The Second Hankel Determinant of Logarithmic Coefficients for Starlike and Convex Functions Involving Four-Leaf-Shaped Domain
Abstract
In this particular research article, we take an analytic function , which makes a four-leaf-shaped image domain. Using this specific function, two subclasses, and , of starlike and convex functions will be defined. For these classes, our aim is to find some sharp bounds of inequalities that consist of logarithmic coefficients. Among the inequalities to be studied here are Zalcman inequalities, the Fekete-Szegö inequality, and the second-order Hankel determinant.
1. Introduction and Definitions
This class evolved as the foundational component of cutting-edge research in this area. In his paper [1], Koebe established the presence of a “covering constant” ζ, demonstrating that if F is holomorphic and Schlicht in with F′(0) = 1 and F(0) = 0, then . Many mathematicians were intrigued by this beautiful result. Within a few years, the wonderful article by Bieberbach [2], which gave rise to the renowned coefficient hypothesis, was published.
In the present paper, our core objective is to find the sharp coefficient type problems of logarithmic functions for the families and Among the inequalities to be studied here are Zalcman inequalities, the Fekete-Szegö inequality, and the second-order Hankel determinant H2,1(JF/2).
2. A Set of Lemmas
The following Lemma consists of the widely used e2 formula [40], the e3 formula [41], and the e4 formula illustrated in [42].
Lemma 1. Let be given in the form (24), then for
The inequalities (28)–(30) are taken from [40, 43] and [26, 44, 45], respectively.
Lemma 5 [46]. Let γ, τ, ψ and ς satisfy that τ, ς ∈ (0, 1) and
3. Coefficient Inequalities for the Class
We start by establishing out the class ’s initial coefficient bounds.
Proof. Let Then, Schwarz function u may therefore be used to express (16) as
From the use of Schwarz function u and if , we have
Using (1), we attain
By some calculation and using the series expansion of (37), we get
Now, by comparing (38) and (39), we get
Utilizing (40) and (10), (11), (12), and (13), we have
From (44), using triangle inequality and (29), we get
Also, from (45), application (30), and triangle inequality, we get
By rearranging (46), we have
By Lemma 4 and triangle inequality, we obtain
By rearranging (47), we have
Comparing the equation of (52) right side with
Thus, Lemma 5’s requirements are all met. Hence,
These are sharp outcomes. Equality is determined by using (10)–(13) and
Theorem 7. If then
The above stated inequality is best possible.
Proof. By utilizing (44) and (45), we have
Implementation of (28) and triangle inequality, we get
Equality is determined by using (10), (11), and
Corollary 8. If then
This inequality is sharp and can be obtained by using (10), (11), and
Theorem 9. Let F be the expansion (1) and if then
The above stated result is the best possible.
Proof. From (44)–(46), we easily attain
By using Lemma 4 and triangle inequality, we obtain
Equality is determined by using (10), (11), (12), and
Proof. From the use (45) and (47), we get
Comparing the right side of (68) with
Thus, Lemma 5’s requirements are all met. Hence,
Equality is determined by using (11), (13), and
Proof. We can write the H2,1(JF/2) as
Using (25) and (26) to express e2 and e3 in terms of e1 and also e1 = e, with 0 ≤ e ≤ 2, we obtain
By changing |δ| ≤ 1 and |x| = c, where c ≤ 1 and utilizing triangle inequality and pickings e ∈ [0, 2], so
Differentiate with respect to c, we have
It is easy exercise to show that Ξ/(e, c) ≥ 0 on [0, 1], so that Ξ(e, c) ≤ Ξ(e, 1). Putting c = 1, we get
As Θ/(e) ≤ 0, so Θ(e) is a decreasing function, so that it gives a maximum value at e = 0
Equality is determined by using (10), (11), (12), and
4. Coefficient Inequalities for the Class
For the function of class , we start this portion by determining the absolute values of the first four initial logarithmic coefficients.
Proof. Let Then, (17) can be written in the form of Schwarz function as
Using (1), we obtain
Now, by comparing (84) and (39), we get
Utilizing (85) and (10), (11), (12), and (13) we have
From (86), using triangle inequality and (29), we get
Also, from (87), application (30), and triangle inequality, we get
By rearranging (88), we have
By Lemma 4 and triangle inequality, we obtain
By rearranging (89), we have
Comparing the right side of (94) with
These are sharp outcomes. Equality is determined by using (10), (11), (12), and (13) along with (22).
Proof. By utilizing (86) and (87), we have
Implementation of (28) and triangle inequality, we get
For λ = 1, we get the below corollary.
Corollary 14. Let , and it has the form (1). Then,
This inequality is sharp and can be obtained by using (10), (11), and (22).
Proof. By using (86)–(88), we obtain
By using Lemma 4 and triangle inequality, we obtain
Proof. By using (87) and (89), we obtain
Comparing the right side of (68) with
Thus, all the conditions of Lemma 5 are satisfied. Hence, we have
Proof. We can write the H2,1(FF/2) as;
Using (25) and (26) to express e2 and e3 in terms of e1 and also e1 = e, with 0 ≤ e ≤ 2, we obtain
By replacing |δ| ≤ 1 and |x| = c, where c ≤ 1 and using triangle inequality and taking e ∈ [0, 2], so
Differentiate with respect to c, we have
It is a simple exercise to show that Ω′(e, c) ≥ 0 on [0, 1], so that Ω(e, c) ≤ Ω(e, 1). Putting c = 1 gives
As Θ′(e) ≤ 0, so Θ(e) is a decreasing function, so that it gives a maximum value at e = 0
Conflicts of Interest
The authors declare that there are no conflicts of interest regarding the publication of this article.
Acknowledgments
This research was supported by the Princess Nourah Bint Abdulrahman University Researchers Supporting Project number (PNURSP2022R183), Princess Nourah Bint Abdulrahman University, P.O. Box 84428, Riyadh 11671, Saudi Arabia.
Open Research
Data Availability
The numerical data used to support the findings of this study are included within the article.