Coefficient Estimates for Inverses of Starlike Functions of Positive Order

Lemma 2. Let the function $f(z) = z + \sum_{n=2}^{\infty} a_n z^n$ be in the class $S^*(\alpha)$ and $a_{-n+\beta}^{(-n)}$ be given by (2.2). Then, for $\alpha \in I_k(n)$ , $$\left|a_{-n+\beta}^{(-n)}\right| \leq \begin{cases} \frac{\Gamma(2n(1-\alpha)+1)}{\Gamma(\beta+1)\Gamma(2n(1-\alpha)+1-\beta)} &, \beta=1,\dots,n-k\\ \frac{\Gamma(2n(1-\alpha)+1)}{\beta\Gamma(n-k)\Gamma(2n(1-\alpha)+1-(n-k))} &, \beta=n-k+1,\dots \end{cases}$$ (2.4) In particular, if $\alpha \in I_{n-1}(n)$ , then $$\left| a_{-n+\beta}^{(-n)} \right| \le \frac{2n(1-\alpha)}{g}, \quad \mathcal{G} = 1, 2, \cdots$$ $$(2.6)$$ The estimates (2.4) and (2.6) are sharp. Remark. By allowing k to vary from 0 to n-1, the estimates (2.4), (2.5) and (2.6) give estimates on the all the coefficients $\left|a_{-n+\theta}^{(-n)}\right|$ for a function $f \in S^*(\alpha)$ , $0 \le \alpha < 1$ . Proof . By a direct calculation, $\left(z(1/(f(z))^n)'\right)/\left(-n(1/(f(z))^n)\right) = z f'(z)/f(z)$ . Thus, $\left(z(1/(f(z)^n)'\right)/\left(-n(1/f(z)^n)\right) = \left(1+(1-2\alpha)w(z)\right)/\left(1-w(z)\right)$ , $z \in D$ , for a function $w(z) = \sum_{m=1}^{\infty} w_m z^m$ analytic in D and satisfying the conditions of Schwarz Lemma. Equivalently, $z\left(1/(f(z))^n\right)' + \left(n/(f(z))^n\right) = \left[z\left(1/(f(z))^n\right)' - \left(n(1-2\alpha)/(f(z))^n\right)\right]w(z)$ . The Substitution of the corresponding series expansions of the functions in this identity and a simplification gives $$\sum_{m=1}^{\infty} m \ a_{-n+m}^{(-n)} \ z^m = \left[ -2n(1-\alpha) + \sum_{m=1}^{\infty} \left\{ m - 2n(1-\alpha) \right\} \ a_{-n+m}^{(-n)} \ z^m \right] w(z)$$ (2.7) Equating coefficients on both sides of (2.7), it is observed that, for every $\mathcal{G}=1,2,\cdots$ , the coefficient $a_{-n+\mathcal{G}}^{(-n)}$ depends only on $a_{-n+1}^{(-n)},\,a_{-n+2}^{(-n)},\,\cdots,\,a_{-n+\mathcal{G}-1}^{(-n)}$ . Hence (2.7) can be rearranged as $\sum_{m=1}^{\mathcal{G}} m \, a_{-n+m}^{(-n)} \, z^m + \sum_{m=2+1}^{\infty} b_m z^m = \left[ -2n(1-\alpha) + \sum_{m=1}^{3-1} \left\{ m - 2n(1-\alpha) \right\} a_{-n+m}^{(-n)} \, z^m \right] \sum_{n=1}^{\infty} w_p \, z^p, \quad \mathcal{G}=1,2,3,\cdots$ the second sum in the left hand side being convergent in D. The inequality |w(z)| < 1 and Parseval's theorem give $$\sum_{m=1}^{g} m^2 \left| a_{-n+m}^{(-n)} \right|^2 \le 4n^2 (1-\alpha)^2 + \sum_{m=1}^{g-1} \left\{ m - 2n(1-\alpha) \right\}^2 \left| a_{-n+m}^{(-n)} \right|^2 .