On Hankel and Inverse Hankel Determinants of Order Two for Some Subclasses of Analytic Functions

Abstract

In view of the subclass $S{L}^{*}\left(\beta \right),$ which for $\beta =0$ reduces to the class $S{L}^{*},$ two more subclasses $\mathcal{C}L\left(\beta \right)$ and $\mathcal{G}L\left(\beta \right)$ are introduced. For all these three subclasses, we investigate upper bounds of second Hankel and second inverse Hankel determinates. In most of the cases, the results are sharp.

1. Introduction

Let us denote by $\mathcal{A}$ the class of analytic functions f over the unit open disk $E=\{z\in \mathbb{C}:|z|<1\},$ which are normalized as $f\left(0\right)=0=f^{\prime }\left(z\right)-1,$ and hence of the form

$$f\left(z\right)=z+\sum _{k=2}^{\infty }{a}_k{z}^k,$$

(1)

and the subclass of members of $\mathcal{A}$ which are univalent too, by $\mathcal{S}.$ The relation of subordination between analytic functions provides a convenient way to define various subclasses of $\mathcal{A}$ or $\mathcal{S}$. We say that an analytic function f in E is subordinate to an analytic function g in E written as $f\prec g,$ if there is some function $w\left(z\right)$ satisfying $\left|w\right(z\left)\right|<\left|z\right|$ and $w\left(0\right)=0$, known as the Schawarz function, such that $f\left(z\right)=g\left(w\right(z\left)\right)$. More specifically, if g is univalent then the relation $f\prec g$ holds if and only if $f\left(E\right)\subset g\left(E\right)$.

Further, for $f\in \mathcal{A}$ and of the form (1), we denote the Hankel determinant of order q by $H_{q,n}\left(f\right),$ where the integers q and n are such that $q\ge 2$ and $n\ge 1$ and is defined as follows:

$$H_{q,n}\left(f\right)=det\left[\begin{array}{cccc}{a}_n& a_{n+1}& \cdots & a_{n+q-1}\\ a_{n+1}& a_{n+2}& \cdots & a_{n+q}\\ ⋮& ⋮& ⋮& ⋮\\ a_{n+q-1}& a_{n+q}& \cdots & a_{n+2q-2}\end{array}\right].$$

(2)

This was from Pommerenke [1], who first worked on it. Hankel determinants have applications in the theory of singularities [2], and in the study of power series having integer coefficients [3,4,5]. Computing upper bounds of the absolute of Hankel determinants for various subclasses of analytic and univalent functions for larger values of q and n are tough to handle but at the same time interesting enough not to be left untouched. Hence, many authors worked in this direction and obtained upper bounds (sharp in many cases) of second- and third-order Hankel determinants for different subclasses of class $\mathcal{S}$; for example, see [6,7,8,9,10,11,12,13].

Expanding the second-order Hankel determinant $H_{2,2}\left(f\right)$ and third-order Hankel determinant $H_{3,1}\left(f\right)$ for f of the form (1), give

$$H_{2,2}\left(f\right)=\left|\begin{array}{cc}{a}_2& a_3\\ a_3& a_4\end{array}\right|=a_2{a}_4-a_3^2,$$

and

$$H_{3,1}\left(f\right)=\left|\begin{array}{ccc}1& a_2& a_3\\ a_2& a_3& a_4\\ a_3& a_4& a_5\end{array}\right|=a_3(a_2{a}_4-a_3^2)-a_4(a_4-a_2{a}_3)+a_5(a_3-a_2^2).$$

While the Hankel determinant $H_{2,1}\left(f\right)=a_3-a_2^2$ is a special case of $a_3-\mu a_2^2,$ for $\mu =1,$ which is the famous Fekete–Szeg$\ddot{o}$ functional [6]. For work in this direction, see for example [14].

Further, researchers have also been started investigating the second Hankel determinant for $f^{-1},$ that is, inverse functions of the functions f in various subclasses of $\mathcal{S}.$ Since each $f\in \mathcal{S}$ is univalent, so if $w=f\left(z\right),$ then there must be some disk $\left|w\right|<r_0\left(f\right)$ for which $f^{-1}$ exists. More specifically, there will be at least the disk with $r_0=1/4,$ guaranteed by Koebe’s quarter theorem. Let

$$f^{-1}\left(w\right)=w+B_2{w}^2+B_3{w}^3+\dots ,$$

(3)

where, $f\left(f^{-1}\left(w\right)\right)=w,$ so from (1) and (3) by equating coefficients, we have

$$\left\{\begin{array}{c}{B}_2=-a_2,\hfill \\ B_3=-a_3+2a_2^2,\hfill \\ B_4=-a_4+5a_2{a}_3-5a_2^3.\hfill \end{array}\right.$$

(4)

Second and third coefficient estimates of inverse functions belonging to a certain subclass of Bazilevic functions have recently been obtained in [15].

