On coefficient functionals associated with the Zalcman conjecture

Theorem 2.1. If $f \in \overline{co(C)}$ , then $$|\lambda a_n a_m - a_{n+m-1}| \le \lambda - 1,$$ where $n, m = 2, 3, \ldots$ and $\lambda \in [2, \infty)$ . Equality holds for the function l(z) = z/(1-z) and its rotations.

Theorem 2.2. If $f \in \mathcal{R}(\beta)$ , then $$|\lambda a_n a_m - a_{n+m-1}| \le \frac{4\lambda(1-\beta)^2}{nm} - \frac{2(1-\beta)}{n+m-1},$$ where $n, m = 2, 3, \ldots$ and $\lambda \in \left[\frac{nm}{(1-\beta)(n+m-1)}, \infty\right)$ . Equality holds for the function $m(z) = -2(1-\beta)\ln(1-z) - z(1-2\beta)$ and its rotations. 2.1. The class $\mathcal{H}$ . Define the class $$\mathcal{H} = \left\{ f \in \mathcal{A} : f(z) = z + \sum_{n=2}^{\infty} a_n z^n \text{ and } \sum_{n=2}^{\infty} r(n) |a_n| \le 1, r(n) > 0 \text{ for } n \ge 2 \right\}.$$ Here is a partial list of restrictions on r(n) such that $\mathcal{H}$ is a subclass of $\mathcal{S}$ . For example, - If $r(n) = (n \beta)/(1 \beta)$ , then $\mathcal{H} \subset \mathcal{S}^*(\beta) \subset \mathcal{S}$ [21]. In particular, for $\beta = 0$ we have $\mathcal{H} = H$ , the Hurwitz class. - If $r(n) = n(n-\beta)/(1-\beta)$ , then $\mathcal{H} \subset \mathcal{C}(\beta) \subset \mathcal{S}$ [21]. - If r(n) = 3n 2, then $\mathcal{H} \subset \mathcal{UST} \subset \mathcal{S}$ [10]. - If r(n) = n(2n-1), then $\mathcal{H} \subset \mathcal{UCV} \subset \mathcal{S}$ [10]. - If $r(n) = n/(1-\beta)$ , then $\mathcal{H} \subset \mathcal{R}(\beta) \subset \mathcal{S}$ . - If $r(n) = 1 + [(n-1)/(1-\beta)] \sec \nu$ , then $\mathcal{H} \subset \mathcal{S}_p^{\nu}(\beta) \subset \mathcal{S}$ [11]. In all these classes $\beta \in [0, 1)$ . We now state our main result for the class $\mathcal{H}$ .

Theorem 2.3. Let $\lambda > 0$ and $n = 2, 3, \ldots$ For $f \in \mathcal{H}$ , we have $$|\lambda a_n^2 - a_{2n-1}| \le \max\left\{\frac{\lambda}{r(n)^2}, \frac{1}{r(2n-1)}\right\}.$$ Equality holds if and only if $$f(z) = \begin{cases} z + \frac{\alpha}{r(2n-1)} z^{2n-1} & \text{for } \lambda \le \frac{r(n)^2}{r(2n-1)}, \\ z + \frac{\alpha}{r(n)} z^n & \text{for } \lambda \ge \frac{r(n)^2}{r(2n-1)}, \end{cases}$$ where $\alpha$ is a complex number such that $|\alpha| = 1$ . We remark that for the choice r(n) = n, Theorem 2.3 turns into Theorem C. Indeed, our proof is much simpler than the proof of [7, Theorem 6].

Lemma 3.1. Let $\lambda \in \mathbb{C}$ , $\mu(\theta)$ be a probability measure on $[0, 2\pi]$ , and for some function s(n) > 0, write $a_n = s(n) \int_0^{2\pi} e^{i(n-1)\theta} d\mu(\theta) = s(n)b_{n-1}/2$ where $b_n$ is same as in Lemma A. Then $$|\lambda a_n a_m - a_{n+m-1}| \le \left|\lambda - \frac{2s(n+m-1)}{s(n)s(m)}\right| s(n)s(m) + s(n+m-1),$$ for n, m = 2, 3, ...

