# The Second Hankel Determinant of Logarithmic Coefficients for Starlike and Convex Functions Involving Four-Leaf-Shaped Domain

Authors: Azzh Saad Alshehry; Rasool Shah; Abdul Bariq
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this particular research article, we take an analytic functionQ4L=1+5/6z+1/6z5, which makes a four-leaf-shaped image domain. Using this specific function, two subclasses, S4L∗ and C4L, 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.


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## Body

## 1. Introduction and Definitions

To properly comprehend the findings provided in the paper, certain important literature on geometric function theory must first be discussed. In this regard, the lettersS and A stand for the normalized univalent (or schlicht) functions class and the normalized holomorphic (or analytic) functions class, respectively. These primary notions are defined in the disc Ud=z∈ℂ:z<1 by
(1)A=F∈GUd:Fz=∑l=1∞blzl,where GUd expresses holomorphic functions class, and
(2)S=F∈A:FisSchlichtinUd.This classS 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 Ud with F′0=1 and F0=0, then FUd=w:w<ζ. 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.The below expression provided the coefficientsλn of logarithmic function JFz for F∈S(3)JFz=12logFzz=λ1z+λ2z2+λ3z3+⋯,∀z∈Ud.The above coefficients have a considerable impact on the theory of Schlicht functions in many estimations. De Branges [3] achieved that n≥1 in1985,
(4)∑l=1nln−l+1λn2≤∑l=1nn−l+1l,and equality will be achieved if F has the form z/1−eiφz2 for some φ∈ℝ. It is obvious that this inequality provides the most general version of the well-known Bieberbach-Robertson-Milin conjectures concerning the Taylor coefficients of F∈S. We quote [4–6] for further information on the demonstration of de Brange’s conclusion. By taking into account, the logarithmic coefficients, in 2005, Kayumov [7] established Brennan’s conjecture for conformal mappings. The major contributions to study the bounds of logarithmic coefficients for various holomorphic univalent functions are due to Alimohammadi et al. [8], Obradović et al. [9], Ye [10], Deng [11], Girela [12], Roth [13], and Andreev and Duren [14].For the prescribed functionsQ1,Q2∈A, the relation of subordination between Q1 and Q2 is as follows (mathematically as Q1≺Q2), if an holomorphic function u comes in Ud with the limitation uz<1 and u0=0 in a manner that Q1z=Q2uz satisfy. Consequently, the following relation applies if Q2∈S in Ud:
(5)Q1z≺Q2z,z∈Udif and only if
(6)Q10=Q20&Q1Ud⊂Q2Ud.By applying the notion of subordination, Ma and Minda [15] proposed a consolidated version of the set S∗ψ in 1992, and the following is a description of it:
(7)S∗ψ=g∈S:zg′zgz≺ψzforz∈Ud,with the Schlicht function ψ that satisfies
