BOUNDEDNESS, UNIVALENCE AND QUASICONFORMAL EXTENSION OF ROBERTSON FUNCTIONS

Theorem 1. Let $f \in \mathcal{R}(\lambda)$ with $\cos \lambda < 1/\sqrt{2}$ , then f is bounded. In order to prove the above result, the growth theorem of spirallike functions in [18] or [1] is needed. Since those known forms are complicated there, we simplify them as follows.

Lemma 2. Let $f \in \mathcal{SP}(\lambda)$ , then for |z| = r < 1, we have $$\Psi_1(r) \le |f(z)| \le \Psi_2(r)$$ where $$\Psi_1(r) = \left| P_{\lambda}(re^{i\theta_1}) \right| = \frac{r \exp\left(-\sin 2\lambda \arcsin(r \sin \lambda)\right)}{(r \cos \lambda - \sqrt{1 - r^2 \sin^2 \lambda})^{2 \cos^2 \lambda}}$$ and $$\Psi_2(r) = \left| P_{\lambda}(re^{i\theta_2}) \right| = \frac{r \exp\left(\sin 2\lambda \arcsin(r \sin \lambda)\right)}{(r \cos \lambda - \sqrt{1 - r^2 \sin^2 \lambda})^{2 \cos^2 \lambda}}$$ where $$P_{\lambda}(z) = \frac{z}{(1-z)^{1+e^{2i\lambda}}}$$ belongs to $SP(\lambda)$ and $\theta_j$ (j = 1, 2) satisfy $$\sin(\lambda + \theta_i) = r \sin \lambda \quad (j = 1, 2)$$ and $\cos(\lambda + \theta_1) < 0$ and $\cos(\lambda + \theta_2) > 0$ respectively. 2.2. Proof of Theorem 1. Equivalence (1) and Lemma 2 show that $$|f(z)| = \left| \int_0^z f'(\zeta) d\zeta \right| = \left| \int_0^r \frac{z}{r} f'(tz/r) dt \right|$$ $$\leq \int_0^r |f'(tz/r)| dt \leq \int_0^r \frac{\exp(\sin(2\lambda)\arcsin(t\sin\lambda))}{(\sqrt{1-t^2\sin^2\lambda} - t\cos\lambda)^2\cos^2\lambda} dt$$ where 0 < |z| = r < 1. Since the numerator in the above integrand is bounded over [0, 1], it is sufficient to estimate only the denominator. Upon a change in the variable s = 1 - t, we obtain $$\begin{split} \sqrt{1-t^2\sin^2\lambda} - t\cos\lambda &= \sqrt{1-(1-s)^2\sin^2\lambda} - (1-s)\cos\lambda \\ &= \sqrt{\cos^2\lambda + 2s\sin^2\lambda - s^2\sin^2\lambda} - (1-s)\cos\lambda \\ &= \cos\lambda\sqrt{1+2s\tan^2\lambda - s^2\tan^2\lambda} - (1-s)\cos\lambda \\ &= \cos\lambda[1+1/2(2s\tan^2\lambda - s^2\tan^2\lambda) + O(s^2)] - (1-s)\cos\lambda \\ &= \frac{s}{\cos\lambda} + O(s^2) \end{split}$$ when $s \to 0$ . Therefore f(z) is bounded whenever $2\cos^2 \lambda < 1$ , that is, $\cos \lambda < 1/\sqrt{2}$ . The example given by (2) ensures the sharpness of the value $1/\sqrt{2}$ . Remark. Note that our method is not applicable for the case when $\cos \lambda = 1/\sqrt{2}$ . Since the function $f_{\lambda}(z)$ given in (2) is bounded when $\cos \lambda = 1/\sqrt{2}$ , we may expect that $\mathcal{R}(\lambda)$ consists of bounded functions as well in this case.

