# The Sharp Upper Bounds of the Hankel Determinant on Logarithmic Coefficients for Certain Analytic Functions Connected with Eight-Shaped Domains

Authors: Pongsakorn Sunthrayuth; Naveed Iqbal; Muhammad Naeem; Yousef Jawarneh; Sallieu K. Samura
Journal: Journal of Function Spaces
 (2022)

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## Abstract

The present study’s intention is to produce exact estimations of some problems involving logarithmic coefficients for functions belonging to the considered subcollectionBTsin of the bounded turning class. Furthermore, for the class BTsin, we look into the accurate bounds of the Zalcman inequality, Fekete-Szegö inequality along with D2,1Gg/2 and D2,2Gg/2. Importantly, all of these bounds are shown to be sharp.


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## Body

## 1. Introduction and Definitions

To properly understand the findings provided in the article, certain important literature on Geometric Function Theory must first be discussed. In this regard, the lettersS and A stand for the normalized univalent functions class and the normalized holomorphic (or analytic) functions class, respectively. These primary notions are defined in the region Ed=z∈ℂ:z<1 by
(1)A=g∈HEd:gz=z+∑k=2∞bkzkz∈Ed,where HEd symbolizes the holomorphic functions class, and
(2)S=g∈A:gisunivalentinEd.The following formula defines the logarithmic coefficientsβn of g that belong to S(3)Ggz≔loggzz=2∑n=1∞βnznforz∈Ed.In many estimations, these coefficients provide a significant contribution to the concept of univalent functions. In 1985, De Branges [1] proved that
(4)∑k=1nkn−k+1βn2≤∑k=1nn−k+1k∀n≥1,and equality will be achieved if g has the form z/1−eiθz2 for some θ∈ℝ. In its most comprehensive version, this inequality offers the famous Bieberbach-Robertson-Milin conjectures regarding Taylor coefficients of g∈S. We refer to [2–4] for further details on the proof of De Branges’ finding. By considering the logarithmic coefficients, Kayumov [5] was able to prove Brennan’s conjecture for conformal mappings in 2005. For your reference, we mention a few works that have made major contributions to the research of the logarithmic coefficients. Andreev and Duren [6], Alimohammadi et al. [7], Deng [8], Roth [9], Ye [10], Obradović et al. [11], and finally the work of Girela [12] are the major contributions to the study of logarithmic coefficients for different subclasses of holomorphic univalent functions.As stated in the definition, it is simple to determine that forg∈S, the logarithmic coefficients are computed by
(5)β1=12b2,(6)β2=12b3−12b22,(7)β3=12b4−b2b3+13b23,(8)β4=12b5−b2b4+b22b3−12b32−14b24.For givenq,n∈ℕ=1,2,⋯, b1=1, and g∈S with the series expansion (1), the Hankel determinant Dq,ng is represented by
(9)Dq,ng=bnbn+1⋯bn+q−1bn+1bn+2⋯bn+q⋮⋮⋯⋮bn+q−1bn+q⋯bn+2q−2.It was defined by Pommerenke [13, 14]. This determinant has indeed been investigated for a number of univalent function subclasses. In specific, the sharp estimate of the functional D2,2g=b2b4−b32 for the sets C (convex functions), S∗(starlike functions), and R (bounded turning functions) has been effectively established in [15, 16]. Later, numerous scholars published their findings on the upper bounds of D2,2g for various subcollections of holomorphic functions; see [17–23]. However, for the class of close-to-convex functions, the exact estimation of this determinant is yet unknown [24].Analogous to the determinantDq,ng mentioned above, Kowalczyk and Lecko [25, 26] considered to examine the following determinant Dq,nGg/2 with entries from logarithmic coefficients of g(10)Dq,nGg2=βnβn+1⋯βn+q−1βn+1βn+2⋯βn+q⋮⋮⋯⋮βn+q−1βn+q⋯βn+2q−2.It is observed that(11)D2,1Gg2=β1β3−β22,D2,2Gg2=β2β4−β32.For the given functionsG1,G2∈A, the subordination between G1 and G2 (mathematically written as G1≺G2), if we get a Schwarz function v with v0=0 and vz<1 for z∈Ed in a way such that G1z=G2vz hold true. Additionally, the following relation applies if G2 in Ed is univalent:
(12)G1z≺G2z,z∈Ed,if and only if
(13)G10=G20,G1Ed⊂G2Ed.In 1992, Ma and Minda [27] developed a consolidated version of the collection S∗π by using the principle of subordination, and the following is a description of it:
(14)S∗π≔g∈S:zg′zgz≺πz,z∈Ed,where the univalent function π satisfies
(15)π′0>0,Reπ>0.The areaπEd is also symmetric about x-axis and has a star-shaped form around the point π0=1. In recent years, a wide variety of the collection S’s subfamilies have been looked into as particular alternatives for the class S∗π. As an illustration:
(i)
SS∗ξ≡S∗πz with πz=1+z/1−zξ and 0<ξ≤1 (see [28])(ii)
SL∗≡S∗1+z1/2 (see [29]), and Scςr∗≡S∗1+4/3z+2/3z2 (see [30, 31])(iii)
Sρ∗≡S∗1+sinh−1z (see [32]), and Se∗≡S∗ez (see [33, 34])(iv)
Scos∗≡S∗cosz (see [35]), and Scosh∗≡S∗coshz (see [36])(v)
Stanh∗≡S∗1+tanhz (see [37, 38])In [39], Cho et al. developed a novel subfamily of starlike function described by
(16)Ssin∗≔g∈A:zg′zgz≺1+sinzz∈Ed.From the definition of the familySsin∗, the authors [39] deduced that
(17)g∈Ssin∗⇔gz=zexp∫0zut−1tdt,for some uz≺u0z=1+sinz. By substituting
(18)uz=u0z=1+sinzin (17), we acquire the function
(19)g0z=zexp∫0zsinttdt=z+z2+12z3+19z4⋯,which acts as the extremal function in a variety of Ssin∗-family problems. In [40], the authors defined the following subfamily BTsin of holomorphic functions by using (18):
(20)BTsin=g∈S:g′z≺1+sinzz∈Ed.Our primary objective in the current paper is to compute the problems involving the sharp logarithmic coefficients for the classBTsin of bounded turning functions connected to an eight-shaped domain. The sharp bounds of the Zalcman inequality, the Fekete-Szegö type inequality, along with the determinants D2,1Gg/2 and D2,2Gg/2 for the family BTsin are found using logarithmic coefficient entries.


