Improvements of certain results of the class of univalent functions

Theorem 1. Let $f \in \mathcal{S}$ and be given by (1) with $a_2 = 0$ . Then 1 ²⁰²⁰ Mathematics Subject Classification. 30C45, 30C50, 30C55. $Key\ words\ and\ phrases.$ univalent functions, Grunsky coefficients, coefficient differences, Hankel determinant. - (i) $|a_3| \leq 1$ , - (ii) $|a_4| \le \frac{2}{3} = 0.666 \dots$ - (iii) $|a_5| \leq \frac{5}{4} + \frac{1}{\sqrt{15}} = 1.508...,$ - $(iv) |H_2(2)| < 1,$ - $(v) |H_3(1)| \le \frac{41}{20} = 2.05.$ We would like to point out that there is a mistake in the estimate of $|a_5|$ given in Theorem 2.1(iii) from [9]. The estimate given in the case (iii) in the theorem above is the correct one. Similar results are given in [9] for the case when $a_3 = 0$ .

Theorem 2. Let $f \in \mathcal{S}$ and be given by (1), with $a_3 = 0$ . Then - $(i) |a_2| \leq 1,$ - $(ii) |a_4| \le \frac{\sqrt{37} + 13}{12} = 1.59023...,$ - (iii) $|a_5| \le \frac{1}{4} \sqrt{\frac{757}{15}} + 1 = 2.77599...,$ - $(iv) |H_2(2)| \le 1.75088...,$ - $(v) |H_3(1)| \leq 1.114596...$ A long standing problem in the theory of univalent functions is to find sharp upper and lower bounds for $|a_{n+1}| - |a_n|$ , when $f \in \mathcal{S}$ . Since the Keobe function has coefficients $a_n = n$ , it is natural to conjecture that $||a_{n+1}| - |a_n|| \le 1$ . As early as 1933, this was shown to be false even when n = 2, when Fekete and Szegö [2] obtained the sharp bounds $$-1 \le |a_3| - |a_2| \le \frac{3}{4} + e^{-\lambda_0} (2e^{-\lambda_0} - 1) = 1.029 \dots,$$ where $\lambda_0$ is the unique value of $\lambda$ in $0 < \lambda < 1$ , satisfying the equation $4\lambda = e^{\lambda}$ . Hayman [4] showed that if $f \in \mathcal{S}$ , then $||a_{n+1}| - |a_n|| \le C$ , where C is an absolute constant. The exact value of C is unknown, the best estimate to date being C = 3.61... (see Grinspan [3]), which because of the sharp estimate above when n = 2, cannot be reduced to 1. In [9] the authors treated the difference coefficients $|a_4| - |a_3|$ for $f \in \mathcal{S}$ , which improves the previous cited result of Grinspan when n = 3 as follows

Theorem 3. Let $f \in \mathcal{S}$ and be given by (1). Then $$|a_4| - |a_3| \le 2.1033299\dots$$ For the proofs of the previously cited results, the authors mainly used the property of Grunsky coefficients given in the book of N. A.Lebedev ([5]). In the proofs of the results in this paper we also use the same tools but with different approach that brings improved estimates. We proceed with the notations and results to be used. Let $f \in \mathcal{S}$ and let $$\log \frac{f(t) - f(z)}{t - z} = \sum_{p,q=0}^{\infty} \omega_{p,q} t^p z^q,$$ where $\omega_{p,q}$ are called Grunsky's coefficients with property $\omega_{p,q} = \omega_{q,p}$ . For those coefficients in [1, 5] we can find the