Proof of The Generalized Zalcman Conjecture for Initial Coefficients of Univalent Functions

Theorem 4.1. Let $f \in \mathcal{S}$ and be given by (1.1), then $$|a_2 a_3 - a_4| \le 2$$ with equality only for functions of the form <span id="page-5-3"></span> $$\frac{z}{(1-e^{i\theta}z)^2}$$ , where $\theta$ is real.

Theorem 4.2. Let $f \in \mathcal{S}$ and be given by (1.1), then $$|a_2 a_4 - a_5| \le 3$$ with equality only for functions of the form <span id="page-5-4"></span> $$\frac{z}{(1-e^{i\theta}z)^2}$$ , where $\theta$ is real.

Lemma 5.10. If h is an analytic self-map of $\mathbb{D}$ fixing the origin with expansion $h(z) = \sum_{n=1}^{\infty} \alpha_n z^n$ , then for $n \geq 1$ , <span id="page-7-4"></span> $$|\alpha_{n+1}| \le 1 - |\alpha_1|^2,$$ with equality only for h(z) = z. It therefore follows from (5.1), and Lemma 5.10, that for a fixed z (5.11) $$|\mu_n(z)| \le 1 - |\mu_1(z)|^2, \quad n = 2, 3, 4, \dots$$ for almost all $z \in \mathbb{D}$ . We will need a much stronger version of inequality (5.11) as follows. $$(5.12)$$ <span id="page-8-1"></span><span id="page-8-0"></span>It is easy to see that (5.12) can be derived from (5.11) provided (5.13) $$|\mu_1(z)| = \text{ constant a.e. on } \mathbb{D}^*.$$ To establish (5.13), Krushkal [12] proved a lemma known as the Surgery Lemma, which plays a vital role in the proof of our main result.

Lemma 5.14. (Surgery Lemma). Given a function $f \in \mathcal{S}$ with $a_2 \neq 0$ , one can construct a holomorphic motion $f^(z,t)$ on $\mathbb{C}_{\infty} \times \mathbb{D}$ such that for any t, the restriction of the fibre map $f_t^(z)$ to $\mathbb D$ belongs to $\mathcal S$ , and the family $f^*(z,t)$ satisfies the following properties: (i.) $f^*(z,t)$ is also holomorphic in t, and its Beltrami coefficients (5.15) $$\mu_{\tilde{f}_{}^{}(z)} = \mu_{1}^{}(z)t + \mu_{2}^{}(z)t^{2} + \mu_{3}^{*}(z)t^{3} + \cdots$$ satisfy (5.16) $$|\mu_1^(z)| = constant \quad on \mathbb{D}$$ (ii.) $$f_t^*(\infty) = \infty$$ for all $t \in \mathbb{D}$ (ii.) $$f_t^*(\infty) = \infty$$ for all $t \in \mathbb{D}$ (iii.) $a_n(f_t^*) = a_n(f_t) + o_n(t^n), n = 2, 3, \cdots$ We now give the proofs of Theorems 4.1 and 4.2. 5.1. Proof of Theorem 4.1. If $f \in \mathcal{S}$ any function maximizing (4.1), then $a_2(f) \neq$ 0. If not, let $a_2 = 0$ , then $b_0 = 0$ and so $a_2a_3 - a_4 = b_2$ . The sharp inequality $$|b_2| \le \frac{2}{3},$$ was proved by Golusin [9]. Jenkins [11, Corollary 11] gave another proof of this, and Duren [7] has given an elementary proof. Equality in (5.17) holds only for the functions <span id="page-8-2"></span> $$z\left(1+\frac{e^{i\theta}}{z^3}\right)^{\frac{2}{3}}, \quad \theta \text{ real},$$ which map $\{z:|z|>1\}$ onto the whole w-plane except for a cut consisting of three lines, each of length $2^{2/3}$ , originating