Generalized Zalcman conjecture for convex functions of order $α$

Lemma 1. Let $-\frac{1}{2} \leq \alpha < 1$ and $f \in \mathcal{F}(\alpha)$ as in the form (1). Then we have (6) $$f'(z) = \int_0^{2\pi} (1 - e^{i\theta}z)^{2\alpha - 2} d\nu(\theta),$$ where $\nu(\theta)$ is a probability measure on $[0, 2\pi]$ and for $n \geq 2$ , (7) $$|\lambda a_n^2 - a_{2n-1}| \le (\lambda A_n^2 - 2A_{2n-1}) \int_0^{2\pi} \cos^2(n-1)\theta \, d\nu(\theta) + A_{2n-1},$$ where $A_n := A_n(\alpha)$ is given by (3). Proof. Let $f \in \mathcal{F}(\alpha)$ for some $\alpha \in [-1/2, 1)$ . By the well-known Herglotz representation theorem for analytic functions p with positive real part in $\mathbb{D}$ , p(0) = 1, and the analytic characterization of $f \in \mathcal{F}(\alpha)$ given by (2), one has $$\frac{1}{1-\alpha} \left( 1 + \frac{zf''(z)}{f'(z)} - \alpha \right) = p(z) := \int_0^{2\pi} \frac{1 + e^{i\theta}z}{1 - e^{i\theta}z} d\nu(\theta), \quad |z| < 1,$$ where $\nu(\theta)$ is a probability measure on $[0, 2\pi]$ . By a computation, we easily have $$f'(z) = \int_0^{2\pi} (1 - e^{i\theta}z)^{2\alpha - 2} d\nu(\theta)$$ $$= 1 + \sum_{n=2}^{\infty} (-1)^{n-1} \frac{(2\alpha - 2)(2\alpha - 3)\cdots(2\alpha - n)}{(n-1)!} \left( \int_0^{2\pi} e^{i(n-1)\theta} d\nu(\theta) \right) z^{n-1}.$$ By comparing the coefficients of $z^{n-1}$ on both sides of the above equation, we easily have (8) $$a_n = A_n \int_0^{2\pi} e^{i(n-1)\theta} d\nu(\theta) \text{ for } n = 2, 3, ...,$$ where $A_n = A_n(\alpha)$ is given by (3), i.e. $$A_n = \frac{\Gamma(n+1-2\alpha)}{n!\Gamma(2-2\alpha)}, \quad n \ge 2.$$ We observe that the last relation quickly gives the necessary coefficient inequality (3) for functions in $\mathcal{F}(\alpha)$ . Since $|\lambda a_n^2 - a_{2n-1}|$ is invariant under rotations, we can consider instead the problem of maximizing the functional Re $(\lambda a_n^2 - a_{2n-1})$ . Consequently, by (8), we begin to observe that $$\operatorname{Re} \left( \lambda a_n^2 - a_{2n-1} \right) = \lambda A_n^2 \left( \int_0^{2\pi} \cos(n-1)\theta \, d\nu(\theta) \right)^2 - \lambda A_n^2 \left( \int_0^{2\pi} \sin(n-1)\theta \, d\nu(\theta) \right)^2 - A_{2n-1} \int_0^{2\pi} \cos 2(n-1)\theta \, d\nu(\theta).$$ If we apply Cauchy-Schwarz inequality to the first integral above and use the trigonometric identity $\cos 2t = 2\cos^2 t - 1$ in the third integral, we find that $$\operatorname{Re}\left(\lambda a_{n}^{2} - a_{2n-1}\right) \leq \lambda A_{n}^{2} \int_{0}^{2\pi} \cos^{2}(n-1)\theta \, d\nu(\theta) - 2A_{2n-1} \int_{0}^{2\pi} \cos^{2}(n-1)\theta \, d\nu(\theta) + A_{2n-1}$$ $$= (\lambda A_{n}^{2} - 2A_{2n-1}) \int_{0}^{2\pi} \cos^{2}(n-1)\theta \, d\nu(\theta) + A_{2n-1}.$$ The proof is completed. Here is an alternate approach to Lemma 1 which works for a more general setting. If X is a linear topological space, then a subset Y of X is called convex if $tx + (1-t)y \in Y$ whenever $x, y \in Y$ and $0 \le t \le 1$ . The closed convex hull of Y is defined as the intersection of all closed convex sets containing Y. A point $u \in Y$ is called an extremal point of Y if u = tx + (1-t)y, 0 < t < 1 and $x, y \in Y$ , implies that x = y. See [10, 18] for a general reference and for many important results on this topic. In order to solve the generalized Zalcman coefficient inequality problem for the class $\mathcal{F}(\alpha)$ , we need the following lemma. Lemma B. ([10]) Suppose that $F_{\alpha}(z) = (1-z)^{2\alpha-2}$ and $\alpha \in [-1/2, 1/2]$ . If $s(F_{\alpha})$ , $\mathcal{H}s(F_{\alpha})$ and $\mathcal{E}\mathcal{H}s(F_{\alpha})$ denote the set of analytic functions subordinate to $F_{\alpha}$ , the closed convex hull of $s(F_{\alpha})$ and the set of the extremal points of $\mathcal{H}s(F_{\alpha})$ , respectively, then $\mathcal{H}s(F_{\alpha})$ consists of all analytic functions represented by (9) $$F_{\alpha}(z) = \int_{|x|=1} (1 - xz)^{2\alpha - 2} d\mu(x),$$ where $\mu(x)$ is a probability measure on the unit circle $\partial \mathbb{D}$ . Moreover, $\mathcal{EH}s(F_{\alpha})$ consists of the functions given by (10) $$F_{\alpha}(z) = (1 - xz)^{2\alpha - 2},$$ where |x| = 1. If $\mathcal{F} \subset \mathcal{H}$ is convex and $L: \mathcal{H} \to \mathbb{R}$ is a real-valued functional on $\mathcal{A}$ , then we say that L is convex on $\mathcal{F}$ provided that $$L(tg_1 + (1-t)g_2) \le tL(g_1) + (1-t)L(g_2)$$ whenever $g_1, g_2 \in \mathcal{F}$ and $0 \le t \le 1$ . Since $\mathcal{H}s(F_{\alpha})$ is convex, we have a real-valued, continuous and convex functional on $\mathcal{H}s(F_{\alpha})$ .