(8)ψ′0>0&Rψ>0.Various subclasses of the setS have been examined in the past few years as particular choices for family S∗ψ. For instance,
(i)
Scos∗≡S∗cosz (see [16]) and Scosh∗≡S∗coshz (see [17])(ii)
Stanh∗≡S∗1+tanhz (see [18, 19])(iii)
Se∗≡S∗ez (see [20, 21]) and Sρ∗≡S∗1+sinh−1z (see [22])(iv)
SS∗ξ≡S∗ψz with ψz=1+z/1−zξ and 0<ξ≤1 (see [23])(v)
SL∗≡S∗1+z (see [24]) and Scar∗≡S∗1+4/3z+2/3z2 (see [25, 26])For givenq,n∈ℕ=1,2,⋯, b1=1, and F∈S with the series representation (1), the Hankel determinant Hq,nF is expressed by
(9)Hq,nF=bnbn+1⋯bn+q−1bn+1bn+2⋯bn+q⋮⋮⋯⋮bn+q−1bn+q⋯bn+2q−2,and it was established by Pommerenke and Pommerenke [27, 28]. For several subcollections of Schlicht functions, the determinant Hq,nF has been examined. In specific, the sharp estimate of the functional H2,2F=b2b4−b32 for sets Cconvexfunctions,S∗starlikefunctions, and Rboundedturningfunctions were determined in [29, 30]. However, for the class of close-to-convex functions, the exact bounds of this determinant remain open [31]. The researchers were inspired by the works of Babalola [32], Bansal, et al. [33], Zaprawa [34], Kwon et al. [35], Kowalczyk et al. [36], and Lecko et al. [37].It is easy to deduce from equation (2) that, for F∈S, the logarithmic coefficients are computed by
(10)λ1=12b2,(11)λ2=12b3−12b22,(12)λ3=12b4−b2b3+13b23,(13)λ4=12b5−b2b4+b22b3−12b32−14b24.Currently, Lecko and Kowalczyk and Kowalczyk and Lecko [38, 39] studied the following Hankel determinant Hq,nJF/2 of logarithmic coefficients
(14)Hq,nJF2=λnλn+1⋯λn+q−1λn+1λn+2⋯λn+q⋮⋮⋯⋮λn+q−1λn+q⋯λn+2q−2.It has been noted that(15)H2,1JF2=λ1λ3−λ22,H2,2JF2=λ2λ4−λ32.By the virtue of the functionQ4L=1+5/6z+1/6z5, we define the following classes:
(16)S4L∗=F∈S:zF′zFz≺Q4L,z∈Ud,(17)C4L=F∈S:1+zF′′zF′z≺Q4L,z∈Ud.Alternatively,F∈S4L∗ if and only if an analytic function q occurs that satisfies qz≺Q4L in such that
(18)Fz=zexp∫0zqt−1tdt.By takingqz=Q4L in (18), we achieve the following function, which serves as an extremal in many of the class S4L∗ problems.
(19)F0z=zexp∫0z56+16t4dt=z+56z2+⋯.The following Alexander-type connection-related two classes were mentioned above. The above two families are interlinked by the following Alexander-type relation(20)F∈C4L⇔zF′∈S4L∗.From (19) and (20), we easily obtain the following extremal functions in various problems of the class C4L(21)g0z=z+512z2+⋯.Clearly,g0z,g0z2,g0z3, and g0z4 belong to the class C4L. That is,
(22)g1z=g0z=z+512z2+⋯,g2z=g0z2=z+536z3+⋯,g3z=g0z3=z+572z4+⋯,g4z=g0z4=z+5120z5+⋯.In the present paper, our core objective is to find the sharp coefficient type problems of logarithmic functions for the familiesS4L∗ and C4L. Among the inequalities to be studied here are Zalcman inequalities, the Fekete-Szegö inequality, and the second-order Hankel determinant H2,1JF/2.


