# Bernardi Integral Operator and Its Application to the Fourth Hankel Determinant

Authors: Abid Khan; Mirajul Haq; Luminiţa-Ioana Cotîrlă; Georgia Irina Oros
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In recent years, the theory of operators got the attention of many authors due to its applications in different fields of sciences and engineering. In this paper, making use of the Bernardi integral operator, we define a new class of starlike functions associated with the sine functions. For our new function class, extended Bernardi’s theorem is studied, and the upper bounds for the fourth Hankel determinant are determined.


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## Body

## 1. Introduction

LetH be the family of holomorphic (or analytic) functions in D=z∈ℂ:z<1, and An⊂H such that f∈An has the series representation:
(1)fz=z+∑j=n+1∞ajzj.LetS be the subfamily of A1≡A containing univalent functions in D. Despite the fact that function theory was first proposed in 1851, it only became a viable research topic in 1916. Many academics attempted to prove or refute the conjecture an≤n, which was initially proven by de Branges in 1985, and as a result, they identified multiple subfamilies of the class S that are connected to various image domains. The starlike, convex, and close-to-convex functions are among those families which are defined by
(2)S∗=f∈S:Rzf′zfz>0,z∈D,C=f∈S:R1+zf′′zf′z>0,z∈D,K=f∈S:Rzf′zgz>0,gz∈S∗,z∈D.Letf and g be the two analytic functions in D; then, f is subordinate to g, denoted by
(3)fz≺gz,z∈D,if there exists a Schwarz function wz satisfying the conditions:
(4)w0=0,wz<1,z∈D,such that
(5)fz=gwz,z∈D.Letpz=1+c1z+c2z2+⋯ be an analytic and regular function in D with p0=1,Rpz>0, satisfying the criteria:
(6)pz≺1+Az1+Bz,−1≤B<A,−1<A<1.Then, this function is referred to as the Janowski function which is represented byPA,B. Geometrically, pz∈PA,B⟺p0=1, and pUd is inside the domain specified by
(7)ΩA,B=ω:ω−1−AB1−B2<A−B1−B2,having diameter end points:
(8)1−A1−B=p−1,1+A1+B=p1.LetS∗A,B be the class of functions ϰz, where ϰ0=0=ϰ′0−1 are holomorphic in Ud and meet the following requirements:
(9)ϰz∈S∗A,B⟺zϰ′zϰz∈PA,B.Distinct subclasses of analytic functions associated with various image domains have been introduced by many scholars. For example, Cho et al. [1] and Dziok et al. [2] discussed various properties of starlike functions related to Bell numbers and a shell-like curve connected with Fibonacci numbers, respectively. Similarly, Kumar and Ravichandran [3] and Mendiratta et al. [4] investigated subclasses of starlike functions associated with rational and exponential functions, respectively. Kanas and Raducanu [5] and Sharma et al. [6] explored some subclasses of analytic functions related to conic and cardioid domains, respectively. Raina and Sokól [7] investigated some important properties related to a certain class of starlike functions. Sokól and Stankiewicz [8] discussed radius problems of some subclasses of strongly starlike functions. Recently, Cho et al. [9] explored a family of starlike functions related to the sine function, which is defined as follows:
(10)Ss∗=f∈A~zf′zfz≺1+sinz.Theqth Hankel determinant for q≥1 and n≥1 of the functions f is introduced by Noonan and Thomas [10], which is given by
(11)Δq,nf=anan+1⋯an+q−1an+1an+2⋯an+q⋮⋮⋮⋮an+q−1an+q⋯an+2q−1a1=1.The following options are provided for some special choices ofn and q:
(1)
Forq=2,n=1,(12)Δ2,1f=a1a2a2a3=a3−a22,a1=1,is the famed Fekete-Szegő functional.
(2)
Forq=2,n=2,(13)Δ2,2f=a2a3a3a4=a2a4−a32is the second Hankel determinant.There are relatively few findings in the literature in connection with the Hankel determinant for functions belonging to the general familyS. Hayman [11] established the well-known sharp inequality:
(14)Δq,nf≤λn,where λ is the absolute constant. Similarly for the same class S, it was obtained in [12] that
(15)Δ2,2f≤λ,1≤λ≤113.For different subclasses of the setS of univalent functions, the growth of Δq,nf has been estimated many times. For example, Janteng [13] investigated the sharp bounds of Δ2,2f for the classes S∗, C, and R as given below:
(16)Δ2,2f≤1,f∈S∗,18,f∈C,49,f∈K.The sharp bound ofΔ2,2f for the class of close-to-convex functions is unknown. On the other hand, Krishna and Reddy [14] calculated a precision estimate of Δ2,2f for the Bazilevic function class.