next Grunsky's inequality: (4) $$\sum_{q=1}^{\infty} q \left| \sum_{p=1}^{\infty} \omega_{p,q} x_p \right|^2 \le \sum_{p=1}^{\infty} \frac{|x_p|^2}{p},$$ where $x_p$ are arbitrary complex numbers such that last series converges. Further, it is well-known that if $f \in \mathcal{S}$ and has the form (1), then also <span id="page-2-1"></span><span id="page-2-0"></span> $$f_2(z) = \sqrt{f(z^2)} = z + c_3 + c_5 z^5 + \dots$$ belongs to the class S. Then for the function $f_2$ we have the appropriate Grunsky's coefficients of the form $\omega_{2p-1,2q-1}^{(2)}$ and the inequality (4) gets the form (5) $$\sum_{q=1}^{\infty} (2q-1) \left| \sum_{p=1}^{\infty} \omega_{2p-1,2q-1}^{(2)} x_{2p-1} \right|^2 \le \sum_{p=1}^{\infty} \frac{|x_{2p-1}|^2}{2p-1}.$$ As it has been shown in [5, p.57], if f is given by (1) then the coefficients $a_2$ , $a_3$ , $a_4$ , $a_5$ are expressed by Grunsky's coefficients $\omega_{2p-1,2q-1}^{(2)}$ of the function $f_2$ given by (3) in the following way (in the next text we omit upper index "(2)" in $\omega_{2p-1,2q-1}^{(2)}$ ): <span id="page-2-3"></span> $$a_{2} = 2\omega_{11},$$ $$a_{3} = 2\omega_{13} + 3\omega_{11}^{2},$$ $$a_{4} = 2\omega_{33} + 8\omega_{11}\omega_{13} + \frac{10}{3}\omega_{11}^{3}$$ $$a_{5} = 2\omega_{35} + 8\omega_{11}\omega_{33} + 5\omega_{13}^{2} + 18\omega_{11}^{2}\omega_{13} + \frac{7}{3}\omega_{11}^{4}$$ $$0 = 3\omega_{15} - 3\omega_{11}\omega_{13} + \omega_{11}^{3} - 3\omega_{33}$$ $$0 = \omega_{17} - \omega_{35} - \omega_{11}\omega_{33} - \omega_{13}^{2} + \frac{1}{3}\omega_{11}^{4}.$$ We note that in [5] there exists a typing mistake for the coefficient $a_5$ . Namely, instead of the therm $5\omega_{13}^2$ , there is $5\omega_{15}^2$ . Also, from (5) for $x_{2p-1} = 0$ , p = 3, 4, ..., we have <span id="page-2-2"></span>(7) $$|\omega_{11}x_1 + \omega_{31}x_3|^2 + 3|\omega_{13}x_1 + \omega_{33}x_3|^2 + |\omega_{15}x_1 + \omega_{35}x_3|^2 + |\omega_{17}x_1 + \omega_{37}x_3|^2 \le |x_1|^2 + \frac{|x_3|^2}{3}.$$ From (7), having in mind that $\omega_{31} = \omega_{13}$ , for $x_1 = 1$ and $x_3 = 0$ we have the next inequalities $$\begin{aligned} |\omega_{11}|^2 &\leq 1, \\ |\omega_{11}|^2 + 3|\omega_{13}|^2 &\leq 1, \\ |\omega_{11}|^2 + 3|\omega_{13}|^2 + 5|\omega_{15}|^2 &\leq 1, \\ |\omega_{11}|^2 + 3|\omega_{13}|^2 + 5|\omega_{15}|^2 + 7|\omega_{17}|^2 &\leq 1. \end{aligned}$$ From the last inequalities we easily obtain <span id="page-3-0"></span> $$|\omega_{11}| \leq 1,$$ $$|\omega_{13}| \leq \frac{1}{\sqrt{3}} \sqrt{1 - |\omega_{11}|^2},$$ $$|\omega_{15}| \leq \frac{1}{\sqrt{5}} \sqrt{1 - |\omega_{11}|^2 - 3|\omega_{13}|^2},$$ $$|\omega_{17}| \leq \frac{1}{\sqrt{7}} \sqrt{1 - |\omega_{11}|^2 - 3|\omega_{13}|^2 - 5|\omega_{15}|^2}.$$ We note that we get the first inequality from (8) also using the fact $|a_2| = |2\omega_{11}| \le 2$ (see (6)).

