Upper bound of the third Hankel determinant for a class of analytic functions related with lemniscate of Bernoulli

Abstract

In this paper, the upper bound of the Hankel determinant $H_3(1)$ for a subclass of analytic functions associated with right half of the lemniscate of Bernoulli ${(x^2+y^2)}^2-2(x^2-y^2)=0$ is investigated.

MSC:30C45, 30C50.

1 Introduction and preliminaries

Let A be the class of functions f of the form

$$f(z)=z+\stackrel{\mathrm{\infty }}{\sum _{n=2}}{a}_n{z}^n,$$

(1.1)

which are analytic in the open unit disk $E=\{z:|z|<1\}$. A function f is said to be subordinate to a function g, written as $f\prec g$, if there exists a Schwartz function w with $w(0)=0$ and $|w(z)|<1$ such that $f(z)=g(w(z))$. In particular, if g is univalent in E, then $f(0)=g(0)$ and $f(E)\subset g(E)$.

Let P denote the class of analytic functions p normalized by

$$p(z)=1+\stackrel{\mathrm{\infty }}{\sum _{n=1}}{p}_n{z}^n$$

(1.2)

such that $Rep(z)>0$. Let ${\mathit{SL}}^{\ast }$ be the class of functions defined by

$${\mathit{SL}}^{\ast }=\{f\in A:|{\left(\frac{z{f}^{\prime }(z)}{f(z)}\right)}^2-1|<1\},z\in E.$$

Thus a function $f\in {\mathit{SL}}^{\ast }$ is such that $\frac{z{f}^{\prime }(z)}{f(z)}$ lies in the region bounded by the right half of the lemniscate of Bernoulli given by the relation $|w^2-1|<1$. It can easily be seen that $f\in {\mathit{SL}}^{\ast }$ if it satisfies the condition

$$\frac{z{f}^{\prime }(z)}{f(z)}\prec \sqrt{1+z},z\in E.$$

(1.3)

This class of functions was introduced by Sokół and Stankiewicz [1] and further investigated by some authors. For details, see [2, 3].

Noonan and Thomas [4] have studied the q th Hankel determinant defined as

$$H_q(n)=\left|\begin{array}{cccc}{a}_n& a_{n+1}& \cdots & a_{n+q-1}\\ a_{n+1}& a_{n+2}& \cdots & a_{n+q-2}\\ ⋮& ⋮& ⋮& ⋮\\ a_{n+q-1}& a_{n+q-2}& \cdots & a_{n+2q-2}\end{array}\right|,$$

(1.4)

where $n\ge 1$ and $q\ge 1$. The Hankel determinant plays an important role in the study of singularities; for instance, see [[5], p.329] and Edrei [6]. This is also important in the study of power series with integral coefficients [[5], p.323] and Cantor [7]. For the use of the Hankel determinant in the study of meromorphic functions, see [8], and various properties of these determinants can be found in [[9], Chapter 4]. It is well known that the Fekete-Szegö functional $|a_3-a_2^2|=H_2(1)$. This functional is further generalized as $|a_3-\mu a_2^2|$ for some μ (real as well as complex). Fekete and Szegö gave sharp estimates of $|a_3-\mu a_2^2|$ for μ real and $f\in S$, the class of univalent functions. It is a very great combination of the two coefficients which describes the area problems posted earlier by Gronwall in 1914-15. Moreover, we also know that the functional $|a_2{a}_4-a_3^2|$ is equivalent to $H_2(2)$. The q th Hankel determinant for some subclasses of analytic functions was recently studied by Arif et al. [10] and Arif et al. [11]. The functional $|a_2{a}_4-a_3^2|$ has been studied by many authors, see [12–14]. Babalola [15] studied the Hankel determinant $H_3(1)$ for some subclasses of analytic functions. In the present investigation, we determine the upper bounds of the Hankel determinant $H_3(1)$ for a subclass of analytic functions related with lemniscate of Bernoulli by using Toeplitz determinants.

We need the following lemmas which will be used in our main results.

