On the difference of coefficients of univalent functions

Theorem 1. Let $f \in \mathcal{S}$ and be given by (1) with $a_2 = 0$ . Then - $(i) |a_3| \leq 1,$ - $(ii) |a_4| \le \frac{2}{3} = 0.666 \dots,$ - (iii) $|a_5| \le \sqrt{\frac{19}{15}} = 1.67666...,$ - $(iv) |H_2(2)| \leq 1,$ - $(v) |H_3(1)| \le \frac{21}{20} = 1.05.$ Proof. (i) The classical inequality $|a_3 - a_2^2| \le 1$ for f in S when $a_2 = 0$ , gives $|a_3| \le 1$ , which from (8) gives $$(10) |\omega_{13}| \le \frac{1}{2}.$$ (ii) Next choose $x_1 = 0$ and $x_3 = 1$ in (7), which gives $$(11) |\omega_{33}| \le \frac{1}{3}.$$ Also, since $\omega_{11} = 0 \ (\Leftrightarrow a_2 = 0)$ , then from (8) and (11) we obtain $$|a_4| = 2|\omega_{33}| \le \frac{2}{3} = 0.666\dots$$ (iii) Again since $\omega_{11} = 0$ , from (8) we obtain $$|a_5| = |2\omega_{35} + 5\omega_{13}^2|.$$ From (6) with $x_1 = 0$ and $x_3 = 1$ we have $(\omega_{11} = 0)$ $$|\omega_{13}|^2 + 3|\omega_{33}|^2 + 5|\omega_{35}|^2 \le \frac{1}{3}$$ and from here (13) $$|\omega_{35}| \le \frac{1}{\sqrt{15}} \sqrt{1 - 3|\omega_{13}|^2}.$$ From (12) and (13) we have $$|a_5| \le 2|\omega_{35}| + 5|\omega_{13}|^2 \le \frac{1}{\sqrt{15}}\sqrt{1 - 3|\omega_{13}|^2} + 5|\omega_{13}|^2 \le \frac{503}{300} = 1.67666\dots$$ - (iv) Since we are assuming $a_2 = 0$ , (i) shows that $|H_2(2)| \le 1$ is trivial. - (v) When $\omega_{11} = 0$ , from the last relation in (8) we have $\omega_{33} = \omega_{15}$ , and from (9), (14) $$|H_3(1)| = |2\omega_{13}^3 + 4\omega_{13}\omega_{35} - 4\omega_{33}^2| \le 2|\omega_{13}|^3 + 4 + \underbrace{|\omega_{13}\omega_{35} - \omega_{15}^2|}_{E_1}.$$ Now choose $x_1 = -\omega_{15}$ , and $x_3 = \omega_{13}$ , and since $\omega_{33} = \omega_{15}$ , from (6) we obtain $$|\omega_{13}|^4 + 5E_1^2 \le |\omega_{15}|^2 + \frac{|\omega_{13}|^2}{3} \le \frac{1}{5} - \frac{3}{5}|\omega_{13}|^2 + \frac{1}{3}|\omega_{13}|^2,$$ (since by (6) $3|\omega_{13}|^2 + 5|\omega_{15}|^2 \le 1$ for $x_1 = 1$ , $x_3 = 0$ and $\omega_{11} = 0$ ), which implies $5E_1^2 \le \frac{1}{5} - \frac{4}{15}|\omega_{13}|^2 - |\omega_{13}|^4$ , i.e., $E_1 \le \frac{1}{5}$ . Finally from (10) and (14), it follows that $$|H_3(1)| \le 2 \cdot \frac{1}{8} + 4 \cdot \frac{1}{5} = \frac{21}{20} = 1.05.$$ This completes the proof of Theorem 1. We next prove a similar result, this time assuming that $a_3 = 0$ .

