ROTARU ALPHA - CONVEX FUNCTIONS

Lemma 2.1. Let $a \in C$ , $|a-1| < b \le Re(a)$ and p(z) be analytic in E with p(0) = 1. Then |p(z) - a| < b, $z \in E$ , if and only if, there exists a function w analytic in E satisfying w(0) = 0, |w(z)| < 1 for $z \in E$ such that $$p(z) = \frac{1+A \ w(z)}{1+\overline{B} \ w(z)}, \quad z \in E,$$ where $A = (b^2 - |a|^2 + a)/b$ and B = (1 - a)/b. The proof of this lemma follows by an application of Schwarz's lemma as in Rotaru [6]. From now on A and B will be as in lemma 2.1 and also $c = (b^2 - |a-1|^2)/b$ . Next we have the well known Jack's lemma [1].

Lemma 2.2. Let w(z) be regular in E with w(0) = 0. If there exists a $z_0 \in E$ such that $\max_{|z| \le |z_0|} |w(z)| = |w(z_0)|$ then $z_0 w'(z_0) = kw(z_0)$ for some $k \ge 1$ . It is well known that all $\alpha$ -convex functions are starlike [2]. We now prove an analogous theorem for the class $S^*(\alpha, a, b)$ .

Theorem 2.1. Let $f(z) \in S^(\alpha, a, b)$ , $\alpha > (4/c) |Im(a)|$ . Then $f(z) \in S^(a, b)$ . Proof. Define an analytic function w(z) in E by $$\frac{zf'(z)}{f(z)} = \frac{1 + A \ w(z)}{1 + \overline{B} \ w(z)}.$$ (2.1) Clearly w(0) = 0 and $w(z) \neq -1/\overline{B}$ in E. In view of lemma 2.1, it sufficies to show that |w(z)| < 1. Suppose there exists a $z_0 \in E$ such that $\max_{|z| \leq |z_0|} |w(z)| = |w(z_0)| = 1$ , then by lemma 2.2, $$z_0 w'(z_0) = k w(z_0), \quad k \ge 1.$$ (2.2) From (2.1) and (2.2) we get $$|J(\alpha, f(z_0)) - a| = b \Big| \frac{B + (1 + AB + (\alpha ck)/b)w(z_0) + Aw^2(z_0)}{1 + (A + \overline{B})w(z_0) + A\overline{B}w^2(z_0)} \Big|,$$ where $c = (b^2 - |a-1|^2)/b > 0$ . Now $|J(\alpha, f(z_0)) - a| > b$ , provided $$|B + (1 + AB + (\alpha ck)/b)w(z_0) + Aw^2(z_0)|^2 > |1 + (A + \overline{B})w(z_0) + A\overline{B}w^2(z_0)|^2.$$ (2.3) Condition (2.3) reduces to the following: $$(\alpha ck)/2b + 1 + Re(AB) + Re[(A + \overline{B})w(z_0)] > 0$$ or equivalently $$\frac{\alpha c(k-1)}{2b} + 1 + Re(A) Re(B) - Im(A) Im(B) + Re(A + \overline{B}) Re(w(z_0))$$ $$-Im(A + \overline{B}) Im(w(z_0)) + \frac{\alpha c}{2b} > 0.$$ (2.4) Since $\frac{\alpha c}{2b}(k-1) \ge 0$ and Im(A) Im(B) < 0, (2.4) holds provided $\frac{\alpha c}{2b} \pm Im(A+\overline{B}) > 0$ and 1 + Re(A) $Re(B) \pm Re(A+\overline{B}) > 0$ . Now, since $\alpha > (4/c)$ |Im(a)| we have $$\frac{\alpha c}{2b} \pm Im(A + \overline{B}) > 0$$ and $$1 + Re(A) Re(B) \pm Re(A + \overline{B}) = (1 \pm Re(A))(1 \pm Re(B)) \ge 0,$$ since $|A| \le 1$ and $|B| < 1.$ This means that $f(z_0) \in S^*(\alpha, a, b)$ , a contradiction. Thus the proof is complete.

Corollary 2.1. Let $f(z) \in S^(\alpha, a, b)$ , $\alpha > 0$ . Then $f(z) \in S^(a, b)$ . The above corollary follows by taking $a = \overline{a}$ in theorem 2.1.