Now, for the inverse function $f^{-1}$ of the form (3), the second Hankel determinant is given by

$$H_{2,2}\left(f^{-1}\right)=B_2{B}_4-B_3^2$$

$$=a_2{a}_4-a_3^2-a_2^2(a_3-a_2^2).$$

(5)

Before introducing our three subclasses, it is necessary to have definitions of the classes, $\mathcal{C}\left(\alpha \right),$${\mathcal{S}}^{*}\left(\alpha \right)$ and $\mathcal{G}\left(\alpha \right).$ For $0\le \alpha \le 1,$ the subclass $\mathcal{C}\left(\alpha \right)$ is known as the class of convex functions of order $\alpha$ and ${\mathcal{S}}^{*}\left(\alpha \right)$ the class of starlike functions of order $\alpha ,$ defined, respectively, as

$$\mathcal{C}\left(\alpha \right)=\left\{f\in \mathcal{A}:\Re \left(1+\frac{z{f}^{\prime \prime }\left(z\right)}{f^{\prime }\left(z\right)}\right)>\alpha ,z\in E\right\},$$

and

$${\mathcal{S}}^{*}\left(\alpha \right)=\left\{f\in \mathcal{A}:\Re \left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)>\alpha ,z\in E\right\}.$$

Since convex functions as well as starlike functions are univalent, both $\mathcal{C}\left(\alpha \right),$${\mathcal{S}}^{*}\left(\alpha \right)$ are subclasses of the class $\mathcal{S}$ and $\mathcal{C}\left(\alpha \right)\subset {\mathcal{S}}^{*}\left(\alpha \right).$ Also for $\alpha =0,$$\mathcal{C}\left(\alpha \right)$ and ${\mathcal{S}}^{*}\left(\alpha \right)$ become the well-known subclasses $\mathcal{C}$ and ${\mathcal{S}}^{*}$, respectively, of convex and starlike functions. For $f\in \mathcal{C}\left(\alpha \right),$ a non-sharp bound of $H_{2,2}\left(f\right)$ was given by V. Krishna et al. [7], which was further improved by Thomas et al. [8], and finally the best possible result was obtained by Y. J. Sim et al. [9]. While for $f\in S^{*}\left(\alpha \right)$, the sharp bound for $H_{2,2}\left(f\right)$ was obtained by N. E. Cho et al. [10]. In the paper [9], they are also successful in obtaining the sharp upper bounds of $H_{2,2}\left(f^{-1}\right)$ for the inverse function $f^{-1}$ when $f\in \mathcal{C}\left(\alpha \right)$, as well as when $f\in {\mathcal{S}}^{*}\left(\alpha \right),$ and as such the struggle for finding sharp upper bounds of $H_{2,1}\left(f\right)$ and $H_{2,2}\left(f^{-1}\right)$ for these two subclasses came to an end.

Further, the class $\mathcal{G}\left(\alpha \right),$ when $0<\alpha \le 1,$ is defined as

$$\mathcal{G}\left(\alpha \right)=\left\{f\in \mathcal{A}:\Re \left(1+\frac{z{f}^{\prime \prime }\left(z\right)}{f^{\prime }\left(z\right)}\right)<1+\frac{\alpha }{2},z\in E\right\},$$

Ozaki [12] considered the class $\mathcal{G}\left(\alpha \right)$ for $\alpha =1,$ denoted by $\mathcal{G}$, and proved that $\mathcal{G}\subset \mathcal{S},$ the class of univalent functions. Later, a general form of this class was discussed by Umezawa [13]; in this paper he showed that each $f\in \mathcal{G}$ is convex in one direction. Apart from this, it has also been proven (see [16,17]) that $\mathcal{G}$ is a subclass of the class ${\mathcal{S}}^{*}$ of starlike functions, and so $\mathcal{G}\left(\alpha \right),$ for $0<\alpha \le 1$ is a subclass of the class of starlike functions ${\mathcal{S}}^{*}.$

Motivated by the work in [18] for the three classes $\mathcal{C}\left(\alpha \right),$${\mathcal{S}}^{*}\left(\alpha \right)$ and $\mathcal{G}\left(\alpha \right)$ and the way the authors in [19] defined the class $S{L}^{*}\left(\beta \right)$ as

$$\mathcal{S}{L}^{*}\left(\beta \right)=\left\{f\in \mathcal{A}:\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\prec \beta +(1-\beta )\sqrt{1+z},z\in E\right\},$$

which for $\beta =0$ reduces to the class $S{L}^{*},$ defined by J. Sokól et al. in [20], we define the two subclasses $\mathcal{C}L\left(\beta \right)$ and $\mathcal{G}L\left(\beta \right)$ for $0\le \beta <1$ and $0<\beta <1$, respectively, by

$$\mathcal{C}L\left(\beta \right)=\left\{f\in \mathcal{A}:1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\prec \beta +(1-\beta )\sqrt{1+z},z\in E\right\},$$

and

$$\mathcal{G}L\left(\beta \right)=\left\{f\in \mathcal{A}:1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\prec 1+\frac{\beta }{2}-\frac{\beta }{2}\sqrt{1+z},z\in E\right\},$$

We need the following lemmas in the sequel.

Lemma 1 

([21]). Let $w\left(z\right)=c_{1}z+c_2{z}^2+c_3{z}^3+c_4{z}^4+\dots ,$ be the Schwarz function with $w\left(0\right)=0$ and $\left|w\right(z\left)\right|<1,$ for all $z\in E$ and $\nu ,$ be any complex number, then

$$|c_2-\nu c_1^2|\le max\{1,\left|\nu \right|\}.$$

Lemma 2 

([11]). For an analytic function of the form

$$w\left(z\right)=\sum _{k=1}^{\infty }{c}_k{z}^k$$

(6)

in the open unit disk $E,$ with $w\left(0\right)=0$ and $\left|w\right(z\left)\right|<1,$ known as Schwarz function, the following inequalities hold:

$$|c_1|\le 1,|c_2|\le 1-|c_1{|}^2,and|c_3|\le 1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|}.$$

2. Second Hankel Determinant

First, we investigate upper bounds for second Hankel determinant, namely $H_{2,2}\left(f\right)$ as well as for $H_{2,1}\left(f\right)$, in all the above-mentioned three subclasses one by one.