Lemma 1.1 [16]

Let $p\in P$ and of the form (1.2). Then

$$|p_2-v{p}_1^2|\le \{\begin{array}{ll}-4v+2,& v<0,\\ 2,& 0\le v\le 1,\\ 4v-2,& v>1.\end{array}$$

When $v<0$ or $v>1$, the equality holds if and only if $p(z)$ is $\frac{1+z}{1-z}$ or one of its rotations. If $0<v<1$, then the equality holds if and only if $p(z)=\frac{1+z^2}{1-z^2}$ or one of its rotations. If $v=0$, the equality holds if and only if $p(z)=(\frac{1}{2}+\frac{\eta }{2})\frac{1+z}{1-z}+(\frac{1}{2}-\frac{\eta }{2})\frac{1-z}{1+z}$ ($0\le \eta \le 1$) or one of its rotations. If $v=1$, the equality holds if and only if p is the reciprocal of one of the functions such that the equality holds in the case of $v=0$. Although the above upper bound is sharp, when $0<v<1$, it can improved as follows:

$$|p_2-v{p}_1^2|+v|p_1{|}^2\le 2(0<v\le 1/2)$$

and

$$|p_2-v{p}_1^2|+(1-v)|p_1{|}^2\le 2(1/2<v\le 1).$$

Lemma 1.2 [16]

If $p(z)=1+p_{1}z+p_2{z}^2+\cdots$ is a function with positive real part in E, then for v a complex number

$$|p_2-v{p}_1^2|\le 2max(1,|2v-1|).$$

This result is sharp for the functions

$$p(z)=\frac{1+z^2}{1-z^2},p(z)=\frac{1+z}{1-z}.$$

Lemma 1.3 [17]

Let $p\in P$ and of the form (1.2). Then

$$2p_2=p_1^2+x(4-p_1^2)$$

for some x, $|x|\le 1$, and

$$4p_3=p_1^3+2(4-p_1^2)p_{1}x-(4-p_1^2)p_1{x}^2+2(4-p_1^2)(1-|x{|}^2)z$$

for some z, $|z|\le 1$.

2 Main results

Although we have discussed the Hankel determinant problem in the paper, the first two problems are specifically related with the Fekete-Szegö functional, which is a special case of the Hankel determinant.

Theorem 2.1 Let $f\in {\mathit{SL}}^{\ast }$ and of the form (1.1). Then

$$|a_3-\mu a_2^2|\le \{\begin{array}{ll}\frac{1}{16}(1-4\mu ),& \mu <-\frac{3}{4},\\ \frac{1}{4},& -\frac{3}{4}\le \mu \le \frac{5}{4},\\ \frac{1}{16}(4\mu -1),& \mu >\frac{5}{4}.\end{array}$$

Furthermore, for $-\frac{3}{4}<\mu \le \frac{1}{4}$,

$$|a_3-\mu a_2^2|+\frac{1}{4}(4\mu +3)|a_2{|}^2\le \frac{1}{4},$$

and for $\frac{1}{4}<\mu \le \frac{5}{4}$,

$$|a_3-\mu a_2^2|+\frac{1}{4}(5-4\mu )|a_2{|}^2\le \frac{1}{4}.$$

These results are sharp.

Proof If $f\in {\mathit{SL}}^{\ast }$, then it follows from (1.3) that

$$\frac{z{f}^{\prime }(z)}{f(z)}\prec \varphi (z),$$

(2.1)

where $\varphi (z)=\sqrt{1+z}$. Define a function

$$p(z)=\frac{1+w(z)}{1-w(z)}=1+p_{1}z+p_2{z}^2+\cdots .$$

It is clear that $p\in P$. This implies that

$$w(z)=\frac{p(z)-1}{p(z)+1}.$$

From (2.1), we have

$$\frac{z{f}^{\prime }(z)}{f(z)}=\varphi (w(z)),$$

with

$$\varphi (w(z))={\left(\frac{2p(z)}{p(z)+1}\right)}^{\frac{1}{2}}.$$

Now

$${\left(\frac{2p(z)}{p(z)+1}\right)}^{\frac{1}{2}}=1+\frac{1}{4}{p}_{1}z+[\frac{1}{4}{p}_2-\frac{5}{32}{p}_1^2]z^2+[\frac{1}{4}{p}_3-\frac{5}{16}{p}_1{p}_2+\frac{13}{128}{p}_1^3]z^3+\cdots .$$

Similarly,

$$\frac{z{f}^{\prime }(z)}{f(z)}=1+a_{2}z+[2a_3-a_2^2]z^2+[3a_4-3a_2{a}_3+a_2^3]z^3+\cdots .$$

Therefore

$$a_2=\frac{1}{4}{p}_1,$$

(2.2)

$$a_3=\frac{1}{8}{p}_2-\frac{3}{64}{p}_1^2,$$

(2.3)

$$a_4=\frac{1}{12}{p}_3-\frac{7}{96}{p}_1{p}_2+\frac{13}{768}{p}_1^2.$$

(2.4)