Theorem 4. Let $f \in \mathcal{S}$ and be given by (1) with $a_2 = 0$ . Then (i) $$|a_5| \le \frac{3}{4} + \frac{1}{\sqrt{7}} = 1.12796...,$$ $$(ii) |H_3(1)| \le 1.026...$$ 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 (6) gives $$(9) |\omega_{13}| \le \frac{1}{2}.$$ <span id="page-3-3"></span><span id="page-3-1"></span>Since $\omega_{11} = 0 \ (\Leftrightarrow a_2 = 0)$ , from the relation for $a_5$ in (6) and last relation in it, we obtain (10) $$|a_5| = |2\omega_{35} + 5\omega_{13}^2|$$ and (11) $$\omega_{35} = \omega_{17} - \omega_{13}^2.$$ <span id="page-3-2"></span>Further, using the relations (10) and (11) we have <span id="page-3-6"></span> $$|a_{5}| = |2\omega_{17} + 3\omega_{13}^{2}|$$ $$\leq 2|\omega_{17}| + 3|\omega_{13}|^{2}$$ $$\leq \frac{2}{\sqrt{7}}\sqrt{1 - 3|\omega_{13}|^{2}} + 3|\omega_{13}|^{2}$$ $$\leq \frac{3}{4} + \frac{1}{\sqrt{7}} = 1.12796...,$$ <span id="page-3-4"></span>since by (6) $(\omega_{11} = 0)$ , (13) $$|\omega_{17}| \le \frac{1}{\sqrt{7}} \sqrt{1 - 3|\omega_{13}|^2 - 5|\omega_{15}|^2} \le \frac{1}{\sqrt{7}} \sqrt{1 - 3|\omega_{13}|^2}$$ and $|\omega_{13}| \le \frac{1}{2}$ by (9). (ii) When $\omega_{11} = 0$ , the fifth relation in (6) gives $\omega_{33} = \omega_{15}$ , and using (11), from (3) we have $$|H_3(1) = | -8\omega_{13}^3 - 4\omega_{33}^2 + 2(2\omega_{35} + 5\omega_{13}^2)\omega_{13}|$$ $$= | -8\omega_{13}^3 - 4\omega_{15}^2 + 4\omega_{13}(\omega_{17} - \omega_{13}^2) + 10\omega_{13}^3|$$ $$= | -2\omega_{13}^3 - 4\omega_{15}^2 + 4\omega_{13}\omega_{17}|$$ $$\leq 2|\omega_{13}|^3 + 4|\omega_{15}|^2 + 4|\omega_{13}||\omega_{17}|$$ $$\leq 2|\omega_{13}|^3 + \frac{4}{5}(1 - 3|\omega_{13}|^2) + \frac{4}{\sqrt{7}}|\omega_{13}|\sqrt{1 - 3|\omega_{13}|^2}$$ $$=: F_1(|\omega_{13}|),$$ where $$F_1(y) = 2y^3 + \frac{4}{5}(1 - 3y^2) + \frac{4}{\sqrt{7}}y\sqrt{1 - 3y^2}, \quad 0 \le y \le \frac{1}{2}.$$ Here we used the relations (10) and (13). Now, using the first derivative test we conclude that the function $F_1$ attains its maximum for $y_0 = 0.286667...$ with $F_1(y_0) = 1.026...$ Finally, we note that the results (i) and (iv) in Theorem 1 are the best possible as the function $f(z) = \frac{z}{1-z^2}$ shows. We next prove a similar result, this time assuming that $a_3 = 0$ .

