On convex and starlike functions in a sector

Theorem 1. Let f be analytic in D with /(0) = /'(O) - 1 = 0 and 0 < p < 1. Then for z e D, $$1 + \frac{zf''(z)}{f'(z)} \prec \left(\frac{1+z}{1-z}\right)^{\alpha(\beta)}$$ implies (1) $$\frac{zf'(z)}{f(z)} \prec \left(\frac{1+z}{1-z}\right)^{\beta},$$ where (2) $$\alpha(\beta) = \frac{2}{\pi} \arctan \left[ \tan \frac{\beta \pi}{2} + \frac{\beta}{(1+\beta)^{(1+\beta)/2} (1-\beta)^{(1-\beta)/2} \cos(\beta \pi/2)} \right]$$ Also for y > 0, (3) $$\frac{zf'(z)}{f(z)} \prec \left(\frac{1+z}{1-z}\right)^{\beta(\gamma)}$$ implies $$\frac{f(z)}{z} \prec \left(\frac{1+z}{1-z}\right)^{\gamma},$$ where (5) $$\beta(\gamma) = \frac{2}{\pi} \arctan \gamma.$$ Furthermore oe(fi) is the largest number such that (1) holds and fi(y) is the largest number such that (4) holds. We shall use the following lemma (see for example [4]). LEMMA. Let F be analytic in D and G be analytic and univalent in D, with F(0) = G(0). // F jk. G, then there is a point z0 € D and £0 € dD such that F{\z\ < |zo|) C G(D), F(z0) = Gttb) andz0F'(z0) = PROOF OF THEOREM 1. Write $$p(z) = \frac{zf'(z)}{f(z)},$$ so that p is analytic in D and p(0) = 1. Thus we need to show that $$p(z) + \frac{zp'(z)}{p(z)} \prec \left(\frac{1+z}{1-z}\right)^{\alpha}$$ implies $$p(z) \prec \left(\frac{1+z}{1-z}\right)^{\beta},$$ whenever a = a (ft). Now let/i (2) = [(l + 2)/(l-z)fand^(2) = [(l + z)/(l-z)f .Then |arg/i(z)| < an/2 and | arg^(z)| < fin/2. Suppose that p -£ q, then from the Lemma there is a point z0 e D and £0 € 3D such that p(z0) = q(£0) and p(\z\ < \zo\) c q(D). We next note that p(z) ^ 0 for z e D, since otherwise we can write p(z) = (z — Zi)k pi(z) for some k > 1 and pi analytic in £), such that P\{z\) ^ 0. Then (6) $$p(z) + \frac{zp'(z)}{p(z)} = \frac{zp'_1(z)}{p_1(z)} + \frac{kz}{z - z_1} + (z - z_1)^k p_1(z).$$ Thus choosing z —> zi, suitably the argument of the right-hand side of (6) can take any value between 0 and In, which contradicts the hypotheses of the Theorem. Since $p(z_0) = q(\zeta_0) \neq 0$ , it follows that $\zeta_0 \neq \pm 1$ . Thus we can write $ri = (1 + \zeta_0)/(1 - \zeta_0)$ for $r \neq 0$ . Then the Lemma gives $$p(z_0) + \frac{z_0 p'(z_0)}{p(z_0)} = q(\zeta_0) + \frac{m\zeta_0 q'(\zeta_0)}{q(\zeta_0)}$$ $$= (ri)^{\beta} + \frac{m\beta(1+r^2)i}{2r}.$$ Next assume that r > 0. (If r < 0, the proof is similar.) Since $m \ge 1$ , an elementary argument shows that $$\arg\left(p(z_0) + \frac{z_0 p'(z_0)}{p(z_0)}\right) = \arctan\left[\tan\frac{\beta\pi}{2} + \frac{m\beta(1+r^2)}{2r^{\beta+1}\cos(\beta\pi/2)}\right]$$ $$\geq \arctan\left[\tan\frac{\beta\pi}{2} + \frac{\beta(1+r^2)}{2r^{\beta+1}\cos(\beta\pi/2)}\right]$$ $$\geq \arctan\left[\tan\frac{\beta\pi}{2} + \frac{\beta}{(1+\beta)^{(1+\beta)/2}(1-\beta)^{(1-\beta)/2}\cos(\beta\pi/2)}\right]$$ $$= \alpha(\beta)\pi/2.