Hankel and Toeplitz determinants of logarithmic coefficients of Inverse functions for certain classes of univalent functions

Lemma 2.1. [9] Let $w(z) = c_1 z + c_2 z^2 + c_3 z^3 + ...$ be a Schwarz function. Then $$|c_1| \le 1$$ , $|c_2| \le 1 - |c_1|^2$ , and $|c_3| \le 1 - |c_1|^2 - \frac{|c_2|^2}{1 + |c_1|}$ . <span id="page-5-1"></span>2.1. Sharp bound of $|H_{2,1}(F_{f^{-1}}/2)|$ for $f \in \mathcal{S}_S$ : We obtain the following sharp bound for $H_{2,1}(F_{f^{-1}}/2)$ for functions in the class $\mathcal{S}_S$ . <span id="page-5-2"></span>Theorem 2.1. Let $f \in \mathcal{S}_S^*$ . Then $$|H_{2,1}(F_{f^{-1}}/2)| \le \frac{1}{4}.$$ The inequality is sharp with the extremal function $$h_1(z) = \frac{z}{1 - z^2} = z + z^3 + z^5 + \cdots, \quad z \in \mathbb{D}.$$ An example is presented below to demonstrate that the strict inequality in Theorem 2.1 remains valid. Example 2.1. Consider the function $$h_2(z) = \frac{z}{1-z} = z + z^2 + z^3 + \cdots, \quad z \in \mathbb{D}.$$ It is easy to see that $$\operatorname{Re}\left(\frac{zh_2'(z)}{h_2(z) - h_2(-z)}\right) = \frac{1}{2}\operatorname{Re}\left(\frac{1+z}{1-z}\right) > 0.$$ Hence, the function $h_2 \in \mathcal{S}_S^*$ . We can easily compute and find that $$|H_{2,1}(F_{h_2^{-1}}/2)| = \frac{1}{12} < \frac{1}{4}.$$ Proof of Theorem 2.1. Let $f \in \mathcal{S}_S^*$ be of the form $f(z) = z + \sum_{n=2}^{\infty} a_n z^n$ , $z \in \mathbb{D}$ . Then by the definition of subordination there exits a Schwarz function $w(z) = \sum_{n=1}^{\infty} c_n z^n$ such that <span id="page-5-3"></span>(2.1) $$\frac{2zf'(z)}{f(z) - f(-z)} = \frac{1 + w(z)}{1 - w(z)}.$$ By equating the coefficients on each side of (2.1), we get <span id="page-6-0"></span>(2.2) $$\begin{cases} a_2 = c_1, \\ a_3 = c_2 + c_1^2, \\ a_4 = \frac{1}{2}(c_3 + 3c_1c_2 + 2c_1^3). \end{cases}$$ In view of (1.8) and (2.2), a simple computation shows that $$H_{2,1}(F_{f^{-1}}/2) = \frac{1}{48} \left( 13a_2^4 - 12a_2^2 a_3 - 12a_3^2 + 12a_2 a_4 \right)$$ $$= \frac{1}{48} \left( c_1^4 - 18c_1^2 c_2 - 12c_2^2 + 6c_1 c_3 \right).$$ (2.3) <span id="page-6-1"></span>Using the Lemma 2.1 into (2.3), we obtain <span id="page-6-2"></span> $$(2.4) 48|H_{2,1}(F_{f^{-1}}/2)| \le |c_1|^4 + 18|c_1|^2|c_2| + 12|c_2|^2 + 6|c_1| \left(1 - |c_1|^2 - \frac{|c_2|^2}{1 + |c_1|}\right).$$ Suppose that $x = |c_1|$ and $y = |c_2|$ . Then it follows from (2.4) that <span id="page-6-3"></span> $$(2.5) 48|H_{2,1}(F_{f^{-1}}/2)| \le M(x,y),$$ where $$M(x,y) := x^4 + 18x^2y + 12y^2 + 6x\left(1 - x^2 - \frac{y^2}{1+x}\right).$$ In view of Lemma 2.1, the region of variability of a pair (x, y) coincides with the set <span id="page-6-4"></span>(2.6) $$\Omega = \{(x,y) : 0 \le x \le 1, 0 \le y \le 1 - x^2\}.$$ The goal is to establish the maximum value of M(x,y) in the region $\Omega$ . Therefore, the critical point of M(x,y) satisfies the conditions $$\frac{\partial M}{\partial x} = 4x^3 - 18x^2 + 36xy + 6 - \frac{6y^2}{(1+x)} + \frac{6xy^2}{(1+x)^2} = 0$$ and $$\frac{\partial M}{\partial y} = 18x^2 + 24y - \frac{12xy}{(1+x)} = 0.