Theorem 5. Let $f \in \mathcal{S}$ and be given by (1), with $a_3 = 0$ . Then - $(i) |a_2| \leq 1,$ - (ii) $|a_4| \le \frac{1}{4}\sqrt{\frac{21}{5}} + \frac{5}{8} = 1.1373...,$ - $(iii) |a_5| \le 1.674896577...,$ - $(iv) |H_2(2)| \le 1.1373...,$ - $(v) |H_3(1)| \leq 0.6647958756...$ Proof. (i) We imitate the proof of Theorem 4(i) and from $|a_3 - a_2^2| \le 1$ , $a_3 = 0$ , we receive $|a_2^2| \le 1$ , i.e., $|a_2| \le 1$ . Further we will use that from (6) follows $a_3 = 2\omega_{13} + 3\omega_{11}^2 = 0$ , and then (14) $$\omega_{13} = -\frac{3}{2}\omega_{11}^2 \quad \left( \Leftrightarrow \ \omega_{11}^2 = -\frac{2}{3}\omega_{13} \right),$$ and also, from $|a_2| = |2\omega_{11}| \le 1$ , <span id="page-4-0"></span> $$|\omega_{11}| \le \frac{1}{2}$$ and $|\omega_{13}| \le \frac{3}{8}$ (by (14). (ii) By using (6) and (14), we obtain $$(15) |a_4| = \left| 2\omega_{33} + 8\omega_{11} \left( -\frac{3}{2}\omega_{11}^2 \right) + \frac{10}{3}\omega_{11}^3 \right| = \left| 2\omega_{33} - \frac{26}{3}\omega_{11}^3 \right|.$$ <span id="page-5-0"></span>On the other hand, using the fifth relation in (6), and (14), we have <span id="page-5-1"></span>(16) $$\omega_{33} = \omega_{15} - \omega_{11}\omega_{13} + \frac{1}{3}\omega_{11}^3 = \omega_{15} - \omega_{11}(-\frac{3}{2}\omega_{11}^2) + \frac{1}{3}\omega_{11}^3 = \omega_{15} + \frac{11}{6}\omega_{11}^3.$$ Combining (15) and (16) we obtain <span id="page-5-4"></span>(17) $$|a_4| = |2\omega_{15} - 5\omega_{11}^3| \le 2|\omega_{15}| + 5|\omega_{11}|^3$$ $$\le \frac{2}{\sqrt{5}}\sqrt{1 - |\omega_{11}|^2 - \frac{27}{4}|\omega_{11}|^4} + 5|\omega_{11}|^3$$ $$=: F_2(|\omega_{11}|),$$ where <span id="page-5-3"></span> $$F_2(x) = \frac{2}{\sqrt{5}}\sqrt{1 - x^2 - \frac{27}{4}x^4} + 5x^3, \quad 0 \le x \le \frac{1}{2}$$ and where we used (8) and (14) to obtain (18) $$|\omega_{15}| \le \frac{1}{\sqrt{5}} \sqrt{1 - |\omega_{11}|^2 - \frac{27}{4} |\omega_{11}|^4}.$$ Finally, $F_2(x)$ is strictly increasing function on the interval [0, 1/2], attaining its maximal value $\frac{1}{4}\sqrt{\frac{21}{5}} + \frac{5}{8} = 1.1373...$ for x = 1/2. (iii) From the last relation in (6), using (14) and (16), after simple calculation we receive (19) $$\omega_{35} = \omega_{17} - \omega_{11}\omega_{15} - \frac{15}{4}\omega_{11}^4.$$ <span id="page-5-2"></span>Using the relations (6) and (19), we get <span id="page-5-5"></span> $$|a_{5}| = \left| 2\omega_{17} + 6\omega_{11}\omega_{15} - \frac{25}{4}\omega_{11}^{4} \right|$$ $$\leq 2|\omega_{17}| + 6|\omega_{11}||\omega_{15}| + \frac{25}{4}|\omega_{11}|^{4}$$ $$\leq \left( \frac{2}{\sqrt{7}} + \frac{6}{\sqrt{5}}|\omega_{11}| \right) \sqrt{1 - |\omega_{11}|^{2} - \frac{27}{4}|\omega_{11}|^{4}} + \frac{25}{4}|\omega_{11}|^{4}$$ $$=: F_{3}(x),$$ where $$F_3(x) = \left(\frac{2}{\sqrt{7}} + \frac{6}{\sqrt{5}}x\right)\sqrt{1 - x^2 - \frac{27}{4}x^4} + \frac{25}{4}x^4, \quad 0 \le x \le \frac{1}{2},$$ and where we used the estimate given in (18), the estimate from (8), and <span id="page-5-6"></span>(21) $$|\omega_{17}| \leq \frac{1}{\sqrt{7}} \sqrt{1 - |\omega_{11}|^2 - 3|\omega_{13}|^2 - 5|\omega_{15}|^2} \\ \leq \frac{1}{\sqrt{7}} \sqrt{1 - |\omega_{11}|^2 - 3|\omega_{13}|^2}.