$$ Hence $$\frac{\alpha(\beta)\pi}{2} \leq \arg\left(p(z_0) + \frac{z_0 p'(z_0)}{p(z_0)}\right) \leq \frac{\pi}{2},$$ which contradicts the fact that $|\arg h(z)| < \alpha(\beta)\pi/2$ provided that (2) holds. To show that (2) is exact, let $$p(z) = \left(\frac{1+z}{1-z}\right)^{\beta}.$$ Then from the minimum principle for harmonic functions, it follows that $$\inf_{|z|<1} \arg \left( p(z) + \frac{zp'(z)}{p(z)} \right)$$ is attained at some point $z = e^{i\theta}$ for $0 < \theta < 2\pi$ . Thus $$p(z) + \frac{zp'(z)}{p(z)} = \left(\frac{\sin\theta}{1 - \cos\theta}\right)^{\beta} e^{\beta\pi i/2} + \frac{i\beta}{\sin\theta},$$ and so writing $t = \cos \theta$ , we obtain $$\arg\left(p(z) + \frac{zp'(z)}{p(z)}\right) = \arctan\left[\tan\frac{\beta\pi}{2} + \frac{\beta}{(1+t)^{(1+\beta)/2}(1-t)^{(1-\beta)/2}\cos(\beta\pi/2)}\right]$$ and an elementary calculation shows that the minimum of this expression is attained when $t = \beta$ . We next show that (3) implies (4). Write p(z) = f(z)/z, so that again p is analytic in D and p(0) = 1. Thus this time we need to show that $$1 + \frac{zp'(z)}{p(z)} \prec \left(\frac{1+z}{1-z}\right)^{\beta(\gamma)}$$ implies $$p(z) \prec \left(\frac{1+z}{1-z}\right)^{\gamma}$$ , whenever (5) holds. As before, let $h(z) = [(1+z)/(1-z)]^{\beta}$ and $q(z) = [(1+z)/(1-z)]^{\gamma}$ and assume that $p \not\prec q$ . Then there exists $z_0 \in D$ and $\zeta_0 \in \partial D$ such that $p(z_0) = q(\zeta_0)$ and $p(|z| < |z_0|) \subset q(D)$ . Since $p(z_0) = q(\zeta_0) \neq 0$ , the Lemma gives $$1 + \frac{z_0 p'(z_0)}{p(z_0)} = 1 + \frac{m\gamma(1+r^2)i}{2r}.$$ The result now follows using the same arguments as before. To show that $\beta(\gamma)$ is the largest number such that (4) holds we again let $p(z) = [(1+z)/(1-z)]^{\beta}$ and $z = e^{i\theta}$ . Then $$\arg\left(1+\frac{zp'(z)}{p(z)}\right)=\arctan\left[\frac{\beta}{\sin\theta}\right],$$ which has a minimum when $\theta = \pi/2$ . The following is an immediate consequence of Theorem 1. COROLLARY. For $\gamma > 0$ , $$1 + \frac{zf''(z)}{f'(z)} \prec \left(\frac{1+z}{1-z}\right)^{\alpha(\gamma)}$$ implies (7) $$\frac{f(z)}{z} \prec \left(\frac{1+z}{1-z}\right)^{\gamma},$$ where $$\alpha(\gamma) = \frac{2}{\pi} \arctan \left[ \tan \frac{\beta(\gamma)\pi}{2} + \frac{\beta(\gamma)}{(1+\beta(\gamma))^{(1+\beta(\gamma))/2} (1-\beta(\gamma))^{(1-\beta(\gamma))/2} \cos(\beta(\gamma)\pi/2)} \right]$$ and $\beta(\gamma)$ is given by (5). Also $\alpha(\gamma)$ is the largest number such that (7) holds. For a > 0, denote by R(a) the class of functions / analytic in D with /(0) = /'(0) - 1 = 0, satisfying f(z) arg — ~ 2 ' for z e D. We now restate the results in terms of C(a), S\(a) and R(a)* as follows.

Theorem 2. For y > 0, $$C(\alpha(\beta(\gamma))) \subset S^*(\beta(\gamma)) \subset R(\gamma),$$ where a(/}) anrf ^(y) are given by (2) a/w/ (5) respectively. Furthermore, a(P(y)) and P(y) are the largest numbers such that the inclusion holds.