symmetrically from the origin. Thus we obtain the following sharp inequality $$|b_2| \le \frac{2}{3},$$ and so it follows from (1.3) that $|a_2a_3-a_4|=|b_2|\leq 2/3$ . This contradicts our hypothesis, since for the Koebe function, $|a_2a_3-a_4|=2$ , which is always greater than 2/3 Hence it is now clear that any function f maximizing (4.1) has second Taylor series coefficient $a_2 \neq 0$ , which leads us to use the Surgery Lemma. So now we consider the holomorphic motion $$\widetilde{F}(z,t) = \frac{1}{\widetilde{f}(\frac{1}{z},t)}$$ with Beltrami coefficient $$\mu_{\widetilde{F}(z,t)} = \mu_{\widetilde{f}_t(\frac{1}{z},t)} \frac{\overline{z}^2}{z^2} = \sum_{n=1}^{\infty} \mu_n \left(\frac{1}{z}\right) t^n \frac{\overline{z}^2}{z^2} = \widetilde{\mu}_1(z)t + \widetilde{\mu}_2(z)t^2 + \widetilde{\mu}_3(z)t^3 + \cdots$$ generated by $\widetilde{f}$ . From the proof of the Surgery Lemma (see [12]), we can assume that $$\widetilde{\mu_1}(z) = \alpha \widetilde{\mu}^0(z),$$ where $$\widetilde{\mu}^0(z) = \frac{|z|}{\overline{z}},$$ and from [12] it is clear that for small |t|, we have the representation $$\widetilde{F}_t(z) = w^{\sigma} \circ K_{\alpha t}(z) = z + b_0 t + b_1 t^2 z^{-1} + \dots,$$ where $$K_{\alpha t}(z) = \frac{1}{\tilde{k}_{\alpha t}\left(\frac{1}{z}\right)} = \begin{cases} z - 2\alpha t + \alpha^2 t^2 z^{-1}, & |z| > 1, \\ z - 2\alpha t |z| + \alpha^2 t^2 \overline{z}, & |z| < 1, \end{cases}$$ has the Beltrami coefficient $-t\alpha\widetilde{\mu}^0,\,||\sigma(z,t)||_\infty\leq |t|^2,$ and the map $$w^{\sigma}(\zeta) = \zeta + \sum_{n=0}^{\infty} b'_n \zeta^{-n} \quad (b'_n = b'_n(t))$$ is conformal in the domain $K_{\alpha t}(\mathbb{D}^*) \supset \{|\zeta| > 1 + O(t)\}.$ We now use the estimates of the coefficients of $w^{\sigma}$ mentioned in [12], and obtain <span id="page-9-0"></span> $$(5.18) b'_0 = O(t^2), b'_1 = O(t^3), b'_2 = O(t^4) and |b'_3| \le \frac{(1 - |\alpha|^2)^2}{\sqrt{3}(1 - |t|)^2} |t|^4 + O(t^6),$$ which gives $$b_0'b_1' = O(t^5).$$ Since $$\widetilde{F}_t(z) = z - (2\alpha t - b_0') + \frac{\alpha^2 t^2 + b_1'}{z} + \frac{2\alpha t b_1' + b_2'}{z^2} + \frac{3\alpha^2 t^2 b_1' + 4\alpha t b_2' + b_3'}{z^3} + \cdots,$$ we have (5.19) $$b_0 = \frac{b'_0 - 2\alpha t}{t}$$ $$b_1 = \frac{\alpha^2 t^2 + b'_1}{t^2}$$ $$b_2 = \frac{2\alpha t b'_1 + b'_2}{t^3}$$ $$b_3 = \frac{3\alpha^2 t^2 b'_1 + 4\alpha t b'_2 + b'_3}{t^3},$$ and so from (1.4), we obtain $$|a_2a_3 - a_4| = |-b_0b_1 + b_2|.$$ Using (5.18), after an easy computation, we obtain $$|a_2a_3 - a_4| \le 2|\alpha|^3$$ , where $0 < \alpha \le 1$ , which implies that $$|a_2a_3 - a_4| \le 2,$$ with equality only if $|\alpha| = 1$ , i.e., for the Koebe function and its rotations. This completes the proof of Theorem 4.1. 