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Theorem 1.1
Theorem 1.1. If f: D → C is analytic in the convex domain D and Re,, then f is univalent in D. The authors kindly acknowledge the support…
Theorem 1.1. If f : D → C is analytic in the convex domain D and
Re
$$f'(z) > 0$$
, $z \in D$ ,
then f is univalent in D.
The authors kindly acknowledge the support from CNCSIS research grant PNII-IDEI 209. c 2010 Australian Mathematical Publishing Association Inc. 0004-9727/2010 \$16.00
In the present paper we introduce the constant K(f, D) associated with a function $f: D \to \mathbb{C}$ analytic in a domain D, which is a measure of the 'degree of univalence' of f (see Proposition 2.1 and the remark following it).
Using the constant K(f, D) thus introduced, in Theorem 2.4 we obtain a sufficient condition for univalence, which shows that a small perturbation of a univalent function is again univalent. As a theoretical consequence of this result, it follows that a univalent function has a neighbourhood consisting entirely of univalent functions (see Remark 2.8).
The Theorem 2.4 is sharp, in the sense that we cannot replace the upper bound appearing in the hypothesis of this theorem by a larger one, as shown in Example 2.9.
For the particular choice of a linear function in Theorem 2.4, we obtain a simple sufficient condition for univalence (Corollary 2.6), which is shown to be equivalent to the Noshiro–Warschawski–Wolff univalence criterion. The main result in Theorem 2.4 can be viewed therefore as a generalization of this classical result, in which the linear function is replaced by a general univalent function.
The paper concludes with another application of the main result in the case of analytic functions defined in the unit disk. Thus, in Theorem 2.11 and the corollary following it, we obtain sufficient conditions for the univalence of an analytic function defined in the unit disk in terms of the coefficients of its Taylor series representation, which might be of independent interest.$
Corollary 2.12 · coeff
Corollary 2.12. Let be an analytic univalent function with Taylor series representation (2.14) If the coefficients satisfy the inequality…
Corollary 2.12. Let $g: U \to \mathbb{C}$ be an analytic univalent function with Taylor series representation
$$g(z) = \sum_{n=0}^{\infty} b_n z^n, \quad z \in U.$$
(2.14)
If the coefficients $a_0, a_1, \ldots \in \mathbb{C}$ satisfy the inequality
$$|a_n - b_n| < K(g, U) \frac{\zeta(p)}{n^{p+1}}, \quad n = 1, 2, \dots,$$
(2.15)
for some p > 1 ( $\zeta$ denotes the Riemann zeta function), then the function $f: U \to \mathbb{C}$ defined by
$$f(z) = \sum_{n=0}^{\infty} a_n z^n, \quad z \in U,$$
(2.16)
is analytic and univalent in U.
EXAMPLE 2.13. Considering the function $g(z) = z/(1-z) = \sum_{n=1}^{\infty} z^n$ defined in Example 2.10, which is analytic and univalent in U and has K(g, U) = 1, from the previous theorem it follows that the function $f: U \to \mathbb{C}$ defined by $f(z) = \sum_{n=0}^{\infty} a_n z^n$ is analytic and univalent in U if the coefficients $a_n$ satisfy the inequality
$$\sum_{n=1}^{\infty} n|a_n - 1| < 1.$$
Using, for example, the fact that $\zeta(2) = \pi^2/6 \approx 1.645$ , from the previous corollary it follows that the function f is also analytic and univalent in U if the coefficients $a_n$ satisfy the inequality
$$|a_n - 1| \le \frac{\pi^2}{6n^3} \approx \frac{1.645}{n^3}, \quad n = 1, 2, \dots$$