$$ There are no solutions of M(x,y) inside the interior of $\Omega$ , hence it is not possible for the function to attain a maximum value within this region. Since M(x,y) is a continuous function on a compact set $\Omega$ , its maximum value must occur at some point on the boundary of $\Omega$ . On the boundary of $\Omega$ , a simple computation shows that $$M(x,0) = 6x - 6x^3 + x^4 \le 2.437828...$$ for $0 \le x \le 1$ , $M(0,y) = 12y^2 \le 12$ for $0 \le y \le 1$ and $$M(x, 1 - x^2) = 12 - 11x^4 \le 12$$ for $0 \le x \le 1$ . Therefore, we see that $\max_{(x,y)\in\Omega} M(x,y) = 12$ and it fllows from (2.5) that <span id="page-7-1"></span> $$(2.7) |H_{2,1}(F_{f^{-1}}/2)| \le \frac{1}{4}.$$ To prove the equality in (2.7) sharp, we consider the function $$h_1(z) = \frac{z}{1 - z^2} = z + z^3 + z^5 + \cdots, \quad z \in \mathbb{D}.$$ It is easy to see that $h_1 \in \mathcal{S}_S^*$ and $|H_{2,1}(F_{h_1^{-1}}/2)| = 1/4$ . Hence equality holds in (2.7). This completes the proof <span id="page-7-0"></span>2.2. Sharp bound of $|H_{2,1}(F_{f^{-1}}/2)|$ for $f \in \mathcal{K}_S$ : By considering functions from the class $\mathcal{K}_S$ , we obtain the following sharp bound of $H_{2,1}(F_{f^{-1}}/2)$ . <span id="page-7-2"></span>Theorem 2.2. Let $f \in \mathcal{K}_S$ . Then $$|H_{2,1}(F_{f^{-1}}/2)| \le \frac{1}{36}.$$ The extremal function creates a precise inequality $$h_3(z) = \frac{1}{2} \log \left( \frac{1+z}{1-z} \right) = z + \frac{z^3}{3} + \frac{z^5}{5} \cdots, \quad z \in \mathbb{D}.$$ An example of a function in the class $\mathcal{K}_S$ is provided to demonstrate Theorem 2.2. The example satisfies the conditions of the theorem and shows that the inequality is strict also. Example 2.2. Consider the function $$h_4(z) = -\log(1-z) = z + \frac{1}{2}z^2 + \frac{1}{3}z^3 + \frac{1}{4}z^4 + \cdots, z \in \mathbb{D}.$$ It is easy to see that $$\operatorname{Re}\left(\frac{(zh_4'(z))'}{(h_4(z)-h_4(-z))'}\right) = \frac{1}{2}\operatorname{Re}\left(\frac{1+z}{1-z}\right) > 0.$$ Therefore, the function $h_4 \in \mathcal{K}_S$ . A simple computation shows that $$|H_{2,1}(F_{h_4^{-1}}/2)| = \frac{11}{576} < \frac{1}{36}.$$ Proof of Theorem 2.2. Let $f \in \mathcal{K}_S$ be of the form $f(z) = z + \sum_{n=2}^{\infty} a_n z^n$ , $z \in \mathbb{D}$ . Then by the definition of subordination there exits a Schwarz function $w(z) = \sum_{n=1}^{\infty} c_n z^n$ such that <span id="page-7-3"></span>(2.8) $$\frac{2(zf'(z))'}{(f(z)-f(-z))'} = \frac{1+w(z)}{1-w(z)}.$$ Comparing the coefficients on both side of (2.8), we obtain <span id="page-7-4"></span>(2.9) $$a_2 = \frac{1}{2}c_1, a_3 = \frac{1}{3}(c_2 + c_1^2), \text{ and } a_4 = \frac{1}{8}(c_3 + 3c_1c_2 + 2c_1^3).$$ By utilizing (1.8) and (2.9), an easy calculation gives that <span id="page-8-0"></span> $$H_{2,1}(F_{f^{-1}}/2) = \frac{1}{48} \left( 13a_2^4 - 12a_2^2a_3 - 12a_3^2 + 12a_2a_4 \right)$$ $$= \frac{1}{2304} \left( -c_1^4 - 68c_1^2c_2 - 64c_2^2 + 36c_1c_3 \right).$$ <span id="page-8-1"></span>Applying the triangle inequality to (2.10) and using Lemma 2.1, we obtain (2.11) $$2304|H_{2,1}(F_{f^{-1}}/2)| \le |c_1|^4 + 68|c_1|^2|c_2| + 64|c_2|^2 + 36|c_1| \left(1 - |c_1|^2 - \frac{|c_2|^2}{1 + |c_1|}\right).