Theorem 1. 

Let $f\in \mathcal{C}L\left(\beta \right)$, and is of the form $\left(1\right)$, then

$$|H_{2,1}\left(f\right)|\le \frac{1}{12}(1-\beta ),$$

and

$$|H_{2,2}\left(f\right)|\le \frac{1}{144}{(1-\beta )}^2.$$

These results are sharp.

Proof. 

If $f\in \mathcal{C}L\left(\beta \right)$, and has the form $\left(1\right)$, then we have

$$1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\prec \beta +(1-\beta )\sqrt{1+z},$$

and so by definition of subordination, there is some Schwarz function $w\left(z\right)$ of the form $\left(6\right),$ with $w\left(0\right)=0$ and $\left|w\right(z\left)\right|<1,$ for all $z\in E,$ such that

$$1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}=\beta +(1-\beta )\sqrt{1+w\left(z\right)}.$$

(7)

Using values from Equations (1) and (6) into the above Equation $\left(7\right)$, then equating the coefficients of $z,z^2,z^3,$ after some simplification we have:

$$a_2=\frac{1}{4}(1-\beta )c_1,$$

(8)

$$a_3=\frac{1}{12}(1-\beta )\left[c_2+\frac{1}{4}(1-2\beta )c_1^2\right],$$

(9)

$$a_4=\frac{1}{384}(1-\beta )\left[16c_3+4(1-3\beta )c_1{c}_2+(2{\beta }^2-\beta +1)c_1^3\right].$$

(10)

By using Equations (8) and (9), we have

$$\begin{array}{ccc}\hfill H_{2,1}\left(f\right)& =& a_3-a_2^2\hfill \\ & =& \frac{1}{12}(1-\beta )\left[c_2+\frac{1}{4}(1-2\beta )c_1^2\right]-\frac{1}{16}{(1-\beta )}^2{c}_1^2\hfill \\ & =& \frac{1}{12}(1-\beta )\left[c_2+\frac{1}{4}(1-2\beta )c_1^2-\frac{3}{4}(1-\beta )c_1^2\right]\hfill \\ & =& \frac{1}{12}(1-\beta )\left[c_2+\frac{1}{4}\left\{(1-2\beta )-3(1-\beta )\right\}{c}_1^2\right],\hfill \end{array}$$

that is,

$$H_{2,1}\left(f\right)=\frac{1}{12}(1-\beta )\left[c_2-\frac{1}{4}(2-\beta )c_1^2\right].$$

(11)

Taking modulus and using Lemma 1, we have

$$|a_3-a_2^2|\le \frac{1}{12}(1-\beta )max\{1,\frac{1}{4}|2-\beta |\}.$$

Hence,

$$|a_3-a_2^2|\le \frac{1}{12}(1-\beta ).$$

This result is sharp for $w\left(z\right)=z^2,$ in Equation $\left(7\right)$ or for $c_2=1$ and $c_1=0$ into Equation $\left(11\right)$ we obtain the equality. Further, by using values from Equations $\left(8\right)$–$\left(10\right),$ into $H_{2,2}\left(f\right)=a_2{a}_4-a_3^2,$ after some computation, we have

$$H_{2,2}\left(f\right)=\frac{1}{2304}{(1-\beta )}^2\left[24c_1{c}_3-16c_2^2-2(1+\beta )c_1^2{c}_2-\frac{1}{2}(2{\beta }^2-5\beta -1)c_1^4\right].$$

(12)

where, $2{\beta }^2-5\beta -1<0,$ for $0\le \beta <1.$ Taking the absolute and applying the triangle inequality, we have

$$|H_{2,2}\left(f\right)|\le \frac{1}{2304}{(1-\beta )}^2\left[24|c_1||c_3|+16|c_2{|}^2+2(1+\beta )|c_1{|}^2|c_2|-\frac{1}{2}(2{\beta }^2-5\beta -1){|c_1|}^4\right].$$

Now, utilizing the results of Lemma 2

$$|H_{2,2}\left(f\right)|\le \frac{1}{2304}{(1-\beta )}^2{\displaystyle [}24|c_1|(1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|})+16|c_2{|}^2$$

$$+2(1+\beta )|c_1{|}^2(1-|c_1{|}^2)-\frac{1}{2}(1+5\beta -2{\beta }^2){|c_1|}^4]$$

$$=\frac{1}{2304}{(1-\beta )}^2{\displaystyle [}24|c_1|-24|c_1{|}^3+(16-\frac{24|c_1|}{1+|c_1|}){|c_2|}^2$$

$$+2(1+\beta )|c_1{|}^2-2(1+\beta )|c_1{|}^4-\frac{1}{2}(1+5\beta -2{\beta }^2){|c_1|}^4]$$

Using the result $|c_2|\le 1-|c_1{|}^2,$ combining coefficients of the same powers of $|c_1|,$ and simplifying, we have

$$|H_{2,2}\left(f\right)|\le \frac{1}{2304}{(1-\beta )}^2\left[16-2(3-\beta )|c_1{|}^2-\frac{1}{2}(19-\beta +{\beta }^2){|c_1|}^4\right].$$

Clearly, the maximum of the above inequality can be obtained at $|c|=0$; thus, we have

$$|H_{2,2}\left(f\right)|\le \frac{1}{2304}{(1-\beta )}^2\left(16\right).$$

This gives the required result, which is sharp for $w\left(z\right)=z^2$, in Equation $\left(7\right)$ or substituting $c_2=1$ and placing all the remaining $c_i^{\prime }s$ equal to zero into Equation $\left(12\right),$ yields the equality. Hence the result. □

Theorem 2. 