Lemma 2. Suppose that $g(z) = 1 + \sum_{n=2}^{\infty} b_n z^{n-1}$ is analytic in $\mathbb{D}$ and $$J(g) = \lambda \frac{(\operatorname{Re} b_n)^2}{n^2} - \frac{\operatorname{Re} b_{2n-1}}{2n-1},$$ where $\lambda > 0$ . Then J is a real-valued, continuous and convex functional on $\mathcal{H}s(F_{\alpha})$ .

Lemma 3. We have $$\max\{J(f): f \in \mathcal{H}s(F_{\alpha})\} = \max\{J(f): f \in s(F_{\alpha})\} = \max\{J(f): f \in \mathcal{EH}s(F_{\alpha})\}.$$

Lemma 4. Let $-\frac{1}{2} \le \alpha < \frac{1}{2}$ , $\lambda > 0$ and $f \in \mathcal{F}(\alpha)$ be as in the form (1). Then f has the form (6) and for $n \ge 2$ , we have $$|\lambda a_n^2 - a_{2n-1}| \le (\lambda A_n^2 - 2A_{2n-1})\cos^2(n-1)\theta + A_{2n-1},$$ where $A_n$ is given by (3).

Lemma 5. For $-\frac{1}{2} \le \alpha < 1$ and $n \ge 3$ , we define (13) $$C_n(\alpha) = \frac{2A_{2n-1}(\alpha)}{A_n^2(\alpha)},$$ where $A_n = A_n(\alpha)$ is given by (3). Then for fixed $\alpha$ , we have - (1) $C_n(\alpha)$ is monotonically decreasing with respect to n and $C_n(\alpha) \leq C_3(\alpha)$ if $-\frac{1}{2} \leq \alpha < 0$ : - (2) $C_n(\alpha)$ is monotonically increasing with respect to n and $C_n(\alpha) \ge C_3(\alpha)$ if $0 < \alpha < 1$ , where $$C_3(\alpha) = \frac{3}{5} \frac{(2\alpha - 4)(2\alpha - 5)}{(2\alpha - 2)(2\alpha - 3)}.$$

Theorem 1. For $\mathcal{F}(0) := \mathcal{K}$ , let $f \in \overline{Co}(\mathcal{K})$ as in the form (1). Then we have - (1) $|\lambda a_n^2 a_{2n-1}| \le \lambda 1$ for $n \ge 3$ and $\lambda \ge 2$ . The equality is attained for the function $f_0(z) = \frac{z}{1-z}$ . - (2) $|\lambda a_n^2 a_{2n-1}| \le 1$ for $n \ge 3$ and $0 < \lambda < 2$ . The equality is attained by convex combination of convex functions in K, namely, for the functions f in the form $$f(z) = \sum_{k=0}^{2n-3} \alpha_k \frac{z}{1 - e^{i\theta_k} z},$$ where $0 \le \alpha_k \le 1$ , $$\theta_k = \frac{(2k+1)\pi}{2n-2}$$ and $\sum_{m=0}^{n-2} \alpha_{2m} = \sum_{m=0}^{n-2} \alpha_{2m+1} = \frac{1}{2}$ .