## 2. Preliminary 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:
(21)P=p∈HEd:p0=1&Rep>0,z∈Ed.That is, ifp∈P, then it has the series representation
(22)pz=1+∑n=1∞pnznz∈Ed.Lemma 1 (see [41]).
Letp∈P and has the series form (22). Then for x,δ,ρ∈Ed¯=Ed∪11(23)2p2=p12+x4−p12,(24)4p3=p13+24−p12p1x−p14−p12x2+24−p121−x2δ,(25)8p4=p14+4−p12xp12x2−3x+3+4x−44−p121−x2,(26)px−1δ+x¯δ2−1−δ2ρ.Lemma 2.
Ifp∈P and has the expansion (22), then
(27)pn≤2n≥1,and if Q∈0,1 and Q2Q−1≤R≤Q, then
(28)p3−2Qp1p2+Rp13≤2.
Also,(29)pn+k−μpnpk≤2max1,2μ−1=21,for0≤μ≤1,2μ−1,otherwise..
The inequalities (27), (28) and (29) are taken from [42, 43], and [44], respectively.Lemma 3 (see [45]).
Letτ,ψ,ρ, and ς satify the inequalities 0<τ<1,0<ς<1 and
(30)8ς1−ςτψ−2ρ2+τς+τ−ψ2+τ1−τψ−2ςτ2≤4ςτ21−τ21−ς.
Ifp∈P has the form (22), then
(31)ρp14+ςp22+2τp1p3−32ψp12p2−p4≤2.


