Meromorphic univalent functions with positive coefficients

Theorem 1. Let $$f(z) = \frac{1}{z} + \sum_{n=1}^{\infty} a_n z^n$$ be regular in D . If (2.1) $$\sum_{n=1}^{\infty} [(1+\beta)n + (2\alpha-1)\beta + 1] |a_n| \le 2\beta(1-\alpha) ,$$ $0 \le \alpha < 1$ , $0 < \beta \le 1$ , then $f \in \Sigma^*(\alpha, \beta)$ . Proof. It is easy to see that (2.1) implies $f(z) \neq 0$ in D. Suppose (2.1) holds for all admissible values of $\alpha$ and $\beta$ . Consider the expression (2.2) $$H(f,f') = |zf'(z) + f(z)| - \beta |zf'(z) + (2\alpha-1)f(z)|$$ Replacing f and f' by their series expansions, we have for $0 < \left|z\right| = r < 1$ , $$H(f,f') = \Big| \sum_{n=1}^{\infty} (n+1) a_n z^n \Big| - \beta \Big| 2(\alpha-1) \frac{1}{z} + \sum_{n=1}^{\infty} (n+2\alpha-1) a_n z^n \Big|$$ or $$rH(f,f') \leq \sum_{n=1}^{\infty} (n+1) \left| a_n \right| r^{n+1} - \beta \{ 2(1-\alpha) - \sum_{n=1}^{\infty} (n+2\alpha-1) \left| a_n \right| r^{n+1} \}$$ $$= \sum_{n=1}^{\infty} [(1+\beta)n + (2\alpha-1)\beta+1] |a_n| r^{n+1} - 2\beta(1-\alpha) .$$ Since the above inequality holds for all $\, r$ , 0 < r < 1 , letting $\, r \, \to \, 1$ , we have $$H(f,f') \leq \sum_{n=1}^{\infty} \left[ (1+\beta)n + (2\alpha-1)\beta \right] \left| a_n \right| - 2\beta(1-\alpha)$$ $$\leq 0$$ by (2.1). Hence it follows that $\left|\frac{zf'(z)}{f(z)}+1\right|<\beta\left|\frac{zf'(z)}{f(z)}+2\alpha-1\right|$ , so that $f\in\Sigma^*(\alpha,\beta)$ . Hence the theorem. Let us write $\Sigma_P^\star(\alpha,\beta) = \Sigma_P \cap \Sigma^\star(\alpha,\beta)$ where $\Sigma_P$ is the class of functions of the form (1.8) that are regular and univalent in D. Our next theorem shows that condition (2.1) is both necessary and sufficient for an f to be in $\Sigma_D^\star(\alpha,\beta)$ .

Theorem 2. Let $f(z)=\frac{1}{z}+\sum\limits_{n=1}^{\infty}a_nz^n$ , $a_n\geq 0$ , be regular in D . Then $f\in \Sigma_{\mathcal{D}}^{\star}(\alpha,\beta)$ if and only if (2.1) is satisfied. Proof. In view of Theorem 1 it is sufficient to show the 'only if' part. Let us assume that $f(z)=\frac{1}{z}+\sum\limits_{n=1}^{\infty}a_{n}z^{n}$ , $a_{n}\geq0$ , is in $\sum\limits_{p}^{\star}(\alpha,\beta)$ . Then $$\left| \frac{\frac{zf'(z)}{f(z)} + 1}{\frac{zf'(z)}{f(z)} + 2\alpha - 1} \right| = \left| \frac{\sum_{n=1}^{\infty} (n+1) \, \alpha_n z^n}{2(1-\alpha)\frac{1}{z} - \sum_{n=1}^{\infty} (n+2\alpha - 1) \, \alpha_n z^n} \right| < \beta$$ for all $z \in \mathcal{D}$ . Using the fact that $\operatorname{Re}(z) \leq |z|$ for all z , it follows that (2.3) $$\operatorname{Re} \left\{ \frac{\sum_{n=1}^{\infty} (n+1) \, a_n z_n}{\sum_{n=1}^{\infty} (n+2\alpha-1)} \right\} < \beta , z \in U.$$ Now choose the values of z on the real axis so that $\frac{zf'(z)}{f(z)}$ is real. Upon clearing the denominator in (2.3) and letting $z \to 1$ through positive values, we obtain $$\sum_{n=1}^{\infty} (n+1) \quad \alpha_n \le \beta (2(1-\alpha) - \sum_{n=1}^{\infty} (n+2\alpha-1) \alpha_n)$$ or $$\sum_{n=1}^{\infty} [(1+\beta)n + (2\alpha-1)\beta+1] a_n \le 2\beta(1-\alpha) .$$ Hence the result follows. COROLLARY. If $$f(z) = \frac{1}{z} + \sum_{n=1}^{\infty} a_n z^n$$ , $a_n \ge 0$ , is in $\Sigma_p^*(\alpha, \beta)$ , then (2.4) $$a_n \leq \frac{2\beta(1-\alpha)}{(1+\beta)n + (2\alpha-1)\beta+1}, n = 1,2,...,$$ with equality for each n, for functions of the form (2.5) $$f_n(z) = \frac{1}{z} + \frac{2\beta(1-\alpha)}{(1+\beta)n + (2\alpha-1)\beta+1} z^n.$$ REMARK. The coefficient estimates for functions of the class $\Sigma_P^*(\alpha)$ determined in [7] follow from Corollary 1 by taking $\beta=1$ .