This implies that

$$|a_3-\mu a_2^2|=\frac{1}{8}|p_2-\frac{1}{8}(4\mu +3)p_1^2|.$$

Now, using Lemma 1.1, we have the required result. □

The results are sharp for the functions $K_i(z)$, $i=1,2,3,4$, such that

$$\begin{array}{c}\frac{z{K}_1^{\prime }(z)}{K_1(z)}=\sqrt{1+z}\text{if }\mu <-\frac{3}{4}\text{ or }\mu >\frac{5}{4},\hfill \\ \frac{z{K}_2^{\prime }(z)}{K_2(z)}=\sqrt{1+z^2}\text{if }-\frac{3}{4}<\mu <\frac{5}{4},\hfill \\ \frac{z{K}_3^{\prime }(z)}{K_3(z)}=\sqrt{1+\mathrm{\Phi }(z)}\text{if }\mu =-\frac{3}{4},\hfill \\ \frac{z{K}_4^{\prime }(z)}{K_4(z)}=\sqrt{1-\mathrm{\Phi }(z)}\text{if }\mu =\frac{5}{4},\hfill \end{array}$$

where $\mathrm{\Phi }(z)=\frac{z(z+\eta )}{1+\eta z}$ with $0\le \eta \le 1$.

Theorem 2.2 Let $f\in {\mathit{SL}}^{\ast }$ and of the form (1.1). Then for a complex number μ,

$$|a_3-\mu a_2^2|\le \frac{1}{4}max\{1;|\mu -\frac{1}{4}\left|\right\}.$$

Proof Since

$$|a_3-\mu a_2^2|=\frac{1}{8}|p_2-\frac{1}{8}(4\mu +3)p_1^2|,$$

therefore, using Lemma 1.2, we get the result. This result is sharp for the functions

$$\frac{z{f}^{\prime }(z)}{f(z)}=\sqrt{1+z}$$

or

$$\frac{z{f}^{\prime }(z)}{f(z)}=\sqrt{1+z^2}.$$

 □

For $\mu =1$, we have $H_2(1)$.

Corollary 2.3 Let $f\in {\mathit{SL}}^{\ast }$ and of the form (1.1). Then

$$|a_3-a_2^2|\le \frac{1}{4}.$$

Theorem 2.4 Let $f\in {\mathit{SL}}^{\ast }$ and of the form (1.1). Then

$$|a_2{a}_4-a_3^2|\le \frac{1}{16}.$$

Proof From (2.2), (2.3) and (2.4), we obtain

$$\begin{array}{rcl}{a}_2{a}_4-a_3^2& =& \frac{1}{48}(p_1{p}_3-\frac{7}{8}{p}_1^2{p}_2+\frac{13}{64}{p}_1^4)-{(\frac{1}{8}{p}_2-\frac{3}{64}{p}_1^2)}^2\\ =& \frac{1}{48}{p}_1{p}_3-\frac{1}{64}{p}_2^2-\frac{5}{768}{p}_1^2{p}_2+\frac{25}{12\text{,}288}{p}_1^4\\ =& \frac{1}{12\text{,}288}(256p_1{p}_3-192p_2^2-80p_1^2{p}_2+25p_1^4).\end{array}$$

Putting the values of $p_2$ and $p_3$ from Lemma 1.3, we assume that $p>0$, and taking $p_1=p\in [0,2]$, we get

$$\begin{array}{rl}|a_2{a}_4-a_3^2|=& \frac{1}{12\text{,}288}|64p_1\{p_1^3+2(4-p_1^2)p_{1}x-(4-p_1^2)p_1{x}^2+2(4-p_1^2)(1-|x{|}^2)z\}\\ -48{\{p_1^2+x(4-p_1^2)\}}^2-40p_1^2\{p_1^2+x(4-p_1^2)\}+25p_1^4|.\end{array}$$

After simple calculations, we get

$$\begin{array}{rl}|a_2{a}_4-a_3^2|=& \frac{1}{12\text{,}288}|41p^4-8(4-p^2)p^{2}x-128(4-p^2)(1-|x{|}^2)z\\ +x^2(4-p^2)(64p^2+48)(4-p^2)|.\end{array}$$

Now, applying the triangle inequality and replacing $|x|$ by ρ, we obtain

$$\begin{array}{rcl}|a_2{a}_4-a_3^2|& \le & \frac{1}{12\text{,}288}[41p^4+128(4-p^2)+8(4-p^2)p^2\rho +{\rho }^2(4-p^2)(16p^2+64)]\\ =& F(p,\rho )\text{(say).}\end{array}$$

Differentiating with respect to ρ, we have

$$\frac{\partial F(p,\rho )}{\partial \rho }=\frac{1}{12\text{,}288}[8(4-p^2)p^2+2\rho (4-p^2)(16p^2+64)].$$

