it follows that for each arbitrary fixed t > 0, there
exists θ = θ(t) ∈ R such that
(3.7)
|w(z, t)| < max |w(ξ, t)| = |w(eiθ , t)|,
|ξ|=1
We denote u = e−t · eiθ . Then |u| = e−t < 1 and from (3.5) we get
2 0
u f (u)
1 − |u|2α u2 f 0 (u)
iθ
2α
w(e , t) =
− 1 |u| + 2
−1
f 2 (u)
α
f 2 (u)
2 0
(1 − |u|2α )2
u f (u)
f (u)
+
− 1 + (1 − α)
−1 .
α2 |u|2α
f 2 (u)
u
Since u ∈ U , the inequality (3.2) implies that |w(eiθ , t)| ≤ 1 and from (3.6) and (3.7) we
conclude that |w(z, t)| < 1 for all z ∈ U and t ≥ 0.
From Theorem 2.1 it results that the function L(z, t) has an analytic and univalent extension
to the whole disk U for each t ∈ I, in particular L(z, 0). But L(z, 0) = Fα (z). Therefore the
function Fα (z) defined by (3.3) is analytic and univalent in U .
If in Theorem 3.1 we take α = 1 we obtain the following corollary which is just Theorem
1.1, namely Ozaki-Nunokawa’s univalence criterion.
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 117, 4 pp.
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4
H ORIANA T UDOR
Corollary 3.2. Let f ∈ A. If for all z ∈ U , the inequality (3.1) holds true, then the function f
is univalent in U .
Proof. For α = 1 we have F1 (z) = f (z) and the inequality (3.2) becomes
2 0
z f (z)
(1 − |z|2 )2 z 2 f 0 (z)
1 2
2
(3.8) − 1 |z| + 2(1 − |z| ) +
= f 2 (z) − 1 · |z|2 ≤ 1.
f 2 (z)
|z|2
It is easy to check that if the inequality (3.1) is true, then the inequality (3.8) is also true. Indeed,
the function g,
z 2 f 0 (z)
g(z) = 2
−1
f (z)
is analytic in U , g(z) = b2 z 2 + b3 z 3 + · · · , which shows that g(0) = g 0 (0) = 0. In view of (1.1)
we have |g(z)| < 1 and using Schwarz’s lemma we get |g(z)| < |z|2 .
Example 3.1. Let n be a natural number, n ≥ 2, and the function
z
(3.9)
f (z) =
n+1 .
1− zn
Then f is univalent in U and F n+1 is analytic and univalent in U , where
2
2
n+1
Z
n + 1 z n−1 0
(3.10)
F n+1 (z) =
u 2 f (u)du
.
2
2
0
Proof. We have
(3.11)
z 2 f 0 (z)
− 1 = z n+1
f 2 (z)
and
f (z)
z n+1
−1=
.
z
n − z n+1
It is clear that condition (3.1) of Theorem 3.1 is satisfied, and the function f is univalent in U .
Taking into account (3.11) and (3.12), condition (3.2) of Theorem 3.1 becomes
2(n+1)
4
4
|z|
+
|z|n+1 (1 − |z|n+1 ) +
(1 − |z|n+1 )2
n+1
(n + 1)2
2(1 − n)
1
n+1 2
+
(1 − |z| )
2
n+1
(n + 1)
n − |z| 1
≤
(n + 1)2 |z|2(n+1) + 4(n + 1)(1 − |z|n+1 ) + 6(1 − |z|n+1 )2
2
(n + 1)
2
1
=
(n − 2n + 3)|z|2(n+1) + (4n − 8)|z|n+1 + 6 ≤ 1,
2
(n + 1)
because the greatest value of the function
g(x) = (n2 − 2n + 3)x2 + (4n − 8)x + 6,
for x ∈ [0, 1], n ≥ 2 is taken for x = 1 and is g(1) = (n + 1)2 . Therefore the function F n+1 is
2
analytic and univalent in U .
(3.12)
R EFERENCES
[1] S. OZAKI AND M. NUNOKAWA, The Schwartzian derivative and univalent functions, Proc. Amer.
Math. Soc., 33(2) (1972), 392–394.
[2] Ch. POMMERENKE, Univalent Functions, Vandenhoech Ruprecht, Göttingen, 1975.
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 117, 4 pp.
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