A subclass of univalent functions

Lemma 2. If $$g(z) = \sum_{k=0}^{\infty} d_k z^k \prec_q G(z) = \sum_{k=0}^{\infty} D_k z^k$$ , then $$(2.2) \qquad \sum_{k=0}^{n} |d_k|^2 \leq \sum_{k=0}^{n} |D_k|^2 \qquad (n = 0, 1, 2, ...).$$ This lemma is due to Robertson [14]. In particular (2.2) holds also when - (i) $g(z) \prec G(z)$ , - (ii) $g(z) \ll G(z)$ . By the application of Lemma 2, we establish

Lemma 3. For $$f \in R(A, B)$$ , if $f'(z) = P(z) = 1 + \sum_{k=1}^{\infty} p_k z^k$ , then (2.3) $|p_n| \le (A - B)$ , $n \ge 1$ . The bounds are sharp. PROOF. From (1.5) we have $$\sum_{k=1}^{\infty} p_k z^k = w(z) \left[ (A - B) - B \sum_{k=1}^{\infty} p_k z^k \right].$$ By the application of Lemma 2(2.2), we get $$\sum_{k=1}^{n} |p_k|^2 \le (A - B)^2 + B^2 \sum_{k=1}^{n-1} |p_k|^2$$ or $$|p_n|^2 \le (A-B)^2 - (1-B^2) \sum_{k=1}^{n-1} |p_k|^2 \le (A-B)^2.$$ This yields (2.3). Equality signs in (2.3) are attained for the functions $P_n(z)$ defined by $$P_n(z) = \frac{1 + A\delta z^n}{1 + B\delta z^n}, \quad |\delta| = 1.$$ In order to determine the radius of convexity, we need the following two lemmas.

Lemma 4. For $w \in U$ and |z| = r, we have $$|zw'(z) - w(z)| \le \frac{r^2 - |w(z)|^2}{1 - r^2}.$$ This result is due to Singh and Goel proved in [15].