$$ Suppose that $x = |c_1|$ and $y = |c_2|$ . Then it follows from (2.11) that <span id="page-8-2"></span> $$(2.12) 2304|H_{2,1}(F_{f^{-1}}/2)| \le N(x,y),$$ where $$N(x,y) := x^4 + 68x^2y + 64y^2 + 36x\left(1 - x^2 - \frac{y^2}{1+x}\right).$$ In view of Lemma 2.1, the region of variability of a pair (x, y) coincides with the set $\Omega$ as defined in (2.6). Given this information, we proceed to find the maximum value of N(x, y) within the region $\Omega$ . Therefore, the critical point of N(x, y) satisfies the conditions $$\frac{\partial N}{\partial x} = 4x^3 - 108x^2 + 136xy + 36 - \frac{36y^2}{(1+x)} + \frac{36xy^2}{(1+x)^2}$$ and $$\frac{\partial N}{\partial y} = 68x^2 + 128y - \frac{72xy}{(1+x)}.$$ By applying the analogous reasoning as in the proof of Theorem 2.1, we establish that the maximum of N(x, y) is obtained on the boundary of $\Omega$ . Specifically, on the boundary, it can be seen that $$N(x,0) = 36x - 36x^3 + x^4 \le 13.969963...$$ for $0 \le x \le 1$ , $N(0,y) = 64y^2 \le 64$ for $0 \le y \le 1$ and $$N(x, 1 - x^2) = 64 - 24x^2 - 39x^4 \le 64$$ for $0 \le x \le 1$ . Therefore, it is clear that $\max_{(x,y)\in\Omega} N(x,y) = 64$ and from (2.12), we easily obtain <span id="page-8-3"></span> $$(2.13) |H_{2,1}(F_{f^{-1}}/2)| \le \frac{1}{36}.$$ In order to show the equality in (2.13) sharp, we consider the function $$h_3(z) = \frac{1}{2} \log \left( \frac{1+z}{1-z} \right) = z + \frac{z^3}{3} + \frac{z^5}{5} \cdots, \quad z \in \mathbb{D}.$$ which belongs to the class $\mathcal{K}_S$ . A simple computation shows that $|H_{2,1}(F_{h_3^{-1}}/2)| = 1/36$ . This shows that the equality holds in (2.12). This completes the proof. <span id="page-9-0"></span>2.3. Sharp bound of $|T_{2,1}(F_{f^{-1}}/2)|$ for $f \in \mathcal{S}_S$ : We derive the subsequent sharp bound of $T_{2,1}(F_{f^{-1}}/2)$ concerning functions within the class $\mathcal{S}_S$ . <span id="page-9-1"></span>Theorem 2.3. Let $f \in \mathcal{S}_S^*$ . Then $$|T_{2,1}(F_{f^{-1}}/2)| \le \frac{5}{16}.$$ The inequality is sharp with the extremal function $$h_5(z) = \frac{z}{1 - iz} = z + iz^2 - z^3 - iz^4 + z^5 + \dots, \ z \in \mathbb{D}.$$ We present a function as an illustrative example to bolster the claim of the strict inequality established in Theorem 2.3. Example 2.3. Consider the function $$h_6(z) = \frac{z}{1-z} = z + z^2 + z^3 + z^4 + \dots, \ z \in \mathbb{D}$$ which belongs to the class $\mathcal{S}_{S}^{*}$ . A simple computation shows that $$|T_{2,1}(F_{h_6^{-1}}/2)| = \frac{3}{16} < \frac{5}{16}.$$ Proof of Theorem 2.3. Let $f \in \mathcal{S}_S^*$ be of the form $f(z) = z + \sum_{n=2}^{\infty} a_n z^n$ , $z \in \mathbb{D}$ . Then by the definition of subordination there exits a Schwarz function $w(z) = \sum_{n=1}^{\infty} c_n z^n$ such that <span id="page-9-2"></span>(2.14) $$\frac{2zf'(z)}{f(z) - f(-z)} = \frac{1 + w(z)}{1 - w(z)}.$$ Now, comparing the coefficients on both side of (2.14), we see that <span id="page-9-3"></span> $$(2.15) a_2 = c_1 \text{ and } a_3 = c_2 + c_1^2.