$$ Now, the first derivative test shows that on the interval [0, 1/2], the function $F_3$ has maximal value 1.674896577... attained for x = 0.43957885... (iv) Since $a_3 = 0$ using (2) we have $H_2(2) = a_2 a_4$ and then from the relation (15) and estimation (18), we obtain $$|H_{2}(2)| = |a_{2}a_{4}| = |2\omega_{11}(2\omega_{15} - 5\omega_{11}^{3})$$ $$\leq 4|\omega_{11}||\omega_{15}| + 10|\omega_{11}|^{4}$$ $$\leq \frac{4}{\sqrt{5}}|\omega_{11}|\sqrt{1 - |\omega_{11}|^{2} - \frac{27}{4}|\omega_{11}|^{4}} + 10|\omega_{11}|^{4}$$ $$=: F_{4}(|\omega_{11}|),$$ where $$F_4(x) = \frac{4}{\sqrt{5}}x\sqrt{1 - x^2 - \frac{27}{4}x^4} + 10x^4, \quad 0 \le x \le \frac{1}{2}.$$ Again, the first derivative test shows that $F_4$ is strictly increasing on the interval [0, 1/2] with maximal value 1.1373... attained for x = 1/2. (v) After applying $a_3 = 0$ in the definition (3) we receive $$(22) H_3(1) = -a_4^2 - a_5 a_2^2.$$ <span id="page-6-0"></span>Recall that in (17) and (20) we found $a_4 = 2\omega_{15} - 5\omega_{11}^3$ and $a_5 = 2\omega_{17} + 6\omega_{11}\omega_{15} - \frac{25}{4}\omega_{11}^4$ , respectively. From these facts and (22), after some calculations we obtain (23) $$H_3(1) = -\omega_{15}^2 - 4\omega_{11}^3\omega_{15} - 8\omega_{11}^2\omega_{17}.$$ So, using (23) and estimates given in (18) and (21) we get <span id="page-6-1"></span> $$\begin{aligned} |H_3(1)| &= |\omega_{15}|^2 + 4|\omega_{11}|^3|\omega_{15}| + 8|\omega_{11}|^2|\omega_{17}| \\ &\leq \frac{1}{5} \left( 1 - |\omega_{11}|^2 - \frac{27}{4}|\omega_{11}|^4 \right) \\ &+ \left( \frac{4}{\sqrt{5}}|\omega_{11}|^3 + \frac{8}{\sqrt{7}}|\omega_{11}|^2 \right) \sqrt{1 - |\omega_{11}|^2 - \frac{27}{4}|\omega_{11}|^4} \\ &=: F_5(|\omega_{11}|), \end{aligned}$$ where $$F_5(x) = \frac{1}{5} \left( 1 - x^2 - \frac{27}{4} x^4 \right) + \left( \frac{4}{\sqrt{5}} x^3 + \frac{8}{\sqrt{7}} x^2 \right) \sqrt{1 - x^2 - \frac{27}{4} x^4},$$ $0 \le x \le \frac{1}{2}$ . The first derivative test shows that the function $F_5(x)$ when $0 \le x \le 1/2$ , has maximal value 0.6647958756... attained for x = 0.458573... \\\\\\\* DTS

Theorem 6. Let $f \in \mathcal{S}$ and be given by (1). - (i) Then $|a_4| |a_3| \le 1.75185...$ - (ii) If f is an odd function, then $|a_5| |a_3| \le \frac{2}{\sqrt{7}} = 0.7559...$ Proof. $\neg$ <span id="page-7-0"></span>(i) Using (6) and $|\omega_{11}| \leq 1$ , we have (24) $$|a_4| - |a_3| \le |a_4| - |\omega_{11}| |a_3| \le |a_4 - \omega_{11} a_3| = 2 \left| \omega_{33} + 3\omega_{11}\omega_{13} + \frac{1}{6}\omega_{11}^3 \right|$$ . From the fifth relation in (6) we obtain (25) $$\omega_{33} = \omega_{15} - \omega_{11}\omega_{13} + \frac{1}{3}\omega_{11}^3,$$ and using the relations (24) and (25), after some simple calculations we have <span