Lemma 5. Let $$p(z) = \frac{1 + Bw(z)}{1 + Aw(z)}, \quad w \in U.$$ Then for |z| = r < 1, (2.5) $$\operatorname{Re}\left(Ap(z) + \frac{B}{p(z)}\right) + \frac{r^2|Ap(z) - B|^2 - |1 - p(z)|^2}{(1 - r^2)|p(z)|}$$ $$\leq \begin{cases} \frac{AB(A+B)r^2 - 4ABr + (A+B)}{(1 - Ar)(1 - Br)}, & R_1 \leq R_0, \\ \frac{2}{(1 - r^2)}\left[(1 - ABr^2) - ((1 - A)(1 - B)(1 + Ar^2)(1 + Br^2))^{1/2}\right], \\ R_1 \geq R_0, A \neq 1. \end{cases}$$ where (2.6) $$R_1 = \frac{1 - Br}{1 - Ar}, \qquad R_0^2 = \frac{(1 - B)(1 + Br^2)}{(1 - A)(1 + Ar^2)}.$$ The bounds are sharp. PROOF. It is easy to see that the transformation $$p(z) = \frac{1 + Bw(z)}{1 + Aw(z)}$$ maps $|w(z)| \le r$ onto the circle $|p(z) - a| \le d$ , where $$a = \frac{(1 - ABr^2)}{(1 - A^2r^2)}$$ and $d = \frac{(A - B)r}{(1 - A^2r^2)}$ . Putting $p(z) = Re^{i\theta}$ $\left(-\frac{\pi}{2} < \theta < \frac{\pi}{2}\right)$ and denoting the left hand side of (2.5) by $T(R, \theta)$ , we get $$T(R,\theta) = (AR + B/R)\cos\theta + \frac{2(1 - ABr^2)\cos\theta}{(1 - r^2)}$$ $$-\frac{(1 - A^2r^2)R}{(1 - r^2)} - \frac{(1 - B^2r^2)}{(1 - r^2)R}.$$ For extreme values of $T(R, \theta)$ , $\partial T/\partial R = 0 = \partial T/\partial \theta$ which yield respectively (2.7) $$\cos \theta = \frac{(1 - A^2 r^2) - (1 - B^2 r^2)/R^2}{(1 - r^2)(A - B/R^2)}, \qquad \left(-\frac{\pi}{2} < \theta < \frac{\pi}{2}\right),$$ and $$(2.8) L(R)\sin\theta = 0,$$ where (2.9) $$L(R) = (AR + B/R) + \frac{2(1 - ABr^2)}{(1 - r^2)}.$$ Now we prove that L(R) remains positive. If $B \ge 0$ , A > 0, then L(R) > 0. We now consider the case when B < 0. The following cases arise. Case I. B < 0, $A \ge 0$ . Using the fact that $0 < \cos \theta \le 1$ , it follows from (2.7) that (2.10) $$\frac{1 - B^2 r^2}{1 - A^2 r^2} < R^2 \le \frac{(1 - B)(1 + Br^2)}{(1 - A)(1 + Ar^2)}.$$ $L'(R) = (A - B/R^2) > 0$ and hence L(R) attains its minimum value at (2.11) $$R = \left(\frac{1 - B^2 r^2}{1 - A^2 r^2}\right)^{1/2} = R_2, \quad \text{say}.$$ Now $$L(R_2) = \left[ (1 - r^2)(A + B) + 2(1 - A^2r^2)^{1/2}(1 - B^2r^2)^{1/2} \right] \times \frac{(1 - ABr^2)}{(1 - r^2)\left[ (1 - A^2r^2)(1 - B^2r^2) \right]^{1/2}}$$ which is positive provided $$(2.12) (1-r^2)(A+B) + 2((1-A^2r^2)(1-B^2r^2))^{1/2} > 0.$$ If $(A + B) \ge 0$ , there is nothing to prove, so we assume (A + B) < 0. (2.12) will hold if $$4(1-A^2r^2)(1-B^2r^2)-(A+B)^2(1-r^2)^2>0$$ or if $$[(1+B)(1+Ar^2)(1+A)(1+Br^2)][2(1+ABr^2)-(A+B)(1+r^2)]>0$$ which is always true. Case II. B < 0, A < 0. Consider the case when $L'(R) = (A - B/R^2) < 0$ . Since $0 < \cos \theta \le 1$ , it follows from (2.7) that $$\frac{(1-B)(1+Br^2)}{(1-A)(1+Ar^2)} \le R^2 < \frac{1-B^2r^2}{1-A^2r^2}.$$ An easy computation would show that this does not hold. Now, consider the case when $(A - B/R^2) \ge 0$ . If $A - B/R^2 = 0$ , then from (2.7), we have $$R^2 = \frac{(1 - B^2 r^2)}{(1 - A^2 r^2)} = \frac{B}{A}$$ and it implies $(A - B)(1 + ABr^2) = 0$ which is evidently not possible. Thus the only case needed to be considered is when $(A - B/R^2) > 0$ . Therefore, by (2.10), $$\frac{1 - B^2 r^2}{1 - A^2 r^2} < R^2 < B/A.$$ The minimum value of L(R) occurs at $R = R_2$ and $L(R_2) > 0$ if $$(1-r^2)(A+B)+2[(1-A^2r^2)(1-B^2r^2)]^{1/2}>0$$ which holds as proved in Case I (when A + B is negative). For extreme values, from (2.8) and (2.7), we get $$\theta = 0, \pi;$$ $R^2 = \frac{(1-B)(1+Br^2)}{(1-A)(1+Ar^2)} = R_0^2, \text{ say.}$ It can be easily verified that $T(R, \theta)$ attains its maximum value at $(\theta = 0, R = R_0)$ . So $$T(R,\theta) \le T(R_0,0)$$ $$= \frac{2}{(1-r^2)} \Big[ (1-ABr^2) - \big( (1-A)(1-B)(1+Ar^2)(1+Br^2) \big)^{1/2} \Big].$$ It is easy to see that $R_0 \ge a - d = (1 + Br)/(1 + Ar)$ . But $R_0$ is not always less than or equal to a + d. In case $R_0 \notin [a = d, a + d]$ , the maximum of T(R, 0) is attained at $$R = R_1 = (a + d) = \frac{1 - Br}{1 - Ar}$$ and equals $$T(R_1,0) = \frac{AB(A+B)r^2 - 4ABr + (A+B)}{(1-Ar)(1-Br)}, \qquad R_1 \le R_0.$$ If $R_1 \le R_0$ , equality sign in (2.5) holds for the function $$p(z)=\frac{1+Bz}{1+Az}.$$ If $R_1 \ge R_0$ , $A \ne 1$ , equality sign in (2.5) holds for the function $$p_0(z) = \frac{1 - (1 + B)z\cos\theta + Bz^2}{1 - (1 + A)z\cos\theta + Az^2}$$ where (2.13) $$R_0 = \frac{1 - (1+B)r\cos\theta + Br^2}{1 - (1+A)r\cos\theta + Ar^2}.$$ Hence the lemma is established.