Corollary 3.3. If $f \in \overline{co(C)}$ , then $$|\lambda a_n^2 - a_{2n-1}| \le \lambda - 1,$$ where $n=2,3,\ldots$ and $\lambda\in[2,\infty)$ . Equality holds for the function l(z)=z/(1-z) and its rotations. Remark 3.4. We can also prove Corollary 3.3 by using the same technique as in [7]. Indeed, from the proof of [7, Theorem 3] we have $$|\lambda a_n^2 - a_{2n-1}| \le (\lambda - 2) \int_0^{2\pi} \cos^2((n-1)\theta) d\mu(\theta) + 1 \le \lambda - 1,$$ where the second inequality follows from the fact that $\cos \theta < 1$ for $0 < \theta < 2\pi$ . Proof of Theorem 2.2. By the Herglotz representation theorem for functions with positive real part [6, 1.9], there is a unique probability measure $\mu$ on $[0, 2\pi]$ such that $$\frac{f'(z) - \beta}{1 - \beta} = \int_0^{2\pi} \frac{1 + e^{i\theta}z}{1 - e^{i\theta}z} d\mu(\theta)$$ or, equivalently, $$1 + \sum_{n=2}^{\infty} n a_n z^{n-1} = 1 + (1 - \beta) \sum_{n=2}^{\infty} 2 \int_0^{2\pi} e^{in\theta} d\mu(\theta) z^n.$$ Comparing the coefficients, we obtain $$a_n = \frac{2(1-\beta)}{n} \int_0^{2\pi} e^{i(n-1)\theta} d\mu(\theta), \quad n \ge 2.$$ From Lemma 3.1, we can see that $s(n) = 2(1 - \beta)/n$ and hence we get $$|\lambda a_n a_m - a_{n+m-1}| \leq \left| \lambda - \frac{nm}{(1-\beta)(n+m-1)} \right| \frac{4(1-\beta)^2}{nm} + \frac{2(1-\beta)}{n+m-1}$$ $$= \frac{4\lambda(1-\beta)^2}{nm} - \frac{2(1-\beta)}{n+m-1}.$$ The sharpness can easily be verified using the given function m(z) stated in the hypothesis of the theorem. In particular when m=n, Theorem 2.2 leads to

Corollary 3.5. If $f \in \mathcal{R}(\beta)$ and $\lambda \in \left[\frac{n^2}{(2n-1)(1-\beta)}, \infty\right)$ , then $$|\lambda a_n^2 - a_{2n-1}| \le \frac{4\lambda(1-\beta)^2}{n^2} - \frac{2(1-\beta)}{2n-1},$$ where $n = 2, 3, \ldots$ Equality holds for the function $m(z) = -2(1-\beta) \ln(1-z) - z(1-2\beta)$ and its rotations. Remark 3.6. Alternative proof of Corollary 3.5 can be done by the same technique as in [7]. Indeed, from the proof of [7, Theorem 4] we have $$|\lambda a_n^2 - a_{2n-1}| \leq \left(\frac{4\lambda(1-\beta)^2}{n^2} - \frac{4(1-\beta)}{2n-1}\right) \int_0^{2\pi} \cos^2((n-1)\theta) d\mu(\theta) + \frac{2(1-\beta)}{2n-1}$$ $$\leq \frac{4\lambda(1-\beta)^2}{n^2} - \frac{2(1-\beta)}{2n-1},$$ where the second inequality follows from the fact that $\cos \theta \leq 1$ for $0 \leq \theta \leq 2\pi$ . Remark 3.7. From Remark 3.6, it is clear that for $f \in \mathcal{R}(\beta)$ and for $0 < \lambda \le 4/3(1-\beta)$ , $$|\lambda a_n^2 - a_{2n-1}| \le \frac{2(1-\beta)}{2n-1}.$$ Equality holds for the function m(z) and its rotations. To prove the generalized Zalcman problem for $\mathcal{H}$ , we need the following lemma which is in a similar form of [7, Lemma 5].

Lemma 3.8. Let $\lambda > 0, n \ge 2, q(n), q(2n-1) > 0$ , and consider the triangular region $$P = \{(u, v) \in \mathbb{R}^2 : u, v \ge 0, q(n)u + q(2n - 1)v \le 1\}$$ in the uv-plane. Then $$\max_{(u,v)\in P} (\lambda u^2 + v) = \max\left\{\frac{\lambda}{q(n)^2}, \frac{1}{q(2n-1)}\right\},\,$$ and the maximum attain only at (u, v) = (0, 1/q(2n-1)) and (u, v) = (1/q(n), 0).