## 3. Coefficient Inequalities for the ClassBTsin

Theorem 4.
Ifg∈BTsin and has the series representation (1), then
(32)β1≤14,β2≤16,β3≤18,β4≤110.
These bounds are sharp and can be obtained from the following extremal functions(33)g0z=∫0z1+sintdt=z+12z2+⋯,g1z=∫0z1+sint2dt=z+13z3+⋯,g2z=∫0z1+sint3dt=z+14z4+⋯,g3z=∫0z1+sint4dt=z+15z5+⋯.Proof.
Letg∈BTsin. Consequently, (20) may be expressed using the Schwarz function as
(34)g′z=1+sinwz,z∈Ed.
The Schwarz functionw may be used to express it if p∈P as follows
(35)pz=1+wz1−wz=1+p1z+p2z2+p3z3+p4z4+⋯,equivalently,
(36)wz=pz−1pz+1=p1z+p2z2+p3z3+⋯2+p1z+p2z2+p3z3+⋯.
From (1), we obtain
(37)g′z=1+2a2z+3a3z2+4a4z3+⋯.
By simplification and using the series expansion of (36), we get
(38)1+sinwz=1+12p1z+−14p12+12p2z2+−12p1p2+548p13+12p3z3+12p4−132p14+516p12p2−12p1p3−14p22z4+⋯.
Comparing (37) and (38), we obtain
(39)a2=14p1,(40)a3=−112p12+16p2,(41)a4=−18p1p2+5192p13+18p3,(42)a5=110p4−1160p14+580p12p2−110p1p3−120p22.
Putting (42) in (5), (6), (7), and (8), we obtain
(43)β1=18p1,(44)β2=−11192p12+112p2,(45)β3=−112p1p2+5192p13+116p3,(46)β4=120p4−103392160p14+17288p12p2−23720p22−21320p1p3.
Forβ1, using (27), in (43), we obtain
(47)β1≤14.
Forβ2, putting (29) in (44), we obtain
(48)β2≤16.
Forβ3, we can rewrite (45) as
(49)β3=116p3−43p1p2+512p13.
Using (28) we get
(50)β3≤18.
Forβ4, we can rewrite (46) as
(51)β4=−12010334608p14+2336p22+22132p1p3−3285108p12p2−p4.
Comparing the right side of (51) with
(52)ϱp14+ςp22+2τp1p3−32ψp12p2−p4,where
(53)ϱ=10334608,ς=2336,τ=2132,ψ=85108.
It follows that(54)8ς1−ςτψ−2ρ2+τς+τ−ψ2+τ1−τψ−2ςτ2=0.01647,4ςτ21−τ21−ς=0.04696.
Using (30) we deduce that
(55)β4≤110.Theorem 5.
Ifg has the series form (1) and belongs to BTsin, then
(56)β2−ηβ12≤max16,1483+3η.
Equality will be attained by using (5), (6), and
(57)g1z=∫0z1+sint2dt=z+13z3+⋯.Proof.
From (43) to (44), we get
(58)β2−ηβ12=−11192p12+112p2−η64p12.
Using (29), we have
(59)β2−ηβ12≤112max2,2211+3η16−1.
After the simplification, we get(60)β2−ηβ12≤max16,1483+3η.Theorem 6.
Ifg has the series expansion (1) and belongs to BTsin, then
(61)β1β2−β3≤18.
Equality can be attained by applying (5), (6), (7), and
(62)g2z=∫0z1+sint3dt=z+14z4+⋯.Proof.
From (43), (44), and (45), we obtain
(63)β1β2−β3=−17512p13+332p1p2−116p3.
After the simplification, we obtain(64)β1β2−β3=116p3−32p1p2+1732p13.
Using (28), we have
(65)β1β2−β3≤18.Theorem 7.
Ifg∈BTsin has given by (1), then
(66)β4−β22≤110.
This result is sharp and equality can be achieved by applying (6), (8), and
(67)g3z=∫0z1+sint4dt=z+15z5+⋯.Proof.
From (44) to (46), we obtain
(68)β4−β22=−7180p22+120p4+791152p12p2−2671184320p14−21320p1p3.
After the simplification, we obtain(69)β4−β22=−12026719216p14+79p22+22132p1p3−32395432p12p2−p4.
Comparing the right side of (69)with
(70)ϱp14+ςp22+2τp1p3−32ψp12p2−p4,where
(71)ϱ=26719216,ς=79,τ=2132,ψ=395432.
It follows that(72)8ς1−ςτψ−2ρ2+τς+τ−ψ2+τ1−τψ−2ςτ2=0.004121,4ςτ21−τ21−ς=0.03518.
Using (30) we deduce that
(73)β4−β22≤110.


