Applications of Livingston-type inequalities to the generalized Zalcman functional

Lemma 1. Let $a, b \in \mathbb{C}$ be arbitrary and C, M > 0. Then (1) $$|a + \lambda b| \le M \max\{C, |\lambda|\}, \quad \text{for all } \lambda \in \mathbb{C}$$ if and only if $$(2) |a| + |b|C \le MC.$$ Equality holds in (1) if and only if it holds in (2) if and only if $|\lambda| = C$ and $\arg \lambda = \arg a - \arg b$ .

Proposition 2. If $g \in \mathcal{P}$ , $g(z) = 1 + \sum_{n=1}^{\infty} p_n z^n$ and $1 \le k \le n-1$ then (4) $$\left| p_n - \frac{1}{2} p_k p_{n-k} \right| + \frac{1}{2} |p_k p_{n-k}| \le 2.$$ The inequality is sharp.

Theorem 3. (a) If $f \in \mathcal{H}$ and $n \ge 2$ then the following inequality holds for the coefficients of f: (5) $$n^2|a_n^2| + (2n-1)|a_{2n-1}| \le 1.$$ This single inequality is equivalent to (6) $$|\lambda a_n^2 - a_{2n-1}| \le \max\left\{\frac{|\lambda|}{n^2}, \frac{1}{2n-1}\right\}, \quad for \ all \ \lambda \in \mathbb{C}.$$ Equality holds if and only if $$f(z) = \begin{cases} z + \frac{\alpha}{2n-1} z^{2n-1}, & \text{for } |\lambda| \le \frac{n^2}{2n-1} \\ z + \frac{\alpha}{n} z^n, & \text{for } |\lambda| \ge \frac{n^2}{2n-1}, \end{cases}$$ where $\alpha$ is a complex number of modulus one. (b) If $f \in \mathcal{H}$ , then for any two distinct values $m, n \ge 2$ we have (7) $$4mn|a_m a_n| + (m+n-1)|a_{m+n-1}| \le 1.$$ The last inequality is equivalent to (8) $$|\lambda a_m a_n - a_{m+n-1}| \le \max \left\{ \frac{|\lambda|}{4mn}, \frac{1}{m+n-1} \right\}, \quad \text{for all } \lambda \in \mathbb{C}.$$ Equality holds if and only if $$f(z) = \begin{cases} z + \frac{\alpha}{m+n-1} z^{m+n-1}, & \text{for } |\lambda| \le \frac{4mn}{m+n-1} \\ z + \frac{\alpha}{2m} z^m + \frac{\beta}{2n} z^n, & \text{for } |\lambda| \ge \frac{4mn}{m+n-1}, \end{cases}$$ where $\alpha$ and $\beta$ are complex numbers such that $|\alpha| = |\beta| = 1$ .

Theorem 4. Let $f \in \mathcal{R}$ and $m, n \ge 2$ . Then the following inequality holds for the coefficients of f: $$\left| \frac{mn}{2(m+n-1)} a_m a_n - a_{m+n-1} \right| + \frac{mn|a_m a_n|}{2(m+n-1)} \le \frac{2}{m+n-1}.$$ This is equivalent to $$|\lambda a_m a_n - a_{m+n-1}| \le \frac{2}{m+n-1} \max \left\{ 1, \left| 1 - 2\lambda \frac{m+n-1}{mn} \right| \right\} \quad \text{for all} \quad \lambda \in \mathbb{C}.$$ Equality holds in both inequalities for the function (9) $$f(z) = 2\log\frac{1}{1-z} - z$$ when $\left|1-2\lambda \frac{m+n-1}{mn}\right| \ge 1$ and for $$f(z) = \int_{[0,z]} \frac{1 + \zeta^{m+n-2}}{1 - \zeta^{m+n-2}} d\zeta$$ (meaning integration over the segment from 0 to z) when $\left|1-2\lambda \frac{m+n-1}{mn}\right| < 1$ .

Theorem 5. Let f be in $\overline{\text{co}(C)}$ and $m, n \ge 2$ . Then $$|a_m a_n - a_{m+n-1}| + |a_m a_n| \le 1$$ . This is equivalent to the following statement: $$|\lambda a_m a_n - a_{m+n-1}| \le \max\{1, |1 - \lambda|\}, \quad \text{for all} \quad \lambda \in \mathbb{C}.$$ Equality holds in both inequalities for the function given by $$(10) f(z) = \frac{z}{1-z}$$ when $|1 - \lambda| \ge 1$ and for (11) $$f(z) = \frac{z}{1 - z^{m+n-2}}$$ when $|1 - \lambda|$ < 1.