Theorem 3. If $f(z) \in \Sigma_{\mathcal{D}}^{\star}(\alpha,\beta)$ , then for 0 < |z| = r < 1, $$(3.1) \qquad \frac{1}{r} - \frac{\beta(1-\alpha)}{1+\alpha\beta} r < |f(z)| \le \frac{1}{r} + \frac{\beta(1-\alpha)}{1+\alpha\beta} r$$ where equality holds for the function (3.2) $$f_1(z) = \frac{1}{z} + \frac{\beta(1-\alpha)}{1+\alpha\beta} z$$ at $z = ir$ , $r$ . Proof. Suppose f(z) is in $\Sigma_p^\star(\alpha,\beta)$ . In view of Theorem 2, we have (3.3) $$\sum_{n=1}^{\infty} a_n \leq \frac{\beta(1-\alpha)}{1+\alpha\beta}.$$ Thus, for 0 < |z| = r < 1, $$|f(z)| = \left|\frac{1}{z} + \sum_{n=1}^{\infty} a_n z^n\right| \le \left|\frac{1}{z}\right| + \sum_{n=1}^{\infty} a_n |z|^n \le \frac{1}{r} + r \sum_{n=1}^{\infty} a_n$$ $$\le \frac{1}{r} + \frac{\beta(1-\alpha)}{1+\alpha\beta} r , \text{ by } (3.3) .$$ This gives the right hand inequality of (3.1). Also $$|f(z)| \ge \frac{1}{r} - \sum_{n=1}^{\infty} a_n r^n \ge \frac{1}{r} - r \sum_{n=1}^{\infty} a_n$$ $$\ge \frac{1}{r} - \frac{\beta(1-\alpha)}{1+\alpha\beta} r$$ which gives the left hand side of (3.1) . It can be easily seen that the function $f_1(z)$ defined by (3.2) is extremal for the theorem.

