AN EXTENSION OF OZAKI AND NUNOKAWA’S UNIVALENCE CRITERION Ozaki and Nunokawa’s Univalence Criterion HORIANA TUDOR Faculty of Mathematics and Informatics "Transilvania" University 2200 Braşov, ROMANIA EMail: htudor@unitbv.ro Horiana Tudor vol. 9, iss. 4, art. 117, 2008 Title Page Contents Received: 11 May, 2008 Accepted: 14 October, 2008 Communicated by: N.E. Cho 2000 AMS Sub. Class.: 30C55. Key words: Univalent functions, Univalence criteria. Abstract: 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. JJ II J I Page 1 of 11 Go Back Full Screen Close Contents 1 Introduction 3 2 Preliminaries 4 3 Main Results 5 Ozaki and Nunokawa’s Univalence Criterion Horiana Tudor vol. 9, iss. 4, art. 117, 2008 Title Page Contents JJ II J I Page 2 of 11 Go Back Full Screen Close 1. Introduction 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, Ozaki and Nunokawa’s Univalence Criterion Horiana Tudor vol. 9, iss. 4, art. 117, 2008 then the function f is univalent in U . Title Page Contents JJ II J I Page 3 of 11 Go Back Full Screen Close 2. Preliminaries 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 ∂t Ozaki and Nunokawa’s Univalence Criterion Horiana Tudor ∀z ∈ Ur , 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 . vol. 9, iss. 4, art. 117, 2008 Title Page Contents JJ II J I Page 4 of 11 Go Back Full Screen Close 3. Main Results 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 Ozaki and Nunokawa’s Univalence Criterion Horiana Tudor vol. 9, iss. 4, art. 117, 2008 Title Page Contents JJ II J I Page 5 of 11 Go Back Full Screen Close g2 (z, t) = z α · 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) 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, ∀z ∈ Ur4 , t ∈ I, 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 , Ozaki and Nunokawa’s Univalence Criterion Horiana Tudor vol. 9, iss. 4, art. 117, 2008 Title Page Contents JJ II J I Page 6 of 11 Go Back Full Screen Close 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 p(z, t) − 1 w(z, t) = 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 (3.5) w(z, t) −2t 2 0 −t e z f (e z) 1 − e−2αt e−2t z 2 f 0 (e−t z) −2αt = −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 Ozaki and Nunokawa’s Univalence Criterion Horiana Tudor vol. 9, iss. 4, art. 117, 2008 Title Page Contents JJ II J I Page 7 of 11 Go Back Full Screen Close 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) 2α iθ w(e , t) = − 1 |u| + 2 −1 f 2 (u) α f 2 (u) 2 0 u f (u) f (u) (1 − |u|2α )2 − 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. 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 2 2 z f (z) (1 − |z| ) 2 2 (3.8) − 1 |z| + 2(1 − |z| ) + f 2 (z) |z|2 2 0 z f (z) 1 ≤ 1. = − 1 · f 2 (z) |z|2 Ozaki and Nunokawa’s Univalence Criterion Horiana Tudor vol. 9, iss. 4, art. 117, 2008 Title Page Contents JJ II J I Page 8 of 11 Go Back Full Screen Close 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, g(z) = z 2 f 0 (z) −1 f 2 (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 f (z) = (3.9) z n+1 . 1− zn Ozaki and Nunokawa’s Univalence Criterion Horiana Tudor vol. 9, iss. 4, art. 117, 2008 Title Page Contents Then f is univalent in U and F n+1 is analytic and univalent in U , where 2 (3.10) F n+1 (z) = 2 n+1 2 Z z u n−1 2 f 0 (u)du 2 n+1 . 0 JJ II J I Page 9 of 11 Proof. We have (3.11) z 2 f 0 (z) − 1 = z n+1 2 f (z) Full Screen Close and (3.12) Go Back z n+1 f (z) −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| ) (n + 1)2 n − |z|n+1 1 2 2(n+1) n+1 n+1 2 ≤ (n + 1) |z| + 4(n + 1)(1 − |z| ) + 6(1 − |z| ) (n + 1)2 2 1 2(n+1) n+1 = (n − 2n + 3)|z| + (4n − 8)|z| + 6 ≤ 1, (n + 1)2 because the greatest value of the function 2 Ozaki and Nunokawa’s Univalence Criterion Horiana Tudor vol. 9, iss. 4, art. 117, 2008 Title Page 2 g(x) = (n − 2n + 3)x + (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 analytic and univalent in U . 2 Contents JJ II J I Page 10 of 11 Go Back Full Screen Close References [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. Ozaki and Nunokawa’s Univalence Criterion Horiana Tudor vol. 9, iss. 4, art. 117, 2008 Title Page Contents JJ II J I Page 11 of 11 Go Back Full Screen Close