Theorem 2. Let $f \in \mathcal{S}$ and be given by (1), with $a_3 = 0$ . Then - $(i) |a_2| \leq 1,$ - $(ii) |a_4| \le \frac{\sqrt{37} + 13}{12} = 1.59023 \dots,$ (iii) $$|a_5| \le \frac{1}{4} \sqrt{\frac{757}{15}} + \frac{85}{64} = 3.10412...,$$ $$(iv) |H_2(2)| \le \frac{13+\sqrt{37}}{12} = 1.59023...,$$ $$(v) |H_3(1)| \le \frac{24+\sqrt{645}}{30} = 1.64656\dots$$ Proof. (i) Since $|a_3 - a_2^2| \le 1$ and $a_3 = 0$ , then $|a_2^2| \le 1$ , i.e., $|a_2| \le 1$ . Also, since by $(8), a_3 = 2\omega_{13} + 3\omega_{11}^2 = 0$ , it follows that (15) $$\omega_{13} = -\frac{3}{2}\omega_{11}^2 \quad \left( \Leftrightarrow \ \omega_{11}^2 = -\frac{2}{3}\omega_{13} \right).$$ Because $|a_2| = |2\omega_{11}| \le 1$ , we have (16) $$|\omega_{11}| \le \frac{1}{2}$$ and $|\omega_{13}| \le \frac{3}{8}$ (by (15). (ii) By using (8) and (15), we obtain $$|a_4| = \left| 2\omega_{33} + 8\omega_{11} \left( -\frac{3}{2}\omega_{11}^2 \right) + \frac{10}{3}\omega_{11}^3 \right|$$ $$= \left| 2\omega_{33} - \frac{26}{3}\omega_{11}^3 \right|$$ $$\leq 2|\omega_{33}| + \frac{26}{3}|\omega_{11}|^3.$$ From (6), using $x_1 = 0$ and $x_3 = 1$ , we have $$|\omega_{13}|^2 + 3|\omega_{33}|^2 \le \frac{1}{3},$$ which implies (with $\omega_{13} = -\frac{3}{2}\omega_{11}^2$ , see (15)) $$|\omega_{33}| \le \sqrt{\frac{1}{9} - \frac{3}{4}|\omega_{11}|^4}.$$ Combining (17) and (18) we obtain (19) $$|a_4| \le 2\sqrt{\frac{1}{9} - \frac{3}{4}|\omega_{11}|^4} + \frac{26}{3}|\omega_{11}|^3 =: \varphi(|\omega_{11}|),$$ where $\varphi(t) = 2\sqrt{\frac{1}{9} - \frac{3}{4}t^4} + \frac{26}{3}t^3$ , $0 \le t = |\omega_{11}| \le \frac{1}{2}$ (by (16)). Since $\varphi$ is increasing function on [0, 1/2], $$\varphi(t) \le \varphi(1/2) = \frac{\sqrt{37} + 13}{12},$$ which, together with (19), gives the desired result. (iii) From the last relation in (8), using (15) we have $\omega_{33} = \omega_{15} + \frac{11}{6}\omega_{11}^3$ , which with the expression for $a_5$ in (8), gives (20) $$|a_{5}| = |2\omega_{35} + 8\omega_{11}\omega_{15} + 5\omega_{13}^{2} - 10\omega_{11}^{4}|$$ $$\leq 2\underbrace{|\omega_{35} + 4\omega_{11}\omega_{15}|}_{C_{7}^{}} + \underbrace{5|\omega_{13}|^{2} + 10|\omega_{11}|^{4}}_{C_{7}^{}}.$$ Once again, using (6) choosing $x_1 = 4\omega_{11}$ , $x_3 = 1$ and $\omega_{13} = -\frac{3}{2}\omega_{11}^2$ , we have $$(C_1^*)^2 = |4\omega_{11}\omega_{15} + \omega_{35}|^2 \le -\frac{5}{4}|\omega_{11}|^4 + \frac{16}{5}|\omega_{11}|^2 + \frac{1}{15} \le \frac{757}{64 \cdot 15},$$ since $|\omega_{11}| \leq \frac{1}{2}$ . Thus $$C_1^* \le \frac{1}{8} \sqrt{\frac{757}{15}}.