## 2. A Set of Lemmas

We must first create the classP in the below set-builder form in order to declare the Lemmas that are employed in our primary findings.
(23)P=q∈GUd:q0=1&Req>0,z∈Ud.That is, ifq∈P, then q has the below series expansion
(24)qz=∑n=0∞enzn,z∈Ud.The following Lemma consists of the widely usede2 formula [40], the e3 formula [41], and the e4 formula illustrated in [42].Lemma 1.
Letq∈P be given in the form (24), then for ,ρ,δ∈Ud¯=Ud∪1.Lemma 2.
Letq∈P be of the form (24), then for x,δ,ρ∈Ud¯=Ud∪1(25)2e2=e12−e12−4x,(26)4e3=e13−2e12−4e1x+e1e12−4x2−2e12−41−x2ρ,(27)8e4=e14−e12−4xe12x2−3x+3+4x+4e12−41−x2ex−1ρ+x¯ρ2−1−ρ2δ.Lemma 3.
Letq∈P and has the expansion (24). Then,
(28)en+l−μenel≤2max1,2μ−1,(29)en≤2forn≥1,(30)en+l−μenel≤2,0≤μ≤1.The inequalities (28)–(30) are taken from [40, 43] and [26, 44, 45], respectively.Lemma 4 (see [40]).
Ifq∈P has the representation (24), then
(31)12Je13−Ke1e2+Le3≤J+K−2J+K−J+L.Lemma 5 [46].
Letγ,τ,ψ and ς satisfy that τ,ς∈0,1 and
(32)81−ςςτς+τ−ψ2+τψ−2γ2+τψ−2ςτ21−τ≤4τ2ς1−ς1−τ2.Ifq∈P has the expansion (24), then
(33)γe14+ςe22+2τe1e3−32ψe12e2−e4≤2.


