# The Third Logarithmic Coefficient for Certain Close-to-Convex Functions

Authors: Najla M. Alarifi
Journal: Journal of Mathematics
 (2022)

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## Abstract

The logarithmic coefficientsγn of a normalized analytic functions f are defined by logfz/z=2∑n=1∞γnzn. For certain close-to-convex functions fz=z+a2z2+⋯, Cho et al. (on the third logarithmic coefficient in some subclasses of close-to-convex functions) has obtained the upper bound of the third logarithmic coefficient γ3 when the second coefficient a2 is real. In the present paper, the upper bound of the third logarithmic coefficient γ3 is computed with no restriction on the second coefficient a2.


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## Body

## 1. Introduction and Preliminaries

LetD:=z∈ℂ:z<1 be the open unit disk in the complex plane ℂ and let A be the set of all analytic normalized functions f:D⟶ℂ of the form(1)fz=z+∑n=2∞anzn,z∈D.LetS be its subclass consisting of functions that are univalent in D. Given a function f∈S, the coefficients γn are defined by(2)logfzz=2∑n=1∞γnzn,z∈D\0,log1:=0.For example (see Figure1), for the Koebe function k given by kz=z/1−z2, the logarithmic coefficients γn=1/n are as follows(3)logkzz=2∑n=1∞1nzn.Figure 1
Plot oflogfz/z,fz=z/1−z2.The Milin conjecture ([1] and ([2] p. 155)) gives an inequality satisfied by the logarithmic coefficients. For f∈S, the logarithmic coefficients satisfy(4)∑m=1n∑k=1mkγk2−1k≤0.The Milin conjecture was confirmed (e.g., ([2] p. 37), by Branges [3] and implies the famous Bieberbach conjecture that an≤n for f∈S. Sharp estimates for the class S are known only for the first two coefficients:(5)γ1≤1,γ2≤12+1e=0.635….Note that Obradović and Tuneski [4] obtained an upper bound of γ3 for the class S. The problem of estimating the modulus of the first three logarithmic coefficients is significantly studied for the subclasses of S, and in some cases, sharp bounds are obtained. For instance, sharp estimates for the class of starlike functions S∗ are given by the inequality γn≤1/n holds for n∈ℕ ([5], p. 42).Furthermore, forf∈SS∗, the class of strongly starlike function of the order β, 0≤β≤1, it holds that γn≤β/nn∈ℕ [6]. The bounds of γn for functions in subclasses of S have been widely studied in recent years. Sharp estimates for different subclasses are given in [6, 7] and ([5], p. 116) and [8], respectively, while nonsharp estimates for the class of Bazilevic and close-to-convex are given in [9–11], respectively.Letℱ1,ℱ2,ℱ3 be the subclasses of S satisfying, respectively, the next conditions:(6)ℜ1−zf′z>0,z∈D,ℜ1−z2f′z>0,z∈D,ℜ1−z+z2f′z>0,z∈D.Note that each class defined above is the subclass of the well-known class of close-to-convex functions; consequently, familiesℱi, i=1,2,3, contain only univalent functions ([2], Vol. II, p. 2). The sharp bounds of γ1,γ2 and partial results for γ3 of the subclasses ℱ1,ℱ2,ℱ3 of S were determined by Pranav Kumar and Vasudevarao [12].Moreover, Cho et al. [13] computed the sharp upper bounds for the third logarithmic coefficient γ3 of f when a2 is a real number. Differentiating (1) and comparing the coefficients with (2), we get γ1=1/2a1, γ2=1/2a3−1/2a2, and(7)γ3=12a4−a2a3+13a23.The main aim of this paper is to determine the upper bound of the third logarithmic coefficient in the general case ofa2. The following lemma is needed to prove our main results.Lemma 1 (see [14]).
Letwz=c1z+c2z2+⋯ be a Schwarz function. Then(8)c1≤1,c2≤1−c12,c3≤1−c12−c221+c1.