Theorem 4. If f(z) is in $\Sigma_P^*(\alpha,\beta)$ , then f(z) is meromorphically convex of order $\delta(0 \le \delta < 1)$ in $|z| < r = r(\alpha,\beta,\delta)$ , where $$r(\alpha,\beta,\delta) = \inf_{n} \left\{ \frac{(1-\delta)((1+\beta)n + (2\alpha-1)\beta+1)}{2\beta(1-\alpha)n(n+2-\delta)} \right\}^{\frac{1}{n+1}}, n = 1,2,...$$ The bound for |z| is sharp for each n, with the extremal function being of the form (2.5). Proof. Let $f(z) \in \Sigma_P^\star(\alpha,\beta)$ . Then, by Theorem 2 (3.5) $$\sum_{n=1}^{\infty} \frac{[(1+\beta)n + (2\alpha-1)\beta+1]}{2\beta(1-\alpha)} a_n \leq 1 .$$ In view of (1.3), it is sufficient to show that $$\left|2 + \frac{zf''(z)}{f'(z)}\right| \le 1 - \delta \quad \text{for} \quad |z| < r(\alpha, \beta, \delta)$$ or, equivalently, to show that 1 6 ⁸ M.L. Mogra, T.R. Reddy and O.P. Juneja (3.6) $$\left| \frac{f'(z) + (zf'(z))'}{f'(z)} \right| \leq 1 - \delta \quad \text{for} \quad |z| < r(\alpha, \beta, \delta) ,$$ where r(a,S,6) is as specified in the statement of the theorem. Substituting the series expansions for f (z) and {zf'(z))' in the left hand side of (3.6) we have $$\begin{vmatrix} \sum_{n=1}^{\infty} n(n+1) \alpha_n z^{n-1} \\ -\frac{1}{z^2} + \sum_{n=1}^{\infty} n \alpha_n z^{n-1} \end{vmatrix} \le \frac{\sum_{n=1}^{\infty} n(n+1) \alpha_n |z|^{n+1}}{1 - \sum_{n=1}^{\infty} n \alpha_n |z|^{n+1}}$$ Thi s wil l b e bounde d b y 1- 6 i f (3.7) $$\sum_{n=1}^{\infty} \frac{n(n-2-\delta)}{1-\delta} a_n |z|^{n+1} \le 1.$$ In view of (3.5), it follows that (3.7) is true if $$\frac{n(n+2-\delta)}{1-\delta} |z|^{n+1} \le \frac{(1+\beta)n + (2\alpha-1)\beta+1}{2\beta(1-\alpha)}, n = 1,2,...$$ (3.8) $$|z| \le \left\{ \frac{(1-\delta)((1+\beta)n + (2\alpha-1)\beta+1)}{2\beta(1-\alpha)n(n+2-\delta)} \right\}^{\frac{1}{n+1}}, n = 1,2,...$$ Settin g \z\ =r(a,B,6 ) in (3.8), the resul t follows. Sharpness can be verifie d easily .

Theorem 5. Let $$f_O(z) = \frac{1}{z}$$ and $$f_n(z) = \frac{1}{z} + \frac{2\beta(1-\alpha)}{(1+\beta)n + (2\alpha-1)\beta+1} z^n, n = 1,2,...$$ Then $f(z) \in \Sigma_{\mathcal{D}}^{\star}(\alpha,\beta)$ if and only if it can be expressed in the form $$f(z) = \sum_{n=0}^{\infty} \lambda_n f_n(z)$$ where $\lambda_n \ge 0$ and $\sum_{n=0}^{\infty} \lambda_n = 1$ . Proof. Let $$f(z) = \sum_{n=0}^{\infty} \lambda_n f_n(z)$$ with $\lambda_n \ge 0$ and $\sum_{n=0}^{\infty} \lambda_n = 1$ . Then $$\begin{split} f(z) &= \sum_{n=0}^{\infty} \lambda_n f_n(z) = \lambda_0 f_0(z) + \sum_{n=1}^{\infty} \lambda_n f_n(z) \\ &= \left(1 - \sum_{n=1}^{\infty} \lambda_n\right) f_0(z) + \sum_{n=1}^{\infty} \lambda_n f_n(z) \\ &= \left(1 - \sum_{n=1}^{\infty} \lambda_n\right) \frac{1}{z} + \sum_{n=1}^{\infty} \lambda_n \left(\frac{1}{z} + \frac{2\beta(1-\alpha)}{(1+\beta)n + (2\alpha-1)\beta+1} z^n\right) \\ &= \frac{1}{z} + \sum_{n=1}^{\infty} \lambda_n \frac{2\beta(1-\alpha)}{(1+\beta)n + (2\alpha-1)\beta+1} z^n \ . \end{split}$$ Since $$\sum_{n=1}^{\infty} \frac{(1+\beta)n + (2\alpha-1)\beta+1}{2\beta(1-\alpha)} \lambda_n \frac{2\beta(1-\alpha)}{(1+\beta)n + (2\alpha-1)\beta+1} = \sum_{n=1}^{\infty} \lambda_n = 1 - \lambda_0 \le 1 ,$$ by Theorem 2, $f(z) \in \Sigma_P^*(\alpha,\beta)$ . Conversely, suppose f(z) $\epsilon$ $\Sigma_P^{\star}(\alpha,\beta)$ . Since $$a_n \le \frac{2\beta(1-\alpha)}{(1+\beta)n + (2\alpha-1)\beta+1}$$ , $n = 1, 2, ...$ , setting $$\lambda_n = \frac{(1+\beta)n + (2\alpha-1)\beta+1}{2\beta(1-\alpha)} a_n$$ , $n = 1, 2, \ldots$ , and $\lambda_0 = 1 - \sum_{n=1}^{\infty} \lambda_n$ , it follows that $f(z) = \sum_{n=0}^{\infty} \lambda_n f_n(z)$ . This completes the proof of the theorem.