Let $f\in \mathcal{S}{L}^{*}\left(\beta \right),$ and be of the form (1), then

$$|H_{2,1}\left(f\right)|\le \frac{1}{4}(1-\beta ),$$

and

$$|H_{2,2}\left(f\right)|\le \frac{1}{16}{(1-\beta )}^2.$$

Both these results are sharp.

Proof. 

If $f\in \mathcal{S}{L}^{*}\left(\beta \right),$ then by the definition of this class, we have

$$\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\prec \beta +(1-\beta )\sqrt{1+z},$$

and so by the definition of subordination, there exists a Schwarz function $w\left(z\right)$ of the form $\left(6\right)$ with $w\left(0\right)=0$ and $|w(z)|<1,$ for $z\in E,$ such that

$$\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}=\beta +(1-\beta )\sqrt{1+w\left(z\right)},$$

(13)

where,

$$\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}=1+a_{2}z+\left(2a_3-a_2^2\right)z^2+\left(3a_4-3a_2{a}_3+a_2^3\right)z^3+\dots ,$$

and

$$\beta +(1-\beta )\sqrt{1+w\left(z\right)}=1+\frac{1}{2}(1-\beta )c_{1}z+\frac{1}{2}(1-\beta )\left(c_2-\frac{1}{4}{c}_1^2\right)z^2$$

$$+\frac{1}{2}(1-\beta )\left(c_3-\frac{1}{2}{c}_1{c}_2+\frac{1}{8}{c}_1^3\right)z^3+\dots .$$

Substituting values from the last two equations into Equation $\left(13\right)$, then by comparing coefficients of $z,$$z^2,$$z^3$ with a little simplification, we have

$$a_2=\frac{1}{2}(1-\beta )c_1,$$

(14)

$$a_3=\frac{1}{4}(1-\beta )\left[c_2+\frac{1}{4}(1-2\beta )c_1^2\right],$$

(15)

$$a_4=\frac{1}{6}(1-\beta )\left[c_3+\frac{1}{4}(1-3\beta )c_1{c}_2+\frac{1}{16}(2{\beta }^2-\beta +1)c_1^3\right].$$

(16)

Using Equations $\left(13\right)$ and $\left(14\right)$ into $H_{1,2}\left(f\right)=a_3-a_2^2$ and simplifying, we have

$$H_{2,1}\left(f\right)=\frac{1}{4}(1-\beta )\left[c_2-\frac{3}{4}(1-\frac{2}{3}\beta )c_1^2\right].$$

(17)

Taking the absolute and using Lemma 1, we have

$$|H_{2,1}\left(f\right)|\le \frac{1}{4}(1-\beta )max\left\{1,\left|\frac{3}{4}(1-\frac{2}{3}\beta )\right|\right\}.$$

where, $|\frac{3}{4}(1-\frac{2}{3}\beta )|\le 1$; therefore, we obtain

$$|H_{2,1}\left(f\right)|\le \frac{1}{4}(1-\beta ),$$

which is sharp for $w\left(z\right)=z^2$ in Equation $\left(13\right),$ or $c_1=0$ and $c_2=1$ in Equation $\left(17\right)$ gives the equality.

Next, we calculate $H_{2,2}\left(f\right)=a_2{a}_4-a_3^2,$ by using Equations $\left(14\right)$–$\left(16\right).$ After some simplification, we have

$$H_{2,2}\left(f\right)=\frac{1}{192}{(1-\beta )}^2\left[16c_1{c}_3-12c_2^2-2c_1^2{c}_2-\frac{1}{4}(4{\beta }^2-8\beta -1)c_1^4\right].$$

(18)

Observe that $4{\beta }^2-8\beta -1<0,$ for $0\le \beta <1.$ Taking the absolute and applying the triangle inequality, we have

$$|H_{2,2}\left(f\right)|\le \frac{1}{192}{(1-\beta )}^2\left[16|c_1||c_3|+12|c_2{|}^2+2|c_1{|}^2|c_2|-\frac{1}{4}(4{\beta }^2-8\beta -1){|c_1|}^4\right].$$