Corollary 1. ([14, Theorem 3.3]) Let $f \in \mathcal{F}(-1/2)$ and f be of the form (1). Then we have (1) $|\lambda a_n^2 - a_{2n-1}| \le \frac{(n+1)^2}{4} \lambda - n$ for $n \ge 3$ and $\lambda \ge \frac{3}{2}$ . The equality is attained for the function $f_{-1/2}(z)$ given by (14) $$f_{-1/2}(z) = \frac{z - z^2/2}{(1-z)^2} = z + \sum_{n=2}^{\infty} \frac{1+n}{2} z^n.$$ - (2) $|\lambda a_n^2 a_{2n-1}| \le \frac{(n+1)^2}{4}\lambda n$ for $0 < \lambda < \frac{3}{2}$ and $n > \frac{4-\lambda+2\sqrt{4-2\lambda}}{\lambda}$ . The equality is attained for the function $f_{-1/2}(z)$ given by (14). - (3) $|\lambda a_n^2 a_{2n-1}| \le n$ for $0 < \lambda < \frac{3}{2}$ and $3 \le n \le \frac{4-\lambda+2\sqrt{4-2\lambda}}{\lambda}$ . The equality is attained for functions f in the following form $$f(z) = \sum_{k=0}^{2n-3} \alpha_k \frac{2z - e^{i\theta_k} z^2}{2(1 - e^{i\theta_k} z)^2} = \sum_{k=0}^{2n-3} \alpha_k e^{-i\theta_k} f_{-1/2}(ze^{i\theta_k}),$$ where $0 \le \alpha_k \le 1$ , $$\theta_k = \frac{(2k+1)\pi}{2n-2}$$ and $\sum_{m=0}^{n-2} \alpha_{2m} = \sum_{m=0}^{n-2} \alpha_{2m+1} = \frac{1}{2}$ . Proof. Set $\alpha = -1/2$ in Lemma 1 or Lemma 4, or apply Lemma 5 directly. Then, because $A_n(-1/2) = (n+1)/2$ for all $n \ge 2$ , it is clear from (7) or (11) that $$|\lambda a_n^2 - a_{2n-1}| \le \begin{cases} \left(\lambda \frac{(n+1)^2}{4} - 2n\right) + n & \text{if } \lambda \ge \frac{8n}{(n+1)^2} \\ n & \text{if } 0 < \lambda \le \frac{8n}{(n+1)^2}. \end{cases}$$ Clearly, $\frac{8n}{(n+1)^2} \leq \frac{3}{2}$ if and only if $(3n-1)(n-3) \geq 0$ and thus, the Case (1) follows. For $0 < \lambda < \frac{3}{2}$ , we see that $\lambda \geq \frac{8n}{(n+1)^2}$ if and only if $\varphi(n) := \lambda n^2 - 2n(4-\lambda) + \lambda \geq 0$ . Since $$\varphi(n) = \lambda \left[ n - \left( \frac{4 - \lambda + 2\sqrt{4 - 2\lambda}}{\lambda} \right) \right] \left[ n - \left( \frac{4 - \lambda - 2\sqrt{4 - 2\lambda}}{\lambda} \right) \right],$$ Case (2) follows. Finally, in the last case the range of n shows that $\lambda < \frac{8n}{(n+1)^2}$ and thus, Case (3) is clear. Setting $\lambda = 1$ and $\frac{3}{2}$ in Corollary 1, we obtain the following.

Corollary 2. Let $f \in \mathcal{F}(-1/2)$ as in the form (1). Then we have - (1) $|a_n^2 a_{2n-1}| \le \frac{(n-1)^2}{4}$ for n > 5. The equality is attained for the function $f_{-1/2}(z)$ given by (14). - (2) $|a_n^2 a_{2n-1}| \le n \text{ for } 3 \le n \le 5.$ - (3) $\left|\frac{3}{2}a_n^2 a_{2n-1}\right| \le \frac{3n^2 2n + 3}{8}$ for n > 3. The equality is attained for the function $f_{-1/2}(z)$ given by (14). - $(4) \left| \frac{3}{2}a_3^2 a_5 \right| \le 3.$