Theorem 3.1. Let $f \in R(A, B)$ then $$|a_n| \le \frac{(A-B)}{n}, \qquad n \ge 2.$$ The bounds are sharp for the functions $f_{(n-1)}(z)$ defined by (3.2) $$f_{(n-1)}(z) = \int_0^z \left( \frac{1 + A\delta t^{n-1}}{1 + B\delta t^{n-1}} \right) dt, \quad |\delta| = 1.$$ PROOF. (3.1) follows on equating the coefficients of $z^n$ in (1.4) and then using (2.3).

Theorem 3.2. If $f \in R(A, B)$ and if $\mu$ is a complex number, then (3.3) $$|a_3 - \mu a_2^2| \leq \frac{(A-B)}{3} \max \left\{ 1, |B + \frac{3(A-B)}{4} \mu| \right\}.$$ The estimate is sharp. PROOF. On equating the coefficients of $z^2$ and $z^3$ in (1.5), we get (3.4) $$c_1 = \frac{2a_2}{(A-B)}, ...$$ (3.5) $$c_2 = \frac{3}{(A-B)} \left[ a_3 + \frac{4B}{3(A-B)} a_2^2 \right].$$ Also $$|c_2| \le 1 - |c_1|^2$$ . Therefore, for every complex number v, we have (3.6) $$|c_2 - vc_1^2| \le |c_2| + |v| |c_1|^2$$ $$\le 1 + (|v| - 1) |c_1|^2$$ $$\le \max\{1, |v|\},$$ since | c, |< 1. The estimate (3.6) is sharp for w(z) = z and w(z) = z 2 respectively for | v | < 1 and | v \ < 1. From (3.4) and (3.5) we have (3.7) $$|a_3 - \mu a_2^2| = \frac{(A-B)}{3} |c_2 - vc_1^2|,$$ $$\mu = \frac{4}{3(A-B)}(v-B)$$ or (3.8) $$v = B + \frac{3(A-B)\mu}{4}$$ (3.7) in conjunction with (3.6) and (3.7), yields (3.3). (3.3) is sharp, being attained for the function/,(z) and/2(z) defined, respectively, by $$f_1'(z) = \frac{1+Az}{1+Bz}$$ and $f_2'(z) = \frac{1+Az^2}{1+Bz^2}$