## 4. Hankel Determinant with Logarithmic Coefficients

Theorem 8.
Letg∈BTsin and be of the form (1). Then
(74)D2,1Gg2=β1β3−β22≤136.
The above stated result is sharp. Equality can be attained with the use of (5), (6), (7), and
(75)g1z=∫0z1+sint2dt=z+13z3+⋯.Proof.
Employing (43), (44), and (45), we obtain
(76)D2,1Gg2=1128p1p3−136864p14−1144p22−11152p12p2.
Using (23) and (24) along with the assumption that p1=p,p∈0,2, we get
(77)D2,1Gg2=−15764−p22x2−14096p4−1512p24−p2x2+12564−p21−x2pδ.
Applying triangle inequality and assumingδ≤1,x=J,J≤1 and also setting p∈0,2, we have
(78)D2,1Gg2≤15764−p22J2+14096p4+1512p24−p2J2+12564−p21−J2p≔ϕp,J.
A little exercise can verify thatϕ′p,J≥0 in 0,1, and this implies ϕp,J≤ϕp,1. Thus, by choosing J=1, we achieve
(79)D2,1Gg2≤15764−p22+14096p4+1512p24−p2≔ϕp,1.
Now, sinceϕ′p,1<0, we see that ϕp,1 is a decreasing function, and so its maximum value appears at the lowest point p=0, which is
(80)D2,1Gg2≤136.Theorem 9.
Ifg∈BTsin and has the form (1), then
(81)D2,2Gg2≤164.
The inequality is sharp and can be obtained by using (6), (7), (8), and
(82)g2z=∫0z1+sint3dt=z+14z4+⋯.Proof.
The determinantD2,2Gg/2 can be written as
(83)D2,2Gg2=β2β4−β32.
Putting (44), (45), and (46), with p1=p, we obtain
(84)D2,2Gg2=117694720432p4p2−3456p2p22−637p6−50688p2p4+8928p3p3−69120p32−47104p23+73728p2p4+87552pp2p3.
Letv=4−p2 in (23), (24), and (25). Now, applying the simplest version of the given lemma, we get
(85)432p4p2=216p6+p4vx,3456p2p22=1728p4vx+864p2v2x2+864p6,50688p2p4=6336p6−25344p3v1−x2δx−25344p2v1−x2x¯δ2+25344vp2x2+25344p3v1−x2δ+19008p4xv+19008p4xv+25344p2v1−x21−δ2ρ+6336p4vx3,8928p3p3=4464p4xv−2232p4vx2+2232p6+4464p3v1−x2δ,69120p32=−17280x2v21−x2pδ−8640p4vx2+4320p6+4320x4v2p2+17280p3v1−x2δ+17280v21−x22δ2+17280p2x2v2+17280p4xv+34560pxv21−x2δ−17280x3v2p2,47104p23=5888v3x3+p6+17664p4vx+p2v2x2,73728p2p4=18432x3v2+4608p6+18432vp2x2+4608p4vx3+4608x4v2p2+18432p4xv+18432pxv21−x2δ−13824p4vx2−13824x3v2p2−18432xv2x¯1−x2x¯δ2−18432p2v1−x2x¯δ2+13824p2x2v2−18432p3v1−x2δx+18432p3v1−x2δ+18432xv21−x2,1−δ2ρ+18432p2v1−x21−δ2ρ−18432x2v21−x2pδ,87552pp2p3=21888pxv21−x2δ−10944x3v2p2+32832p4xv−10944p4vx2+21888p3v1−x2δ+10944p6+21888p2x2v2.
Putting the above expressions in (84), we get,
(86)D2,2Gg2=117694720−6912vp2x2−5888x3v3−45p6+2160p3v1−x2δ+264p4xv−1728p4vx3+288x4v2p2−96p2x2v2+18432x3v2−7488x3v2p2+6912p3v1−x2δx+6912p2v1−x2x¯δ2+5760pxv21−x2δ+648p4vx2+18432xv21−x21−δ2ρ−1152x2v21−x2pδ−18432xv21−x2x¯δ2−17280v21−x22δ2−6912p2v1−x21−δ2ρ.
Sincev=4−p2,(87)D2,2Gg2=117694720q1p,x+q2p,xδ+q3p,xδ2+q4p,x,δρ,where ρ,x,δ∈Ud¯, and
