Third Hankel Determinant for a Subfamily of Holomorphic Functions Related with Lemniscate of Bernoulli

Abstract

The main goal of this investigation is to obtain sharp upper bounds for Fekete-Szegö functional and the third Hankel determinant for a certain subclass ${\mathcal{SL}}^{\ast }\left(u,v,\alpha \right)$ of holomorphic functions defined by the Carlson-Shaffer operator in the unit disk. Finally, for some special values of parameters, several corollaries were presented.

1. Introduction and Definitions

Denote by $\mathcal{A}$ the family of holomorphic functions defined in the unit disk $\Omega =\left\{\varsigma \in \mathbb{C}:\left|\varsigma \right|<1\right\},$ with expansion

$$l\left(\varsigma \right)=\varsigma +\sum _{k=2}^{\infty }{m}_k{\varsigma }^k$$

(1)

and let $\mathcal{S}$ be the subset of $\mathcal{A}$, consisting of functions which are univalent in $\Omega$.

Let $\mathcal{P}$ be a family of the holomorphic functions t of the form

$$t\left(\varsigma \right)=1+\sum _{k=1}^{\infty }{t}_k{\varsigma }^k,\left(\varsigma \in \Omega \right)$$

(2)

satisfying $\mathrm{Re}\left(t\right(\varsigma \left)\right)>0$ in $\Omega .$ The family of starlike functions in $\Omega$ are represented by the symbol ${\mathcal{S}}^{\ast }$, which satisfies

$$\frac{\varsigma l^{\prime }\left(\varsigma \right)}{l\left(\varsigma \right)}\in \mathcal{P},\left(\mathrm{for}\mathrm{all}\varsigma \in \Omega \right).$$

In addition, the symbol ${\mathcal{SL}}^{\ast }$ represents the family of functions that satisfy

$$\left|{\left(\frac{\varsigma l^{\prime }\left(\varsigma \right)}{l\left(\varsigma \right)}\right)}^2-1\right|<1,\left(\varsigma \in \Omega \right).$$

As a result, $l\in {\mathcal{SL}}^{\ast }$ can be expressed by

$$\left|w^2-1\right|<1.$$

if and only if $\frac{\varsigma l^{\prime }\left(\varsigma \right)}{l\left(\varsigma \right)}$ is the inside region bounded by the right half of the Bernoulli lemniscate.

This class was introduced by Sokól [1] and Sokól et al. [2]. If there is a Schwarz function w that is holomorphic in $\Omega$, with $w\left(0\right)=0,$ $\left|w\left(\varsigma \right)\right|<1$, such that $l\left(\varsigma \right)=h\left(w\right(\varsigma \left)\right),$ $\varsigma \in \Omega$, then the function l is subordinate to $h,$ denoted by the notation $l\prec h$. If the function h is univalent in $\Omega$, then $l\prec h$ if

$$l\left(0\right)=h\left(0\right)\mathrm{and}l(\Omega )\subset h(\Omega ).$$

A function $l\in \mathcal{A}$ is said to be starlike of order $\alpha$ if and only if

$$\mathrm{Re}\left\{\frac{\varsigma l^{\prime }\left(\varsigma \right)}{l\left(\varsigma \right)}\right\}>\alpha ,\left(\varsigma \in \Omega \right)$$

for some $\alpha \left(0\le \alpha <1\right)$. We denote the class of all starlike functions of order $\alpha$ by ${\mathcal{S}}^{\ast }\left(\alpha \right)$. We also note that ${\mathcal{S}}^{\ast }\left(0\right)={\mathcal{S}}^{\ast }$ is the well-known class of all normalized starlike functions in $\Omega$.

Now, the function

$${\mathcal{K}}_{\alpha }\left(\varsigma \right)=\frac{\varsigma }{{\left(1-\varsigma \right)}^{2\left(1-\alpha \right)}}$$

(3)

is a well known extremal function for the class ${\mathcal{S}}^{\ast }\left(\alpha \right),$ (see [3,4,5]).

Setting

$$\psi (\alpha ,k)=\frac{{\Pi }_{k=2}^k\left(k-2\alpha \right)}{\left(k-1\right)!},(k\ge 2),$$

(4)

the function ${\mathcal{K}}_{\alpha }$ can be written in the form as follows:

$${\mathcal{K}}_{\alpha }\left(\varsigma \right)=\varsigma +\sum _{k=2}^{\infty }\psi (\alpha ,k){\varsigma }^k.$$

(5)

We denote by $\mathcal{F}(\alpha ,k,\psi )$ the class of functions ${\mathcal{K}}_{\alpha }.$ Then, we note that $\psi (\alpha ,k)$ is a decreasing function in $\alpha$ and satisfies

$$\underset{k\to \infty }{lim}\psi (\alpha ,k)=\left\{\begin{array}{cc}\infty & \left(\alpha <\frac{1}{2}\right)\\ 1& \left(\alpha =\frac{1}{2}\right)\\ 0& \left(\alpha >\frac{1}{2}\right)\end{array}.\right.$$

Let $(l\ast h)\left(\varsigma \right)$ be the Hadamard product (or convolution) of two functions l and h, that is, if l given by (1) and h is given by

$$h\left(\varsigma \right)=\varsigma +\sum _{k=2}^{\infty }{n}_k{\varsigma }^k.$$

Then,

$$(l\ast h)\left(\varsigma \right)=\varsigma +\sum _{k=2}^{\infty }{m}_k{n}_k{\varsigma }^k=(h\ast l)\left(\varsigma \right),\left(\varsigma \in \Omega \right).$$

(6)

Let $\Theta (u,v,\varsigma )$ be defined by

$$\Theta \left(u,v,\varsigma \right)=\varsigma +\sum _{k=2}^{\infty }\frac{{\left(u\right)}_{k-1}}{{\left(v\right)}_{k-1}}{\varsigma }^k,(u\in \mathbb{C},v\in \mathbb{C}\setminus {\mathbb{Z}}_0^{-},{\mathbb{Z}}_0^{-}=\{\dots -2,-1,0\};\varsigma \in \Omega ).$$