Theorem 3. Let $f \in \mathcal{A}$ , $k \in [0,1)$ and $\lambda \in (-\pi/2, \pi/2)$ , q > -1 be related by <span id="page-3-0"></span> $$0 < \cos \lambda \le \begin{cases} k/2, & \text{if } -1 < q \le 0, \\ k/(2+4q), & \text{if } 0 < q. \end{cases}$$ (3) $\it If f satisfies$ <span id="page-3-2"></span> $$\operatorname{Re}\left\{e^{-i\lambda}\left(1 + \frac{zf''(z)}{f'(z)} + q\frac{zf'(z)}{f(z)}\right)\right\} > 0 \tag{4}$$ for all $z \in \mathbb{D}$ , then $f \in \mathcal{S}(k)$ . If, in addition, f''(0) = 0, (3) can be replaced by <span id="page-3-3"></span> $$0 < \cos \lambda \le \begin{cases} k, & \text{if } -1 < q \le 0, \\ k/(1+2q), & \text{if } 0 < q. \end{cases}$$ (5) We note that when q=0 Theorem 3 claims quasiconformal extension of $\lambda$ -Robertson functions which can be stated explicitly as follows; <span id="page-4-0"></span>Corollary 4. Let $f \in \mathcal{R}(\lambda)$ with $\lambda \in (-\pi/2, \pi/2)$ satisfying $$0 < \cos \lambda \le k/2$$ , then $f \in \mathcal{S}(k)$ . If, in addition, f''(0) = 0 and (4) can be replied by $$0 < \cos \lambda \le k$$ . then $f \in \mathcal{S}(k)$ . We note here that the second case in Corollary 4 also implies that function $f \in \mathcal{R}(\lambda)$ with f''(0) = 0 for arbitrary $\lambda \in (-\pi/2, \pi/2)$ is univalent which was proved by Singh and Chichra [19] by means of Ahlfors's criterion for univalence as well. 3.2. Preliminaries. The following several results will be used later in our arguments. Here, set $$H_s(z) = s \left( 1 + \frac{zf''(z)}{f'(z)} \right) + (1 - s) \frac{zf'(z)}{f(z)}.$$ <span id="page-4-2"></span>Theorem B ([7]). Let a > 0, $b \in \mathbb{R}$ , s = a + ib, $k \in [0, 1)$ and $f \in \mathcal{A}$ . Assume that for a constant $c \in \mathbb{C}$ and all $z \in \mathbb{D}$ $$|c|z|^2 + s - a(1 - |z|^2)H_s(z)| \le M$$ with $$M = \begin{cases} ak|s| + (a-1)|s+c|, & if \quad 0 < a \le 1, \\ k|s|, & if \quad a > 1, \end{cases}$$ then $f \in \mathcal{S}(l)$ , where $$l = \frac{2ka + (1 - k^2)|b|}{(1 + k^2)a + (1 - k^2)|s|} < 1.$$ We note that in the above theorem l = k if and only if b = 0 ([7]). <span id="page-4-1"></span>Lemma C (e.g. [17]). Let $p(z) = 1 + a_n z^n + \cdots$ be analytic and $\operatorname{Re} p(z) > 0$ on $\mathbb{D}$ . Then $$\left| p(z) - 1 - \frac{2|z|^{2n}}{1 - |z|^{2n}} \right| \le \frac{2|z|^{2n}}{1 - |z|^{2n}}$$ for all $z \in \mathbb{D}$ . 3.3. Proof of Theorem 3. Let s = 1/(1+q) and $f(z) = z + \sum_{n=2}^{\infty} a_n z^n$ , then for $$p(z) = \frac{e^{-i\lambda}H_s(z) + i\sin\lambda}{\cos\lambda}$$ $$= 1 + \frac{e^{-i\lambda}}{\cos\lambda}(s+1)a_2z + \cdots.$$ we have p'(0)=0 if and only if f''(0)=0. Condition (4) implies that p(z) is analytic in $\mathbb D$ and fulfills $\operatorname{Re} p(z)>0$ for all $z\in \mathbb D$ . With $(c+s)=\frac{2}{n}se^{i\lambda}\cos\gamma,\ n=1,2$ , we obtain from Lemma C that $$\left| \frac{(c+s)|z|^2}{1-|z|^2} - s(H_s(z)-1) \right| \\ \leq s|\cos\lambda| \left\{ \left| \frac{2|z|^{2n}}{1-|z|^{2n}} - (p(z)-1) \right| + \left| \frac{2|z|^{2n}}{1-|z|^{2n}} - \frac{2}{n} \frac{|z|^2}{1-|z|^2} \right| \right\} \\ \leq \frac{2s}{n} \frac{|\cos\lambda|}{1-|z|^2}.$$ Therefore by Theorem B f can be extended to a k-quasiconformal automorphism of $\mathbb C$ whenever $$\frac{2}{n}s|\cos\lambda| \le \begin{cases} ks^2 + \frac{2}{n}s|\cos\lambda|(s-1), & if \quad 0 < s \le 1, \\ ks, & if \quad 1 < s, \end{cases}$$ which is equivalent to (3) if n = 1 and to (5) if n = 2.