Theorem 4.1. LetfG R(A, B), then for \ z |= r < 1, $$(4.1) |f'(z)| \leq \frac{1+Ar}{1+Br};$$ (4.2) $$\operatorname{Re} f'(z) \ge \frac{1 - Ar}{1 - Br};$$ $$(4.3) |f(z)| \leq \begin{cases} \frac{A}{B} \left[ r + \left( \frac{1}{A} - \frac{1}{B} \right) \log(1 + Br) \right], & B \neq 0, \\ r + \frac{A}{2} r^2, & B = 0, \end{cases}$$ $$|f(z)| \geqslant \begin{cases} \frac{A}{B} \left[ r - \left( \frac{1}{A} - \frac{1}{B} \right) \log(1 - Br) \right], & B \neq 0, \\ r - \frac{A}{2} r^2, & B = 0. \end{cases}$$ All the estimates are sharp. PROOF. From (1.5), it is easy to establish (4.1) and (4.2). Using (4.1), $$|f(z)| \le \int_0^r |f'(te^{i\beta})| dt \le \int_0^r \frac{1+At}{1+Bt} dt$$ which yields (4.3). Again using (4.2), $$|f(z)| \ge \int_0^r \operatorname{Re} f'(te^{i\beta}) dt \ge \int_0^r \frac{1 - At}{1 - Bt} dt$$ which gives (4.4). (4.1) and (4.3) are sharp, being attained for the function $$f_1(z) = \begin{cases} \frac{A}{B} \left[ z + \left( \frac{1}{A} - \frac{1}{B} \right) \log(1 + Bz) \right], & B \neq 0, \\ z + \frac{A}{2} z^2, & B = 0. \end{cases}$$ (4.2) and (4.4) are sharp for the function $$f_{(1)}(z) = \begin{cases} \frac{A}{B} \left[ z - \left( \frac{1}{A} - \frac{1}{B} \right) \log(1 - Bz) \right], & B \neq 0, \\ z - \frac{A}{2} z^2, & B = 0. \end{cases}$$ Let W be any complex number such that $$|W| < \begin{cases} \frac{A}{B} \left[ r - \left( \frac{1}{A} - \frac{1}{B} \right) \log(1 - Br) \right], & B \neq 0, \\ r - \frac{A}{2} r^2, & B = 0. \end{cases}$$ By Rouche's Theorem it follows that/(z) and/(z) — W have the same number of zeros in | z |< r, that is, precisely one. Hence we have the following: COROLLARY. Every function f(z) in R(A, B) maps E onto a domain which covers the disc $$|W| < \begin{cases} \frac{A}{B} \left[ 1 - \left( \frac{1}{A} - \frac{1}{B} \right) \log(1 - B) \right], & B \neq 0, \\ 1 - \frac{A}{2}, & B = 0. \end{cases}$$

Theorem 5.1. If f £ R(A, 5) then (5.1) $$|\arg f'(z)| \le \sin^{-1} \frac{(A-B)r}{1-ABr^2}, \quad |z|=r.$$ The result is sharp. PROOF. It is easy to show that f'(z) = (1 + Aw(z))/(1 + Bw(z)) maps $|w(z)| \le r$ onto the circle (5.2) $$\left| f'(z) - \frac{(1 - ABr^2)}{1 - B^2 r^2} \right| \le \frac{(A - B)r}{(1 - B^2 r^2)}.$$ (5.1) is an immediate consequence of (5.2). The result is sharp, being attained for the function $f_0(z)$ defined by $$f_0'(z) = \frac{1 + A\delta z}{1 + B\delta z},$$ where $$\delta = \frac{z}{r} \left[ \frac{-(A+B)r + i((1-A^2r^2)(1-B^2r^2))^{1/2}}{1+ABr^2} \right].$$

Theorem 6.1. If f and $h \in R(A, B)$ , then $$\lambda f + (1 - \lambda)h \in R(A, B), \quad (0 \le \lambda \le 1).$$ PROOF. By definition, $$(6.1) f'(z) < \frac{1+Az}{1+Bz},$$ (6.2) $$h'(z) < \frac{1 + Az}{1 + Bz}.$$ Since (1 + Az)/(1 + Bz) is convex univalent in E, it follows by a result due to Bernardi [2, page 57, Example 2] that $$\lambda f'(z) + (1-\lambda)h'(z) < \frac{1+Az}{1+Bz}.$$ Hence $$\lambda f' + (1 - \lambda)h' \in R(A, B).$$