The function, $\Theta \left(u,v,\varsigma \right)$ is known as the incomplete beta function. The term ${\left(\varkappa \right)}_k$ is the Pochhammer symbol that can be expanded in Gamma functions as

$${\left(\varkappa \right)}_k=\frac{\Gamma (\varkappa +k)}{\Gamma \left(\varkappa \right)}=\left\{\begin{array}{cc}1,& k=0\\ \varkappa \left(\varkappa +1\right)\left(\varkappa +2\right)\dots \left(\varkappa +k-1\right),& k\in \mathbb{N}=\{1,2,3,\cdots \}.\end{array}\right.$$

Corresponding to the $\Theta \left(u,v,\varsigma \right)$ Carlson-Shaffer function [6], an operator $\mathcal{L}\left(u,v\right)$ is introduced for $l\in \mathcal{A}$ using the Hadamard product as follows:

$$\mathcal{L}\left(u,v\right)l\left(\varsigma \right)=\Theta \left(u,v,\varsigma \right)\ast l\left(\varsigma \right)=\varsigma +\sum _{k=2}^{\infty }\frac{{\left(u\right)}_{k-1}}{{\left(v\right)}_{k-1}}{m}_k{\varsigma }^k(\varsigma \in \Omega ).$$

Further, for the function $\mathcal{L}\left(u,v\right)l\left(\varsigma \right)$

$$\tau \left(\varsigma \right)=\mathcal{L}\left(u,v\right)l\left(\varsigma \right)\ast {\mathcal{K}}_{\alpha }\left(\varsigma \right)=\varsigma +\sum _{k=2}^{\infty }\frac{{\left(u\right)}_{k-1}}{{\left(v\right)}_{k-1}}\psi \left(\alpha ,k\right)m_k{\varsigma }^k$$

(7)

where $\mathcal{L}\left(u,v\right)$ is called the Carlson-Shaffer operator [6], and the operator ∗ stands for the Hadamard product (or convolution product) of two power series as given by (6). We will show by $\tilde{\mathcal{F}}(\alpha ,k,\psi )$ the family of functions $\tau \left(\varsigma \right).$

Definition 1.

We consider that${\mathcal{SL}}^{\ast }\left(u,v,\alpha \right)$is the family of holomorphic functions given by

$${\mathcal{SL}}^{\ast }\left(u,v,\alpha \right)=\left\{\tau \left(\varsigma \right)\in \tilde{\mathcal{F}}(\alpha ,k,\psi ):\left|{\left(\frac{\varsigma {\tau }^{\prime }\left(\varsigma \right)}{\tau \left(\varsigma \right)}\right)}^2-1\right|<1\right\},$$

(8)

$$\frac{\varsigma {\tau }^{\prime }\left(\varsigma \right)}{\tau \left(\varsigma \right)}\prec \sqrt{1+\varsigma },\left(\varsigma \in \Omega \right),$$

(9)

where

$$\tau \left(\varsigma \right)=\varsigma +\sum _{k=2}^{\infty }\frac{{\left(u\right)}_{k-1}}{{\left(v\right)}_{k-1}}\psi \left(\alpha ,k\right)m_k{\varsigma }^k.$$

(10)

Hankel matrices arise naturally in a wide range of applications in science, engineering, and other related areas, such as signal processing and control theory. For a survey of Hankel matrices and polynomials, the reader is referred to [7,8] and the references therein.

The Hankel determinant ${\mathcal{H}}_{q,k}\left(l\right)$ $(q,k\in \mathbb{N})$ for a function $l\in \mathcal{S}$ of the form (1) was defined by Pommerenke (see [9,10]) as

$${\mathcal{H}}_{q,k}\left(l\right)=\left|\begin{array}{cccc}{m}_k& m_{k+1}& \cdots & m_{k+q-1}\\ m_{k+1}& m_{k+2}& \cdots & m_{k+q}\\ ⋮& ⋮& & ⋮\\ m_{k+q-1}& m_{k+q}& \cdots & m_{k+2q-2}\end{array}\right|(m_1=1).$$

For fixed integer q and k, the growth of ${\mathcal{H}}_{q,k}\left(l\right)$ has been studied for different subfamilies of univalent functions. These studies focus on the main subclasses of certain holomorphic functions. In fact, the majority of papers discuss the determinants ${\mathcal{H}}_{2,2}\left(l\right)$ and ${\mathcal{H}}_{3,1}\left(l\right).$ Case ${\mathcal{H}}_{2,1}\left(l\right)=m_3-m_2^2$ is also very well known. In the year 1933, Fekete and Szegö (see [11]) obtained a sharp bound of the function $m_3-\mu m_2^2$ with real $\mu \in \mathbb{R}$ for a univalent function $l.$ For $\mu \in \mathbb{C}$ this functional was generalized as $\left|m_3-\mu m_2^2\right|.$ Estimating for the upper bound of $\left|m_3-\mu m_2^2\right|$ is known as the Fekete-Szegö problem, (see [12,13,14]). The second Hankel determinant ${\mathcal{H}}_{2,2}\left(l\right)$ is given by ${\mathcal{H}}_{2,2}\left(l\right)=m_2{m}_4-m_3^2.$ In recent years, the research on Hankel determinants has focused on the estimation of $\left|{\mathcal{H}}_{2,2}\left(l\right)\right|.$ Several authors obtained results for different classes of univalent functions. For example, the sharp bounds for the second Hankel determinant ${\mathcal{H}}_{2,2}\left(l\right)$ were obtained for the classes of starlike and convex functions in [15,16,17,18]. Lee et al. [19] established the sharp bound for $\left|{\mathcal{H}}_{2,2}\left(l\right)\right|$ by generalizing their classes by means of the principle of subordination between holomorphic functions. Our main focus in this investigation is for the class ${\mathcal{SL}}^{\ast }\left(u,v,\alpha \right)$ on the Hankel determinant ${\mathcal{H}}_{3,1}\left(l\right)$. The calculation of $\left|{\mathcal{H}}_{3,1}\left(l\right)\right|$ is far more challenging compared to finding the bound of $\left|{\mathcal{H}}_{2,2}\left(l\right)\right|.$ Further, in this work, we find the sharp bounds for $\left|{\mathcal{H}}_{2,2}\left(l\right)\right|,$ when $l\in {\mathcal{SL}}^{\ast }\left(u,v,\alpha \right),$ $\alpha \in \left[0,1\right),$ together with the sharp bound of the functional

$$\mathcal{Z}=\left|m_2{m}_3-m_4\right|,$$

when $l\in {\mathcal{SL}}^{\ast }\left(u,v,\alpha \right)$ and $\alpha \in \left[0,1\right)$.