(88)q1p,x=4−p24−p2288x4p2−1600x3p2−5120x3−96x2p2−1728x3p4+264xp4−6912x2p2+648x2p4−45p6,q2p,x=4−p21−x24−p25760xp−1152x2p+2160p3+6912xp3,q3p,x=4−p21−x24−p2−17280−1152x2+6912x¯p2,q4p,x,δ=4−p21−x21−δ218432x4−p2−6912p2.
Now, by the virtue ofδ=y,x=x, and ρ≤1, we get
(89)D2,2Gg2≤117694720q1p,x+q2p,xy+q3p,xy2+q4p,x,δ≤Tp,x,y17694720,where
(90)Tp,x,y=m1p,x+m2p,xy+m3p,xy2+m4p,x1−y2,with
(91)m1p,x=4−p24−p2288x4p2+1600x3p2+5120x3+96x2p2+1728x3p4+264xp4+6912x2p2+648x2p4+45p6,m2p,x=4−p21−x24−p25760xp+1152x2p+2160p3+6912xp3,m3p,x=4−p21−x24−p217280+1152x2+6912xp2,m4p,x=4−p21−x218432x4−p2+6912p2.
To illustrate the sharp bounds of the given problem, we must maximizeTp,x,y in the closed cuboid Y:0,2×0,1×0,1.
(1)
Interior points of cuboidY
Let us choosep,x,y∈0,2×0,1×0,1. Then simple calculation yields
(92)∂T∂y=1444−p21−x216yx−16p2+x−154−p2+8pxx+54−p2+p26x+158.
Putting∂T/∂y=0, we obtain
(93)y=8pxx+54−p2+p26x+15/816x−1−6p2+15−x4−p2=y0.
Ify0∈Y is a critical point, then y0∈0,1, and it is applicable only if
(94)8pxx+54−p2+p348x+15+161−x15−x4−p2<96p21−x.(95)p2>415−x21−x.
To check critical points existence, we must find solutions that fulfill both constraints (94) and (95).
Letkx=415−x/21−x. As k′x<0 for all x∈0,1, it is evident that kx is decreasing in 0,1. Hence p2>14/5. It is easy to showcase that the inequality (94) does not hold in this scenario for all x∈2/5,1. As a result, Tp,x,y does not have a critical point in 0,2×2/5,1×0,1. Assume a critical point p~,x~,y~ of T exists inside the interior of the cuboid Y, it must unquestionably fulfil that x~<2/5.
From the arguments above, it is undeniable thatp~2≥292/103 and y~∈0,1. Now let us establish that Tp~,x~,y~<276480. For p,x,y∈292/1031/2,2×0,2/5×0,1, by invoking x<2/5 and 1−x2<1, it is not hard to observe that
(96)m1p,x≤4−p24−p2288254p2+1600253p2+5120253+96252p2+1728253p4+26425p4+6912252p2+648252p4+45p6=4−p2799232625p2+121712625p4+819200625+45p6≔Θ1p,m2p,x≤4−p24−p2576025p+1152252p+2160p3+691225p3=4−p224883225p+6091225p3≔Θ2p,m3p,x≤4−p24−p217280+1152252+691225p2=4−p2174643225−36748825p2≔Θ3p,m4p,x≤4−p218432254−p2+6912p2=4−p21474565−23045p2≔Θ4p.
Therefore, we have(97)Tp,x,y≤Θ1p+Θ4p+Θ2py+Θ3p−Θ4py2≔Γp,y.
Obviously, it can be seen that(98)∂Γ∂y=Θ2p+2yΘ3p−Θ4p,∂2Γ∂y2=2Θ3p−Θ4p=24−p2100915225−35596825p2.
SinceΘ3p−Θ4p≤0 for p∈292/1031/2,2, we obtain that ∂2Γ/∂y2≤0 for y∈0,1, and thus, it follows that
(99)∂Γ∂y≥∂Γ∂yy=1=4−p2−71193625p2+6091225p3+201830425+24883225p≥0.
Therefore, we have(100)Γp,y≤Γp,1=Θ1p+Θ2p+Θ3p≔ιp.
It is easy to be calculated thatιp attains its maximum value 74510.30 at p≈1.68373. Thus, we have
