Univalence Criteria and Quasiconformal Extensions

Theorem 2.1. Let $L(z,t) = a_1(t)z + a_2(t)z^2 + ...$ , $a_1(t) \neq 0$ be analytic in $\mathcal{U}_r$ for all $t \in I$ , locally absolutely continuous in I, and locally uniform with respect to $\mathcal{U}_r$ . For almost all $t \in I$ , suppose that <span id="page-1-0"></span>(2.1) $$z\frac{\partial L(z,t)}{\partial z} = p(z,t)\frac{\partial L(z,t)}{\partial t}, \ \forall z \in \mathcal{U}_r$$ where p(z,t) is analytic in $\mathcal{U}$ and satisfies the condition $\Re p(z,t) > 0$ for all $z \in \mathcal{U}$ , $t \in I$ . If $|a_1(t)| \to \infty$ for $t \to \infty$ and $\{L(z,t) \not = a_1(t)\}$ forms a normal family in $\mathcal{U}_r$ , then for each $t \in I$ , the function L(z,t) has an analytic and univalent extension to the whole disk $\mathcal{U}$ . Let k be constant in [0,1). Then a homeomorphism f of $G \subset \mathbb{C}$ is said to be k-quasiconformal, if $\partial_z f$ and $\partial_{\overline{z}} f$ in the distributional sense are locally integrable on G and fulfill the inequality $|\partial_{\overline{z}} f| \leq k |\partial_z f|$ almost everywhere in G. If we do not need to specify k, we will simply call that f is quasiconformal. The method of constructing quasiconformal extension criteria is based on the following result due to Becker (see [4], [5] and also [6]).

Theorem 2.2. Suppose that L(z,t) is a Loewner chain. Consider $$w(z,t) = \frac{p(z,t) - 1}{p(z,t) + 1}, \ z \in \mathcal{U}, \ t \ge 0$$ where p(z,t) is given in (2.1). If $$|w(z,t)| \le k, \ 0 \le k < 1$$ for all $z \in \mathcal{U}$ and $t \geq 0$ , then L(z,t) admits a continuous extension to $\overline{\mathcal{U}}$ for each $t \geq 0$ and the function $F(z,\overline{z})$ defined by $$F(z, \overline{z}) = \begin{cases} \mathcal{L}(z, 0), & \text{if } |z| < 1\\ \mathcal{L}(\frac{z}{|z|}, \log|z|), & \text{if } |z| \ge 1 \end{cases}$$ is a k-quasiconformal extension of L(z,0) to $\mathbb{C}$ . Examples of quasiconformal extension criteria can be found in [1], [3], [7], [17], [24] and more recently in [9], [13]-[15], [30].

Theorem 3.1. Consider $f \in A$ and g be an analytic function in $\mathcal{U}$ , $g(z) = 1 + b_1 z + ...$ . Let $\alpha, \beta, A$ and B complex numbers such that $\Re(\alpha) > \frac{1}{2}$ , $A + B \neq 0$ , |A - B| < 2, $|A| \leq 1$ and $|B| \leq 1$ . If the inequalities $$\left| \frac{1}{\alpha} \left( \frac{f'(z)}{g(z) - \beta} - 1 \right) \right| < \frac{|A + B|}{2 - |A - B|}$$ and <span id="page-2-0"></span> $$\left| \left( \frac{f'(z)}{g(z) - \beta} - 1 \right) |z|^2 + \left( 1 - |z|^2 \right) \left[ \left( \frac{1 - \alpha}{\alpha} \right) \frac{zf'(z)}{f(z)} + \frac{zg'(z)}{g(z) - \beta} \right] - \frac{\left( \overline{A} - \overline{B} \right) (A + B)}{4 - |A - B|^2} \right| \le \frac{2 |A + B|}{4 - |A - B|^2}$$ are satisfied for all $z \in \mathcal{U}$ , then the function f is univalent in $\mathcal{U}$ .