(3)
Forq=3,n=1,(17)Δ3,1f=a1a2a3a2a3a4a3a4a5is the third Hankel determinant.The calculations in (17) represent that estimating Δ3,1f is significantly more difficult than estimating the bound of Δ2,2f. In the first paper of Babalola [15] on Δ3,1f, he obtained the upper bound of Δ3,1f for the classes S∗, C, and K . Later, some more contributions have been made by different authors to calculate the bounds of Δ3,1f for different subclasses of analytic and univalent functions. Zaprawa [16] enhanced the results of Babalola [15] and demonstrated that
(18)Δ3,1f≤1,f∈S∗,49540,f∈C,4160,f∈K.He also observed that the bounds are still not sharp.(4)
Forq=4,n=1,(19)Δ4,1f=a1a2a3a4a2a3a4a5a3a4a5a6a4a5a6a7is the fourth Hankel determinant.Sincef∈S and a1=1, thus Δ4,1f=a7Λ1−a6Λ2+a5Λ3−a4Λ4, where
(20)a3a2a4−a32−a4a4−a2a3+a5a3−a22=Λ1,a3a2a5−a3a4−a4a5−a2a4+a6a3−a22=Λ2,a3a3a5−a42−a5a5−a2a4+a6a4−a2a3=Λ3,a4a3a5−a42−a5a2a5−a3a4+a6a4−a2a3=Λ4.In the last few years, many articles have been published to investigate the upper bounds for the second-order Hankel determinantΔ2,2f, third-order Hankel determinant Δ3,1f, and fourth Hankel determinant Δ4,1f. For the functions with bounded turning, Arif et al. [17, 18] estimated the bound for the fourth- and fifth-order Hankel determinants. Khan et al. [19] also addressed this issue and derived upper bounds for the third- and fourth-order Hankel determinants for a class of functions with bounded turning that are related to sine functions. For more study about the Hankel determinant, we refer to [19–29].In geometric function theory (GFT), especially in the category of univalent functions, integral and differential operators are extremely helpful and important. Convolution of certain analytic functions has been used to introduce certain differential and integral operators. This approach is developed to facilitate further exploration of geometric features of analytic and univalent functions. Libera and Bernardi were the ones who investigated the classes of starlike, convex, and close-to-convex functions by introducing the idea of integral operators. Recently, some researchers have shown a keen interest in this field and developed various features of the integral and differential operators. Srivastava et al. [30] investigated a new family of complex-order analytic functions by using the fractional q-calculus operator. Mahmood et al. [31] looked at a group of analytic functions that were defined using q-integral operators. Using the q-analogue of the Ruscheweyh-type operator, Arif et al. [32] constructed a family of multivalent functions. Srivastava [33] presented a review on basic (or q-) calculus operators, fractional q-calculus operators, and their applications in GFT and complex analysis. This review article has been proven very helpful to investigate some new subclasses from different viewpoints and perspectives [34–40].Inspired from the above recent developments, in this study, we investigate the inclusion of the Bernardi integral operator in the class of starlike function associated with sine function inD. The Bernardi integral operator Jz:A⟶A was defined by Bernardi [41], which is given by the following relation:
(21)Jz=γ+1zγ∫0ztγ−1gtdt,γ>−1.In the first part of the study, we extend Bernardi’s theorem to a certain classSs∗ of univalent starlike functions in D. Particularly, we prove that if g∈Ss∗, then Jz∈Ss∗. In the second part of the study, we investigate the upper bounds for the fourth-order Hankel determinant Δ4,1f with respect to the function class Ss∗ associated with the sine function.


## 2. Main Results

In order to obtain our desired results, we first need the following lemmas.Lemma 1.
LetM and N be holomorphic functions in Ud such that N maps Ud onto many sheeted starlike regions with M0=N0=0 and M′0=N′0=1. If M′z/N′z∈Ss∗, then Mz/Nz∈Ss∗.Proof.