Using inequalities of Lemma 2, we have

$$|H_{2,2}\left(f\right)|\le \frac{1}{192}{(1-\beta )}^2{\displaystyle [}16|c_1|(1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|})+12|c_2{|}^2+2|c_1{|}^2(1-|c_1{|}^2)$$

$$+\frac{1}{4}(1+8\beta -4{\beta }^2){|c_1|}^4]$$

$$=\frac{1}{192}{(1-\beta )}^2{\displaystyle [}16|c_1|-16|c_1{|}^3+(12-\frac{16|c_1|}{1+|c_1|})|c_2{|}^2+2|c_1{|}^2-2{|c_1|}^4$$

$$+\frac{1}{4}(1+8\beta -4{\beta }^2){|c_1|}^4].$$

Using $|c_2|\le 1-|c_1{|}^2$ from Lemma 2, and gathering coefficients of like powers of $|c_1|,$ after some computation, we have

$$|H_{2,2}\left(f\right)|\le \frac{1}{192}{(1-\beta )}^2\left[12-6|c_1{|}^2-\frac{1}{4}(4{\beta }^2-8\beta +23){|c_1|}^4\right],$$

where, clearly $(4{\beta }^2-8\beta +23)>0,$ for $0\le \beta <1,$ hence the maximum of the above inequality occurs at $|c_1|=0.$ Therefore, we obtain the required result for $|c|=0.$ Also, the inequality is the best possible for $w\left(z\right)=z^2$ in Equation $\left(13\right)$ or for $c_2=1$ and all the remaining zeros in Equation $\left(18\right).$

For $\beta =0,$ we have $|H_{2,2}\left(f\right)|\le 1/16,$ which agrees with the result obtained by J. Sokól et al. in [22]. □

Theorem 3. 

Let $f\in \mathcal{G}L\left(\beta \right),$$0<\beta <1$, and be of the form $\left(1\right),$ then

$$|H_{2,1}\left(f\right)|\le \frac{1}{24}\beta ,$$

and

$$|H_{2,2}\left(f\right)|\le \frac{1}{576}\beta .$$

Both the above inequalities are sharp.

Proof. 

If $f\in \mathcal{G}L\left(\beta \right)$ and is of the form $\left(1\right),$ then by definition of this class, we have

$$1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\prec 1+\frac{\beta }{2}-\frac{\beta }{2}\sqrt{1+z}.$$

Hence, by the definition of subordination, there is a Schwarz function $w\left(z\right)$ of the form (6), with $w\left(0\right)=0,$ and $|w(z)|<1$ for all $z\in E,$ such that

$$1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}=1+\frac{\beta }{2}-\frac{\beta }{2}\sqrt{1+w\left(z\right)}.$$

(19)

After simplification, and comparing coefficients of like powers of $z,$ we have

$$a_2=-\frac{1}{8}\beta c_1,$$

(20)

$$a_3=-\frac{1}{24}\beta \left[c_2-\frac{1}{4}(1+\beta )c_1^2\right],$$

(21)

$$a_4=-\frac{1}{48}\beta \left[c_3-\frac{1}{2}(1+\frac{3}{4}\beta )c_1{c}_2+\frac{1}{32}({\beta }^2+3\beta +4)c_1^3\right].$$

(22)

Now, by utilizing Equations $\left(20\right)$ and $\left(21\right),$ we compute $H_{2,1}\left(f\right)$ as follows:

$$H_{2,1}\left(f\right)=-\frac{1}{24}\beta \left[c_2-\frac{1}{8}(2-\beta )c_1^2\right].$$

(23)

Applying the absolute and using Lemma 1, we have

$$|H_{2,1}|\le \frac{1}{24}\beta max\{1,\frac{1}{8}|2-\beta |\}.$$

Since $(2-\beta )/8<1$ for $0<\beta <1,$ therefore, we have

$$|H_{2,1}\left(f\right)|\le \frac{1}{24}\beta .$$

This result is sharp for $w\left(z\right)=z^2$ in Equation $\left(19\right)$, or for $c_1=0$ and $c_2=1$ in Equation $\left(23\right).$

Next, we compute $H_{2,2}\left(f\right)$ by utilizing Equations $\left(20\right)$–$\left(22\right).$ After some simplification, we have

$$H_{2,2}\left(f\right)=\frac{1}{36864}{\beta }^2\left[96c_1{c}_3-64c_2^2-4(4+\beta )c_1^2{c}_2+(8+\beta -{\beta }^2)c_1^4\right].$$

(24)

Taking the modulus, and applying the triangle inequality, we have

$$|H_{2,2}\left(f\right)|\le \frac{1}{36864}{\beta }^2\left[96|c_1||c_3|+64|c_2{|}^2+4(4+\beta )|c_1{|}^2|c_2|+(8-\beta +{\beta }^2){|c_1|}^4\right].$$

Applying results from Lemma 2, we have

$$|H_{2,2}\left(f\right)|\le \frac{1}{36864}{\beta }^2{\displaystyle [}96|c_1|(1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|})+64|c_2{|}^2+4(4+\beta )|c_1{|}^2(1-|c_1{|}^2)$$

$$+(8-\beta +{\beta }^2){|c_1|}^4]$$

$$=\frac{1}{36864}{\beta }^2{\displaystyle [}96|c_1|-96|c_1{|}^3+(64-\frac{96|c_1|}{1+|c_1|})|c_2{|}^2+4(4+\beta ){|c_1|}^2$$

$$-4(4+\beta )|c_1{|}^4+(8-\beta +{\beta }^2){|c_1|}^4].$$

Utilizing $|c_2|\le 1-|c_1{|}^2$ from Lemma 2, and combining and simplifying coefficients of like powers of $|c_1|,$ we have

$$|H_{2,2}\left(f\right)|\le \frac{1}{36864}{\beta }^2\left[64-4(4-\beta )|c_1{|}^2-({\beta }^2+3\beta +40){|c_1|}^4\right]$$

$$\le \frac{1}{576}{\beta }^2.$$

For $w\left(z\right)=z^2$ in Equation (19), or $c_2=1$ and putting all other $c_i^{\prime }s$ equal to zero in Equation (24) proves the sharpness of this result. □

3. Second Hankel Determinant of Inverse Functions

Studying Hankel determinants of inverse functions provides valuable insights into different aspects of analytic functions and their geometry, including univalence and distortion, boundary behavior, geometric properties under inversion, analytic continuation and coefficient relationships of the analytic functions and its inverses. Researchers in Geometric Function Theory explore these properties to understand the intricate relationship between analytic functions, their inverses and the geometric implications within the complex plane.