Corollary 3.2. Consider $f \in A$ and g be an analytic function in $\mathcal{U}$ , $g(z) = 1 + b_1 z + ...$ Let $\alpha, \beta$ and A are complex numbers such that $\Re(\alpha) > \frac{1}{2}$ , $A \neq 0$ , $|A| \leq 1$ . If the inequalities $$\left| \frac{1}{\alpha} \left( \frac{f'(z)}{g(z) - \beta} - 1 \right) \right| < |A|$$ and $$\left| \left( \frac{f'(z)}{g(z) - \beta} - 1 \right) |z|^2 + \left( 1 - |z|^2 \right) \left[ \left( \frac{1 - \alpha}{\alpha} \right) \frac{zf'(z)}{f(z)} + \frac{zg'(z)}{g(z) - \beta} \right] \right| \le |A|$$ are satisfied for all $z \in \mathcal{U}$ , then the function f is univalent in $\mathcal{U}$ . If we choose $\alpha = 1$ in Theorem 3.1, we obtain the following univalence criterion.

Corollary 3.3. Consider $f \in A$ and g be an analytic function in U, $g(z) = 1 + b_1 z + ...$ Let $\beta$ , A and B complex numbers such that $A + B \neq 0$ , |A - B| < 2, $|A| \leq 1$ and $|B| \leq 1$ . If the inequalities $$\left| \frac{f'(z)}{g(z) - \beta} - 1 \right| < \frac{|A + B|}{2 - |A - B|}$$ and $$\left| \left( \frac{f'(z)}{g(z) - \beta} - 1 \right) |z|^2 + \left( 1 - |z|^2 \right) \frac{zg'(z)}{g(z) - \beta} - \frac{\left( \overline{A} - \overline{B} \right) (A + B)}{4 - |A - B|^2} \right| \le \frac{2|A + B|}{4 - |A - B|^2}$$ are satisfied for all $z \in \mathcal{U}$ , then the function f is univalent in $\mathcal{U}$ .

Corollary 3.4. Consider $f \in A$ and g be an analytic function in $\mathcal{U}$ , $g(z) = 1 + b_1 z + ...$ Let A and B complex numbers such that $A + B \neq 0$ , |A - B| < 2, $|A| \leq 1$ and $|B| \leq 1$ . If the inequalities $$\left|\frac{f'(z)}{g(z)} - 1\right| < \frac{|A+B|}{2 - |A-B|}$$ and <span id="page-4-2"></span> $$\left| \left( \frac{f'(z)}{g(z)} - 1 \right) |z|^2 + \left( 1 - |z|^2 \right) \frac{zg'(z)}{g(z)} - \frac{\left( \overline{A} - \overline{B} \right) (A+B)}{4 - |A-B|^2} \right| \le \frac{2|A+B|}{4 - |A-B|^2}$$ are satisfied for all $z \in \mathcal{U}$ , then the function f is univalent in $\mathcal{U}$ .

Corollary 3.5. Consider $f \in A$ . Let A and B complex numbers such that $A + B \neq 0$ , |A - B| < 2, $|A| \leq 1$ and $|B| \leq 1$ . If the inequality $$(3.20) \left| \left( 1 - |z|^2 \right) \frac{zf''(z)}{f'(z)} - \frac{\left( \overline{A} - \overline{B} \right) (A+B)}{4 - |A-B|^2} \right| \le \frac{2|A+B|}{4 - |A-B|^2}$$ is satisfied for all $z \in \mathcal{U}$ , then the function f is univalent in $\mathcal{U}$ . <span id="page-5-0"></span>Corollary 3.6. Consider $f \in A$ . Let $\beta$ , A and B complex numbers such that $A+B \neq 0$ , |A-B| < 2, $|A| \leq 1$ and $|B| \leq 1$ . If the inequalities <span id="page-5-1"></span> $$\left| \frac{\beta}{f'(z) - \beta} \right| < \frac{|A + B|}{2 - |A - B|}$$ and <span id="page-5-2"></span> $$\left| \frac{\beta |z|^2 + \left(1 - |z|^2\right) z f''(z)}{f'(z) - \beta} - \frac{\left(\overline{A} - \overline{B}\right) (A + B)}{4 - |A - B|^2} \right| \le \frac{2 |A + B|}{4 - |A - B|^2}$$ are satisfied for all $z \in \mathcal{U}$ , then the function f is univalent in $\mathcal{U}$ .