$$ Next, since $\omega_{13} = -\frac{3}{2}\omega_{11}^2$ and $|\omega_{11}| \leq \frac{1}{2}$ , we have $$C_2^* = 5 \cdot \frac{9}{4} \cdot |\omega_{11}|^4 + 10|\omega_{11}|^4 = \frac{85}{4}|\omega_{11}|^4 \le \frac{85}{4} \cdot \frac{1}{16} = \frac{85}{64}$$ since $|\omega_{11}| \leq \frac{1}{2}$ . Finally from (20) we have $$|a_5| \le \frac{1}{4} \sqrt{\frac{757}{15}} + \frac{85}{64} = 3.10412\dots$$ (iv) By using (9), (8) and (15), we have (21) $$H_2(2) = 4\omega_{11}\omega_{33} + 4\omega_{11}^2\omega_{13} - 4\omega_{13}^2 - \frac{7}{3}\omega_{11}^4$$ $$= 4\omega_{11}\omega_{33} - \frac{52}{3}\omega_{11}^4$$ and from here $$(22) |H_2(2)| \le 4|\omega_{11}||\omega_{33}| + \frac{52}{3}|\omega_{11}|^4.$$ From (18) and (22) we have $$|H_2(2)| \le 4|\omega_{11}|\sqrt{\frac{1}{9} - \frac{3}{4}|\omega_{11}|^4} + \frac{52}{3}|\omega_{11}|^4 =: \varphi_1(|\omega_{11}|, |\omega_{11}|)$$ where $$\varphi_1(t) = 4t\sqrt{\frac{1}{9} - \frac{3}{4}t^4} + \frac{52}{3}t^4,$$ with $0 \le t = |\omega_{11}| \le \frac{1}{2}$ . Finally, it can be checked that $\varphi_1$ is an increasing function on the interval (0, 1/2), and so $$|H_2(2)| \le \varphi_1(1/2) = \frac{13 + \sqrt{37}}{12} = 1.59023....$$ (v) By using the last relation from (8) with $\omega_{13} = -\frac{3}{2}\omega_{11}^2$ , it follows that $\omega_{33} = \omega_{15} + \frac{11}{6}\omega_{11}^3$ , and so using (9), after some calculations we obtain $$H_3(1) = -12\omega_{11}^2 \left(\omega_{11}\omega_{15} + \frac{2}{3}\omega_{35}\right) - 4\omega_{15}^2 - 30\omega_{11}^6,$$ which gives (23) $$|H_3(1)| \leq \underbrace{12|\omega_{11}|^2 \left|\omega_{11}\omega_{15} + \frac{2}{3}\omega_{35}\right|}_{D_2} + \underbrace{4|\omega_{15}|^2 + 30|\omega_{11}|^6}_{D_2}.$$ Now choose $x_1 = \omega_{11}$ and $x_3 = \frac{2}{3}$ in (6), then (since $\omega_{13} = -\frac{3}{2}\omega_{11}^2$ ), $$\left|\omega_{11}\omega_{15} + \frac{2}{3}\omega_{35}\right| \le \sqrt{\frac{1}{5}\left(|\omega_{11}|^2 + \frac{4}{27}\right)},$$ and so (24) $$D_{1} \leq 12|\omega_{11}|^{2} \sqrt{\frac{1}{5} \left(|\omega_{11}|^{2} + \frac{4}{27}\right)}$$ $$\leq 12 \cdot \frac{1}{4} \sqrt{\frac{1}{5} \left(\frac{1}{4} + \frac{4}{27}\right)} = \sqrt{\frac{43}{60}} = \frac{\sqrt{645}}{30} = 0.84656 \dots,$$ since $|\omega_{11}| \leq \frac{1}{2}$ . Also, as in the proof of (iii), we have $$|5|\omega_{15}|^2 \le 1 - |\omega_{11}|^2 - 3|\omega_{13}|^2 = 1 - |\omega_{11}|^2 - \frac{27}{4}|\omega_{11}|^4$$ where we have once again used $\omega_{13} = -\frac{3}{2}\omega_{11}^2$ . Now $$D_2 \le \frac{4}{5} - \frac{4}{5}|\omega_{11}|^2 - \frac{27}{5}|\omega_{11}|^4 + 30|\omega_{11}|^6 =: \varphi_2(|\omega_{11}|^2),$$ where $$\varphi_2(t) = \frac{1}{5} \left( 4 - 4t - 27t^2 + 150t^3 \right),\,$$ and $0 \le t = |\omega_{11}|^2 \le \frac{1}{4}$ . Since $\varphi_2$ attains its maximum at $t_0 = 0$ , (25) $$D_2 \le \varphi_2(0) = \frac{4}{5}.$$ Finally, by using (23), (24) and (25) we obtain $$|H_3(1)| \le D_1 + D_2 \le \frac{24 + \sqrt{645}}{30} = 1.64656...$$

Theorem 3. Let $f \in \mathcal{S}$ and be given by (1). Then $$|a_4| - |a_3| < 2.1033299...$$