Theorem 6. The class E£(a,6) is closed under convex linear combinations. Proof. Suppose $$f^{(i)}(z) = \frac{1}{z} + \sum_{n=1}^{\infty} a_n^{(i)} z^n$$ , $(i = 1, 2)$ are in E\*(a,6) . Let ^P $$f(z) = (1-t)f^{(1)}(z) + tf^{(2)}(z), 0 \le t \le 1.$$ Then $$f(z) = \frac{1}{z} + \sum_{n=1}^{\infty} [(1-t) \ a_n^{(1)} + ta_n^{(2)}] z^n$$ In view of Theorem 2, we have $$\sum_{n=1}^{\infty} [(1+\beta)n + (2\alpha-1)\beta+1]((1-t)\alpha_n^{(1)} + t\alpha_n^{(2)})$$ $$= (1-t)\sum_{n=1}^{\infty} [(1+\beta)n + (2\alpha-1)\beta+1]\alpha_n^{(1)} + t\sum_{n=1}^{\infty} [(1+\beta)n + (2\alpha-1)\beta+1]\alpha_n^{(2)}$$ $$\leq (1-t)2\beta(1-\alpha) + t 2\beta(1-\alpha) = 2\beta(1-\alpha)$$ which shows that f e Z\*(a,3) . Hence the theorem.

Theorem 7. If f{z) is in E\*(a,6) , then the integral transforms $$F_c(z) = c \int_0^1 u^c f(uz) du , 0 < c < \infty ,$$ are in Z\*(6) , where (5.2) $$\delta = \delta(\alpha, \beta, c) = \frac{(1+\alpha\beta)(c+2) - \beta c(1-\alpha)}{\beta c(1-\alpha) + (1+\alpha\beta)(c+2)}.$$ result is best possible for the function f(z) = — + S(1~") s . 3 1+aB Proof. Suppose $f(z) = \frac{1}{z} + \sum_{n=1}^{\infty} a_n z^n \in \Sigma_P^*(\alpha, \beta)$ . We have $$F_c(z) = c \int_0^1 u^c f(uz) du = \frac{1}{z} + \sum_{n=1}^{\infty} \frac{c a_n}{n + c + 1} z^n$$ . It is sufficient to show that (5.3) $$\sum_{n=1}^{\infty} \frac{n+\delta}{1-\delta} \frac{ca_n}{n+c+1} \le 1.$$ Since $f(z) \in \Sigma_P^{\star}(\alpha,\beta)$ , we have (5.4) $$\sum_{n=1}^{\infty} \frac{(1+\beta)n + (2\alpha-1)\beta+1}{2\beta(1-\alpha)} a_n \le 1.$$ Thus (5.3) will be satisfied if $$\frac{(n+\delta)\,c}{(1-\delta)\,(n+c+1)} \leq \frac{(1+\beta)\,n\,+\,(2\alpha-1)\,\beta+1}{2\beta\,(1-\alpha)} \quad \text{for each} \quad n \ ,$$ or (5.5) $$\delta \leq \frac{\left((1+\beta)n + (2\alpha-1)\beta+1\right)(n+c+1) - 2\beta(1-\alpha)cn}{2\beta(1-\alpha)c + \left((1+\beta)n + (2\alpha-1)\beta+1\right)(n+c+1)}.$$ Since the right hand side of (5.5) is an increasing function of n, putting n=1 in (5.5) we get $$\delta \leq \frac{\left(1+\alpha\beta\right)\left(2+c\right) \; - \; \beta\left(1-\alpha\right)c}{\left(1+\alpha\beta\right)\left(2+c\right) \; + \; \beta\left(1-\alpha\right)c} \ .$$ Hence the theorem. REMARK. It is interesting to note that for c=1 and $(\alpha,\beta)=(0,1)$ , Theorem 7 gives that if $f\in \Sigma_P^(0,1)$ then $F_1(z)=\int_0^1 uf(uz)\,du \quad \text{is in} \quad \Sigma_P^(\frac{1}{2})\ .$