## 3. Coefficient Inequalities for the ClassS4L∗

We start by establishing out the classS4L∗’s initial coefficient bounds.Theorem 6.
LetF be the series form (1) and if F∈S4L∗, then
(34)λ1≤512,λ2≤524,λ3≤536,λ4≤548.
These bounds are sharp.Proof.
LetF∈S4L∗. Then, Schwarz function u may therefore be used to express (16) as
(35)zF′zFz=1+56uz+16uz5=αz.
From the use of Schwarz functionu and if q∈P, we have
(36)qz=1+uz1−uz≔1+e1z+e2z2+⋯,and by simple computation, we get
(37)uz=12e1z+12e2−14e12z2+18e13−12e1e2+12e3z3+12e4−12e1e3−14e22−116e14+38e12e2z4+⋯.
Using (1), we attain
(38)zF′zFz≔1+b2z+−b22+2b3z2+−3b2b3+3b4+b23z3+−2b32+4b5−4b2b4+4b22b3−b24z4+⋯.
By some calculation and using the series expansion of (37), we get
(39)αz=1+512e1z+512e2−524e12z2+548e13−512e1e2+512e3z3+512e4−596e14+516e12e2−512e1e3−524e22z4+⋯.
Now, by comparing (38) and (39), we get
(40)b2=512e1,(41)b3=524e2−5288e12,(42)b4=536e3+3510368e13−596e1e2,(43)b5=548e4−455497664e14+1156912e12e2−351152e22−5108e1e3.
Utilizing (40) and (10), (11), (12), and (13), we have
(44)λ1=524e1,(45)λ2=548e2−596e12,(46)λ3=5288e13−572e1e2+572e3,(47)λ4=596e4−5768e14+5128e12e2−596e1e3−5192e22.
From (44), using triangle inequality and (29), we get
(48)λ1≤512.
Also, from (45), application (30), and triangle inequality, we get
(49)λ2≤524.
By rearranging (46), we have
(50)λ3=5288e13−4e1e2+4e3.
By Lemma4 and triangle inequality, we obtain
(51)λ3≤536.
By rearranging (47), we have
(52)λ4=−59612e22+18e14−34e2e12+e1e3−e4.
Comparing the equation of (52) right side with
(53)γe14+ςe22+2τe1e3−32ψe12e2−e4,we get γ=1/8,ς=1/2,τ=1/2,ψ=1/2, and
(54)81−ςςτς+τ−ψ2+τψ−2γ2+τψ−2ςτ21−τ≤4τ2ς1−ς1−τ2.
Thus, Lemma5’s requirements are all met. Hence,
(55)λ4≤5962=548.
These are sharp outcomes. Equality is determined by using (10)–(13) and
(56)F1z=zexp∫0z56+16t4dt=z+56z2+⋯,F2z=zexp∫0z56t+16t9dt=z+512z3+⋯,F3z=zexp∫0z56t2+16t14dt=z+518z4+⋯,F4z=zexp∫0z56t3+16t19dt=z+524z5+⋯·Theorem 7.
IfF∈S4L∗, then
(57)λ2−μλ12≤max524,5485μ3.
The above stated inequality is best possible.Proof.
By utilizing (44) and (45), we have
(58)λ2−μλ12=548e2−e1226+5μ6.
Implementation of (28) and triangle inequality, we get
(59)λ2−μλ12≤max524,5485μ3.
Equality is determined by using (10), (11), and
(60)F2z=zexp∫0z56t+16t9dt=z+512z3+⋯.Corollary 8.
IfF∈S4L∗, then
(61)λ2−λ12≤524.
This inequality is sharp and can be obtained by using (10), (11), and
(62)F2z=zexp∫0z56t+16t9dt=z+512z3+⋯.Theorem 9.
LetF be the expansion (1) and if F∈S4L∗, then
(63)λ1λ2−λ3≤536.
The above stated result is the best possible.Proof.
From (44)–(46), we easily attain
(64)λ1λ2−λ3=652304−e13+4213e1e2−3213e3.
By using Lemma4 and triangle inequality, we obtain
(65)λ1λ2−λ3≤6523046413=536.
Equality is determined by using (10), (11), (12), and
(66)F3z=zexp∫0z56t2+16t14dt=z+518z4+⋯.Theorem 10.
LetF be the expansion (1) and if F∈S4L∗, then
(67)λ4−λ22≤548.
The last stated inequality is the finest.Proof.
From the use (45) and (47), we get
(68)λ4−λ22=−5961724e22−2344e2e12+1796e14+e1e3−e4.
Comparing the right side of (68) with
(69)γe14+ςe22+2τe1e3−32ψe12e2−e4,we get γ=17/96,ς=17/24,τ=1/2, ψ=23/36, and
(70)81−ςςτς+τ−ψ2+τψ−2γ2+τψ−2ςτ21−τ=0.0051909,4τ2ς1−ς1−τ2=0.051649.
Thus, Lemma5’s requirements are all met. Hence,
(71)λ4−λ22≤5962=548.
Equality is determined by using (11), (13), and
(72)F4z=zexp∫0z56t3+16t19dt=z+524z2+⋯.Theorem 11.
LetF∈S4L∗ be the representation (1). Then,
(73)H2,1JF/2≤25576.
This result is sharp.Proof.
We can write theH2,1JF/2 as
(74)H2,1JF/2=λ1λ3−λ22.
From (44)–(46), we have
(75)λ1λ3−λ22=2527648e14−4e12e2+16e1e3−12e22.
Using (25) and (26) to express e2 and e3 in terms of e1 and also e1=e, with 0≤e≤2, we obtain
(76)λ1λ3−λ22=2527648−4e2x24−e2+8e1−x24−e2δ−3x24−e22.
By changingδ≤1 and x=c, where c≤1 and utilizing triangle inequality and pickings e∈0,2, so
(77)λ1λ3−λ22≤25276484e2c24−e2+8e1−c24−e2+3c24−e22≔Ξe,c.
Differentiate with respect toc, we have
(78)∂Ξe,c∂c=2527648−2ce4+16ce3−16ce2−64ce+96c.
It is easy exercise to show thatΞ/e,c≥0 on 0,1, so that Ξe,c≤Ξe,1. Putting c=1, we get
(79)λ1λ3−λ22≤25276484e24−e2+34−e22≔Θe.
AsΘ/e≤0, so Θe is a decreasing function, so that it gives a maximum value at e=0(80)H2,1JF2≤252764848=25576.
Equality is determined by using (10), (11), (12), and
(81)F2z=zexp∫0z56t+16t9dt=z+512z3+⋯.