5.2. Proof of Theorem 4.2. If $f \in \mathcal{S}$ is a function maximizing (4.2), then $a_2(f) \neq 0$ . If not, let $a_2 = 0$ , then $b_0 = 0$ , and so $a_2a_4 - a_5 = b_3 - b_1^2$ . We need to find the sharp bound for $|b_3 - b_1^2|$ , and to do this we use the method of Jenkins [11]. Jenkins [11, Corollary 13] proved that, if $g \in \Sigma$ , then for $\psi$ real and $0 \le \sigma \le 2$ , $$\Re\left\{e^{-4i\psi}\left(b_3 + \frac{1}{2}b_1^2 - \sigma^2 e^{2i\psi}b_1\right)\right\} \ge \begin{cases} -\frac{1}{2} - \frac{3}{16}\sigma^4 + \frac{1}{8}\sigma^4\log\left(\frac{\sigma^2}{4}\right), & 0 < \sigma \le 2\\ -\frac{1}{2}, & \sigma = 0. \end{cases}$$ Using the above inequality, we now find the sharp bound for $|b_3 - b_1^2|$ . Without loss in generality, choose $\psi$ real so that $$-\Re\left\{e^{-4i\psi}(b_3-b_1^2)\right\} = |b_3-b_1^2|,$$ and $$-\Re\left\{e^{-2i\psi}b_1\right\} \ge 0.$$ Then it follows that $$|b_3 - b_1^2| \le \frac{1}{2} + \frac{3}{16}\sigma^4 - \frac{\sigma^4}{8} - \log\frac{\sigma^4}{4} + \frac{3}{2}\Re(e^{-4i\psi}b_1^2) - \sigma^2\Re(e^{-2i\psi}b_1), \quad 0 < \sigma \le 2$$ $$\le \frac{1}{2}, \quad \sigma = 0.$$ It is easy to see that $$\Re\left\{e^{-4i\psi}b_1^2\right\} \le \left(\Re\left\{e^{-2i\psi}b_1\right\}\right)^2$$ with the strict inequality unless $$\Im\left\{e^{-2i\psi}b_1\right\} = 0.$$ Therefore $$|b_3 - b_1^2| \le \frac{1}{2} + \frac{3}{16}\sigma^4 - \frac{\sigma^4}{8} - \log\frac{\sigma^4}{4} + \frac{3}{2}\left(\Re\left\{e^{-2i\psi}b_1\right\}\right)^2 - \sigma^2\Re(e^{-2i\psi}b_1), \quad 0 < \sigma \le 2$$ $$\le \frac{1}{2}, \quad \sigma = 0.$$ If $\Re\left\{e^{-2i\psi}b_1\right\}=0$ , then taking $\sigma=0$ we obtain $$|b_3 - b_1^2| \le \frac{1}{2}.$$ If $\Re\left\{e^{-2i\psi}b_1\right\} > 0$ then we can find $\sigma$ with $0 < \sigma \le 2$ , so that $$\Re\left\{e^{-2i\psi}b_1\right\} = \frac{1}{4}\sigma^2\left(1 - \log\left(\frac{\sigma^2}{4}\right)\right),\,$$ and using this we obtain $$|b_3 - b_1^2| \le \frac{1}{2} + \frac{1}{32}\sigma^4 - \frac{1}{16}\sigma^4 \log \frac{\sigma^2}{4} + \frac{3}{32}\sigma^2 \log \frac{\sigma^2}{4} = \Theta(\sigma).$$ Clearly the function $\Theta(\sigma)$ is continuous on $0 < \sigma \le 2$ , and further <span id="page-11-0"></span> $$|b_3 - b_1^2| \le \max_{0 < \sigma \le 2} \Theta(\sigma).$$ It is easy to see that the function $\Theta$ is increasing on $[0, \sigma]$ up to the point $\sigma = 2e^{-\frac{1}{6}}$ and is decreasing after $\sigma = 2e^{-\frac{1}{6}}$ . Hence the maximum of $\Theta$ occurs at this value of $\sigma$ and so $$|b_3 - b_1^2| \le \frac{1}{2} + e^{-\frac{2}{3}}.