Ruscheweyh's Univalence Criterion and Quasiconformal Extensions

Theorem 1. Let s = a + ib, a > 0, $b \in \mathbb{R}$ , $k \in [0,1)$ and $f \in \mathcal{A}$ . Assume that for a constant $c \in \mathbb{C}$ and all $z \in \mathbb{D}$ $$\left| c|z|^2 + s - a(1 - |z|^2) \left\{ s \left( 1 + \frac{zf''(z)}{f'(z)} \right) + (1 - s) \frac{zf'(z)}{f(z)} \right\} \right| \le M \tag{3}$$ with $$M = \begin{cases} ak|s| + (a-1)|s+c|, & if \quad 0 < a \le 1, \\ k|s|, & if \quad 1 < a, \end{cases}$$ then f has an l-quasiconformal extension to $\mathbb{C}$ , where $$l = \frac{2ka + (1 - k^2)|b|}{(1 + k^2)a + (1 - k^2)|s|} < 1.$$ (4) Remark 1.1. If $f \in \mathcal{A}$ , then it is easy to verify that there exists a sequence $\{z_n\} \subset \mathbb{D}$ with $|z_n| \to 1$ such that for each $s \in \{z \in \mathbb{C} : \operatorname{Re} z > 0\}$ $$\sup_{n} \left| s \left( 1 + \frac{z_n f''(z_n)}{f'(z_n)} \right) + (1 - s) \frac{z_n f'(z_n)}{f(z_n)} \right| < \infty$$ which shows that (3) implies the inequality $$|c+s| \le M. \tag{5}$$ This inequality is needed for proving that f(z) has no zeros in 0 < |z| < 1 (see Lemma 7). In [8], it is mentioned that (3) implies $f(z) \neq 0$ , 0 < |z| < 1, without proof. The part of (5) can be found in [8]. Remark 1.2. A similar argument to Remark 1.1 is also valid for Theorem 1. It follows that the assumption $|c| \le k$ is embedded in the inequality (1). The next application follows from Theorem 1. Let $\alpha > 0$ and $\beta \in \mathbb{R}$ . It follows from a result of Sheil-Small [9, Theorem 2] that $$\operatorname{Re}\left\{1 + \frac{zf''(z)}{f'(z)} + (\alpha + i\beta - 1)\frac{zf'(z)}{f(z)}\right\} > 0 \qquad (z \in \mathbb{D})$$ (6) is sufficient for $f \in \mathcal{A}$ to be a Bazilevič function of type $(\alpha, \beta)^{-1}$ (see also [5]). Here, a function $f \in \mathcal{A}$ is called Bazilevič of type $(\alpha, \beta)$ if $$f(z) = \left[ (\alpha + i\beta) \int_0^z g(\zeta)^\alpha h(\zeta) \zeta^{i\beta - 1} d\zeta \right]^{1/(\alpha + i\beta)}$$ ^&lt;sup>1</sup>The author would like to thank Professor Yong Chan Kim for this remark. for a starlike univalent function $g \in \mathcal{A}$ and an analytic function h with h(0) = 1 satisfying $Re(e^{i\lambda}h) > 0$ in $\mathbb{D}$ for some $\lambda \in \mathbb{R}$ . Together with this fact, the next theorem follows;

Theorem 2. Let $\alpha > 0, \beta \in \mathbb{R}$ and $k \in [0, 1)$ . If $f \in \mathcal{A}$ satisfies $$\left|1 + \frac{zf''(z)}{f'(z)} + (\alpha + i\beta - 1)\frac{zf'(z)}{f(z)} - \frac{\alpha^2 + \beta^2}{\alpha}\right| \le M \tag{7}$$ for all $z \in \mathbb{D}$ with $$M = \begin{cases} k & \text{if } \alpha < \alpha^2 + \beta^2, \\ k(\alpha^2 + \beta^2)/\alpha & \text{if } \alpha^2 + \beta^2 \le \alpha, \end{cases}$$ then f is a Bazilevič function of type $(\alpha, \beta)$ and can be extended to a $\tilde{k}$ -quasiconformal automorphism of $\mathbb{C}$ , where $$\tilde{k} = \frac{2k\alpha + (1 - k^2)|\beta|}{(1 + k^2)\alpha + (1 - k^2)\sqrt{\alpha^2 + \beta^2}}.$$ Next, we shall discuss quasiconformal extensibility of functions $g(z)=z+\frac{d}{z}+\cdots$ analytic in $\mathbb{D}^*$ .