2. Preliminary Lemmas

Some preliminary results required in the following section are now listed.

Lemma 1

([20]). Suppose that $\mathcal{P}$ denotes the family of holomorphic functions t normalized by

$$t\left(\varsigma \right)=1+t_1\varsigma +t_2{\varsigma }^2+\dots$$

(11)

and satisfying the condition $Re\left(t\left(\varsigma \right)\right)>0,$ $\varsigma \in \Omega .$ Then, for any $\eta \in \mathbb{R}$,

$$\left|t_2-\eta t_1^2\right|\le \left\{\begin{array}{cc}-4\eta +2,& \eta <0\\ 2,& 0\le \eta \le 1\\ 4\eta -2,& \eta \ge 1\end{array}\right.$$

(12)

The equality holds true in (12) if and only if

$$t\left(\varsigma \right)=\frac{1+\varsigma }{1-\varsigma }$$

or one of its rotations, when$\eta <0$or$\eta >1$. If$0<\eta <1$, then the equality holds true in (12) if and only if

$$t\left(\varsigma \right)=\frac{1+{\varsigma }^2}{1-{\varsigma }^2}$$

or one of its rotations. If $\eta =0$, the equality holds true in (12) if and only if

$$t\left(\varsigma \right)=\left(\frac{1+\delta }{2}\right)\frac{1+\varsigma }{1-\varsigma }+\left(\frac{1-\delta }{2}\right)\frac{1-\varsigma }{1+\varsigma },0\le \delta \le 1$$

or one of its rotations. If $\eta =1$, then the equality in (12) holds true if$t\left(\varsigma \right)$is a reciprocal of one of the functions, such that the equality holds true in the case when$\eta =0$.

Lemma 2

([21]). Assume that $t\in \mathcal{P}$ is the form Equation (2), and $\eta \in \mathbb{C}$, we have

$$\left|t_2-\eta t_1^2\right|\le 2max\left\{1,\left|1-2\eta \right|\right\}.$$

Lemma 3

([22,23]). If $t\in \mathcal{P}$ and has the form (11) then

$$2t_2=t_1^2+x(4-t_1^2)$$

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

$$4t_3=t_1^3+2(4-t_1^2)t_{1}x-(4-t_1^2)t_1{x}^2+2(4-t_1^2)(1-{\left|x\right|}^2)\varsigma$$

for some $\varsigma ,$ $\left|\varsigma \right|\le 1.$

Lemma 4

([24]). If $t\in \mathcal{P}$ and has the form (11), then

$$\left|t_k\right|\le 2\left(k\in \mathbb{N}\right)$$

and the inequality is sharp.

3. Main Results

In the remainder of this work, we will assume that $u\ge v>0$ until explicitly stated otherwise.

We now prove our first result asserted by Theorem 1 below.

Theorem 1.

If the function$l,$given by (1) belongs to the class${\mathcal{S}}^{\ast }\left(u,v,\alpha \right)$, then$\mu \in \mathbb{R},$we have

$$\left|m_3-\mu m_2^2\right|\le \left\{\begin{array}{cc}\frac{1}{16}\left(\frac{v\left(v+1\right)}{u\left(u+1\right)\left(1-\alpha \right)\left(3-2\alpha \right)}-\mu \frac{v^2}{u^2{\left(1-\alpha \right)}^2}\right)& \mu \le -\frac{3u\left(v+1\right)\left(1-\alpha \right)}{v\left(u+1\right)\left(3-2\alpha \right)}\\ \frac{1}{4}\frac{v\left(v+1\right)}{u\left(u+1\right)\left(1-\alpha \right)\left(3-2\alpha \right)},& -\frac{3u\left(v+1\right)\left(1-\alpha \right)}{v\left(u+1\right)\left(3-2\alpha \right)}\le \mu \le \frac{5u\left(v+1\right)\left(1-\alpha \right)}{v\left(u+1\right)\left(3-2\alpha \right)}\\ \frac{1}{16}\left(-\frac{v\left(v+1\right)}{u\left(u+1\right)\left(1-\alpha \right)\left(3-2\alpha \right)}+\mu \frac{v^2}{u^2{\left(1-\alpha \right)}^2}\right),& \mu \ge \frac{5u\left(v+1\right)\left(1-\alpha \right)}{v\left(u+1\right)\left(3-2\alpha \right)}\end{array}\right..$$

Proof. 

From Equation (9), it follows that

$$\frac{\varsigma {\tau }^{\prime }\left(\varsigma \right)}{\tau \left(\varsigma \right)}\prec \Phi \left(\varsigma \right).$$

Define the function first,

$$t\left(\varsigma \right)=1+\sum _{k=1}^{\infty }{t}_k{\varsigma }^k=\frac{1+w\left(\varsigma \right)}{1-w\left(\varsigma \right)}.$$

Since $t\in \mathcal{P}$,

$$w\left(\varsigma \right)=\frac{t\left(\varsigma \right)-1}{t\left(\varsigma \right)+1}.$$

Using Equation (9), we have

$$\frac{\varsigma {\tau }^{\prime }\left(\varsigma \right)}{\tau \left(\varsigma \right)}=\Phi \left(w\left(\varsigma \right)\right).$$

Now as

$${\left[\frac{2t\left(\varsigma \right)}{1+t\left(\varsigma \right)}\right]}^{\frac{1}{2}}={\left[2-\frac{2}{1+t\left(\varsigma \right)}\right]}^{\frac{1}{2}},$$

so, we have

$$\begin{array}{ccc}\hfill {\left[\frac{2t\left(\varsigma \right)}{1+t\left(\varsigma \right)}\right]}^{\frac{1}{2}}& =& 1+\frac{1}{4}{t}_1\varsigma +\left(\frac{1}{4}{t}_2-\frac{5}{32}{t}_1^2\right){\varsigma }^2+\left(\frac{1}{4}{t}_3-\frac{5}{16}{t}_1{t}_2+\frac{13}{128}{t}_1^3\right){\varsigma }^3\hfill \\ & & +\left(\frac{19}{8}{t}_1{t}_3-\frac{3}{2}{t}_4+\frac{361}{512}{t}_1^4+\frac{9}{8}{t}_2^2-\frac{34}{16}{t}_1^2{t}_2\right){\varsigma }^4+\dots \hfill \end{array}$$