## 4. Coefficient Inequalities for the ClassC4L

For the function of classC4L, we start this portion by determining the absolute values of the first four initial logarithmic coefficients.Theorem 12.
LetF be given by (1) and if F∈C4L, then
(82)λ1≤524,λ2≤572,λ3≤5144,λ4≤148.
These bounds are sharp.Proof.
LetF∈C4L. Then, (17) can be written in the form of Schwarz function as
(83)1+zF′′zF′z=1+56uz+16uz5=ψz.
Using (1), we obtain
(84)1+zF′′zF′z≔1+2b2z+6b3−4b22z2+8b23−18b2b3+12b4z3+20b5−16b24+48b22b3−32b2b4−18b32z4+⋯.
Now, by comparing (84) and (39), we get
(85)b2=524e1,b3=572e2−5864e12,b4=5144e3+3541472e13−5384e1e2,b5=148e4−91497664e14+236912e12e2−71152e22−1108e1e3.
Utilizing (85) and (10), (11), (12), and (13) we have
(86)λ1=524e1,(87)λ2=548e2−596e12,(88)λ3=5288e13−572e1e2+572e3,(89)λ4=596e4−5768e14+5128e12e2−596e1e3−5192e22.
From (86), using triangle inequality and (29), we get
(90)λ1≤524.
Also, from (87), application (30), and triangle inequality, we get
(91)λ2≤572.
By rearranging (88), we have
(92)λ3=5288748e13−1924e1e2+e3.
By Lemma4 and triangle inequality, we obtain
(93)λ3≤5144.
By rearranging (89), we have
(94)λ4=−1961127e22+13109248832e14−23535184e2e12+1924e1e3−e4.
Comparing the right side of (94) with
(95)γe14+ςe22+2τe1e3−32ψe12e2−e4,we get γ=13109/248832,ς=11/27,τ=19/48, and ψ=2353/7776. Thus, all the conditions of Lemma 5 are satisfied. Hence, we have
(96)λ4≤1962=148.
These are sharp outcomes. Equality is determined by using (10), (11), (12), and (13) along with (22).Theorem 13.
LetF∈C4L be the series form (1). Then,
(97)λ2−μλ12≤max572,5727+15μ12,forμ∈ℂ.
This inequality is sharp.Proof.
By utilizing (86) and (87), we have
(98)λ2−μλ12=5144e2−e12219+15μ24.
Implementation of (28) and triangle inequality, we get
(99)λ2−μλ12≤max572,5727+15μ12.
Equality is determined by using (10), (11), and (22).Forλ=1, we get the below corollary.Corollary 14.
LetF∈C4L, and it has the form (1). Then,
(100)λ2−λ12≤572.
This inequality is sharp and can be obtained by using (10), (11), and (22).Theorem 15.
LetF be the form (1) and if F∈C4L, then
(101)λ1λ2−λ3≤5144.
This result is sharp.Proof.
By using (86)–(88), we obtain
(102)λ1λ2−λ3=5288−2631152e13+e1e2−e3.
By using Lemma4 and triangle inequality, we obtain
(103)λ1λ2−λ3≤52882=5144.
Equality is determined by using (10), (11), (12), and (22).Theorem 16.
LetF be the form (1) and F∈C4L. Then,
(104)λ4−λ22≤148.
This result is sharp.Proof.
By using (87) and (89), we obtain
(105)λ4−λ22=−196113216e22−7071296e2e12+35243497664e14+1924e1e3−e4.
Comparing the right side of (68) with
(106)γe14+ςe22+2τe1e3−32ψe12e2−e4,we get γ=35243/497664,ς=113/216,τ=19/24,ψ=707/1944, and
(107)81−ςςτς+τ−ψ2+τψ−2γ2+τψ−2ςτ21−τ=0.00062010,4τ2ς1−ς1−τ2=0.057070.
Thus, all the conditions of Lemma5 are satisfied. Hence, we have
(108)λ4−λ22≤1962=148.
Equality is determined by using (11), (13), and (22).Theorem 17.
LetF be given the form (1) and F∈C4L. Then,
(109)H2,1JF2≤25576.
This result is sharp.Proof.
We can write theH2,1FF/2 as;
(110)H2,1JF2=λ1λ3−λ22.
From (86)–(88), we have
(111)λ1λ3−λ22=1477757443575e14−22800e12e2+86400e1e3−57600e22.
Using (25) and (26) to express e2 and e3 in terms of e1 and also e1=e, with 0≤e≤2, we obtain
(112)λ1λ3−λ22=147775744−21600e2x24−e2+43200e1−x24−e2δ−14400x24−e22+3000e2x4−e2−625e4.
By replacingδ≤1 and x=c, where c≤1 and using triangle inequality and taking e∈0,2, so
(113)λ1λ3−λ22≤14777574421600e2c24−e2+43200e1−c24−e2+14400c24−e22+3000e2c4−e2+625e4≔Ωe,c.
Differentiate with respect toc, we have
(114)∂Ωe,c∂c=147775744−600e−2e+224ce2−144ce+5e2+192c.
It is a simple exercise to show thatΩ′e,c≥0 on 0,1, so that Ωe,c≤Ωe,1. Putting c=1 gives
(115)λ1λ3−λ22≤14777574424600e24−e2+625e4+144004−e22≔Θe.
AsΘ′e≤0, so Θe is a decreasing function, so that it gives a maximum value at e=0(116)H2,1JF2≤147775744230400=255184.
Equality is determined by using (10), (11), (12), and (22).



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