It is clear that $\frac{\partial F(p,\rho )}{\partial \rho }>0$, which shows that $F(p,\rho )$ is an increasing function on the closed interval $[0,1]$. This implies that maximum occurs at $\rho =1$. Therefore $maxF(p,\rho )=F(p,1)=G(p)$ (say). Now

$$G(p)=\frac{1}{12\text{,}288}[17p^4-96p^2+768].$$

Therefore

$$G^{\prime }(p)=\frac{1}{12\text{,}288}[68p^3-192p]$$

and

$$G^{″}(p)=\frac{1}{12\text{,}288}[204p^2-192]<0$$

for $p=0$. This shows that maximum of $G(p)$ occurs at $p=0$. Hence, we obtain

$$\begin{array}{rcl}|a_2{a}_4-a_3^2|& \le & \frac{768}{12\text{,}288}\\ =& \frac{1}{16}.\end{array}$$

This result is sharp for the functions

$$\frac{z{f}^{\prime }(z)}{f(z)}=\sqrt{1+z}$$

or

$$\frac{z{f}^{\prime }(z)}{f(z)}=\sqrt{1+z^2}.$$

 □

Theorem 2.5 Let $f\in {\mathit{SL}}^{\ast }$ and of the form (1.1). Then

$$|a_2{a}_3-a_4|\le \frac{1}{6}.$$

Proof Since

$$\begin{array}{c}{a}_2=\frac{1}{4}{p}_1,\hfill \\ a_3=\frac{1}{8}{p}_2-\frac{3}{64}{p}_1^2,\hfill \\ a_4=\frac{1}{12}{p}_3-\frac{7}{96}{p}_1{p}_2+\frac{13}{768}{p}_1^2.\hfill \end{array}$$

Therefore, by using Lemma 1.3, we can obtain

$$|a_2{a}_3-a_4|\le \frac{1}{768}\{7p^3+8p\rho (4-p^2)+32(4-p^2)+16{\rho }^2(p-2)(4-p^2)\}.$$

Let

$$F_1(p,\rho )=\frac{1}{768}\{7p^3+8p\rho (4-p^2)+32(4-p^2)+16{\rho }^2(p-2)(4-p^2)\}.$$

(2.5)

We assume that the upper bound occurs at the interior point of the rectangle $[0,2]\times [0,1]$. Differentiating (2.5) with respect to ρ, we get

$$\frac{\partial F_1}{\partial \rho }=\frac{1}{768}\{8p(4-p^2)+32\rho (p-2)(4-p^2)\}.$$

For $0<\rho <1$ and fixed $p\in (0,2)$, it can easily be seen that $\frac{\partial F_1}{\partial \rho }<0$. This shows that $F_1(p,\rho )$ is a decreasing function of ρ, which contradicts our assumption; therefore, $max{F}_1(p,\rho )=F_1(p,0)=G_1(p)$. This implies that

$$G_1^{\prime }(p)=\frac{1}{768}\{21p^2-64p\}$$

and

$$G_1^{\prime \prime }(p)=\frac{1}{768}\{42p-64\}<0$$

for $p=0$. Therefore $p=0$ is a point of maximum. Hence, we get the required result. □

Lemma 2.6 If the function $f(z)=\stackrel{\mathrm{\infty }}{\sum _{n=1}}{a}_n{z}^n$ belongs to the class ${\mathit{SL}}^{\ast }$, then

$$|a_2|\le 1/2,|a_3|\le 1/4,|a_4|\le 1/6,|a_5|\le 1/8.$$

These estimations are sharp. The first three bounds were obtained by Sokół [3]and the bound for $|a_5|$ can be obtained in a similar way.

Theorem 2.7 Let $f\in {\mathit{SL}}^{\ast }$ and of the form (1.1). Then

$$|H_3(1)|\le \frac{43}{576}.$$

Proof Since

$$H_3(1)=a_3(a_2{a}_4-a_3^2)-a_4(a_4-a_2{a}_3)+a_5(a_1{a}_3-a_2^2).$$

Now, using the triangle inequality, we obtain

$$|H_3(1)|\le |a_3||a_2{a}_4-a_3^2|+|a_4||a_2{a}_3-a_4|+|a_5||a_1{a}_3-a_2^2|.$$

Using the fact that $a_1=1$ with the results of Corollary 2.3, Theorem 2.4, Theorem 2.5 and Lemma 2.6, we obtain

$$|H_3(1)|\le \frac{43}{576}.$$

 □