Similarly,

$$\begin{array}{ccc}\hfill \frac{\varsigma {\tau }^{\prime }\left(\varsigma \right)}{\tau \left(\varsigma \right)}& =& 1+Q_2{m}_2\varsigma +\left(2Q_3{m}_3-Q_2^2{m}_2^2\right){\varsigma }^2+\left(3Q_4{m}_4+Q_2^3{m}_2^3-3Q_2{Q}_3{m}_2{m}_3\right){\varsigma }^3\hfill \\ & & +\left(4Q_5{m}_5-4Q_2{Q}_4{m}_2{m}_4-2Q_3^2{m}_3^2-Q_2^4{m}_2^4+4Q_2^2{Q}_3{m}_2^2{m}_3\right){\varsigma }^4+\dots \hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill Q_2& =& \frac{u}{v}\left(2-2\alpha \right),\hfill \\ \hfill Q_3& =& \frac{u\left(u+1\right)}{2v\left(v+1\right)}\left(2-2\alpha \right)\left(3-2\alpha \right),\hfill \\ \hfill Q_4& =& \frac{u\left(u+1\right)\left(u+2\right)}{6v\left(v+1\right)\left(v+2\right)}\left(2-2\alpha \right)\left(3-2\alpha \right)\left(4-2\alpha \right),\hfill \\ \hfill Q_5& =& \frac{u\left(u+1\right)\left(u+2\right)\left(u+2\right)}{24v\left(v+1\right)\left(v+2\right)\left(v+2\right)}\left(2-2\alpha \right)\left(3-2\alpha \right)\left(4-2\alpha \right)\left(5-2\alpha \right).\hfill \end{array}$$

Thus,

$$m_2=\frac{v}{4u\left(2-2\alpha \right)}{t}_1,$$

(13)

$$m_3=\frac{v\left(v+1\right)}{u\left(u+1\right)\left(2-2\alpha \right)\left(3-2\alpha \right)}\left[\frac{1}{4}{t}_2-\frac{3}{32}{t}_1^2\right],$$

(14)

$$m_4=\frac{v\left(v+1\right)\left(v+2\right)}{u\left(u+1\right)\left(u+2\right)\left(2-2\alpha \right)\left(3-2\alpha \right)\left(4-2\alpha \right)}\left[\frac{1}{2}{t}_3-\frac{7}{16}{t}_1{t}_2+\frac{13}{128}{t}_1^3\right]$$

(15)

and

$$m_5=\frac{v\left(v+1\right)\left(v+2\right)\left(v+3\right)}{u\left(u+1\right)\left(u+2\right)\left(u+3\right)\left(2-2\alpha \right)\left(3-2\alpha \right)\left(4-2\alpha \right)\left(5-2\alpha \right)}\left[\frac{19}{8}{t}_1{t}_3-\frac{3}{2}{t}_4+\frac{361}{512}{t}_1^4+\frac{9}{8}{t}_2^2-\frac{34}{16}{t}_1^2{t}_2\right].$$

(16)

We now have the following using the Equations (13) and (14):

$$m_3-\mu m_2^2=\frac{v\left(v+1\right)}{u\left(u+1\right)\left(2-2\alpha \right)\left(3-2\alpha \right)}\left[\frac{1}{4}{t}_2-\frac{3}{32}{t}_1^2\right]-\mu \frac{v^2}{16u^2{\left(2-2\alpha \right)}^2}{t}_1^2,$$

$$\left|m_3-\mu m_2^2\right|\le \frac{v\left(v+1\right)}{8u\left(u+1\right)\left(1-\alpha \right)\left(3-2\alpha \right)}\left|t_2-\frac{1}{8}\left(\mu \frac{v\left(u+1\right)\left(3-2\alpha \right)}{u\left(1-\alpha \right)\left(v+1\right)}+3\right)t_1^2\right|.$$

(17)

We obtained the required result by applying Lemma 1 to Equation (17). This completes the proof of Theorem 1.  □

Theorem 2.

If the function$l,$given by (1), belongs to the class${\mathcal{SL}}^{\ast }\left(u,v,\alpha \right)$, then$\mu \in \mathbb{C},$we have

$$\left|m_3-\mu m_2^2\right|\le \frac{v\left(v+1\right)}{4u\left(u+1\right)\left(1-\alpha \right)\left(3-2\alpha \right)}max\left\{1,\left|\frac{1}{4}\mu \frac{v\left(u+1\right)\left(3-2\alpha \right)}{u\left(v+1\right)\left(1-\alpha \right)}-\frac{1}{4}\right|\right\}.$$

Proof. 

By making use of Equations (13) and (14), we have

$$m_3-\mu m_2^2=\frac{v\left(v+1\right)}{u\left(u+1\right)\left(2-2\alpha \right)\left(3-2\alpha \right)}\left[\frac{1}{4}{t}_2-\frac{3}{32}{t}_1^2\right]-\mu \frac{v^2}{16u^2{\left(2-2\alpha \right)}^2}{t}_1^2,$$

$$\left|m_3-\mu m_2^2\right|\le \frac{v\left(v+1\right)}{8u\left(u+1\right)\left(1-\alpha \right)\left(3-2\alpha \right)}\left|t_2-\frac{1}{8}\left(\mu \frac{v\left(u+1\right)\left(3-2\alpha \right)}{u\left(v+1\right)\left(1-\alpha \right)}+3\right)t_1^2\right|$$

therefore, using Lemma 2, we obtain the result,

$$\left|m_3-\mu m_2^2\right|\le \frac{v\left(v+1\right)}{4u\left(u+1\right)\left(2-2\alpha \right)\left(3-2\alpha \right)}max\left\{1,\left|\frac{1}{4}\mu \frac{v\left(u+1\right)\left(3-2\alpha \right)}{u\left(v+1\right)\left(1-\alpha \right)}-\frac{1}{4}\right|\right\}.$$

Thus, the proof of Theorem 2 is completed.  □

For the case $\mu \in \mathbb{C}$ and $u=v$ in Theorem 2, this reduces to the following result.