We know that(22)M′zN′z∈Ss∗⟺M′zN′z=1+sinwz≺1+sinz.
Also,σz=1+sinz maps z<r onto the disc σz−1<sin1. But M′z/N′z takes values in the same disc, and therefore,
(23)M′zN′z−1<sin1,z<r,0<r<1.
ChooseΛz so that
(24)N′zΛz=M′z−N′z.
Then,Λz<sin1. Fix z0 in Ud. Let L be the segment joining 0 and Nz0, which lies in one sheet of the starlike image of Ud by N. Let L−1 be the preimage of L under N. Then,
(25)Mz0−Nz0=∫0z0M′t−N′tdt=∫L−1N′tΛtdt<sin1∫LdNt=sin1Nz0.
That is,(26)MzNz−1<sin1.
This implies that(27)MzNz≺1+sinz,and hence,
(28)MzNz∈Ss∗.Lemma 2 (see [12]).
LetMz and Nz be regular in D and Nz map D onto many sheeted starlike regions:
(29)M0=N0=0,M′0N′0=1,RM′zN′z>0.
Then,(30)RMzNz>0.Lemma 3.
Letg∈Ss∗ such that
(31)Jz=∫0ztγ−1gtdt.
Then,J is 1+γ-valent starlike for γ=1,2,3,⋯, in D.Proof.
The proof is analogous to the one given in [41] and hence omitted.Lemma 4.
Letg∈Ss∗ and Jz=1+γ/zγJz for γ=1,2,3,⋯, where Jz is given by (31) in Lemma 3. Then, Jz∈Ss∗.Proof.
Let(32)J′z=1+γzgz−γzJz.
Then,(33)zJ′zJz=zJ′zJz.zγzγ=zγgz−γJzJz=MzNz,where
(34)Mz=zγgz−γJz,Nz=Jz.
By Lemma3, N≪J is 1+γ-valent starlike for γ=1,2,3,⋯ in D:
(35)M′0N′0=zg′0g0=1,and since g∈Ss∗,
(36)M′zN′z=zg′zgz∈Ss∗.
From Lemma1, we can get the conclusion:
(37)MzNz=zg′zgz∈Ss∗.Lemma 5 (see [42]).
Ifpz∈P, then there exists some x,z with x≤1,z≤1, such that
(38)2c2=c12+4−c12,4c3=c13+2c1x4−c12−c1x24−c12+24−c121−x2z.Lemma 6 (see [43]).
Letpz∈P; then,
(39)c14+c22+2c1c3−3c12c2−c4≤2,c15+3c1c22+3c12c3−4c13c2−2c1c4−2c2c3+c5≤2,c16+6c12c22+4c13c3+2c1c5+2c2c4+c32−c23−5c14c2−3c12c4−6c1c2c3−c6≤2,cn≤2,forn=1,2,3,⋯.Lemma 7 (see [44]).
Letpz∈P; then,
(40)c2−c122≤2−c122,cn+k−μcnck<2,0≤μ≤1,cn+2k−μcnck2≤21+2μ.Now, we are in position to present our main results.Theorem 8.
If the functionfz∈Ss∗ and is of the form (1), then
(41)a2≤1β2,(42)a3≤12β3,(43)a4≤0.344β4,(44)a5≤38β5,(45)a6≤67120β6,(46)a7≤558710800β7.Proof.
SinceJz∈Ss∗, according to the definition of subordination, there exists a Schwarz function wz with w0=0 and wz<1 such that
(47)zJ′zJz=1+sinwz.
Now,(48)zJ′zJz=1+β2a2z+2β3a3−β22a22z2+β23a23−3β2β3a2a3+3β4a4z3+4β5a5−β24a24+4β22β3a22a3−4β2β4a2a4−2β24a23z4+5β6a6−5β2β5a2a5+β25a25−5β3β4a3a4−5β23β3a23a3+5a22a4β22β4+5β2β32a2a32z5+6β7a7−6β2β6a2a6+6β22β5a22a5−6β3β5a3a5+12β2β3β4a2a3a4−β26a26−6β23β4a23a4−3β42a42+2β33a33−9β22β32a22a32+6β24β3a24a3z6+⋯,where βn=n+γ/1+γ. We define a function:
(49)pz=1+wz1−wz=1+c1z+c2z2+⋯.