Theorem 4. 

Let $f\in \mathcal{C}L\left(\beta \right),$ and be of the form $\left(1\right),$ then

$$H_{2,2}\left(f^{-1}\right)=\left\{\begin{array}{cc}\frac{1}{144}{(1-\beta )}^2\left[1+\frac{{(3-5\beta )}^2}{8(31-7\beta -4{\beta }^2)}\right],\hfill & if0\le \beta \le \frac{3}{5},\hfill \\ \frac{1}{144}{(1-\beta )}^2,\hfill & if\frac{3}{5}\le \beta <1.\hfill \end{array}\right.$$

The second inequality is sharp.

Proof. 

Consider Equation $\left(5\right),$ that is

$$H_{2,2}\left(f^{-1}\right)=a_2{a}_4-a_3^2-a_2^2(a_3-a_2^2),$$

and as $f\in \mathcal{C}L\left(\beta \right),$ so using Equations $\left(8\right)$–$\left(10\right)$ and a little simplification yields

$$H_{2,2}\left(f^{-1}\right)=\frac{1}{1152}{(1-\beta )}^2\left[12c_1{c}_3-8c_2^2-(7-5\beta )c_1^2{c}_2+\frac{1}{4}(4{\beta }^2-13\beta +13)c_1^4\right].$$

(25)

Taking the absolute and applying the triangle inequality, we have

$$|H_{2,2}\left(f^{-1}\right)|\le \frac{1}{1152}{(1-\beta )}^2\left[12|c_1|c_3|+8|c_2{|}^2+(7-5\beta )|c_1{|}^2|c_2|+\frac{1}{4}(4{\beta }^2-13\beta +13){|c_1|}^4\right].$$

Using results from Lemma 2, we have

$$|H_{2,2}\left(f^{-1}\right)|\le \frac{1}{1152}{(1-\beta )}^2{\displaystyle [}12|c_1|(1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|})+8|c_2{|}^2+(7-5\beta )|c_1{|}^2(1-|c_1{|}^2)$$

$$+\frac{1}{4}(4{\beta }^2-13\beta +13){|c_1|}^4]$$

$$=\frac{1}{1152}{(1-\beta )}^2[12|c_1|-12|c_1{|}^3+(8-\frac{12|c_1|}{1+|c_1|})|c_2{|}^2+(7-5\beta ){|c_1|}^2$$

$$-(7-5\beta )|c_1{|}^4+\frac{1}{4}(4{\beta }^2-13\beta +13){|c_1|}^4].$$

Utilizing the result $|c_2|\le 1-|c_1{|}^2$ from Lemma 2, and computing the coefficients of the same powers of $|c_1|,$ we have

$$|H_{2,2}\left(f^{-1}\right)|\le \frac{1}{1152}{(1-\beta )}^2\left[8+(3-5\beta )|c_1{|}^2-\frac{1}{4}(31-7\beta -4{\beta }^2){|c_1|}^4\right].$$

(26)

Here, $(31-7\beta -4{\beta }^2)>0,$ for $0\le \beta <1.$ □

Case-I: Here, $0\le \beta <\frac{3}{5}.$ For this interval of $\beta ,$$(3-5\beta )>0.$

Let $|c_1|=x,$ then by Lemma 2, $0\le x\le 1.$ Also, if

$$g\left(x\right)=8+(3-5\beta )x^2-\frac{1}{4}(31-7\beta -4{\beta }^2)|x^4.$$

Then, the inequality $\left(26\right)$ becomes

$$|H_{2,2}\left(f^{-1}\right)|\le \frac{1}{1152}{(1-\beta )}^{2}g\left(x\right).$$

Differentiating $g\left(x\right)$ twice with respect to x, we have

$$g^{\prime }\left(x\right)=2(3-5\beta )x-(31-7\beta -4{\beta }^2)x^3,$$

and

$$g^{\prime \prime }\left(x\right)=2(3-5\beta )-3(31-7\beta -4{\beta }^2)x^2.$$

setting $g^{\prime }\left(x\right)=0,$ gives

$$x=0,\sqrt{\frac{2(3-5\beta )}{31-7\beta -4{\beta }^2}}.$$

Since for $x=0,$$g^{\prime \prime }\left(0\right)=2(3-5\beta )>0,$ and for $x^2=\frac{2(3-5\beta )}{31-7\beta -4{\beta }^2},$ we have $g^{\prime \prime }\left(x\right)=-4(3-5\beta )<0,$ therefore

$$max\left(g\left(x\right)\right)=g\left(\sqrt{\frac{2(3-5\beta )}{31-7\beta -4{\beta }^2}}\right).$$