Corollary 1.

Let$\alpha \in [0,1)$and$\mu \in \mathbb{C}$. If the function l, given by (1), belongs to the class${\mathcal{SL}}^{\ast }\left(u,u,\alpha \right)={\mathcal{SL}}^{\ast }\left(u,\alpha \right)$, then

$$\left|m_3-\mu m_2^2\right|\le \frac{1}{4\left(1-\alpha \right)\left(3-2\alpha \right)}max\left\{1,\left|\frac{1}{4}\mu \frac{\left(3-2\alpha \right)}{\left(1-\alpha \right)}-\frac{1}{4}\right|\right\}$$

and the inequality is sharp.

4. The Hankel Determinant ${\mathcal{H}}_{\mathbf{2}\mathbf{,}\mathbf{2}}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$

In this section, we find the sharp bound for the modulus of the second Hankel determinant ${\mathcal{H}}_{2,2}\left(l\right)=m_2{m}_4-m_3^2,$ when $l\in {\mathcal{SL}}^{\ast }\left(u,v,\alpha \right)$.

Theorem 3.

If the function$l,$given by (1), belongs to the class${\mathcal{SL}}^{\ast }\left(u,v,\alpha \right)$, then

$$\left|m_2{m}_4-m_3^2\right|\le \frac{v^2{\left(v+1\right)}^2}{16u^2{\left(u+1\right)}^2{\left(1-\alpha \right)}^2{\left(3-2\alpha \right)}^2}.$$

Proof. 

Using the Equations (13)–(15), we obtain the following

$$\begin{array}{cc}\hfill m_2{m}_4-m_3^2=& \left(\frac{v}{4u\left(2-2\alpha \right)}{t}_1\right)\left\{\frac{v\left(v+1\right)\left(v+2\right)}{u\left(u+1\right)\left(u+2\right)\left(2-2\alpha \right)\left(3-2\alpha \right)\left(4-2\alpha \right)}\left[\frac{1}{2}{t}_3-\frac{7}{16}{t}_1{t}_2+\frac{13}{128}{t}_1^3\right]\right\}\hfill \\ & -{\left[\frac{v\left(v+1\right)}{u\left(u+1\right)\left(2-2\alpha \right)\left(3-2\alpha \right)}\left(\frac{1}{4}{t}_2-\frac{3}{32}{t}_1^2\right)\right]}^2.\hfill \end{array}$$

After simplification, we have

$$\begin{array}{cc}\hfill m_2{m}_4-m_3^2=& \frac{v^2\left(v+1\right)}{12,288u^2\left(u+1\right){\left(2-2\alpha \right)}^2\left(3-2\alpha \right)}\left\{\frac{1536\left(v+2\right)}{\left(u+2\right)\left(4-2\alpha \right)}{t}_1{t}_3-\frac{768\left(v+1\right)}{\left(u+1\right)\left(3-2\alpha \right)}{t}_2^2\right.\hfill \\ & \left.+\left(\frac{576\left(v+1\right)}{\left(u+1\right)\left(3-2\alpha \right)}-\frac{1344\left(v+2\right)}{\left(u+2\right)\left(4-2\alpha \right)}\right)t_1^2{t}_2+\left(\frac{312\left(v+2\right)}{\left(u+2\right)\left(4-2\alpha \right)}-\frac{108\left(v+1\right)}{\left(u+1\right)\left(3-2\alpha \right)}\right)t_1^4\right\}.\hfill \end{array}$$

By substituting values of $t_2$ and $t_3$ from Lemma 3, after some simplification, we arrive at

$$\begin{array}{ccc}\hfill m_2{m}_4-m_3^2& =& \frac{v^2\left(v+1\right)}{12,288u^2\left(u+1\right){\left(2-2\alpha \right)}^2\left(3-2\alpha \right)}\left\{\left(\frac{312\left(v+2\right)}{\left(u+2\right)\left(4-2\alpha \right)}-\frac{108\left(v+1\right)}{\left(u+1\right)\left(3-2\alpha \right)}\right)t_1^4\right.\hfill \\ & & +\frac{384\left(v+2\right)}{\left(u+2\right)\left(4-2\alpha \right)}{t}_1\left[t_1^3+2t_1\left(4-t_1^2\right)x-t_1\left(4-t_1^2\right)x^2+2\left(4-t_1^2\right)\left(1-{\left|x\right|}^2\varsigma \right)\right]\hfill \\ & & +\left(\frac{288\left(v+1\right)}{\left(u+1\right)\left(3-2\alpha \right)}-\frac{672\left(v+2\right)}{\left(u+2\right)\left(4-2\alpha \right)}\right)t_1^2\left[t_1^2+\left(4-t_1^2\right)x\right]\hfill \\ & & \left.-\frac{192\left(v+1\right)}{\left(u+1\right)\left(3-2\alpha \right)}{\left[t_1^2+\left(4-t_1^2\right)x\right]}^2\right\}.\hfill \end{array}$$

Now, taking the module and replacing $\left|x\right|$ by $\rho$ and $t_1$ by t, we have

$$\begin{array}{ccc}\hfill \left|m_2{m}_4-m_3^2\right|& \le & \frac{v^2\left(v+1\right)}{12,288u^2\left(u+1\right){\left(2-2\alpha \right)}^2\left(3-2\alpha \right)}\left\{\left(\frac{12\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}-\frac{12\left(v+1\right)}{\left(u+1\right)\left(3-2\alpha \right)}\right)t^4\right.\hfill \\ & & +\left(\frac{96\left(v+1\right)}{\left(u+1\right)\left(3-2\alpha \right)}-\frac{48\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}\right)t^2\left(4-t^2\right)\rho +\frac{384\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}t\left(4-t^2\right)\hfill \\ & & \left.+\left(\frac{192\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}{t}^2+\frac{384\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}t+\frac{192\left(v+1\right)}{\left(u+1\right)\left(3-2\alpha \right)}\left(4-t^2\right)\right){\rho }^2\left(4-t^2\right)\right\}\hfill \\ & =& F\left(t,\rho \right).\hfill \end{array}$$