$$ Equivalently, $$\mathcal{G}^{2} \left| a_{-n+\mathcal{G}}^{(-n)} \right|^{2} \le 4n(1-\alpha) \left[ n(1-\alpha) + \sum_{m=1}^{\mathcal{G}-1} \left\{ n(1-\alpha) - m \right\} \left| a_{-n+m}^{(-n)} \right|^{2} \right]$$ (2.8) The sign of each term inside the summation symbol on the right hand side of (2.8) depends on the sign of the expression $n(1-\alpha)-m$ , $m=1,\dots, g-1$ . To determine the sign of this expression, we need to partition the interval $0 \le \alpha < 1$ into n semi-closed intervals $I_k(n)$ , $k=0,1,\dots, n-1$ . For any fixed k, if $\alpha \in I_k(n)$ , $k=0,1,\dots, n-1$ , then $n-k-1 < n(1-\alpha) \le n-k$ so that $n(1-\alpha)-m>0$ if $m=1,\dots, n-k-1$ and $n(1-\alpha)-m\le 0$ if $m=n-k,\dots$ . Considering only nonnegative contributions in the right hand summation in (2.8), it follows by using the above inequalities that, for $g=1,\dots, n-k$ , $$\mathcal{G}^{2} \left| a_{-n+\mathcal{G}}^{(-n)} \right|^{2} \leq 4n (1-\alpha) \left[ n (1-\alpha) + \sum_{m=1}^{9-1} \left\{ n (1-\alpha) - m \right\} \left| a_{-n+m}^{(-n)} \right|^{2} \right]$$ (2.9) while, if $\vartheta = n-k+1,....$ , then $$\mathcal{9}^{2}\left|a_{-n+\mathcal{9}}^{(-n)}\right|^{2} \leq 4n \ (1-\alpha) \left[n \ (1-\alpha) + \sum_{m=1}^{n-k-1} \left\{n \ (1-\alpha) - m\right\} \left|a_{-n+m}^{(-n)}\right|^{2}\right] + \sum_{m=n-k}^{9-1} \left\{n \ (1-\alpha) - m\right\} \left|a_{-n+m}^{(-n)}\right|^{2}$$ $$\leq 4n (1-\alpha) \left[ n (1-\alpha) + \sum_{m=1}^{n-k-1} \left\{ n (1-\alpha) - m \right\} \left| a_{-n+m}^{(-n)} \right|^2 \right]$$ (2.10) We now use induction on $\mathcal{G}$ . For $\mathcal{G}=1$ , it follows from (2.9) that $\left|a_{-n+1}^{(-n)}\right| \leq 2n \ (1-\alpha)$ , giving the estimate (2.4) in this case. Now let, for $\mathcal{G}=1,2,\cdots,n-k-1$ , the estimate $$\left| a_{-n+9}^{(-n)} \right| \le \prod_{j=0}^{g-1} \frac{2n \ (1-\alpha) - j}{j+1} \tag{2.11}$$ hold. Then, using (2.9), (2.11) and Lemma 1, it follows that, for $\theta = 1, \dots, n-k$ , $$\mathcal{G}^{2} \left| a_{-n+9}^{(-n)} \right|^{2} \leq 4n \ (1-\alpha) \left[ n \ (1-\alpha) + \sum_{m=1}^{g-1} \left\{ n \ (1-\alpha) - m \right\} \left( \prod_{j=0}^{m-1} \frac{2n \ (1-\alpha) - j}{j+1} \right)^{2} \right] \\ = \frac{1}{\left( (g-1)! \right)^{2}} \prod_{j=0}^{g-1} \left( 2n \ (1-\alpha) - j \right)^{2} \tag{2.12}$$ Thus, for $\theta = 1, \dots, n-k$ , Coefficient estimates for inverses of starlike functions $$\left| a_{-n+\theta}^{(-n)} \right| \le \prod_{j=0}^{g-1} \frac{2n \ (1-\alpha) - j}{j+1} = \frac{\Gamma\left(2n \ (1-\alpha) + 1\right)}{\Gamma(g+1) \ \Gamma\left(2n(1-\alpha) + 1 - g\right)}$$ (2.13) This establishes the inequality (2.4). Next, if $\theta = n - k + 1, n - k + 2, \dots$ , using (2.10), the induction hypothesis (2.11) and Lemma 1, we get $$\begin{aligned} \mathcal{G}^{2} \left| a_{-n+\mathcal{G}}^{(-n)} \right|^{2} &\leq 4n \ (1-\alpha) \left[ n \ (1-\alpha) + \sum_{m=1}^{n-k-1} \left\{ n \ (1-\alpha) - m \right\} \left( \prod_{j=0}^{m-1} \frac{2n \ (1-\alpha) - j}{j+1} \right)^{2} \right] \\ &= \frac{1}{\left( (n-k-1)! \right)^{2}} \prod_{j=0}^{n-(k+1)} \left( 2n(1-\alpha) - j \right)^{2} = \left( \frac{\Gamma\left( 2n \ (1-\alpha) + 1 \right)}{\Gamma(n-k) \Gamma\left( 2n \ (1-\alpha) + 1 - (n-k) \right)} \right)^{2} \end{aligned}$$ The above inequality yields the estimate (2.5). For k = n - 1, the estimates (2.4) and (2.5) respectively reduce to $\left| a_{-n+1}^{(-n)} \right| \le 2n \ (1 - \alpha)$ and $\left| a_{-n+\theta}^{(-n)} \right| \le \frac{2n \ (1 - \alpha)}{9}$ for $\theta = 2, 3, \dots$ . Combining the above inequalities, (2.6) follows. Equality holds in (2.4) for every $g = 1, \dots, n-k$ for $(-n)^{th}$ power of the function $K_{\alpha}(z)$ defined in (1.10). On the other hand for each $g = n-k+1, \dots, (-n)^{th}$ power of the function $K_{\alpha,\beta}(z)$ , defined in (1.10), provides the sharpness for the estimate (2.6). This completes the proof of Lemma 2. We also need in the sequel the following result of Jabotinsky [8]:

Lemma 3. If the function f, given by (2.1), is in $\Omega$ then $f^{-1} \in \Omega$ . Further, if $$f^{-1}(w) = \sum_{n=1}^{\infty} A_n w^n$$ then, $$A_n^{(p)} = \frac{p}{n} a_{-p}^{(-n)}, \quad n = 1, 2, \dots; \quad p = \pm 1, \pm 2, \dots$$ (2.14) and $A_0^{(p)}$ is defined by $$\sum_{p=-\infty}^{\infty} A_0^{(p)} z^{-p-1} = \frac{f'(z)}{f(z)}$$ (2.15)

Theorem 1. Let $f \in S^*(\alpha)$ , $0 \le \alpha < 1$ and, for $|w| < \frac{1}{4}$ , $$f^{-1}(w) = w + \sum_{n=2}^{\infty} A_n w^n$$ (3.1) Then, (a) $for \alpha \in I_0(n) \bigcup I_1(n)$ , $$\left| A_n \right| \le \frac{\Gamma\left(2n\left(1-\alpha\right)+1\right)}{\Gamma(n+1)\,\Gamma\left(2n\left(1-\alpha\right)+2-n\right)} \tag{3.2}$$ (b) for $\alpha \in I_k(n)$ , $k = 2, \dots, n-2$ , $$\left| A_n \right| \le \frac{\Gamma\left(2n\left(1-\alpha\right)+1\right)}{n(n-1)\,\Gamma(n-k)\,\Gamma\left(2n\left(1-\alpha\right)+1+k-n\right)} \tag{3.3}$$ (c) for $\alpha \in I_{n-1}$ , $$\left| A_n \right| \le \frac{2(1-\alpha)}{n-1} \tag{3.4}$$ where, $I_k(n) \equiv [\frac{k}{n}, \frac{k+1}{n})$ , k = 0, 1, ..., n-1. The estimates (3.2) and (3.4) are sharp. Proof. It is known (see e.g. [12]) that $A_n = (1/2\pi i n) \int_{|z|=r}^{\infty} (1/(f(z))^n) dz = (1/n) a_{-1}^{(-n)}, 0 < r < 1.$ Therefore, it is sufficient to find suitable estimates for $\left|a_{-1}^{(-n)}\right|$ . To this end, taking $\mathcal{G}=n-1$ in Lemma 2, using the appropriate inequality (2.4), (2.5) or (2.6) for different values of k=0,1,...,n-1 and observing that only for k=0 and k=1 the inequality (2.4) is applicable, the following estimate is obtained for $\alpha \in [0,\frac{2}{n})$ , $$\left| a_{-1}^{(-n)} \right| \le \frac{\Gamma\left(2n(1-\alpha)+1\right)}{\Gamma(n)\,\Gamma\left(2n(1-\alpha)+2-n\right)} \tag{3.5}$$ which gives (a). Similarly, for $\alpha \in I_k(n)$ , k = 2, ..., n-2, the inequality (2.5) yields $$\left| a_{-1}^{(-n)} \right| \le \frac{\Gamma\left(2n\left(1-\alpha\right)+1\right)}{(n-1)\Gamma(n-k)\Gamma\left(2n\left(1-\alpha\right)+1+k-n\right)} . \tag{3.6}$$ This gives (b). Finally, for $\alpha \in I_{n-1}(n)$ , the inequality (2.6) gives $$\left| a_{-1}^{(-n)} \right| \le \frac{2n \ (1-\alpha)}{n-1}.$$ (3.7) Consequently, (c) follows. It is easily verified that equality holds in (3.5) and (3.7) for the $(-n)^{th}$ power of the function $K_n(z)$ and $(-n)^{th}$ power of the function $K_{\alpha,n-1}(z)$ respectively. Thus, the estimates (3.2) and (3.4) are sharp. This completes the proof of Theorem 1. Remark. The sharp coefficient bounds of Krzysz, Libera and Zlotkiewicz [12] for $|A_2|$ and $|A_3|$ follow as a particular case of Theorem 1. Remark: The sharp coefficient bounds of Krzyz, Libera and Zlotkiewicz [12] for $|A_2|$ and $|A_3|$ follow as a particular case of Theorem 1. The sharp coefficient estimates for functions in $\Sigma^*(\alpha)$ , $0 \le \alpha < 1$ , are described by the following:

Theorem 2. Let the function $g \in \Sigma^*(\alpha)$ , $0 \le \alpha < 1$ , be given by the series $$g(z) = z + b_0 + \frac{b_1}{z} + \frac{b_2}{z^2} + \cdots, \quad z \in V.$$ Then, $$|b_m| \le \frac{2(1-\alpha)}{m+1}, \quad m = 0,1...$$ (3.8) The estimate (3.8) is sharp. Proof. The mapping $f(z) \to g(z) = 1/f(1/z)$ establishes a one-to-one correspondence between $S^(\alpha)$ and $\Sigma^(\alpha)$ . Since, (zg'(z)/g(z)) = (z(1/f(1/z))')/(1/f(1/z)) = ((1/z)f'(1/z))/f(1/z), this mapping too is one-to-one from $S^(\alpha)$ onto $\Sigma^(\alpha)$ . We note that the coefficient expansion of 1/f(z) around origin is same as the coefficient expansion of g(z) about infinity. Therefore, $$\max_{g \in \Sigma^(\alpha)} |b_m| = \max_{f \in S^(\alpha)} |a_m^{(-1)}|, \quad m = 0, 1...$$ (3.9) Thus, by Lemma 2, $$\left| a_{-1+\beta}^{(-1)} \right| \le \frac{2(1-\alpha)}{\beta}, \ 0 \le \alpha < 1; \ \beta = 1,...$$ (3.10) The inequality (3.10) can be equivalently expressed as $$\left| a_m^{(-1)} \right| \le \frac{2(1-\alpha)}{m+1}, \quad 0 \le \alpha < 1; \ m = 0, 1, \dots$$ (3.11) and the inequality (3.11) together with (3.9) gives (3. 8). It is easily seen that the function $g(z) = z(1-(1/z^{m+1}))^{2(1-\alpha)/(m+1)}$ belongs to the class $\Sigma^*(\alpha)$ and its $m^{th}$ coefficient equals $2(1-\alpha)/(m+1)$ . Therefore, the estimate (3.8) is sharp. This completes the proof of Theorem 2. The coefficient estimates for the inverse of a function in the class $\Sigma^*(\alpha)$ are found in the following:

Theorem 3. Let the function $g \in \Sigma^*(\alpha)$ , $0 \le \alpha < 1$ , have the series expansion $g^{-1}(w) = w + \sum_{n=0}^{\infty} B_n w^{-n}$ in some neighbourhood of infinity. Then, (a) $$|B_0| \le 2(1-\alpha), \quad 0 \le \alpha < 1$$ (3.12) (b) For $\alpha \in I_k(n)$ , $k = 0, \dots, n-2$ , $$\left|B_{n}\right| \leq \frac{\Gamma\left(2n\left(1-\alpha\right)+1\right)}{n(n+1)\,\Gamma(n-k)\,\Gamma\left(2n\left(1-\alpha\right)+1-(n-k)\right)}\tag{3.13}$$ (c) $For \alpha \in I_{n-1}(n)$ , $$\left|B_n\right| \le \frac{2(1-\alpha)}{n+1} \tag{3.14}$$ where, $I_k(n) = \left[\frac{k}{n}, \frac{k+1}{n}\right]$ , k = 0, 1, ..., n-1. The estimates (3.12) and (3.14) are sharp. Proof. For any $g \in \Sigma^(\alpha)$ , $0 \le \alpha < 1$ , there exists $f \in S^(\alpha)$ such that g(z) = 1/f(1/z). It can be easily verified that $g^{-1}(w) = 1/f^{-1}(1/w)$ . Since the coefficients in the expansion of $1/f^{-1}(w)$ around origin and those of $1/f^{-1}(1/w)$ about infinity are the same, we have $$B_n = A_n^{(-1)}, \quad n = 0, 1, 2, \cdots.$$ (3.15) By (2.15), with $\operatorname{Re} Q(z) > \alpha$ , Q(0) = 1 and $z \in D$ , $$\sum_{p=-\infty}^{\infty} A_0^{(p)} z^{-p-1} = \frac{f'(z)}{f(z)} = \frac{1}{z} Q(z) = \frac{1}{z} [1 + \sum_{n=1}^{\infty} q_n z^n]$$ Now, using the well known sharp estimate $$|q_n| \le 2 (1-\alpha), \quad n = 1, 2, \cdots$$ (3.16) we get, $$\max_{g^{-1} \in \Sigma^{-1}(\alpha)} \left| B_0 \right| = \max_{f^{-1} \in S^{-1}(\alpha)} \left| A_0^{-1} \right| = Max \left| q_1 \right| \le 2 (1 - \alpha),$$ This establishes (3.12). Since the bound in (3.16) is sharp, it follows that the estimate (3.12) is also sharp. For n=1, 2, ..., (3.15) together with (2.14) gives $$\max_{g^{-1} \in \Sigma^{-1}(\alpha)} |B_n| = \max_{f^{-1} \in S^{-1}(\alpha)} |A_n^{(-1)}| = (1/n) \max_{f^{-1} \in S^{*}(\alpha)} |a_1^{(-n)}|$$ (3.17) It is easily verified that $a_1^{(-n)} = a_{-n+(n+1)}^{(-n)}$ . Since the sharp estimate (2.4) in Lemma 2 is not applicable for any value of $k = 0, \dots, n-1$ , in order to get (3.13), the estimate (2.5) with g = n+1 has to be used for $k = 0, 1, \dots, n-2$ . This gives $$\left| a_{1}^{(-n)} \right| \le \frac{\Gamma(2n(1-\alpha)+1)}{(n+1)\Gamma(n-k)\Gamma(2n(1-\alpha)+1-(n-k))}$$ (3.18) The estimate (3.13) now easily follows from (3.17) and (3.18). Similarly, the estimate (2.6) with $\theta = n+1$ gives, $$\left| a_{1}^{(-n)} \right| \le \frac{2n \ (1-\alpha)}{n+1}$$ (3.19) By combining (3.17) and (3.19), the estimate (3.14) follows. Since the inequality (2.6) is sharp, it follows that the estimate (3.14) is sharp. This completes the proof of Theorem 3. Remark. The construction of a suitable example to exhibit the sharpness of inequality (2.5) seems to be quite involved and the sharpness of estimates (3.3) and (3.13) depend on the sharpness of the inequality (2.5).