Corollary 2.2. Let $\alpha > 0$ , b > 1/2 and $f(z) \in S^(\alpha, b, b)$ . Then $f(z) \in S^(b, b)$ . The above result is obtained if, in corollary 2.1, we take a = b, b > 1/2. Remark. Corollary 2.2 improves a result of Miller, Mocanu and Reade [4] who proved it when $b \ge 1$ . Let us choose $a = (1 + \rho - 2\rho\beta)/2(1 - \beta)$ and $b = (1 - \rho)/2(1 - \beta)$ , where $0 \le \rho < 1$ and $0 < \beta \le 1$ . Then the undermentioned corollary follows now from corollary 2.1.

Corollary 2.3. Let $\alpha > 0$ , $0 \le \rho < 1$ , $0 < \beta \le 1$ and $f \in V$ with $f(z)f'(z)/z \ne 0$ for z in E. Then $$\left| \frac{(zf'(z)/f(z)) - 1}{2\beta((zf'(z)/f(z)) - \rho) - ((zf'(z)/f(z)) - 1)} \right| < 1, \quad \text{for } z \text{ in } E.$$ whenever $$\left|\frac{J(\alpha,f(z))-1}{2\beta(J(\alpha,f(z))-\rho)-(J(\alpha,f(z))-1}\right|<1,\quad for\ z\ in\ E,$$ where $J(\alpha, f(z)) = (1-\alpha)(zf'(z)/f(z)) + \alpha(zf'(z))'/f'(z)$ and $\alpha$ a non-negative real number. Remark. For $\rho = (1 - \lambda)/(1 + \lambda)$ and $\beta = (1 + \lambda)/2$ , $0 < \lambda \le 1$ , corollary 2.3 agrees with theorem 1 of Padmanabhan and Bharati [5].

Theorem 2.2. For $(4/c) |Im(a)| < \beta < \alpha, S^(\alpha, a, b) \subset S^(\beta, a, b)$ . Proof. Let $f(z) \in S^*(\alpha, a, b)$ . Form theorem 2.1 and the identity $$J(\beta, f(z)) - a = \frac{\beta}{\alpha} [J(\alpha, f(z)) - a] + (1 - \frac{\beta}{\alpha}) [zf'(z)/f(z)) - a], \quad \beta < \alpha,$$ we arrive at the desired result $f(z) \in S^*(\beta, a, b)$ .

Corollary 2.4. For $\alpha \geq 1 > (4/c) |Im(a)|$ , $S^*(\alpha, a, b) \subset K(a, b)$ . 3. In this section, we obtain an important integral representation for the elements of $S^*(\alpha, a, b)$ .

Theorem 3.1. Let $f(z) \in S^(\alpha, a, b)$ , $\alpha > (4/c) |Im(a)|$ , and if for $(4/c) |Im(a)| < \beta < \alpha$ we choose the branch of $[zf'(z)/f(z)]^{\beta}$ which is equal to 1 when z = 0, then the function $F_{\beta}(z) = f(z) [zf'(z)/f(z)]^{\beta}$ is in $S^(a, b)$ . Proof. A simple calculation yields $$\frac{zF_{\beta}'(z)}{F_{\beta}(z)} = J(\beta, f(z)).$$ Since $f(z) \in S^*(\alpha, a, b)$ , by theorem 2.2, we have $$\left|\frac{zF_{\beta}'(z)}{F_{\beta}(z)} - a\right| = |J(\beta, f(z)) - a| < b,$$ for $(4/c) |Im(a)| < \beta < \alpha$ . Hence $F_{\beta}(z) \in S^*(a,b)$ . Now we consider the converse problem. Given the function $F(z) \in S^*(a,b)$ and $\alpha > (4/c) |Im(a)|$ is the solution $$f(z) = \left\{ \frac{1}{\alpha} \int_0^z F^{1/\alpha}(t) t^{-1} dt \right\}^{\alpha}, \tag{3.1}$$ of the differential equation $$F(z) = f(z)[zf'(z)/f(z)]^{\alpha},$$ (3.2) with boundary condition f(0) = 0, a function in $S^(\alpha, a, b)$ ? The answer is yes, and our solution provides us with an integral representation formula for functions is $S^(\alpha, a, b)$ .