(18)

Upon differentiating both sides (18) with respect to $\rho$, we obtain

$$\begin{array}{ccc}\hfill \frac{\partial F\left(t,\rho \right)}{\partial \rho }& =& \frac{v^2\left(v+1\right)}{12,288u^2\left(u+1\right){\left(2-2\alpha \right)}^2\left(3-2\alpha \right)}\left\{\left(\frac{96\left(v+1\right)}{\left(u+1\right)\left(3-2\alpha \right)}-\frac{48\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}\right)t^2\left(4-t^2\right)\right.\hfill \\ & & \left.+\left(\frac{192\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}{t}^2+\frac{384\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}t+\frac{192\left(v+1\right)}{\left(u+1\right)\left(3-2\alpha \right)}\left(4-t^2\right)\right)2\rho \left(4-t^2\right)\right\}.\hfill \end{array}$$

It is clear that

$$\frac{\partial F\left(t,\rho \right)}{\partial \rho }>0,$$

which show that $F\left(t,\rho \right)$ is an increasing function of $\rho$ on the closed interval $\left[0,1\right]$. This implies that the maximum value occurs at $\rho =1$. This implies that

$$max\left\{F\left(t,\rho \right)\right\}=F\left(t,1\right)=G\left(t\right).$$

We now observe that

$$\begin{array}{ccc}\hfill G\left(t\right)& =& \frac{v^2\left(v+1\right)}{12,288u^2\left(u+1\right){\left(2-2\alpha \right)}^2\left(3-2\alpha \right)}\left\{\left(\frac{12\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}-\frac{12\left(v+1\right)}{\left(u+1\right)\left(3-2\alpha \right)}\right)t^4\right.\hfill \\ & & +\left(\frac{96\left(v+1\right)}{\left(u+1\right)\left(3-2\alpha \right)}-\frac{48\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}\right)t^2\left(4-t^2\right)+\frac{384\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}t\left(4-t^2\right)\hfill \\ & & \left.+\left(\frac{192\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}{t}^2+\frac{384\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}t+\frac{192\left(v+1\right)}{\left(u+1\right)\left(3-2\alpha \right)}\left(4-t^2\right)\right)\left(4-t^2\right)\right\}.\hfill \end{array}$$

(19)

Differentiating (19) with respect to t, we obtain

$$\begin{array}{ccc}\hfill G^{\prime }\left(t\right)& =& \frac{v^2\left(v+1\right)}{12,288u^2\left(u+1\right){\left(2-2\alpha \right)}^2\left(3-2\alpha \right)}\left\{4\left(\frac{84\left(v+1\right)}{\left(u+1\right)\left(3-2\alpha \right)}-\frac{132\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}\right)t^3\right.\hfill \\ & & \left.-\frac{96\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}{t}^2+16\left(\frac{72\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}-\frac{144\left(v+1\right)}{\left(u+1\right)\left(3-2\alpha \right)}\right)t+\frac{3072\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}\right\}.\hfill \end{array}$$

Differentiating again above equation with respect to t, we have

$$\begin{array}{ccc}\hfill G^{\prime \prime }\left(t\right)& =& \frac{v^2\left(v+1\right)}{12,288u^2\left(u+1\right){\left(2-2\alpha \right)}^2\left(3-2\alpha \right)}\left\{12\left(\frac{84\left(v+1\right)}{\left(u+1\right)\left(3-2\alpha \right)}-\frac{132\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}\right)t^2\right.\hfill \\ & & \left.-\frac{192\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}t+16\left(\frac{72\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}-\frac{144\left(v+1\right)}{\left(u+1\right)\left(3-2\alpha \right)}\right)\right\}.\hfill \end{array}$$

For $t=0$, $\left(t\in \left[0,2\right]\right)$ shows that the maximum value of $G\left(t\right)$ occurs at $t=0$. Hence, we obtain,

$$\left|m_2{m}_4-m_3^2\right|\le \frac{v^2{\left(v+1\right)}^2}{16u^2{\left(u+1\right)}^2{\left(1-\alpha \right)}^2{\left(3-2\alpha \right)}^2}.$$

Thus, the proof of Theorem 3 is completed.  □

Upon setting $u=v$ in Theorem 3, we are led to the following results, respectively:

Corollary 2.

Let$\alpha \in [0,1)$. If the function l, given by (1), belongs to the class${\mathcal{SL}}^{\ast }\left(u,u,\alpha \right)={\mathcal{SL}}^{\ast }\left(u,\alpha \right)$, then

$$\left|m_2{m}_4-m_3^2\right|\le \frac{1}{16{\left(1-\alpha \right)}^2{\left(3-2\alpha \right)}^2}$$

and the inequality is sharp.

If we choose $\alpha =0$ in Corollary 2, we obtain the following corollary.

Corollary 3.

Let$\alpha \in [0,1)$. If the function l, given by (1), belongs to the class${\mathcal{SL}}^{\ast }\left(u,u,0\right)={\mathcal{SL}}^{\ast }\left(u\right)$, then

$$\left|m_2{m}_4-m_3^2\right|\le \frac{1}{144}$$

and the inequality is sharp.

5. The Zalcman Functional

In this section, we prove the following theorem on the upper bound estimate of the Zalcman functional $\left|m_2{m}_3-m_4\right|,$ noting that a non-sharp inequality was found in [25,26,27,28,29].

Theorem 4.

If the function$l,$given by (1), belongs to the class${\mathcal{SL}}^{\ast }\left(u,v,\alpha \right)$, then

$$\left|m_2{m}_3-m_4\right|\le \frac{v\left(v+1\right)\left(v+2\right)}{4u\left(u+1\right)\left(u+2\right)\left(1-\alpha \right)\left(2-\alpha \right)\left(3-2\alpha \right)}.$$

Proof. 