Theorem 3. Let $s=a+ib, \ a\geq 1, b\in\mathbb{R}$ and $k\in[0,1)$ which satisfies $|b/s|\leq k$ . Let $g(\zeta)=\zeta+\frac{d}{\zeta}+\cdots$ be analytic in $\mathbb{D}^*$ and fulfill $$\left| ib + (1 - |\zeta|^2) a \left\{ (1 - s) \left( 1 - \frac{\zeta g'(\zeta)}{g(\zeta)} \right) - s \frac{\zeta g''(\zeta)}{g'(\zeta)} \right\} \right| \le ak|s| - |b|(a - 1) \tag{8}$$ for all $\zeta \in \mathbb{D}^*$ . Then g can be extended to an l-quasiconformal automorphism of $\widehat{\mathbb{C}}$ , where $$l = \frac{2ka + (1 - k^2)|b|}{(1 + k^2)a + (1 - k^2)|s|}.$$ The case $k \to 1$ corresponds to a univalence criterion which is due to Ruscheweyh [8]. Theorem 3 yields the following corollary which gives a positive answer to an open problem proposed by Ruscheweyh [8], i.e., whether a function $g(\zeta) = \zeta + d/\zeta + \cdots$ with $(|\zeta|^2 - 1)|1 + (\zeta f''(\zeta)/f'(\zeta)) - (\zeta f'(\zeta)/f(\zeta))| \leq k$ for all $\zeta \in \mathbb{D}^*$ admits a quasiconformal extension to $\mathbb{C}$ ;

Corollary 4. Let $g(\zeta) = \zeta + \frac{d}{\zeta} + \cdots$ be analytic in $\mathbb{D}^*$ . If there exists $k \in [0,1)$ such that $$(|\zeta|^2 - 1) \left| 1 + \frac{\zeta g''(\zeta)}{g'(\zeta)} - \frac{\zeta g'(\zeta)}{g(\zeta)} \right| \le k$$ for all $\zeta \in \mathbb{D}^*$ , then g can be extended to a k-quasiconformal automorphism of $\widehat{\mathbb{C}} - \{0\}$ . From the above corollary we have another extension criterion for analytic functions on $\mathbb{D}$ ;

Corollary 5. Let $f \in A$ with f''(0) = 0. If there exists $k \in [0,1)$ such that $$(1-|z|^2)\left|1+\frac{zf''(z)}{f'(z)}-\frac{zf'(z)}{f(z)}\right| \le k$$ for all $z \in \mathbb{D}$ , then f can be extended to a k-quasiconformal automorphism of $\mathbb{C}$ .