It is easy to see thatpz∈P and
(50)wz=pz+1pz−1=c1z+c2z2+c3z3+⋯2+c1z+c2z2+c3z3+⋯.
On the other hand,(51)1+sinwz=1+c12z+c22−c124z2+5c1348−c1c22+c32z3+5c12c216−c1432−c1c32−c224+c42z4+c52−c1c42−c2c32+5c12c316+c1c2216−c13c28+c153840z5+c62−c2c42−c1c52+5c1c2c38+5c2348−c324−5c16512+c14c2768−3c12c2216+5c12c416−c13c38z6+⋯.
When the coefficients ofz,z2,z3 are compared between the equations (51) and (48), then we get
(52)a2=c12β2,a3=c24β3,a4=1β4c36−c1c224−c13144,(53)a5=1β5c48−c1c324+5c141152−c12c2192−c2232,(54)a6=1β6−3c1c480−7c3c2120−11c154800−43c1c22960+71c13c25760+c510,(55)a7=1β7c12c4480+c1c2c3480+833c16691200−41c12c223840−109c14c211520−c1c530−5c2c496+5c231152+c612+c13c3144.
Using Lemma6, we can simply obtain
(56)a2≤1β2a3≤12β3.(57)Jz=γ+1zγ∫0ztγ−1gtdt=∑n=1∞bnβnzn,
withb1=1 and
(58)βn=n+γ1+γ.
IfJz=∑n=1∞Anzn, then by comparing like powers of z,z2,⋯,zn, we have
(59)A1=1,A2=b2β2,A3=b3β3,⋯,An=bnβn.
For sharpness, if we take(60)gz=1+sinz=1+z−16z3+1120z5−⋯,and thus b2=1, b3=0, and b4=−1/6, then A2=1/β2. This shows that the obtained second coefficient bound is sharp.
Again, by Lemma6,
(61)a4=1β4c36−c13144−c1c224=1β416c3−c1c23+c172c2−c122.
Letc1=c, with c∈0,2; then, by Lemma 7, we can get
(62)a4=1β416c3−c1c23+c172c2−c122≤1β413+c722−c22.
Now, suppose that(63)Fc=1β413+c722−c22.
Then obviously,(64)F′c=1β4136−c248.
SettingF′c=0, we can get c=23/2, and hence, the maximum value of Fc is given by
(65)a4≤F232=1β413+3162=0.344β4.
Also,(66)a5=1β5c48−c1c324+5c141152−c12c2192−c2232=1β518c4−c1c33−c12576c2−c122−c232c2−c122−7c12c2576.
Letc1=c, with c∈0,2; then, again by Lemma 7,
(67)a5≤1β514+5c25762−c22+1162−c22+7c2288.
Assume that(68)Fc=1β514+5c25762−c22+1162−c22+7c2288.
Obviously, we meet the requirement:(69)F′c=1β5−7c144−5c3288≤0.
So the functionFc attains its maximum value at c=0, and it is given by
(70)a5≤F0=38β5.
Next,(71)a6=1β6−3c1c480−7c3c2120−11c154800−43c1c22960+71c13c25760+c510=1β6124c5−9c1c410+11c132400c2−c122−43c1c2960c2−c122+211c13c214400+7c5−c3c2120.
Takec1=c, with c∈0,2; then, according to Lemma 7, we have
(72)a6≤1β6760+112+11c324002−c22+432402−c22+211c37200.
Suppose that(73)Fc=1β6760+112+11c324002−c22+432402−c22+211c37200.
Then obviously,(74)F′c=1β6−c240+277c22400−55c44800,
We see thatF′′0<0, and we get the maximum value at c=0:
(75)a6≤F0=67120β6.
Finally,(76)a7=1β7c12c4480+c1c2c3480+833c16691200−41c12c223840−109c14c211520−c1c530−5c2c496+5c231152+c13c3144+c612=1β7−37c16691200−25c12c225760−c1c530+c12c4−c12480+c1c2c3−c1c2480+c13c3−c1c2144−29c14c2−c12/211520+5c22c2−c12/21152+c6−5/8c2c412.