## 2. Main Results

Our main result is as follows:Theorem 1.
Letf∈ℱ1. Then(9)γ3≤15.7548=0.328125.Proof.
Sincef∈ℱ1, and for analytic function w in D with w0=0 satisfying the formula(10)1−zf′z=1+wz1−wz=1+2wz+2w2z+⋯.
We obtain(11)wz=c1z+c2z2+⋯.
Then, by using (10) along with (11) leads to(12)a2=121+2c1,a3=131+2c1+2c12+2c2,a4=141+2c1+2c2+2c3+2c12+4c1c2+2c13.
From (7) and (12), we obtain(13)γ3=1483+2c1+4c2+12c3+8c1c2+4c13.
In view of Lemma1, we attain(14)48γ3≤3+2c1+4c2+12c3+8c1c2+4c13≤3+2c1+4c2+121−c12−c221+c1+8c1c2+4c13=:f1c1,c2,where(15)f1x,y=3+2x+4y+121−x2−y21+x+8xy+4x3,x,y∈E:0≤x≤1,0≤y≤1−x2.
The system(16)∂f1x,y∂x=2−24x+12y1+x2+8y+12x2=0,∂f1x,y∂y=4−24y1+x+8x=0,has a unique solution x1,y1=1/4,5/16∈E\∂E with(17)f1x1,y1=15.75.
The maximum value off1 is obtained when x,y is a point on the boundary of E. In view of this, we have(18)f1x,0=15+2x−12x2+4x3≤9+10309=15.08580…,f10,y=3+4y+121−y2=15+4y−12y2≤463=15.33…,and(19)f1x,1−x2=7+22x−4x2−16x3≤15.304035….
Using (14) and (17)–(19), we conclude the following outcome:(20)48γ3≤15.75,i.e.,γ3≤0.328125.
This completes the proof.Remark 1.
Iff∈ℱ1, where f″0 is a real number, then we get the result in [13](21)γ3≤128811+1530=0.323466….Theorem 2.
Letf∈ℱ2. Then(22)γ3≤0.258765….Proof.
Sincef∈ℱ2, then there exists an analytic function w in D with w0=0 and(23)1−z2f′z=1+wz1−wz=1+2wz+2w2z+⋯.
The coefficients can be determined by comparing the information in (11) and (23)(24)a2=c1,a3=131+2c2+2c12,a4=12c1+c3+2c1c2+c13.
From (7) and (24), we have the following conclusion:(25)γ3=112c1+3c3+2c1c2+c13.
Moreover, according to Lemma1, we get the following inequality:(26)12γ3≤c1+3c3+2c1c2+c13≤c1+31−c12−c221+c1+2c1c2+c13=:f2c1,c2,where(27)f2x,y=31−x2−y21+x+2xy+x+x3,x,y∈E:0≤x≤1,0≤y≤1−x2.
From the system,(28)∂f2x,y∂x=6x+3y1+x2+2y+1+3x2=0,∂f2x,y∂y=−6y1+x+2x=0,only one solution x2,y2 lies in the interior of E, where(29)x2=4−76=0.22570…,y2=47−147108=0.092217…,and(30)f2x2,y2=3.10518….
On the boundary ofE, we have the next property(31)f2x,0=31−x2+x+x3≤2+496=3.08866,for0≤x≤1,f20,y=31−y2≤3,for0≤y≤1,f2x,1−x2=6x−4x3≤22=2.82842….
Consequently, (26), (30), and (31) yield(32)12γ3≤3.10518…,i.e.,γ3≤0.258765….Remark 2.
Iff∈ℱ2, where f″0 is a real number, then [13](33)γ3≤197295+2346=0.258223….Theorem 3.
Letf∈ℱ3. Then(34)γ3≤17.7548=0.36979….Proof.
Letf∈ℱ3 and an analytic function w in D with w0=0 such that(35)1−z+z2f′z=1+wz1−wz=1+2wz+2w2z+⋯.
Substituting (11) into (35), we have(36)a2=121+2c1,a3=23c1+c2+c12,a4=142c2+2c3+2c12+2c13+4c1c2−1.
By using (7) and (36), we obtain(37)γ3=148−5−2c1+4c2+12c3+8c1c2+4c13.
According to Lemma1, we conclude that(38)48γ3≤5+2c1+4c2+12c3+8c1c2+4c13≤5+2c1+4c2+121−c12−c221+c1+8c1c2+4c13=:f3c1,c2,where(39)f3x,y=5+2x+4y+121−x2−y21+x+8xy+4x3,x,y∈E:0≤x≤1,0≤y≤1−x2.
The system(40)∂f3x,y∂x=2−24x+12y1+x2+8y+12x2=0,∂f3x,y∂y=4−24y1+x+8x=0,admits a unique solution x3,y3=1/4,5/16 in the interior of E such that(41)f1x3,y3=17.75.
On the boundary ofE, the following cases are observed:(42)f3x,0=17+2x−12x2+4x3≤11+10309=17.08580…,f30,y=3+4y+121−y2=17+4y−12y2≤463=17.33…,and(43)f3x,1−x2=9+22x−4x2−20x3≤16.56455….
Equations (38), (41)–(43) show that(44)γ3≤17.7548=0.36979….Remark 3.
Letf∈ℱ3, where f″0 is a real number. Then [13](45)γ3≤17776743+131262=0.368238….



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