Theorem 3.2. Let $F(z) \in S^(a,b)$ and $\alpha > (4/c) |Im(a)|$ . Then f(z) defined by (3.1) belongs to $S^(\alpha,a,b)$ . The proof of this theorem consists of showing that f(z) is well defined, regular in E and is in $S^*(\alpha, a, b)$ . The technique is similar to the one employed in [3, Theorem 5] and is omitted. 4. In this section we obtain an inequality for the coefficients of functions in $S^*(\alpha, a, b)$

Theorem 4.1. Let $f(z) \in S^*(\alpha, a, b)$ and let $s_1 = 0$ , $t_1 = 1 - (A/\overline{B})$ , $s_m = (1 - \alpha)(\beta_m - \alpha_m) + \alpha \gamma_{m-1}$ , $t_m = (\alpha - (A/\overline{B}))\alpha_m + (1 - \alpha)\beta_m + \alpha \gamma_{m-1}$ , $m = 2, 3, 4, \ldots$ , where $\alpha_m$ , $\beta_m$ and $\gamma_m$ are defined by $$\alpha_{m} = \sum_{k=1}^{m} (m - k + 1) a_{k} a_{m-k+1},$$ $$\beta_{m} = \sum_{k=1}^{m} k(m - k + 1) a_{k} a_{m-k+1},$$ $$\gamma_{m} = \sum_{k=1}^{m} k(k + 1) a_{k+1} a_{m-k+1}.$$ (4.1) Then the coefficients $a_n$ satisfy the following inequality: $$\sum_{m=1}^{n} |s_m|^2 \le |B|^2 \sum_{m=1}^{n} |t_m|^2, \quad n = 2, 3, 4, \dots$$ Proof. Since $f(z) \in S^*(\alpha, a, b)$ , $$(1 - \alpha) \frac{z f'(z)}{f(z)} + \alpha \left[ \frac{(z f'(z))'}{f'(z)} \right] = \frac{1 + A w(z)}{1 + \overline{B} w(z)},$$ where w is analytic in E, w(0) = 0 and |w(z)| < 1 in E. This gives $$(1 - \alpha)[z(f'(z))^{2} - f(z)f'(z)] + \alpha z f(z)f''(z)$$ $$= -\overline{B}w(z)[\alpha - \frac{A}{B})f(z)f'(z) + (1 - \alpha)z(f'(z))^{2} + \alpha z f(z)f''(z)].$$ (4.2) Given $f(z) = z + \sum_{n=2}^{\infty} a_n z^n$ , we note that $$f(z)f'(z) = \sum_{m=1}^{\infty} \alpha_m z^m, \ (f'(z))^2 = \sum_{m=1}^{\infty} \beta_m z^{m-1}, \ f(z)f''(z) = \sum_{m=1}^{\infty} \gamma_m z^m,$$ where $\alpha_m$ , $\beta_m$ and $\gamma_m$ are defined in (4.1). Thus (4.2) becomes $$(1-\alpha)\sum_{m=1}^{\infty}(\beta_m-\alpha_m)z^m+\alpha\sum_{m=1}^{\infty}\gamma_mz^{m+1}$$ $$=-\overline{B}w(z)\{(1-\alpha)\sum_{m=1}^{\infty}\beta_mz^m+(\alpha-\frac{A}{\overline{B}})\sum_{m=1}^{\infty}\alpha_mz^m+\alpha\sum_{m=1}^{\infty}\gamma_mz^{m+1}\}$$ which simplifies to $$\sum_{m=1}^{\infty} s_m z^m = -\overline{B}w(z) \left\{ \sum_{m=1}^{\infty} t_m z^m \right\}. \tag{4.3}$$ Now from (4.3) we get $$\Big| \sum_{m=1}^{n} s_m z_m + \sum_{m=n+1}^{\infty} h_m z^m \Big| < |B| \Big| \sum_{m=1}^{n-1} t_m z^m \Big|,$$ where $h_m$ 's are some complex numbers. This yields $$\sum_{m=1}^{n} |s_m|^2 + \sum_{m=n+1}^{\infty} |h_m|^2 \le |B|^2 \sum_{m=1}^{n-1} |t_m|^2$$ or $$\sum_{m=1}^{n} |s_m|^2 \le |B|^2 \sum_{m=1}^{n-1} |t_m|^2.$$ The above result is sharp for the function $$f_{\alpha}(z) = \left\{ \frac{1}{\alpha} \int_{0}^{z} t^{1/\alpha - 1} (1 + \frac{1 - \overline{a}}{b} t) \frac{c}{\alpha (1 - \overline{a})} dt \right\}^{\alpha} \quad \text{if} \quad a \neq 1$$ $$= \left\{ \frac{1}{\alpha} \int_{0}^{z} t^{1/\alpha - 1} e^{\frac{bt}{\alpha}} dt \right\}^{\alpha} \quad \text{if} \quad a = 1.$$