Lemma 6. Under the assumption of Theorem 1, we have $$|aP(e^{-st/|s|}z, t/|s|) + ib| < k|s|$$ (13) for $z \in \mathbb{D}$ and $t \in [0, \infty)$ . Proof. We have $$|aP + ib| \le m_1 + m_2$$ by triangle inequality, where $$m_1 = (1 - e^{-2t/|s|}) \left| \frac{ce^{-2at/|s|} + s}{1 - e^{-2at/|s|}} - aH_s(e^{-st/|s|}z) \right|$$ and $$m_2 = \left| (ce^{-2at/|s|} + s) \frac{1 - e^{-2t/|s|}}{1 - e^{-2at/|s|}} - (ce^{-2t/|s|} + s) \right|.$$ Then it is enough to show that $m_1 + m_2 < k|s|$ . (3) implies $$\left| \frac{c|e^{st/|s|}z|^2 + s}{1 - |e^{st/|s|}z|^2} - aH_s(e^{-st/|s|}z) \right| \le \frac{M}{1 - |e^{st/|s|}z|^2} \le \frac{M}{1 - e^{-2at/|s|}}$$ for $z \in \mathbb{D}$ . Let $q(t) = (1 - e^{-2t/|s|})/(1 - e^{-2at/|s|})$ . Applying the maximum modulus principle to the function $$\frac{ce^{-2at/|s|} + s}{1 - e^{-2at/|s|}} - aH_s(e^{-st/|s|}z)$$ we have $$m_1 < q(t)M$$ . On the other hand $$m_2 \le |c+s||1-q(t)|.$$ Since $1 \le q(t) < 1/a$ if $0 < a \le 1$ and $1/a < q(t) \le 1$ if 1 < a for all $t \in [0, \infty)$ , we conclude that $m_1 + m_2 < k|s|$ which is our desired inequality. We now let $\Delta$ and $\Delta'$ be disks which are defined by replacing P in (12) and (13) to a complex variable w. It remains to find the smallest l so that $\Delta'$ is contained by $\Delta$ . Note that if k = l = 1 then these two disks coincide. The following condition is necessary and sufficient for $\Delta' \subset \Delta$ ; $$\left| \frac{(1+l^2)b}{(1+l^2)a + (1-l^2)|s|} - \frac{b}{a} \right| \le \frac{2l|s|}{(1+l^2)a + (1-l^2)|s|} - \frac{k|s|}{a}. \tag{14}$$ Then we conclude $$l \le \frac{2ka + (1 - k^2)|b|}{(1 + k^2)a + (1 - k^2)\sqrt{a^2 + b^2}}.$$ which is suitable for our purpose. (ii) In order to eliminate the additional assumption that $f(z)/z \neq 0$ in $\mathbb{D}$ , we need a sort of stability of the condition (3);

Lemma 7. If $f \in \mathcal{A}$ satisfies the assumption of Theorem 1, then so does $f_r(z) = \frac{1}{r}f(rz)$ , $r \in (0,1)$ . Proof. It follows from the assumption that $aH_s(rz)$ is contained in the disk $$\Delta = \left\{ w \in \mathbb{C} : \left| w - \frac{cr^2|z|^2 + s}{1 - r^2|z|^2} \right| \le \frac{M}{1 - r^2|z|^2} \right\}.$$ We want to deduce that $aH_s(rz)$ lies in the disk $$\Delta' = \left\{ w \in \mathbb{C} : \left| w - \frac{c|z|^2 + s}{1 - |z|^2} \right| \le \frac{M}{1 - |z|^2} \right\}.$$ Therefore it is enough to see that $\Delta \subset \Delta'$ , that is, $$\left| \frac{c|z|^2 + s}{1 - |z|^2} - \frac{cr^2|z|^2 + s}{1 - r^2|z|^2} \right| \le \frac{M}{1 - |z|^2} - \frac{M}{1 - r^2|z|^2}. \tag{15}$$ In view of the identity $$\frac{|z|^2}{1-|z|^2} - \frac{r^2|z|^2}{1-|z|^2} = \frac{1}{1-|z|^2} - \frac{1}{1-r^2|z|^2},$$ the inequality (15) is equivalent to (5). Now we shall show that the condition $f(z)/z \neq 0$ in $\mathbb{D}$ follows from the assumption of Theorem 1. Suppose, to the contrary, that $f(z_0) = 0$ for some $0 < |z_0| < 1$ . We may assume that $f(z) \neq 0$ for $0 < |z| < |z_0|$ . Then by Lemma 7 we can apply Theorem 1 to the function $f_{r_0}(z) = f(r_0z)/r_0$ , $r_0 = |z_0|$ to conclude that $f_{r_0}$ has a quasiconformal extension to $\mathbb{C}$ . In particular, $f_{r_0}$ is injective on $\overline{\mathbb{D}}$ . This, however, contradicts the relation $f_{r_0}(z_0/r_0) = f_{r_0}(0) = 0$ . Remark 3.1. We can replace |s| in (10) to any positive real value and continue our argument. However, it will be found that |s| gives the smallest l by calculations. Remark 3.2. We have $l \geq k$ , where l = k if and only if b = 0. Indeed, let l = l(k). Then we have l'(k) > 0 and $l''(k) \leq 0$ which imply $l \geq k$ . If we suppose $l = k \neq 0$ , then the right-hand side of (14) is greater than or equal to 0 only if b = 0. In the case l = k = 0 we also have b = 0 by (14). It easily follows from (4) that l = k if b = 0.