Theorem 6. Let $\alpha < 1$ , $f \in \overline{\operatorname{co}(C(\alpha))}$ , $m, n \ge 2$ , and $A_n$ as above. Then $$\left| \frac{a_m a_n}{A_m A_n} - \frac{a_{m+n-1}}{A_{m+n-1}} \right| + \frac{|a_m a_n|}{A_m A_n} \le 1.$$ This is equivalent to the following statement $$|\lambda a_m a_n - a_{m+n-1}| \le \max\{A_{m+n-1}, |\lambda A_m A_n - A_{m+n-1}|\}, \quad for \ all \quad \lambda \in \mathbb{C}.$$ Equality holds in both inequalities above for the function given by $f = f_{\alpha}$ in the case when $|\lambda A_m A_n - A_{m+n-1}| \ge A_{m+n-1}$ and for the function $$f(z) = \frac{1}{m+n-2} \sum_{k=1}^{m+n-2} e^{-\frac{2\pi ki}{m+n-2}} f_{\alpha} \left( e^{\frac{2\pi ki}{m+n-2}} z \right)$$ in the case when $|\lambda A_m A_n - A_{m+n-1}| < A_{m+n-1}$ .

Theorem 7. Let $f \in \overline{\cos(S^*)}$ and $m, n \ge 2$ . Then $$\left|\frac{a_m a_n}{mn} - \frac{a_{m+n-1}}{m+n-1}\right| + \frac{|a_m a_n|}{mn} \le 1.$$ This statement is equivalent to $$|\lambda a_m a_n - a_{m+n-1}| \le (m+n-1) \max \left\{ 1, \left| 1 - \frac{mn}{m+n-1} \lambda \right| \right\}, \quad for \ all \ \lambda \in \mathbb{C}.$$ In both cases, equality holds for the function given by (12) $$f(z) = \frac{z}{(1-z)^2}$$ when $\left|1 - \frac{mn}{m+n-1}\lambda\right| \ge 1$ and for (13) $$f(z) = \frac{z}{1 - z^{m+n-2}} + (m+n-2) \frac{z^{m+n-1}}{(1 - z^{m+n-2})^2}$$ when $$\left|1 - \frac{mn}{m+n-1}\lambda\right| < 1$$ .

Theorem 8. Let $f(z) = z + a_2 z^2 + ...$ be in S, with Hayman index $\alpha$ , and let $\lambda \in \mathbb{C}$ . Then (14) $$\lim_{m,n\to\infty} \frac{|\lambda a_m a_n - a_{m+n-1}|}{|\lambda m n - m - n + 1|} = \alpha^2.$$ Also, if we define $B_{m,n}(\lambda) = \sup_{f \in S} |\lambda a_m a_n - a_{m+n-1}|$ , then $$\lim_{m,n\to\infty}\frac{B_{m,n}(\lambda)}{|\lambda mn-m-n+1|}=1.$$ In both limits, we understand that $(m,n) \to (\infty,\infty)$ unconditionally in $\mathbb{N}^2$ (meaning that $m+n\to\infty$ ).

Corollary 9. If $f \in S$ is not a rotation of the Koebe function, then for every $\delta \in (0, 1 - \alpha^2)$ there exist $m_0$ and $n_0$ in $\mathbb{N}$ (which depend on f) such that $$|\lambda a_m a_n - a_{m+n-1}| \le (1-\delta)|\lambda mn - m - n + 1|,$$ for all $m \ge m_0$ , $n \ge n_0$ . Some equivalent reformulations of the Zalcman conjecture. For the sake of simplicity, we treat only the original conjecture: $|a_n^2 - a_{2n-1}| \le (n-1)^2$ . We first recall that, if assumed true for all n, it easily implies the Bieberbach conjecture (now de Branges' theorem). Since the proof of this implication for one value of n uses the validity of the conjecture for another n, in order to avoid this discussion in the sequel, we shall simply take for granted the Bieberbach conjecture for odd integers: $|a_{2n-1}| \le 2n-1$ . With this in mind, the Zalcman conjecture can be reformulated in several ways.

Theorem 10. Let $f \in S$ be fixed, $f(z) = z + \sum_{k=2}^{\infty} a_k z^k$ , and let $n \ge 2$ be arbitrary. Then the following statements are equivalent: - (a) The Zalcman conjecture holds: $|a_n^2 a_{2n-1}| \le (n-1)^2 = n^2 (2n-1);$ (b) $|a_n^2 ta_{2n-1}| \le n^2 t(2n-1)$ for all $t \in [0,1];$ (c) $|a_n^2 a_{2n-1}| + r|a_{2n-1}| \le (n-1)^2 + r(2n-1)$ for all r > 0;(d) $|a_n^2 wa_{2n-1}| \le (n-1)^2 + |w-1|(2n-1)$ for all $w \in \mathbb{C}$ .

Theorem 11. Assume only a weaker statement than the Bieberbach conjecture, for example, Littlewood's theorem [5, Theorem 2.8]: $|a_n| < en$ , for all $n \ge 2$ . Under these assumptions we have: - (a) The Zalcman conjecture implies Conjecture 3. - (b) Conjecture 3 implies Conjecture 2. - (c) Conjecture 2 implies Conjecture 1 (with t = 1 r). - (d) Conjecture 1 implies the Bieberbach conjecture. (e) All weak conjectures: Conjecture 1, Conjecture 2, and Conjecture 3 are asymptotically true. For example, if f is a function in S with Hayman index $\alpha$ , and $t \in [0,1]$ then (15) $$\lim_{n \to \infty} \frac{|a_n^2 - t a_{2n-1}|}{n^2 - t(2n-1)} = \alpha^2.$$