(101)Tp,x,y<276480,p,x,y∈292103,2×0,25×0,1.
Hence,Tp~,x~,y~<276480. This implies that T is less than 276480 at all the critical points in the interior of Y. Therefore, T has no optimal solution in the interior of Y.(2)
Interior of all the six faces of cuboidY:(i)
On the facep=0,Tp,x,y yields(102)b1x,y=T0,x,y=204891−x216x+x−15x−1y2+40x3,x,y∈0,1.
Differentiatingb1x,y with respect to y, we have
(103)∂b1∂y=36864y1−x2x−15x−1,x,y∈0,1.
Thus,b1x,y has no critical point in the interval 0,1×0,1.(ii)
On the facep=2,Tp,x,y becomes(104)T2,x,y=2880.(iii)
On the facex=0,Tp,x,y reduces to(105)b2p,y=Tp,0,y=4−p26912p2+2160p3y−24192y2p2+69120y2+45p6.
Differentiatingb2p,y partially with respect to y, we have
(106)∂b2∂y=4−p22160p3−48384yp2+138240y.
Solving∂b2/∂y=0, we obtain
(107)y=5p3167p2−20=y1.
For the given range ofy,y1 should belong to 0,1, which is possible only if p>p0,p0≈1.7609. Also derivative of b2p,y partially with respect to p is
(108)∂b2∂p=−4320p4y−13824p3+4−p2−48384y2p+6480yp2+13824p+48384y2p3−138240y2p+270p5.
Putting the value ofy in (108), with ∂b2/∂p=0 and simplifying, we obtain
(109)∂b2∂p=−2749576p7+35p9−385072p5−819200p+983040p3=0.
A calculation gives the solution of (109) in the interval 0,1, that is, p≈1.3851. Thus, b2p,y has no optimal point in the interval 0,2×0,1.(iv)
On the facex=1,Tp,x,y becomes(110)b3p,y=Tp,1,y=45p6+4−p24−p21984p2+5120+6912p2+2640p4.
Then(111)∂b3∂p=−3666p5−28416p3+36864p.
By setting∂b3/∂p=0, we get the critical point p≈1.0639 at which b3p,y attains its maximum value, which is given below
(112)Tp,1,y≤92795.48842.(v)
On the facey=0,Tp,x,y yields(113)b4p,x=Tp,x,0=−128p6x3+288p6x4−552p6x2+19488p4x−264p6x−147456p2x+45p6+4608p2x4+294912x−19200p4x3+1536p2x2+132096p2x3+1824p4x2−6912p4+27648p2−2304p4x4−212992x3.
A numerical computation shows that the solution for the system of equations(114)∂b4∂p=0,∂b4∂x=0does not exists in the interval 0,2×0,1. Hence, b4p,x has no optimal solution in the interval 0,2×0,1.(vi)
On the facey=1,Tp,x,y reduces to(115)b5p,x=Tp,x,1=45p6+288p6x4+9216p3x4+1152p5x3−2160p5−264p6x−1152p5x4+18432p3x3+3312p5x2+6144p4x3−128p6x3−43008p2x3−3456p4x4−552p6x2−1152p5x−138240p2−21216p4x2+276480+13824p2x4+81920x3−5856p4x−92160px3−17856p3x2−18432px4−258048x2+92160px−18432x4−18432p3x+8640p3+18432px2+17280p4+27648p2x+158208p2x2.
As in the above case, we conclude the same result for the facey=0, that is, system of equations
(116)∂b5∂p=0,∂b5∂x=0has no solution in the interval 0,2×0,1.(3)
On the edges of cuboidY:(i)
On the edgex=0 and y=0,Tp,x,y reduces to(117)Tp,0,0=−6912p4+45p6+27648p2=b6p.