Using the values given in (13)–(15) we have

$$\begin{array}{ccc}\hfill m_2{m}_3-m_4& =& \left(\frac{v}{4u\left(2-2\alpha \right)}{t}_1\right)\left[\frac{v\left(v+1\right)}{u\left(u+1\right)\left(2-2\alpha \right)\left(3-2\alpha \right)}\left(\frac{1}{4}{t}_2-\frac{3}{32}{t}_1^2\right)\right]\hfill \\ & & -\left\{\frac{v\left(v+1\right)\left(v+2\right)}{u\left(u+1\right)\left(u+2\right)\left(2-2\alpha \right)\left(3-2\alpha \right)\left(4-2\alpha \right)}\left[\frac{1}{2}{t}_3-\frac{7}{16}{t}_1{t}_2+\frac{13}{128}{t}_1^3\right]\right\}.\hfill \end{array}$$

By substituting values of $t_2$ and $t_3$ from Lemma 3, after some simplification, we have

$$\begin{array}{ccc}\hfill m_2{m}_3-m_4& =& \frac{v\left(v+1\right)}{u\left(u+1\right)\left(2-2\alpha \right)\left(3-2\alpha \right)}\left\{\frac{v}{4u\left(2-2\alpha \right)}\left(\frac{1}{4}{t}_1{t}_2-\frac{3}{32}{t}_1^3\right)\right.\hfill \\ & & \left.-\frac{\left(v+2\right)}{\left(u+2\right)\left(4-2\alpha \right)}\left[\frac{1}{2}{t}_3-\frac{7}{16}{t}_1{t}_2+\frac{13}{128}{t}_1^3\right]\right\}\hfill \\ & =& \frac{v\left(v+1\right)}{u\left(u+1\right)\left(2-2\alpha \right)\left(3-2\alpha \right)}\left\{\frac{1}{32}\frac{v}{u\left(2-2\alpha \right)}{t}_1\left[t_1^2+\left(4-t_1^2\right)x\right]-\frac{3v}{128u\left(2-2\alpha \right)}{t}_1^3\right.\hfill \\ & & -\frac{1}{8}\frac{\left(v+2\right)}{\left(u+2\right)\left(4-2\alpha \right)}\left[t_1^3+2t_1\left(4-t_1^2\right)x-t_1\left(4-t_1^2\right)x^2+2\left(4-t_1^2\right)\left(1-{\left|x\right|}^2\varsigma \right)\right]\hfill \\ & & \left.+\frac{7\left(v+2\right)}{32\left(u+2\right)\left(4-2\alpha \right)}{t}_1\left[t_1^2+\left(4-t_1^2\right)x\right]-\frac{13\left(v+2\right)}{128\left(u+2\right)\left(4-2\alpha \right)}{t}_1^3\right\}.\hfill \end{array}$$

Using Lemma 3, and since $t_1\le 2$ by Lemma 4, let $t_1=t$ and assume, without restriction, that $t\in \left[0,2\right]$. By using the triangle inequality with $\rho =\left|x\right|,$ we arrive at

$$\begin{array}{ccc}\hfill \left|m_2{m}_3-m_4\right|& \le & \frac{v\left(v+1\right)}{768u\left(u+1\right)\left(2-2\alpha \right)\left(3-2\alpha \right)}\left\{\left(\frac{3v}{u\left(1-\alpha \right)}-\frac{3v}{\left(u+2\right)\left(2-\alpha \right)}\right)t^3\right.\hfill \\ & & +\left(\frac{12v}{u\left(1-\alpha \right)}-\frac{12\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}\right)t\left(4-t^2\right)\rho \hfill \\ & & \left.+\left(\frac{96\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}+\frac{96\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}{\rho }^2+\frac{48\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}t{\rho }^2\right)\left(4-t^2\right)\right\}\hfill \\ & =& F_1(t,\rho ).\hfill \end{array}$$

Differentiating $F_1(t,\rho )$ with respect to $\rho$, we have

$$\begin{array}{ccc}\hfill F_1^{\prime }\left(\rho \right)& =& \frac{v\left(v+1\right)}{768u\left(u+1\right)\left(2-2\alpha \right)\left(3-2\alpha \right)}\left\{\left(\frac{12v}{u\left(1-\alpha \right)}-\frac{12\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}\right)t\left(4-t^2\right)\right.\hfill \\ & & \left.+\frac{192\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}\rho \left(4-t^2\right)+\frac{96\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}t\rho \left(4-t^2\right)\right\}\hfill \\ & >& 0.\hfill \end{array}$$

This implies that $F_1(t,\rho )$ is an increasing function of $\rho$ on the closed interval $\left[0,1\right].$

Hence, $F_1\left(\rho \right)\le F_1\left(0\right)$ for all $\rho \in \left[0,1\right]$, that is

$$\begin{array}{ccc}\hfill F_1\left(\rho \right)& =& \frac{v\left(v+1\right)}{768u\left(u+1\right)\left(2-2\alpha \right)\left(3-2\alpha \right)}\left\{\left(\frac{3v}{u\left(1-\alpha \right)}-\frac{3v}{\left(u+2\right)\left(2-\alpha \right)}\right)t^3\right.\hfill \\ & & \left.+\frac{96\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}\left(4-t^2\right)\right\}\hfill \\ & =& G_1\left(t\right).\hfill \end{array}$$

Differentiating $G_1\left(t\right)$ with respect to t, we have

$$\begin{array}{ccc}\hfill G_1^{\prime }\left(t\right)& =& \frac{v\left(v+1\right)}{768u\left(u+1\right)\left(2-2\alpha \right)\left(3-2\alpha \right)}\left\{\left(\frac{9v}{u\left(1-\alpha \right)}-\frac{9v}{\left(u+2\right)\left(2-\alpha \right)}\right)t^2\right.\hfill \\ & & \left.-\frac{192\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}t\right\}.\hfill \end{array}$$

Again, differentiating the above equation with respect to t, we have

$$\begin{array}{ccc}\hfill G_1^{″}\left(t\right)& =& \frac{v\left(v+1\right)}{768u\left(u+1\right)\left(2-2\alpha \right)\left(3-2\alpha \right)}\left\{\left(\frac{18v}{u\left(1-\alpha \right)}-\frac{18v}{\left(u+2\right)\left(2-\alpha \right)}\right)t\right.\hfill \\ & & \left.-\frac{192\left(v+2\right)}{\left(u+2\right)\left(2-\alpha \right)}\right\}\hfill \\ & <& 0.\hfill \end{array}$$

Since $t\in \left[0,2\right]$, by the assumption, it follows that $G_1\left(t\right)$ attains maximum at $t=0$, which corresponds to $\rho =0$, and it is the desired upper bound. Hence, we obtain

$$\left|m_2{m}_3-m_4\right|\le \frac{v\left(v+1\right)\left(v+2\right)}{4u\left(u+1\right)\left(u+2\right)\left(1-\alpha \right)\left(2-\alpha \right)\left(3-2\alpha \right)}.$$

The proof of Theorem 4 is thus completed.  □

If we put $u=v$ in Theorem 4, we have the following results, respectively:

Corollary 4.

