Harmonic mappings, univalence criteria and a theorem of Lehtinen

Theorem 1. For every domain Ω in C that omits an open set we have that $$\sigma_H(\Omega) \le \sigma_H(\mathbb{D}) \le 3/2.$$ 3

Lemma 1. Let $\omega$ be a holomorphic self-mapping of $\mathbb{D}$ . Then $$\left| \frac{(1 - |z|^2)^2 \omega''(z)}{2(1 - |\omega(z)|^2)} - \overline{z}\omega^(z) + \overline{\omega(z)}\omega^(z)^2 \right| \le 1 - |\omega^*(z)|^2, \qquad z \in \mathbb{D}$$

Theorem 2. Let f be a locally univalent harmonic mapping in $\mathbb{D}$ with dilatation $\omega : \mathbb{D} \to \mathbb{D}$ . If $$(1-|z|^2)^2|S_f(z)|+2|\omega^(z)A_f(z)| \le \frac{1}{2}|\omega^(z)|^2, \qquad z \in \mathbb{D},$$ then f is injective.

Theorem 3. Let $f = h + \overline{g}$ be a locally univalent harmonic mapping in $\mathbb{D}$ . If $|Sh(z)| \leq 2c^2$ , where $c \approx 0,6533$ is the first no negative solution of the equation $2x \tan x = 1$ , then f is injective. Proof. Let a and b be complex numbers for which ah''(0) + b = 0, and set $f_1 = af + b = h_1 + \overline{g_1}$ . Then $h''_1(0) = 0$ and $Sh_1 = Sh$ . In view of Theorem 5.2 in [2], the normalization of $h_1$ along with the condition $|Sh_1(z)| \leq 2c^2$ imply that $h_1$ is convex. Hence, $f_1$ is injective by [7]. It follows that f is injective. Acknowledgements. The authors wish to thank María José Martín for drawing their attention to the non-injective harmonic mapping constructed in [19] and suggesting its use in the proof of the second inequality in Theorem 1.