Theorem 2. Let $n \geq 3$ , $-\frac{1}{2} \leq \alpha < 0$ , $f \in \mathcal{F}(\alpha)$ as in the form (1), $A_n = A_n(\alpha)$ and $C_n = C_n(\alpha)$ be given by (3) and (13), respectively. (1) If $n \geq 3$ and $\lambda \geq C_3(\alpha)$ , then we have $$|\lambda a_n^2 - a_{2n-1}| \le \lambda A_n^2 - A_{2n-1},$$ where the equality is attained for the function $f_{\alpha}(z)$ defined by (4). (2) If $0 < \lambda < C_3(\alpha)$ , then there exists a fixed $n_0 > 3$ such that $$C_{n_0-1}(\alpha) > \lambda \ge C_{n_0}(\alpha).$$ If $0 < \lambda < C_3(\alpha)$ and $n \ge n_0$ , then $$|\lambda a_n^2 - a_{2n-1}| \le \lambda A_n^2 - A_{2n-1},$$ where the equality is attained for the function $f_{\alpha}(z)$ defined by (4). (3) If $0 < \lambda < C_3(\alpha)$ and $3 \le n < n_0$ , then $$|\lambda a_n^2 - a_{2n-1}| \le A_{2n-1},$$ where the equality is attained by convex combination of rotations of functions $f_{\alpha} \in \mathcal{F}(\alpha)$ .

Theorem 3. Let $n \geq 3$ , $0 < \alpha < 1$ , $\alpha \neq \frac{1}{2}$ and $f \in \mathcal{F}(\alpha)$ as in the form $(1), A_n = A_n(\alpha)$ and $C_n = C_n(\alpha)$ be given by (3) and (13), respectively. (1) If $n \geq 3$ and $0 < \lambda \leq C_3(\alpha)$ , then $$|\lambda a_n^2 - a_{2n-1}| \le A_{2n-1},$$ where the equality is attained by convex combination of rotations of functions $f_{\alpha} \in$ $\mathcal{F}(\alpha)$ defined by (4). (2) If $\lambda > C_3(\alpha)$ , then there exists a fixed $n_0 > 3$ such that $$C_{n_0-1}(\alpha) < \lambda \le C_{n_0}(\alpha).$$ Furthermore, if $\lambda > C_3(\alpha)$ and $n \geq n_0$ , then $$|\lambda a_n^2 - a_{2n-1}| \le A_{2n-1},$$ where the equality is attained by convex combination of rotations of functions $f_{\alpha} \in$ $\mathcal{F}(\alpha)$ defined by (4). (3) If $\lambda > C_3(\alpha)$ and $3 \leq n < n_0$ , then $$|\lambda a_n^2 - a_{2n-1}| \le \lambda A_n^2 - A_{2n-1},$$ where the equality is attained for the function $f_{\alpha}(z)$ defined by (4). Proof. Case (1). $0 < \lambda \le C_3(\alpha)$ and $n \ge 3$ . In this case, $\lambda \leq C_3(\alpha) \leq C_n(\alpha)$ by Lemma 5 and thus, $\lambda A_n^2 - 2A_{2n-1} \leq 0$ , which by (15) implies the desired inequality $$|\lambda a_n^2 - a_{2n-1}| \le A_{2n-1}.$$ If $\lambda > C_3(\alpha)$ , then there exists a fixed $n_0 > 3$ such that $C_{n_0-1}(\alpha) < \lambda \leq C_{n_0}(\alpha)$ by Lemma 5. Case (2). $\lambda > C_3(\alpha)$ and $n \geq n_0$ . For this case, Lemma 5 yields that $\lambda A_n^2 - 2A_{2n-1} \leq 0$ and the conclusion follows from (15). Case (3). $\lambda > C_3(\alpha)$ and $3 \le n < n_0$ . In this case, $\lambda A_n^2 - 2A_{2n-1} \ge 0$ and the desired inequality follows as before.

Theorem 4. Let $f \in \mathcal{F}(1/2)$ as in the form (1). Then we have - (1) |λa_n² a_2n-1| ≤ 1/(2n-1) for n ≥ 3 and 0 < λ ≤ 18/5, where the equality is attained by convex combination of rotations of functions f_1/2 ∈ F(1/2) defined by (4). (2) |λa_n² a_2n-1| ≤ 1/(2n-1) for λ > 18/5 and n ≥ λ+√(λ²-2λ)/2, where the equality is attained by convex combination of rotations of functions f_1/2 ∈ F(1/2). - (3) $|\lambda a_n^2 a_{2n-1}| \le \frac{\lambda}{n^2} \frac{1}{2n-1}$ for $\lambda > \frac{18}{5}$ and $3 \le n < \frac{\lambda + \sqrt{\lambda^2 2\lambda}}{2}$ . The equality is attained for the function $f_{1/2}(z) = -\log(1-z)$ .