Theorem 8. If $$f(z) = \frac{1}{z} + \sum_{n=1}^{\infty} a_n z^n$$ and $g(z) = \frac{1}{z} + \sum_{n=1}^{\infty} b_n z^n$ are in $$\Sigma_{P}^{\star}(\alpha,\beta)$$ then $f(z) \star g(z) = \frac{1}{z} + \sum_{n=1}^{\infty} a_{n}b_{n}z^{n}$ is in $\Sigma_{P}^{\star}(\gamma,\beta)$ where (6.1) $$\gamma = \frac{(1+\alpha\beta)^2 - \beta(1-\alpha)^2}{(1+\alpha\beta)^2 + \beta^2(1-\alpha)^2}.$$ The result is best possible for the functions $$f(z) = g(z) = \frac{1}{z} + \frac{\beta(1-\alpha)}{1+\alpha\beta} z.$$ Proof. Suppose f(z) and g(z) are in $\Sigma_P^*(\alpha,\beta)$ . In view of Theorem 2, we have (6.2) $$\sum_{n=1}^{\infty} \frac{(1+\beta)n + (2\alpha-1)\beta+1}{2\beta(1-\alpha)} a_n \le 1$$ and $$\sum_{n=1}^{\infty} \frac{(1+\beta)n + (2\alpha-1)\beta+1}{2\beta(1-\alpha)} b_n \le 1.$$ Since f(z) and g(z) are regular in D , so is f(z)\*g(z) . Further, $$\sum_{n=1}^{\infty} \frac{(1+\beta)n + \beta \left[ 2 \frac{(1+\alpha\beta)^{2} - \beta(1-\alpha)^{2}}{(1+\alpha\beta)^{2} + \beta^{2}(1-\alpha)^{2}} - 1 \right] + 1}{2\beta \left[ 1 - \frac{(1+\alpha\beta)^{2} - \beta(1-\alpha)^{2}}{(1+\alpha\beta)^{2} + \beta^{2}(1-\alpha)^{2}} \right]} a_{n}b_{n}$$ $$\leq \sum_{n=1}^{\infty} \left[ \frac{(1+\beta)n + (2\alpha-1)\beta+1}{2\beta(1-\alpha)} \right]^2 a_n b_n$$ $$\leq \left(\sum_{n=1}^{\infty} \frac{(1+\beta)n + (2\alpha-1)\beta+1}{2\beta(1-\alpha)} a_n\right) \left(\sum_{n=1}^{\infty} \frac{(1+\beta)n + (2\alpha-1)\beta+1}{2\beta(1-\alpha)} b_n\right)$$ ≤ 1 in view of (6.2) and (6.3). Hence by Theorem 2, $f(z)g(z) \in \Sigma_P^(\alpha,\beta)$ where $\gamma$ is given by (6.1). This completes the proof of the theorem. Putting $\beta$ = 1 in the above theorem, we get the following result determined in [7]. COROLLARY. If $$f(z) = \frac{1}{z} + \sum_{n=1}^{\infty} a_n z^n$$ and $g(z) = \frac{1}{z} + \sum_{n=1}^{\infty} b_n z^n$ are elements of $\Sigma_P^(\alpha)$ , then $h(z) = f(z) g(z) = \frac{1}{z} + \sum_{n=1}^{\infty} a_n b_n z^n$ is an element of $\Sigma_P^*\left(\frac{2\alpha}{1+\alpha^2}\right)$ . The result is best possible. REMARK. If f(z) and g(z) are in $\Sigma_P^(0)$ , then, according to the corollary, $f(z)^g(z)$ is also in $\Sigma_P^*(0)$ ; the result is best possible. This is in sharp contrast with the corresponding result for starlike functions with negative coefficients (cf. Remark following Theorem 1 of Schild and Silverman [14]). Using similar arguments as in the proof of Theorem 8, we can obtain the following result.