Thus, substituting $|c_1|=\sqrt{\frac{2(3-5\beta )}{31-7\beta -4{\beta }^2}}$ into the inequality $\left(26\right),$ gives

$$H_{2,2}\left(f^{-1}\right)\le \frac{1}{144}{(1-\beta )}^2)\left[1+\frac{{(3-5\beta )}^2}{8(31-7\beta -4{\beta }^2)}\right].$$

Case-II: Here, $\frac{3}{5}\le \beta <1.$ For this range of $\beta ,$ we have $(3-5\beta )\le 0.$ Therefore, the right side of the inequality $\left(26\right),$ attains its maximum at $|c_1|=0,$ which gives

$$H_{2,2}\left(f^{-1}\right)\le \frac{1}{1152}{(1-\beta )}^2\left(8\right).$$

The last inequality implies the required result. This inequality is sharp for $w\left(z\right)=z^2$ in Equation $\left(7\right),$ or for $c_2=1$ and all other coefficients’ zeros in Equation $\left(25\right).$

Theorem 5. 

If $f\in \mathcal{S}{L}^{*}\left(\beta \right),$ and is of the form $\left(1\right),$ then

$$H_{2,2}\left(f^{-1}\right)\le \left\{\begin{array}{cc}\frac{1}{16}{(1-\beta )}^2\left[1+\frac{3{(1-2\beta )}^2}{35+4\beta -20{\beta }^2}\right],\hfill & if0\le \beta <\frac{1}{2},\hfill \\ \frac{1}{16}{(1-\beta )}^2,\hfill & if\frac{1}{2}\le \beta <1.\hfill \end{array}\right.$$

The second inequality is sharp.

Proof. 

If $f\in \mathcal{S}{L}^{*}\left(\beta \right),$ then we have to substitute values from Equation $\left(14\right)$ into Equation $\left(16\right)$ and into Equation $\left(5\right).$ Doing so, after some simplification, we obtain

$$H_{2,2}\left(f^{-1}\right)=\frac{1}{192}{(1-\beta )}^2\left[16c_1{c}_3-12c_2^2-12(\frac{7}{6}-\beta )c_1^2{c}_2+(5{\beta }^2-13\beta +\frac{37}{4})c_1^4\right].$$

(27)

Taking the absolute and applying the triangle inequality, we have

$$|H_{2,2}\left(f^{-1}\right)|\le \frac{1}{192}{(1-\beta )}^2\left[16|c_1||c_3|+12|c_2{|}^2+12(\frac{7}{6}-\beta )|c_1{|}^2|c_2|+(5{\beta }^2-13\beta +\frac{37}{4}){|c_1|}^4\right].$$

Now applying results of Lemma 2, we have

$$|H_{2,2}\left(f^{-1}\right)|\le \frac{1}{192}{(1-\beta )}^2{\displaystyle [}16|c_1|(1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|})+12|c_2{|}^2+12(\frac{7}{6}-\beta )|c_1{|}^2(1-|c_1{|}^2)$$

$$+(5{\beta }^2-13\beta +\frac{37}{4}){|c_1|}^4]$$

$$=\frac{1}{192}{(1-\beta )}^2{\displaystyle [}16|c_1|-16|c_1{|}^3+(12-\frac{16|c_1|}{1+|c_1|})|c_2{|}^2+12(\frac{7}{6}-\beta ){|c_1|}^2$$

$$-12(\frac{7}{6}-\beta )|c_1{|}^4+(5{\beta }^2-13\beta +\frac{37}{4}){|c_1|}^4].$$

Combining coefficients of equal powers of $|c_1|$ and simplifying, we have

$$|H_{2,2}\left(f^{-1}\right)|\le \frac{1}{192}{(1-\beta )}^2\left[12+6(1-2\beta )|c_1{|}^2-\frac{1}{4}(35+4\beta -20{\beta }^2){|c_1|}^4\right].$$

(28)

where, $(35+4\beta -20{\beta }^2)>0,$ for $0\le \beta <1.$

Case-I: Here, $0\le \beta <\frac{1}{2};$ in this case $(1-2\beta )>0,$ therefore, here we let

$$h\left(x\right)=12+6(1-2\beta )x^2-\frac{1}{4}(35+4\beta -20{\beta }^2)x^4,$$

as such, the inequality $\left(28\right)$ becomes

$$|H_{2,2}\left(f^{-1}\right)|\le \frac{1}{192}{(1-\beta )}^{2}h\left(x\right),$$

where, $x=|c_1|,$ and so $0\le x\le 1,$ by Lemma 2. Differentiating $h\left(x\right)$ with respect to x twice, we have

$$h^{\prime }\left(x\right)=12(1-2\beta )x-(35+4\beta -20{\beta }^2)x^3,$$

and

$$h^{\prime \prime }\left(x\right)=12(1-2\beta )-3(35+4\beta -20{\beta }^2)x^2.$$

Setting $h^{\prime }\left(x\right)=0,$ and solving, we have

$$x=0,\sqrt{\frac{12(1-2\beta )}{35+4\beta -20{\beta }^2}}.$$

Further, as $g^{\prime \prime }\left(x\right)>0,$ when $x=0$ and $g^{\prime \prime }\left(x\right)<0,$ when $x=\sqrt{\frac{12(1-2\beta )}{35+4\beta -20{\beta }^2}},$ therefore,

$$max\left(h\left(x\right)\right)=h\left(\sqrt{\frac{12(1-2\beta )}{35+4\beta -20{\beta }^2}}\right).$$