Lemma 8. Let $g(\zeta) = \zeta + \frac{d}{\zeta} + \cdots$ be analytic in $\mathbb{D}^*$ . If g satisfies the same assumption of Theorem 3, then so does $g_R(\zeta) = \frac{1}{R} f(R\zeta), R > 1$ . Proof. We need to prove $$\left| \frac{ib}{|\zeta|^2 - 1} - aG_s(R\zeta) \right| \le \frac{ak|s| - |b|(a-1)}{|\zeta|^2 - 1}$$ by using $$\left| \frac{ib}{R^2 |\zeta|^2 - 1} - aG_s(R\zeta) \right| \le \frac{ak|s| - |b|(a-1)}{R^2 |\zeta|^2 - 1},$$ where $$G_s(\zeta) = (1-s)\left(\frac{\zeta g'(\zeta)}{g(\zeta)} - 1\right) + s\frac{\zeta g''(\zeta)}{g'(\zeta)}.$$ In a similar way to the proof of Lemma 7, it suffices to see that $$\left|\frac{ib}{|\zeta|^2-1} - \frac{ib}{R^2|\zeta|^2-1}\right| \leq \frac{ak|s|-|b|(a-1)}{|\zeta|^2-1} - \frac{ak|s|-|b|(a-1)}{R^2|\zeta|^2-1}.$$ This is equivalent to $|b| \leq k|s|$ . Then we let $$f(1/\zeta,t) = \frac{1}{g(e^{st}\zeta)} \left\{ 1 - (1 - e^{-2t})e^{st}\zeta \frac{g'(e^{st}\zeta)}{g(e^{st}\zeta)} \right\}^{-s}$$ and $$F(1/\zeta, t) = f(1/\zeta, t/|s|).$$ Since $$h(1/\zeta,t) = \frac{\dot{F}(1/\zeta,t)}{(1/\zeta)F'(1/\zeta,t)} = \frac{s}{|s|} \cdot \frac{1 + P(e^{st/|s|}\zeta,t/|s|)}{1 - P(e^{st/|s|}\zeta,t/|s|)}$$ where $$P(\zeta, t) = (e^{2t/|s|} - 1)G_s(\zeta),$$ it is sufficient to see that $$|aP(e^{st/|s|}\zeta, t/|s|) + ib| < k|s| \tag{16}$$ for all $\zeta \in \mathbb{D}^*$ and $t \in [0, \infty)$ under the assumption of the theorem. By triangle inequality we have $$|aP + ib| \le \left| \frac{1 - e^{2t/|s|}}{1 - e^{2at/|s|}} \left( ib + (1 - e^{2at/|s|}) aG_s(e^{st/|s|}\zeta) \right) \right| + \left| ib \left( 1 - \frac{1 - e^{2t/|s|}}{1 - e^{2at/|s|}} \right) \right|$$ for $\zeta \in \mathbb{D}^*$ and $t \in [0, \infty)$ . Following the lines of the proof of Lemma 6, one can obtain that (8) implies (16). Therefore, a similar argument of the proof of Theorem 1 implies our assertion. The case s = 1 follows from a theorem of Becker [2].