$$ In view of (1.9) and (2.15), a simple computation shows that $$T_{2,1}(F_{f^{-1}}/2) = \frac{1}{16}(-9a_2^4 + 4a_2^2 - 4a_3^2 + 12a_2^2a_3)$$ $$= \frac{1}{16}(-c_1^4 + 4c_1^2 - 4c_2^2 + 4c_1^2c_2).$$ (2.16) <span id="page-9-4"></span>In view of the triangle inequality, the equation (2.16) can be written as <span id="page-9-5"></span> $$(2.17) 16|T_{2,1}(F_{f^{-1}}/2)| \le |c_1|^4 + 4|c_1|^2 + 4|c_2|^2 + 4|c_1|^2|c_2|.$$ Suppose that $x = |c_1|$ and $y = |c_2|$ . Then it follows from (2.17) that <span id="page-9-6"></span> $$(2.18) 16|T_{2,1}(F_{f^{-1}}/2)| \le P(x,y),$$ where $$P(x,y) = x^4 + 4x^2 + 4y^2 + 4x^2y.$$ In view of Lemma 2.1, the region of variability of a pair (x, y) coincides with the set $\Omega$ as defined in (2.6). Taking this information into account, we proceed to compute the maximum value of P(x,y) within the bounded area $\Omega$ . The critical point of P(x,y) satisfies the conditions $$\frac{\partial P}{\partial x} = 4x^3 + 8x + 8xy,$$ $$\frac{\partial P}{\partial y} = 8y + 4x^2.$$ The function P(x,y) cannot have a maximum in the interior of $\Omega$ , as it has no solution in the interior of $\Omega$ . In fact, P(x,y) is continuous on a compact set $\Omega$ , the maximum of P(x,y) attains boundary of $\Omega$ . On the boundary, an easy computation yields that $$P(x,0) = x^4 + 4x^2 \le 5 \text{ for } 0 \le x \le 1,$$ $P(0,y) = 4y^2 \le 4 \text{ for } 0 \le y \le 1$ and $$P(x, 1 - x^2) = x^4 + 4 \le 5$$ for $0 \le x \le 1$ . Therefore, we see that $\max_{(x,y)\in\Omega} P(x,y) = 5$ and from (2.18) we obtain <span id="page-10-1"></span> $$|T_{2,1}(F_{f^{-1}}/2)| \le \frac{5}{16}.$$ To prove the equality in (2.19), we consider the function $$h_5(z) = \frac{z}{1 - iz} = z + iz^2 - z^3 - iz^4 + z^5 + \dots, \ z \in \mathbb{D}.$$ It is easy to see that $$\operatorname{Re}\left(\frac{zh_5'(z)}{h_5(z)-h_5(-z)}\right) = \frac{1}{2}\operatorname{Re}\left(\frac{1+iz}{1-iz}\right) > 0, \ z \in \mathbb{D}.$$ and hence the function $h_1$ belongs to the class $\mathcal{S}_S^*$ . Moreover, by a simple computation, it can be shown that $|T_{2,1}(F_{h_5^{-1}}/2)| = 5/16$ and hence equality hods in (2.19). This completes the proof. <span id="page-10-0"></span>2.4. Sharp bound of $|T_{2,1}(F_{f^{-1}}/2)|$ for $f \in \mathcal{K}_S$ : In the context of $\mathcal{K}_S$ functions, we deduce the sharp bound for $T_{2,1}(F_{f^{-1}}/2)$ . <span id="page-10-2"></span>Theorem 2.4. Let $f \in \mathcal{K}_S$ . Then $$|T_{2,1}(F_{f^{-1}}/2)| \le \frac{145}{2304}.$$ The extremal function establishes a sharp inequality $$h_7(z) = \frac{\sqrt{6}}{\sqrt{3\sqrt{145}}} \log \left( \frac{1 + \left(\frac{\sqrt{3\sqrt{145}}}{2\sqrt{6}}\right) z}{1 - \left(\frac{\sqrt{3\sqrt{145}}}{2\sqrt{6}}\right) z} \right) = z + \frac{\sqrt{145}}{24} z^3 + \frac{29}{64} z^5 + \dots \quad z \in \mathbb{D}.$$ An example of a function in the class $\mathcal{K}_S$ is provided to demonstrate Theorem 2.2. The example provided satisfies the prerequisites of the theorem and serves as evidence that the inequality is strictly true.