Theorem 7.1. Let $f \in R(A, B)$ , then (i) for $A_0 \le A \le 1$ , f(z) is convex in $|z| < r_0$ , where $r_0$ is the smallest positive root of $$(7.1) ABr^2 - 2Ar + 1 = 0;$$ (ii) for $-1 < A \le A_0$ , f(z) is convex in $|z| < r_1$ , where $r_1$ is the smallest positive root of (7.2) $$A(1-r^2) - \left[ (1-ABr^2) - \left( (1-A)(1-B)(1+Ar^2)(1+Br^2) \right)^{1/2} \right] = 0;$$ $$A_0 = \frac{(2+B-2B^2) + (20-36B+21B^2-4B^3)^{1/2}}{2(4-2B-B^2)}.$$ The results are sharp. PROOF. Differentiating logarithmically, (1.5) yields (7.3) $$1 + \frac{zf''(z)}{f'(z)} = 1 + (A - B) \frac{zw'(z)}{(1 + Aw(z))(1 + Bw(z))}.$$ (7.3) together with Lemma 4 gives (7.4) $$\operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)}\right) \ge 1 + (A - B)$$ $$\times \left[\operatorname{Re}\frac{w(z)}{(1 + Aw(z))(1 + Bw(z))} - \frac{r^2 - |w(z)|^2}{(1 - r^2)|(1 + Aw(z))(1 + Bw(z))|}\right].$$ Putting p(z) = (1 + Bw(z))/(1 + Aw(z)), and using Lemma 5, (7.5) $$\operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)}\right)$$ $$\begin{cases} \frac{2A}{A - B} - \frac{AB(A + B)r^2 - 4ABr + (A + B)}{(A - B)(1 - Ar)(1 - Br)}, & R_1 \leq R_0, \\ \frac{2A}{A - B} - 2\left[\frac{(1 - ABr^2) - ((1 - A)(1 - B)(1 + Ar^2)(1 + Br^2))^{1/2}}{(1 - r^2)(A - B)}\right], \\ R_1 \geq R_0, A \neq 1. \end{cases}$$ (7.1) and (7.2) follow by equating the right hand sides of (7.5) to zero. The equation $R_0 = R_1$ yields $$(7.6) ABr^4 - 2ABr^3 + [2(A+B) - AB - 1]r^2 - 2r + 1 = 0.$$ Elimination of r between (7.1) and (7.6) leads to $$(7.7) \qquad (4-2B-B^2)A^2-(2+B-2B^2)A-(1-B)^2=0.$$ (7.7), on verification of the signs, yields $$A = \frac{(2+B-2B^2) + (20-36B+21B^2-4B^3)^{1/2}}{2(4-2B-B^2)} = A_0, \text{ say.}$$ The results are sharp, being attained respectively, for the functions $f_1(z)$ and $f_{\theta}(z)$ defined by $$f'_1(z) = \frac{1+Az}{1+Bz}, \qquad f'_{\theta}(z) = \frac{1-(1+A)z\cos\theta+Az^2}{1-(1+B)z\cos\theta+Bz^2},$$ where $\theta$ is defined by (2.13). REMARK 1. Radii of convexity for the classes R, R(1) and $S(\alpha)$ , at once, follow from (7.1). REMARK 2. On taking $A = \alpha$ , $B = -\alpha$ (0 < $\alpha \le 1$ ) in (7.7), we get $\alpha^4 - 4\alpha^3 - 4\alpha^2 + 4\alpha + 1 = 0$ which gives $$\alpha = \frac{(2^{1/2} - 1)(3^{1/2} + 1)}{2^{1/2}} = \alpha_0,$$ say. Hence - (i) for $\alpha_0 \le \alpha \le 1$ , f(z) maps $|z| < (2^{1/2} 1)/\alpha$ onto a convex domain; - (ii) for $0 < \alpha \le \alpha_0$ , f(z) maps $$|z| < \left[ \frac{(\alpha^2 - 1) + ((1 - \alpha^2)(1 + 4\alpha - \alpha^2))^{1/2}}{2\alpha(1 + \alpha)} \right]^{1/2}$$ onto a convex domain. This result was established by Padmanabhan in [13] and also by Capling and Causey in [4].