$$ Equality occurs in (5.20) only for functions of the form $h(z, 2e^{-1/6}, \psi) + k$ , where $\psi$ is real, and k is a constant, where $h(z, \sigma, \psi)$ is defined in [11]. Thus we have the sharp bound for $|b_3 - b_1^2|$ . Thus $$|a_2a_4 - a_5| = |b_3 - b_1^2| \le \frac{1}{2} + e^{-\frac{2}{3}},$$ which is a contradiction, since for the Koebe function, $|a_2a_4 - a_5| = 3$ . Thus it is clear that any function f maximizing (4.2) has second Taylor series coefficient $a_2 \neq 0$ . So again we can use the Surgery Lemma and proceed as before. Using equation (1.4), we obtain $$|a_2a_4 - a_5| = |-b_0b_2 + 2b_1b_0^2 + b_3 - b_1^2|.$$ Using equation (5.18) in above and after an easy computation, we obtain $$|a_2a_4 - a_5| < 3|\alpha|^4$$ , where $0 < \alpha < 1$ . This implies that $$|\lambda a_2 a_4 - a_5| < 3,$$ with equality only if $|\alpha| = 1$ i.e., for the Koebe function and its rotations. This completes the proof of Theorem 4.2. Acknowledgement: Authors thanks Professors T. Sugawa and D. K. Thomas for their careful reading of this manuscript and giving constructive suggestions for improvements to this paper. The first author thanks SERB-CRG, and the second author thanks PMRF-MHRD, Govt. of India for their support.

Definition 3.1. Let E be a subset of $\mathbb{C}_{\infty}$ containing at least three points. A holomorphic motion of E is a function $f: E \times \mathbb{D} \to \mathbb{C}_{\infty}$ such that - (i) for every fixed $z \in E$ , the function $t \mapsto f(z,t) : E \times \mathbb{D} \to \mathbb{C}_{\infty}$ is holomorphic in $\mathbb{D}$ ; - (ii) for every fixed $t \in \mathbb{D}$ , the map $f(z,t) = f_t(z) : E \to \mathbb{C}_{\infty}$ is injective; - (iii) f(z,0) = z for all $z \in E$ . The following $\lambda$ -lemma assures that such a holomorphic motion is a holomorphic family of quasiconformal maps. $\lambda$ -lemma: If $f: E \times \mathbb{D} \to \mathbb{C}_{\infty}$ is a holomorphic motion, then f has an extension $\widetilde{f}: \overline{E} \times \mathbb{D} \to \mathbb{C}_{\infty}$ such that - (i) $\widetilde{f}$ is a holomorphic motion of the closure $\overline{E}$ of E; - (ii) each $\widetilde{f}_t(z) = \widetilde{f}(t,z) : \overline{E} \to \mathbb{C}$ is quasiconformal on the interior of $\overline{E}$ ; - (iii) $\widetilde{f}$ is jointly continuous in (z,t). The obvious question that arises is can we extend a holomorphic motion from any set to the whole sphere? The Slodkowski lifting theorem [23] solves this problem which was posed by Sullivan and Thurston. Extended $\lambda$ -lemma: Any holomorphic motion $f: E \times \mathbb{D} \to \mathbb{C}_{\infty}$ can be extended to a holomorphic motion $\widetilde{f}: \mathbb{C}_{\infty} \times \mathbb{D} \to \mathbb{C}_{\infty}$ , with $\widetilde{f}|_{E \times \mathbb{D}} = f$ . In view of [3, Theorem 2], the function $\phi : \mathbb{D} \to \mathcal{M}(\mathbb{D})$ defined by $$\phi(t) = \mu_{\widetilde{f}_t}(z) = \frac{\partial_{\overline{z}}\widetilde{f}(z,t)}{\partial_z\widetilde{f}(z,t)}$$ is holomorphic, where $\mathcal{M}(\mathbb{D})$ is a unit ball in $\mathcal{L}^{\infty}(\mathbb{D})$ .