Corollary 3.7. Consider $\beta < 0$ in Corollary 3.6. By elementary calculation we obtain that the inequality (3.21) for A = B = 1 is equivalent to $$\Re f'(z) > \frac{1}{2\beta} \left| f'(z) \right|^2, \ z \in \mathcal{U}.$$ If in the last inequality we let $\beta \to -\infty$ we obtain that $$\Re f'(z) > 0.$$ Since (3.22) for A=B=1 and $\beta\to -\infty$ it follows from Corollary 3.6 that the function f is univalent in $\mathcal U$ . Therefore, we can conclude that the univalence criterion due to Alexander-Noshiro-Warshawski [2], [20], [32] is a limit case of Corollary 3.6. Remark 3.8. Some particular cases of Theorem 3.1 are the following: <span id="page-5-3"></span>(i) When $\alpha = 1$ , $\beta = 0$ , A = B = 1 and g(z) = f'(z) inequality (3.2) becomes (3.23) $$\left(1 - |z|^2\right) \left| \frac{zf''(z)}{f'(z)} \right| \le 1, \ z \in \mathcal{U}$$ which is Becker's condition of univalence [4]. (ii) A result due to N. N. Pascu [23] is obtained when $\alpha = 1$ , A = B = 1 and g(z) = f'(z). Remark 3.9. It is worth to notice that the condition (3.2) assures the univalence of an analytic function in more general case than that of condition (3.23). Remark 3.10. If we put $g(z) = \frac{f(z)}{z}$ into (3.19), we have $$\left|\frac{zf'(z)}{f(z)} - 1\right| \le \frac{|A+B|}{2 - |A-B|}, \ z \in \mathcal{U}$$ the class of functions starlike with respect to origin.

Theorem 4.1. Consider $f \in A$ , g be an analytic function in $\mathcal{U}$ , $g(z) = 1 + b_1 z + ...$ and $k \in [0,1)$ . Let $\alpha, \beta, A$ and B complex numbers such that $\Re(\alpha) > \frac{1}{2}$ , $A + B \neq 0$ , k |A - B| < 2, $|A| \leq 1$ and $|B| \leq 1$ . If the inequalities $$\left| \frac{1}{\alpha} \left( \frac{f'(z)}{g(z) - \beta} - 1 \right) \right| < \frac{k |A + B|}{2 - k |A - B|}$$ and (4.2) $$\left| \left( \frac{f'(z)}{g(z) - \beta} - 1 \right) |z|^2 + \left( 1 - |z|^2 \right) \left[ \left( \frac{1 - \alpha}{\alpha} \right) \frac{zf'(z)}{f(z)} + \frac{zg'(z)}{g(z) - \beta} \right] - \frac{k^2 \left( \overline{A} - \overline{B} \right) (A + B)}{4 - k^2 \left| A - B \right|^2} \right| \le \frac{2k \left| A + B \right|}{4 - k^2 \left| A - B \right|^2}$$ are satisfied for all $z \in \mathcal{U}$ , then the function f has a k-quasiconformal extension to $\mathbb{C}$ .

Corollary 4.2. Consider $f \in A$ and $k \in [0,1)$ . Let A and B complex numbers such that $A + B \neq 0$ , k|A - B| < 2, $|A| \le 1$ and $|B| \le 1$ . If the inequality $$\left| \left( 1 - |z|^2 \right) \left( \frac{zf''(z)}{f'(z)} \right) - \frac{k^2 \left( \overline{A} - \overline{B} \right) (A+B)}{4 - k^2 |A-B|^2} \right| \le \frac{2k |A+B|}{4 - k^2 |A-B|^2}$$ is satisfied for all $z \in \mathcal{U}$ , then the function f has a k-quasiconformal extension to $\mathbb{C}$ . For A = B = 1 in Corollary 4.2, we have result of Becker [4].

Corollary 4.3. Consider $f \in A$ and $k \in [0,1)$ . If the inequality $$(4.6) \left(1 - |z|^2\right) \left| \frac{zf''(z)}{f'(z)} \right| \le k$$ is satisfied for all $z \in \mathcal{U}$ , then the function f has a k-quasiconformal extension to $\mathbb{C}$ . Acknowledgement 1. The present investigation was supported by Atatürk University Rectorship under "The Scientific and Research Project of Atatürk University", Project No: 2012/173.