Theorem 8.1. Let $f \in R(A, B)$ and $L_r(f)$ denotes the length of the image of |z| = r under f(z), 0 < r < 1, then (8.1) $$L_{r}(f) \leq \begin{cases} \pi r \left[ \left| \frac{A+B}{B} \right| - \frac{(A-B)}{|B|} - 2\frac{(A-B)}{\pi B} \log \left( \frac{1-Br}{1+Br} \right) \right], & B \neq 0, \\ r \int_{0}^{2\pi} \left| 1 + Are^{i\theta} \right| d\theta, & B = 0. \end{cases}$$ The results are sharp. PROOF. In Lemma 1, set g(z) = f'(z), G(z) = (1 + Az)/(1 + Bz) and $\lambda = 1$ . Then (8.2) $$\int_0^{2\pi} |f'(re^{i\theta})| d\theta \leq \int_0^{2\pi} \left| \frac{1 + Are^{i\theta}}{1 + Bre^{i\theta}} \right| d\theta.$$ Now $$L_r(f) = \int_{|z|=r} |f'(z)| |dz|$$ $$= \int_0^{2\pi} |f'(re^{i\theta})| r d\theta.$$ By (8.2), $$\begin{split} L_{r}(f) &\leq r \int_{0}^{2\pi} \left| \frac{1 + Are^{i\theta}}{1 + Bre^{i\theta}} \right| d\theta \\ &= r \int_{0}^{2\pi} \left| \frac{A + B}{2B} - \frac{(A - B)(1 - B^{2}r^{2})}{2B(1 + 2Br\cos\theta + B^{2}r^{2})} \right| \\ &+ i \frac{(A - B)r\sin\theta}{1 + 2Br\cos\theta + B^{2}r^{2}} \\ &\leq \pi r \left| \frac{A + B}{B} \right| + \frac{(A - B)r}{2|B|} \int_{0}^{2\pi} \frac{(1 - B^{2}r^{2})}{1 + 2Br\cos\theta + B^{2}r^{2}} d\theta \\ &+ (A - B)r \int_{0}^{2\pi} \frac{r|\sin\theta|}{1 + 2Br\cos\theta + B^{2}r^{2}} d\theta \\ &= \pi r \left| \frac{A + B}{B} \right| + \frac{\pi(A - B)r}{|B|} - \frac{(A - B)r}{B} \int_{0}^{\pi} \frac{-2Br\sin\theta}{1 + 2Br\cos\theta + B^{2}r^{2}} d\theta \\ &= \pi r \left| \frac{A + B}{B} \right| + \frac{(A - B)}{|B|} - 2\frac{(A - B)}{\pi B} \log\left(\frac{1 - Br}{1 + Br}\right) \right]. \end{split}$$ For B = 0, result is trivial. The extremal function $f_0(z)$ is given by (8.3) $$f_0'(z) = \frac{1 + A\delta z}{1 + B\delta z}, \quad |\delta| = 1.$$ COROLLARY. For the class $R(\alpha)$ , we deduce, from (8.1), $$L_r(f) \leq 2\pi r + 4r \log\left(\frac{1+\alpha r}{1-\alpha r}\right).$$ This is a result established by Capling and Causey [4].

Theorem 8.2. // / £ R(A, B), and if Ar(f) denotes the area of image of\z\=r under f(z), 0 < r < 1, then (8.4) $$A_r(f) \leq \begin{cases} \pi r^2 \left[ \left( 1 - \frac{(A - B)^2}{B^2} \right) - \frac{(A - B)^2}{B^4 r^2} \log(1 - B^2 r^2) \right], & B \neq 0, \\ \pi r^2 \left[ 1 + \frac{A^2 r^2}{2} \right], & B = 0. \end{cases}$$ The inequalities are sharp. (8.4) are direct consequences of Lemma 1 (X = 2) and interior area theorem. Equality sign is attained for the function fo(z) defined by (8.3). COROLLARY. For the class R(a), we have from (8.4), $$A_r(f) \le \pi r^2 \left[ -3 - \frac{4}{\alpha^2 r^2} \log(1 - \alpha^2 r^2) \right].$$ This is a result due to Capling and Causey [4].

