Convexity theorems for subclasses of univalent functions

Theorem 1. Suppose $p(z) \in P$ , h is defined by (1), and a is defined as in Theorem A. Then $$\operatorname{Re}\left\{\frac{zp'(z)}{p(z)+h}\right\} \ge \begin{cases} -\frac{2r}{(1+r)[(1+h-(1-h)r])} & (0 \le r \le r_{\beta}) \\ 2\sqrt{h^2+ah}-a-2h & (r_{\beta} < r < 1), \end{cases}$$ where $r_{\beta}$ is the unique root of the equation $(1-2\beta)r^3-3(1-2\beta)r^2+3r-1=0$ in the interval (0,1]. This result is sharp.

Theorem 2. Suppose $r_{\beta}$ is the unique root of $$t(r) = (1 - 2\beta)r^3 - 3(1 - 2\beta)r^2 + 3r - 1$$ in the interval (0, 1]. Set $$r(\alpha,\beta) = \frac{1}{2-\alpha-2\beta+\sqrt{\alpha^2-2\alpha+4\beta^2-6\beta+3}}.$$ Then the radius of convexity of $C(\alpha, \beta)$ is $r(\alpha, \beta)$ when $0 < r(\alpha, \beta) \le r_{\beta}$ , and is otherwise the smallest root greater than $r_{\beta}$ of the polynomial equation $$v(r) = [\alpha^{2} - \beta(\alpha^{2} + 2\alpha - 1)]r^{4} - 2(1 - \alpha)(\beta + \alpha\beta - \alpha)r^{3} + [(1 - \alpha)^{2}(1 - \beta) + 2\alpha\beta]r^{2} + 2\beta(1 - \alpha)r - \beta.$$ This result is sharp for all $\alpha$ and $\beta$ .

Theorem 3. If $f(z) \in S$ with $Ref'(z) > \beta$ , then f(z) is convex in a disk of radius $$\begin{cases} \frac{1}{1 - 2\beta + \sqrt{4\beta^2 - 6\beta + 2}} & (0 \le \beta \le \frac{1}{10}) \\ \left(1 + \sqrt{\frac{1 - \beta}{\beta}}\right)^{-\frac{1}{2}} & (\frac{1}{10} < \beta < 1). \end{cases}$$ This result is sharp. Proof. Since $\phi(z) = z$ is the only function in K(1), the class under consideration is $C(1, \beta)$ so that Theorem 2 may be applied. As we saw in the corollary to Theorem 2, $$r(1,\beta) = \frac{1}{1-2\beta+\sqrt{4\beta^2-6\beta+2}} \le r_{\beta} \qquad (0 \le \beta \le \frac{1}{10}),$$ which gives the first part of the theorem. Since $t(r_{\beta}) = 0$ when $\alpha = 1$ and $\beta = \frac{1}{10}$ , the radius of convexity of $C(1, \beta)$ for $\beta > \frac{1}{10}$ is the only positive root of $$(1-2\beta)r^4+2\beta r^2-\beta=0$$ , or $$r^{2} = \frac{-\beta + \sqrt{\beta - \beta^{2}}}{1 - 2\beta} = \frac{1}{1 + \sqrt{\frac{1 - \beta}{\beta}}}.$$ This completes the proof. REMARK. The cases $\beta = 0$ and $\beta = \frac{1}{2}$ were proved, respectively, by MacGregor [6] and Hallenbeck [2].

Theorem 4. If $f(z) \in C(\alpha, \beta)$ , then $$|\arg f'(z)| \le 2(1-\alpha)\sin^{-1}r + \sin^{-1}\left[\frac{2(1-\beta)r}{1+(1-2\beta)r^2}\right].$$ This result is sharp. Proof. We may write $$f'(z) = \phi'(z)q(z)$$ , where $\phi(z) \in K(\alpha)$ and $q(z) \in P_{\beta}$ . Hence $$|\arg f'(z)| \le |\arg \phi'(z)| + |\arg q(z)|.$$ But by a result of Pinchuk [8], $$|\arg \phi'(z)| \le 2(1-\alpha)\sin^{-1} r \qquad (|z| \le r).$$ Since Re $q(z) > \beta$ , the function $$\omega(z) = \frac{(q(z) - \beta) - (1 - \beta)}{(q(z) - \beta) + (1 - \beta)} = \frac{q(z) - 1}{q(z) - (2\beta - 1)}$$ is analytic with $\omega(0) = 0$ and $|\omega(z)| < 1$ in |z| < 1. Thus by Schwarz's lemma, $$\left|\frac{q(z)-1}{q(z)-(2\beta-1)}\right| < |z| \quad \text{for} \quad |z| < 1.$$ Hence the values of q(z) are contained in the circle of Apollonius whose diameter is the line segment from $(1 + (2\beta - 1)r)/(1 + r)$ to $(1-(2\beta-1)r)/(1-r)$ . The circle is centered at the point $(1+(1-2\beta)r^2)/(1-r^2)$ and has radius $(2(1-\beta)r)/(1-r^2)$ . Thus $|\arg q(z)|$ attains its maximum at points where a ray from the origin is tangent to the circle, that is, when (23) $$\arg q(z) = \pm \sin^{-1} \frac{2(1-\beta)r}{1+(1-2\beta)r^2}.$$ Substituting (22) and (23) into (21), the result follows. Equality holds for functions of the form $$f(z) = \int_0^z \frac{1 + (1 - 2\beta)\eta t}{(1 - \epsilon t)^{2(1 - \alpha)}(1 - \eta t)} dt$$ with suitably chosen $\epsilon$ , $\eta$ , where $|\epsilon| = |\eta| = 1$ . REMARK. For $\alpha = \beta = 0$ , this reduces to $$|\arg f'(z)| \le 2\sin^{-1}r + \sin^{-1}\frac{2r}{1+r^2} = 2(\sin^{-1}r + \tan^{-1}r),$$ a result of Krzyz [4].

Theorem 5. Suppose f(z), $g(z) \in C(\alpha, \beta)$ . Then $$\lambda f(z) + (1 - \lambda)g(z)$$ $(0 \le \lambda \le 1)$ is univalent in a disk |z| < r, where r is the smallest positive root of the equation $$2(1-\alpha)\sin^{-1}r + \sin^{-1}\left(\frac{2(1-\beta)r}{1+(1-2\beta)r^2}\right) = \frac{\pi}{2}.$$ This result is sharp.