Theorem 9. If $f(z)\in \Sigma_P^\star(\alpha,\beta)$ and $g(z)\in \Sigma_P^\star(\gamma,\beta)$ , then $f(z)*g(z)\in \Sigma_D^\star(\tau,\beta) \text{ where }$ $$\tau = \frac{(1+\alpha\beta)(1+\beta\gamma) - \beta(1-\alpha)(1-\gamma)}{(1+\alpha\beta)(1+\beta\gamma) + \beta^2(1-\alpha)(1-\gamma)}.$$ The result is best possible for $$f(z) = \frac{1}{z} + \frac{\beta(1-\alpha)}{1+\alpha\beta} z \quad \text{and} \quad g(z) = \frac{1}{z} + \frac{\beta(1-\gamma)}{1+\gamma\beta} z \ .$$

Theorem 10. If $$f(z)=\frac{1}{z}+\sum\limits_{n=1}^{\infty}a_{n}z^{n}\in\Sigma_{P}^{\star}(\alpha,\beta)$$ and $$g(z)=\frac{1}{z}+\sum\limits_{n=1}^{\infty}b_{b}z^{n} \text{ with } |b_{n}|\leq1\text{ , }n=1,2,\ldots\text{ , then }$$ $f(z) g(z) \in \Sigma (\alpha, \beta)$ . Proof. Since $$\sum_{n=1}^{\infty} \frac{(1+\beta)n + (2\alpha-1)\beta+1}{2\beta(1-\alpha)} |a_n b_n| = \sum_{n=1}^{\infty} \frac{(1+\beta)n + (2\alpha-1)\beta+1}{2\beta(1-\alpha)} |a_n| |b_n|$$ $$\leq \sum_{n=1}^{\infty} \frac{(1+\beta)n + (2\alpha-1)\beta+1}{2\beta(1-\alpha)} |a_n| \leq 1,$$ by Theorem 1 it follows that $f(z)g(z) \in \Sigma^(\alpha,\beta)$ . REMARK. g(z) need not be univalent in the above theorem. For example, if $g(z) = \frac{1}{z} - \frac{a}{a+b} z^2$ where 0 < b < a, then $\left| \frac{-a}{a+b} \right| < 1$ but $g'(z) = -\frac{1}{z^2} - \frac{2a}{a+b} z = 0$ for $z = -\left(\frac{a+b}{2a}\right)^{1/3}$ which lies inside D. Thus g(z) is not univalent in D .

Theorem 11. If f(z) and g(z) are in $\Sigma_P^{\star}(\alpha,\beta)$ and $1+3\alpha\beta-2\beta\geq 0$ , then $$F(z) = \frac{1}{z} + \sum_{n=1}^{\infty} (a_n^2 + b_n^2) z^n$$ also belongs to $\Sigma_{\mathcal{D}}^{\star}(\alpha,\beta)$ . Proof. Since $f(z) \in \Sigma_P^{\star}(\alpha,\beta)$ we have $$\sum_{n=1}^{\infty} \frac{(1+\beta)n + (2\alpha-1)\beta+1}{2\beta(1-\alpha)} a_n \le 1.$$ Therefore, $$\sum_{n=1}^{\infty} \left( \frac{(1+\beta)n + (2\alpha-1)\beta+1}{2\beta(1-\alpha)} \right) \alpha_n^2 \le 1.$$ Similarly Meromorphic univalent functions $$\sum_{n=1}^{\infty} \left[ \frac{(1+\beta)n + (2\alpha-1)\beta+1}{2\beta(1-\alpha)} \right]^2 b_n^2 \le 1.$$ Hence $$\sum_{n=1}^{\infty} \frac{1}{2} \left( \frac{(1+\beta)n + (2\alpha-1)\beta+1}{2\beta(1-\alpha)} \right)^2 (a_n^2 + b_n^2) \le 1.$$ In view of Theorem 2, it is sufficient to show that (6.4) $$\sum_{n=1}^{\infty} \left[ \frac{(1+\beta)n + (2\alpha-1)\beta+1}{2\beta(1-\alpha)} \right] (a_n^2 + b_n^2) \le 1 .$$ Thus (6.4) will be satisfied if, for n = 1, 2, ..., $$\frac{(1+\beta)n + (2\alpha-1)\beta+1}{2\beta(1-\alpha)} \leq \frac{1}{2} \left[ \frac{(1+\beta)n + (2\alpha-1)\beta+1}{2\beta(1-\alpha)} \right]^2$$ or if (6.5) $$(1+\beta)n + 6\alpha\beta - 5\beta + 1 \ge 0$$ for $n = 1, 2, ...$ The left hand side of (6.5) is an increasing function of n, hence (6.5) is satisfied for all n if $1+3\alpha\beta-2\beta\geq 0$ which is true by our assumption. Hence the result.