Bounds on the coefficients of certain analytic and univalent functions

Theorem 1. ([2]). Let $$f \in T_n^{\alpha}(\beta)$$ . Define $$B_{m} = \frac{2^{m}(1-\beta)^{m}\alpha^{m(n-1)}\prod_{j=0}^{m-1}(1-j\alpha)}{m!}$$ $$A_{k}(n,\alpha,\beta) = \sum_{m=1}^{k-1}B_{m}Q_{k-1}^{(m)}$$ $$A_{k-1}(n,\alpha,\beta) = \sum_{m=1}^{k-2}B_{m}Q_{k-1}^{(m)}$$ where for each $m = 1, 2, ..., Q_{k-1}^{(m)}$ is defined by the power series $$\left(\sum_{k=1}^{\infty} \frac{z^k}{(\alpha+k)^n}\right)^m = Q_m^{(m)} z^m + Q_{m+1}^{(m)} z^{m+1} + Q_{m+2}^{(m)} z^{m+2} + \dots$$ Also let $$\Omega_1 = \{\alpha | 0 < \alpha < (k-2)^{-1}, k = 2, 3, ...\},$$ $$\Omega_2 = \{\alpha | (k-2)^{-1} \le \alpha \le (k-3)^{-1}, k = 4, 6, ...\},$$ $$\Omega_3 = \{\alpha | (k-2)^{-1} \le \alpha < (k-3)^{-1}, k = 3, 5, ...\}$$ Then $$|a_k| \le \begin{cases} A_k & \text{if } \alpha \in \Omega_1 \cup \Omega_2, \\ A_{k-1} & \text{if } \alpha \in \Omega_3. \end{cases}$$ The inequalities are sharp. Equalities are attained for f(z) satisfying $$\frac{D^n f(z)^{\alpha}}{\alpha^n z^{\alpha}} = \begin{cases} \frac{1 + (1 - 2\beta)z}{1 - z} & \text{if } \alpha \in \Omega_1 \cup \Omega_2, \\ \frac{1 + (1 - 2\beta)z^2}{1 - z^2} & \text{if } \alpha \in \Omega_3. \end{cases}$$ The rough estimate was also given as:

Theorem 2. ([2]). Let $f \in T_n^{\alpha}(\beta), n \in N$ . Suppose $$\frac{f(z)^{\alpha}}{z^{\alpha}} = \sum_{k=0}^{\infty} A_{k+1}(\alpha)z^k, A_1(\alpha) = 1.$$ Then $$|A_{k+1}(\alpha)| < \exp\left\{0.624\alpha^2 + (2\alpha^2 - \frac{1}{2})\sum_{j=1}^k \frac{1}{j}\right\}.$$ From Theorem 1, which is best-possible, it is obvious that the problem has only been solved completely for $a_2$ and $a_3$ for all values of the index $\alpha$ . For $k \geq 4$ , the problem has remained open for all $\alpha > (k-3)^{-1}$ . In this article we proceed with the proof of the sharp bounds on the coefficients of functions in the class $T_n^{\alpha}(\beta)$ , $\alpha > 1$ . Our result is the following:

Theorem 3. Let $f \in T_n^{\alpha}(\beta)$ . If $\alpha > 1$ , we have the sharp inequalities (3) $$|a_k| \le \frac{2(1-\beta)\alpha^{n-1}}{(\alpha+k-1)^n}, \quad k=2, 3, 4, \dots$$ Equalities are attained for f(z) satisfying <span id="page-1-0"></span> $$\frac{D^n f(z)^{\alpha}}{\alpha^n z^{\alpha}} = \frac{1 + (1 - 2\beta)z^{k-1}}{1 - z^{k-1}}.$$ For k=2, 3, the above result is contained in Theorem 1 above. The proof here is however new. Also Singh [6] proved the same result for k=2, 3, 4 for the particular case n=1 and $\beta=0$ . In our proof we combine a method of classical analysis due to Nehari and Netanyau [3] (also used by Singh [6]) with an application of certain integral iteration of the Caratheodory functions [1]. That is presented in Section 3. In the next section we state and prove some preliminary lemmas.

Lemma 1. ([3]). If $p(z) = 1 + \sum_{k=1}^{\infty} b_k z^k$ and $q(z) = 1 + \sum_{k=1}^{\infty} c_k z^k$ belong to P, then $r(z) = 1 + \frac{1}{2} \sum_{k=1}^{\infty} b_k c_k z^k$ also belongs to P.

Lemma 2. ([3]). Let $h(z) = 1 + \sum_{k=1}^{\infty} d_k z^k$ and $1 + G(z) = 1 + \sum_{k=1}^{\infty} b'_k z^k$ be functions in P. Set (6) $$\gamma_m = \frac{1}{2^m} \left[ 1 + \frac{1}{2} \sum_{\mu=1}^m {m \choose \mu} d_\mu \right], \quad \gamma_0 = 1.$$ If $A_k$ is defined by <span id="page-2-0"></span> $$\sum_{m=1}^{\infty} (-1)^{m+1} \gamma_{m-1} G_1^m(z) = \sum_{k=1}^{\infty} A_k z^k,$$ then $$|A_k| < 2, \quad k = 1, 2, \dots$$ If, in the proof of the above lemma (as contained in [3]),we define $h_n(z)$ as the nth iterated integral transform of $h_0(z) = h(z)$ we immediately obtain the following corollary.

Corollary 1. Let $h_n(z)$ be the nth integral iteration of $h_0(z) = 1 + \sum_{k=1}^{\infty} d_k z^k$ with Re $h_n(z) > \beta$ , and $1 + G(z) = 1 + \sum_{k=1}^{\infty} b'_k z^k$ be functions in P. Define $\gamma_m$ as in (6) and <span id="page-3-5"></span>(7) $$\eta_m = \frac{(1-\beta)\alpha^n}{(\alpha+m)^n} \gamma_m, \quad \eta_0 = 1-\beta.$$ If $A_k$ is defined by <span id="page-3-3"></span>(8) $$\sum_{m=1}^{\infty} (-1)^{m+1} \eta_{m-1} G_1^m(z) = \sum_{k=1}^{\infty} A_k z^k,$$ then <span id="page-3-4"></span>(9) $$|A_k| \le \frac{2(1-\beta)\alpha^n}{(\alpha+k)^n}, \quad k=1, 2, \dots$$

Lemma 3. ([2]). Let $G(z) = \sum_{k=0}^{\infty} c_k z^k$ be a power series. Then the mth integer product of G(z) is $G^m(z) = \sum_{k=0}^{\infty} c_k^{(m)} z^k$ where $c_k^{(1)} = c_k$ and $c_k^{(m)} = \sum_{j=0}^{k} c_j c_{k-j}^{(m-1)}$ , $m \ge 2$ . We now turn to the proof of the main result.