Let$\alpha \in [0,1)$. If the function l, given by (1), belongs to the class${\mathcal{SL}}^{\ast }\left(u,u,\alpha \right)={\mathcal{SL}}^{\ast }\left(u,\alpha \right)$, then

$$\left|m_2{m}_3-m_4\right|\le \frac{1}{4\left(1-\alpha \right)\left(2-\alpha \right)\left(3-2\alpha \right)}$$

and the inequality is sharp.

If we choose $\alpha =0$ in Corollary 4, we arrive at the following result.

Corollary 5.

Let$\alpha \in [0,1)$. If the function l, given by (1), belongs to the class${\mathcal{SL}}^{\ast }\left(u,u,0\right)={\mathcal{SL}}^{\ast }\left(u\right)$, then

$$\left|m_2{m}_3-m_4\right|\le \frac{1}{24}$$

and the inequality is sharp.

Theorem 5.

If the function$l,$given by (1), belongs to the class${\mathcal{SL}}^{\ast }\left(u,v,\alpha \right)$, then

$$\begin{array}{ccc}\hfill \left|H_3\left(1\right)\right|& \le & \frac{v^2{\left(v+1\right)}^2}{512u^2{\left(u+1\right)}^2{\left(1-\alpha \right)}^2{\left(3-2\alpha \right)}^2}\hfill \\ & & \times \left\{\frac{2v\left(v+1\right)}{u\left(u+1\right)\left(1-\alpha \right)\left(3-2\alpha \right)}+\frac{50{\left(v+2\right)}^2}{{\left(u+2\right)}^2{\left(2-\alpha \right)}^2}+\frac{169\left(v+2\right)\left(v+3\right)}{\left(u+2\right)\left(u+3\right)\left(2-\alpha \right)\left(5-2\alpha \right)}\right\}.\hfill \end{array}$$

Proof. 

Since

$$\left|H_3\left(1\right)\right|\le \left|m_3\right|\left|m_2{m}_4-m_3^2\right|+\left|m_4\right|\left|m_2{m}_3-m_4\right|+\left|m_5\right|\left|m_3-m_2^2\right|$$

using the fact that $m_1=1$, with Theorems 1, 3 and 4 and Lemma 4, we have the required result

$$\begin{array}{ccc}\hfill \left|H_3\left(1\right)\right|& \le & \left|m_3\right|\left|m_2{m}_4-m_3^2\right|+\left|m_4\right|\left|m_2{m}_3-m_4\right|+\left|m_5\right|\left|m_1{m}_3-m_2^2\right|\hfill \\ & =& \frac{v\left(v+1\right)}{16u\left(u+1\right)\left(1-\alpha \right)\left(3-2\alpha \right)}\left[\frac{1}{16}{\left(\frac{v\left(v+1\right)}{u\left(u+1\right)\left(1-\alpha \right)\left(3-2\alpha \right)}\right)}^2\right]\hfill \\ & & +\frac{25v\left(v+1\right)\left(v+2\right)}{64u\left(u+1\right)\left(u+2\right)\left(1-\alpha \right)\left(3-2\alpha \right)\left(2-\alpha \right)}\left[\frac{1}{4}\left(\frac{v\left(v+1\right)\left(v+2\right)}{u\left(u+1\right)\left(u+2\right)\left(1-\alpha \right)\left(2-\alpha \right)\left(3-2\alpha \right)}\right)\right]\hfill \\ & & +\frac{169v\left(v+1\right)\left(v+2\right)\left(v+3\right)}{128u\left(u+1\right)\left(u+2\right)\left(u+3\right)\left(1-\alpha \right)\left(3-2\alpha \right)\left(2-\alpha \right)\left(5-2\alpha \right)}\left[\frac{1}{4}\frac{v\left(v+1\right)}{u\left(u+1\right)\left(1-\alpha \right)\left(3-2\alpha \right)}\right]\hfill \\ & =& \frac{v^2{\left(v+1\right)}^2}{512u^2{\left(u+1\right)}^2{\left(1-\alpha \right)}^2{\left(3-2\alpha \right)}^2}\hfill \\ & & \times \left\{\frac{2v\left(v+1\right)}{u\left(u+1\right)\left(1-\alpha \right)\left(3-2\alpha \right)}+\frac{50{\left(v+2\right)}^2}{{\left(u+2\right)}^2{\left(2-\alpha \right)}^2}+\frac{169\left(v+2\right)\left(v+3\right)}{\left(u+2\right)\left(u+3\right)\left(2-\alpha \right)\left(5-2\alpha \right)}\right\}.\hfill \end{array}$$

The proof of Theorem 5 is thus completed.  □

If we set $u=v$ in Theorem 5, we establish the below inequality.

Corollary 6.

Let$\alpha \in [0,1)$. If the function l, given by (1), belongs to the class${\mathcal{SL}}^{\ast }\left(u,u,\alpha \right)={\mathcal{SL}}^{\ast }\left(u,\alpha \right)$, then

$$\left|H_3\left(1\right)\right|\le \frac{1}{512{\left(1-\alpha \right)}^2{\left(3-2\alpha \right)}^2}\left\{\frac{2}{\left(1-\alpha \right)\left(3-2\alpha \right)}+\frac{50}{{\left(2-\alpha \right)}^2}+\frac{169}{\left(2-\alpha \right)\left(5-2\alpha \right)}\right\}$$

and the inequality is sharp.

6. Conclusions

In the present investigation, we have estimated smaller upper bounds and more accurate estimations for the functionals $\left|m_3-\mu m_2^2\right|$ and $\left|m_2{m}_4-m_3^2\right|$ for the class ${\mathcal{SL}}^{\ast }\left(u,v,\alpha \right)$ of holomorphic functions associated with the Carlson-Shaffer operator in the unit disk.