Hence, putting $|c_1|=\sqrt{\frac{12(1-2\beta )}{35+4\beta -20{\beta }^2}}$ into the inequality $\left(28\right),$ and simplifying, we have

$$H_{2,2}\left(f^{-1}\right)\le \frac{1}{16}{(1-\beta )}^2\left[1+\frac{3(1-2\beta )}{35+4\beta -20{\beta }^2}\right].$$

Case-II: Here, $\frac{1}{2}\le \beta <1$ and $(1-2\beta )\le 0,$ therefore the maximum of the right side of inequality $\left(28\right)$ occurs at $|c_1|=0,$ and so we obtain

$$H_{2,2}\left(f^{-1}\right)\le \frac{1}{192}{(1-\beta )}^2\left(12\right).$$

This inequality is sharp for $w\left(z\right)=z^2$ in Equation $\left(13\right),$ or for $c_2=1$ and all other coefficients’ zeros in Equation $\left(27\right).$ □

Theorem 6. 

Let $f\in \mathcal{G}L\left(\beta \right),$ and be of the form $\left(1\right),$ then

$$H_{2,2}\left(f^{-1}\right)\le \frac{1}{576}{\beta }^2,$$

This result is sharp.

Proof. 

We remember that $\beta$ in the subclass $\mathcal{G}L\left(\beta \right)$ ranges as $0<\beta <1.$ Further, if f lies in this class, then we have to substitute the values of $a_2,a_3$ and $a_4$ from Equations $\left(20\right)$–$\left(22\right)$, respectively, into Equation $\left(5\right)$ to compute $H_{2,2}\left(f^{-1}\right).$ As such, after a little simplification, we have

$$H_{2,2}\left(f^{-1}\right)=\frac{1}{18432}{\beta }^2\left[48c_1{c}_3-32c_2^2+(10\beta -8)c_1^2{c}_2+\frac{1}{2}(2{\beta }^2-5\beta +8)c_1^4\right].$$

(29)

As before, taking the modulus and applying the triangle inequality, we have

$$|H_{2,2}\left(f^{-1}\right)|\le \frac{1}{18432}{\beta }^2\left[48|c_1||c_3|+32|c_2{|}^2+|10\beta -8||c_1{|}^2|c_2|+\frac{1}{2}(2{\beta }^2-5\beta +8){|c_1|}^4\right].$$

Now, utilizing results of Lemma 2, we have

$$|H_{2,2}\left(f^{-1}\right)|\le \frac{1}{18432}{\beta }^2{\displaystyle [}48|c_1|(1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|})+32|c_2{|}^2+|10\beta -8||c_1{|}^2(1-|c_1{|}^2)$$

$$+\frac{1}{2}(2{\beta }^2-5\beta +8){|c_1|}^4]$$

$$=\frac{1}{18432}{\beta }^2[48|c_1|-48|c_1{|}^3+(32-\frac{48|c_1|}{1+|c_1|})|c_2{|}^2+|10\beta -8||c_1{|}^2$$

$$-|10\beta -8||c_1{|}^4+\frac{1}{2}(2{\beta }^2-5\beta +8){|c_1|}^4].$$

Combining the coefficients of like powers of $|c_1|$ and computing, we have

$$|H_{2,2}\left(f^{-1}\right)|\le \frac{1}{18432}{\beta }^2[32+(|10\beta -8|-16)|c_1{|}^2+\frac{1}{2}\left(2{\beta }^2-5\beta -24-2|10\beta -8|\right){|c_1|}^4].$$

(30)

□

Case-I: Here, $0<\beta \le \frac{8}{10}.$ For this range of $\beta ,$ we have $10\beta -8\le 0.$ Therefore, from the above inequality $\left(30\right),$ we obtain

$$|H_{2,2}\left(f^{-1}\right)|\le \frac{1}{18432}{\beta }^2\left[32-2(4+5\beta )|c_1{|}^2-\frac{1}{2}(40-15\beta -2{\beta }^2){|c_1|}^4\right].$$

Clearly, the maximum occurs at $|c_1|=0.$ Thus, we have

$$|H_{2,2}\left(f^{-1}\right)|\le \frac{1}{18432}{\beta }^2\left(32\right).$$

Case-II: Here, $\frac{8}{10}\le \beta <1.$ For this interval of $\beta$, $10\beta -8\ge 0.$ After some simplification from inequality $\left(30\right),$ we have

$$|H_{2,2}\left(f^{-1}\right)|\le \frac{1}{18432}{\beta }^2\left[32-2(12-5\beta )|c_1{|}^2-\frac{1}{2}(8+25\beta -2{\beta }^2){|c_1|}^4\right].$$

Obviously, the maximum here occurs at $|c_1|=0$ too. As such, we have

$$|H_{2,2}\left(f^{-1}\right)|\le \frac{1}{18432}{\beta }^2\left(32\right)=\frac{1}{576}{\beta }^2.$$

In both the cases the result is the same, and this inequality is sharp for $w\left(z\right)=z^2$ in Equation $\left(19\right)$, or for $c_2=1$ and all other coefficients’ zeros in Equation $\left(29\right).$