It follows that(118)b6′p=−27648p3+270p5+55296p.
We see thatb6′p=0 for the critical point p0≈1.4285 at which b6p obtain its maximum value, which is given by
(119)Tp,0,0≤28018.97.(ii)
On the edgex=0 and y=1,Tp,x,y becomes(120)Tp,0,1=−2160p5−138240p2+17280p4+45p6+8640p3+276480=b7p.
Differentiatingb7p with respect to p, we have
(121)b7′p=−10800p4−276480p+69120p3+270p5+25920p2.
We know thatb7′p<0 in 0,2 follows that b7p is decreasing over 0,2. Therefore, b7p gets its maxima at p=0. Hence
(122)Tp,0,1≤276480.(iii)
On the edgep=0 and x=0,Tp,x,y reduces to(123)T0,0,y=276480y2=b8y.
Noting thatb8′y>0 in 0,1 shows that b8y is increasing over 0,1. Thus, b8y gets its maxima at y=1. Thus, we have
(124)T0,0,y≤276480.(iv)
On the edgesTp,1,1 and Tp,1,0
SinceTp,1,y is free of y, therefore
(125)Tp,1,1=Tp,1,0=−7104p4−611p6+81920+18432p2=b9p.
Then(126)b9′p=−28416p3−3666p5+36864p.
By puttingb9′p=0, we obtain the critical point p0≈1.0639 at which b9p attains its maximum value, which is given by
(127)Tp,1,1=Tp,1,0≤92795.48.(v)
On the edgep=0 and x=1,Tp,x,y becomes(128)T0,1,y=81920.(vi)
On the edgep=2,Tp,x,y reduces to(129)T2,x,y=2880.
T2,x,y is independent of x and y; therefore
(130)T2,x,0=T2,x,1=T2,0,y=T2,1,y=2880.(vii)
On the edgep=0 and y=1,Tp,x,y takes the form(131)T0,x,1=81920x3−18432x4+276480−258048x2=b10x.
It is clear that(132)b10′x=245760x2−73728x3−516096x.
We see thatb10′x<0 in 0,1 shows that b10x is decreasing over 0,1. Thus, b10x gets its maxima at x=0. Hence, we have
(133)T0,x,1≤276480.(viii)
On the edgep=0 and y=0,Tp,x,y yields(134)T0,x,0=294912x−212992x3=b11x.
It follows that(135)b11′x=294912−638976x2.
By takingb11′x=0, we obtain the critical point x0≈0.6793 at which b11x attains its maximum value, which is given by
(136)T0,x,0≤133568.833.
Hence, from the above cases we deduce that(137)Tp,x,y≤276480on0,2×0,1×0,1.
From (89)we have
(138)D2,2Gg2≤Tp,x,y17694720≤164.
Ifg∈BTsin, then sharp bound for this Hankel determinant is determined by
(139)D2,2Gg2=164≈0.0156.
Thus, we have completed the proof.


## 5. Conclusion

In our current investigation, we have considered a classBTsin of bounded turning functions associated with an eight-shaped domain. For such a class, we studied some interesting problems involving logarithmic coefficients. The Zalcman inequality, the Fekete-Szegö inequality, and the determinants D2,2Gg/2 and D2,1Gg/2 for the family BTsin have been studied here in this article. All the obtained results are proven to be the best possible.



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*Source: 2229960-2022-09-09.xml*