id="page-7-1"></span> $$|a_4| - |a_3| \le |2\omega_{15} + 4\omega_{11}\omega_{13} + \omega_{11}^3|$$ $$\le |\omega_{15}| + 4|\omega_{11}||\omega_{13}| + |\omega_{11}|^3$$ $$\le \frac{2}{\sqrt{5}}\sqrt{1 - |\omega_{11}|^2 - 3|\omega_{13}|^2} + 4|\omega_{11}||\omega_{13}| + |\omega_{13}|^3$$ $$=: F_6(|\omega_{11}|, |\omega_{13}|,$$ where $$F_6(x,y) = \frac{2}{\sqrt{5}}\sqrt{1-x^2-3y^2} + 4xy + x^3,$$ and $0 \le x \le 1, \ 0 \le y \le \frac{1}{\sqrt{3}}\sqrt{1-x^2}.$ Now, we need to find maximum of the function $F_6(x,y)$ when $$(x,y) \in \left\{ (x,y) : 0 \le x \le 1, 0 \le y \le \frac{1}{\sqrt{3}} \sqrt{1-x^2} \right\} := D_1.$$ We start the analysis in the interior of $D_1$ . From $$\frac{\partial F_6(x,y)}{\partial x} = -\frac{2x}{\sqrt{5}\sqrt{1 - x^2 - 3y^2}} + 3x^2 + 4y$$ and $$\frac{\partial F_6(x,y)}{\partial y} = 4x - \frac{6y}{\sqrt{5}\sqrt{1-x^2-3y^2}},$$ we receive that the stationary points (it existing) satisfy $$3y\frac{\partial F_6(x,y)}{\partial x} - x\frac{\partial F_6(x,y)}{\partial y} = x^2(9y - 4) + 12y^2 = 0,$$ i.e., on $D_1$ , $$x = \frac{2\sqrt{3}y}{\sqrt{4 - 9y}}.$$ Finally, substituting $\frac{2\sqrt{3}y}{\sqrt{4-9y}}$ for x in $\frac{\partial F_6(x,y)}{\partial x}=0$ , and numerically solving the corresponding equation for y, we receive solution $y_{01}=0.2872\ldots$ , and further $x_{01}=\frac{2\sqrt{3}y_{01}}{\sqrt{4-9y_{01}}}=0.83634\ldots$ It is easy to check that $(x_{01},y_{01})\in D_1$ and that $F_6(x_{01},y_{01})=1.75185\ldots$ - On the edges of $D_1$ we have: $F_6(0,y) = \frac{2}{\sqrt{5}}\sqrt{1-3y^2}$ , with maximal value for $y \ge 0$ , $\frac{2}{\sqrt{5}} = 0.8944...$ ; - $F_6(x,0) = \frac{2}{\sqrt{5}}\sqrt{1-x^2} + x^3$ , with maximal value for $0 \le x \le 1$ equaling to 1.13666... for x = 0.9494...; $F_6\left(x, \frac{\sqrt{1-x^2}}{\sqrt{3}}\right) = \frac{4}{\sqrt{3}}x\sqrt{1-x^2} + x^3$ , with maximal value for $0 \le x \le 1$ - equaling to 1.6496... for x = 0.8628... So, the function $F_6$ attains its maximal value 1.75185... in the interior point $(x_{01}, y_{01})$ of its domain $D_1$ and the conclusion of part (i) follows. (ii) Since f is odd, $a_2 = 0$ , and then using $|a_3 - a_2| \le 1$ , we have $|a_3| = |2\omega_{13}| \le 1$ , i.e., $|\omega_{13}| \le \frac{1}{2}$ . Further, from (6), (11) and (12), we receive $$|a_5| = |2\omega_{17} + 3\omega_{13}^2|.$$ Also, since $a_4=0$ , from (6) follows $\omega_{33}=0$ and $\omega_{15}=0$ . So, since $|a_3|=|2\omega_{13}|\leq 1$ , $$\begin{aligned} |a_5| - |a_3| &\leq |a_5| - 2|\omega_{13}| |a_3| = |2\omega_{17} + 3\omega_{13}^2| - 4|\omega_{13}|^2 \\ &\leq |(2\omega_{17} + 3\omega_{13}^2) - 4\omega_{13}^2| = |2\omega_{17} - \omega_{13}^2| \\ &\leq 2|\omega_{17}| + |\omega_{13}|^2 \leq \frac{2}{\sqrt{7}} \sqrt{1 - 3|\omega_{13}|^2} + |\omega_{13}|^2 \\ &\leq \frac{2}{\sqrt{7}}, \end{aligned}$$ where $|\omega_{13}| \leq \frac{1}{2}$ . Remark 1. The estimate given in Theorem 6(ii) is an improvement of the result $|a_5| - |a_3| < 1$ given in [6, p. 17].