Theorem 9.1. If f(z) = z + 2"=2 anz" and h(z) = z + 2~=2 bnz" belong to the class R(A, B), then so does $$F(z) = z + \frac{1}{2} \sum_{n=2}^{\infty} n a_n b_n z^n$$ PROOF. Since/ e R(A, B), it follows by (1.6) that \f'(z) - 1 |<| A - Bf'(z) | . It is equivalent to $$(9.1) |f'(z) - b| \leq C$$ where b = (1 - AB)/{\ - B² ), C = (A - B)/(\ - B² ). It is easy to see that 1 - b< C < b. We know that if H(z) = l^=Qhnz n is regular for | z |< 1 and | H(z)\\z M,* then, by [11, page 101], (9.2) $$\sum_{n=0}^{\infty} |h_n|^2 \le M^2.$$ Applying (9.2) to (9.1), we get (1 - b)² + 2~ ⁼ ² n2 \an\ < C² or (9.3) $$\sum_{n=2}^{\infty} n^2 |a_n|^2 < \frac{(A-B)^2}{(1-B^2)^2}.$$ Similarly (9.4) $$\sum_{n=2}^{\infty} n^2 |b_n|^2 < \frac{(A-B)^2}{(1-B^2)^2}.$$ Now $$|F'(z) - b|^{2} = \left| (1 - b) + \frac{1}{2} \sum_{n=2}^{\infty} n^{2} a_{n} b_{n} z^{n-1} \right|^{2}$$ $$\leq (1 - b)^{2} + (1 - b) \sum_{n=2}^{\infty} n^{2} |a_{n}| |b_{n}| r^{n-1} + \frac{1}{4} \left| \sum_{n=2}^{\infty} n^{2} a_{n} b_{n} z^{n-1} \right|^{2}$$ $$\leq (1 - b)^{2} + (1 - b) \left( \sum_{n=2}^{\infty} n^{2} |a_{n}|^{2} r^{n-1} \right)^{1/2} \left( \sum_{n=2}^{\infty} n^{2} |b_{n}|^{2} r^{n-1} \right)^{1/2}$$ $$+ \frac{1}{4} \left( \sum_{n=2}^{\infty} n^{2} |a_{n}|^{2} r^{n-1} \right) \left( \sum_{n=2}^{\infty} n^{2} |b_{n}|^{2} r^{n-1} \right)$$ (by Cauchy-Schwarz inequality) $$\leq (1 - b)^{2} + (1 - b) \left( \sum_{n=2}^{\infty} n^{2} |a_{n}|^{2} \right)^{1/2} \left( \sum_{n=2}^{\infty} n^{2} |b_{n}|^{2} \right)^{1/2}$$ $$+ \frac{1}{4} \left( \sum_{n=2}^{\infty} n^{2} |a_{n}|^{2} \right) \left( \sum_{n=2}^{\infty} n^{2} |b_{n}|^{2} \right)$$ $$\leq (1 - b)^{2} + (1 - b) \frac{(A - B)^{2}}{(1 - B^{2})^{3}} + \frac{1}{4} \frac{(A - B)^{4}}{(1 - B^{2})^{4}}$$ [using (9.3) and (9.4)] $F \in R(A, B)$ if $$\frac{B^2(A-B)^2}{(1-B^2)^2} + \frac{B(A-B)^3}{(1-B^2)^3} + \frac{1}{4} \frac{(A-B)^4}{(1-B^2)^4} < \frac{(A-B)^2}{(1-B^2)^2}.$$ This gives on simplification, (A + B) < 2 which is true. Hence $F \in R(A, B)$ . $=\frac{B^2(A-B)^2}{(1-B^2)^2}+\frac{B(A-B)^3}{(1-B^2)^3}+\frac{1}{4}\frac{(A-B)^4}{(1-B^2)^4}.$ REMARK. The first author [6] proved this theorem for the class $S(\alpha)$ .