Again, takingc1=c, with c∈0,2, and using the result of Lemma 7, we can obtain
(77)a7≤1β716+c2240+9c120+25c21440+29c42−c2/211520+c372+52−c2/2288+37c6691200.
Let(78)Fc=1β716+c2240+9c120+25c21440+29c42−c2/211520+c372+52−c2/2288+37c6691200.
Then obviously,F′c≥0. As a result, the function Fc attains its maximum value at c=2. Hence,
(79)a7≤F2≤558710800β7.Theorem 9.
If the functionfz∈Ss∗ and is of the form (1), then we have
(80)a3−a22≤12β3.Proof.
From (52), we can write
(81)a3−a22=c24β3−c124β22.
Using Lemma5, we get
(82)a3−a22=β22−2β3c128β3β22−x4−c128β3.
We suppose thatx=t∈0,1, and c1=c∈0,2. Also, if we apply the triangle inequality to the above equation, then we get
(83)a3−a22≤β22−2β3c28β3β22+t4−c28β3.
Assume that(84)Fc,t=a3−a22≤β22−2β3c28β3β22+t4−c28β3.
Obviously, we can write(85)∂F∂t=4−c28β3≥0,
Fc,t is increasing on 0,1. Therefore, at t=1, the function Fc,t will obtain its maximum value:
(86)maxFc,t=Fc,1=β22−2β3c28β3β22+4−c28β3.
Let us take(87)Gc=β22−2β3c28β3β22+4−c28β3,G′c=β22−2β3c4β3β22−c4β3=−c2β22,G′′c=−12β22<0.
It is clear thatGc is decreasing on 0,2. So at c=0, the function Gc will obtain its maximum value:
(88)a3−a22≤G0=12β3.
This complete the proof.Theorem 10.
If the functionfz∈Ss∗ and is of the form (1), then we have
(89)a2a3−a4≤13β4.Proof.
From (52), we can write
(90)a2a3−a4=c13144β4+c1c224β4+c1c28β2β3−c36β4.
From Lemma5, we can deduce that
(91)a2a3−a4=9β4−2β2β3c13144β2β3β4+c1x24−c1224β4−4−c121−x2z12β4.
We suppose thatx=t∈0,1, and c1=c∈0,2. Once again, if we apply the triangle inequality to the above equation, then we get
(92)a2a3−a4≤9β4−2β2β3c3144β2β3β4+ct24−c224β4+4−c21−t2z12β4.
Suppose that(93)Fc,t=9β4−2β2β3c3144β2β3β4+ct24−c224β4+4−c21−t2z12β4.
Then, we get(94)∂F∂t=c−2t4−c212β4<0.
The above expression shows thatFc,t is a decreasing function about t on the closed interval 0,1. This implies that Fc,t will attain its maximum value at t=0, which is
(95)maxFc,t=Fc,0=9β4−2β2β3c3144β2β3β4+4−c212β4.
Now define(96)Gc=9β4−2β2β3c3144β2β3β4+4−c212β4,G′c=9β4−2β2β3c248β2β3β4−c6β4,G′′c=−16β4.
SinceG′′c<0, the function Gc has maximum value at c=0. That is,
(97)a2a3−a4=Gc=G0≤13β4,and this completes the proof.Theorem 11.
If the functionfz∈Ss∗ and is of the form (1), then we have
(98)a2a4−a32≤13β2β4+14β32.Proof.
Again from (52), we can write
(99)a2a4−a32=−c12c248β2β4−c14288β2β4+c1c312β2β4−c2216β32.
Using the result of Lemma5, we can obtain
(100)a2a4−a32=c1c3−c13/2412β2β4−c12c248β2β4−c2216β32.
Also, by Lemma7, we have
(101)a2a4−a32≤c6β2β4+c224β2β4+14β32,=16β2β4c+c24+14β32=16β2β4Hc+14β32,where
(102)Hc=c+c24.
ClearlyHc is an increasing function about c on the closed interval 0,2. This means that Hc will attain its maximum value at c=2, which is Hc≤3. Thus,
(103)a2a4−a32≤12β2β4+14β32.Theorem 12.
If the functionfz∈Ss∗ and is of the form (1), then we have
(104)a2a5−a3a4≤736β3β4+0.81562β2β5.Proof.
From (52) and (53), we have
(105)a2a5−a3a4=c24β3β4c13144+c1c224−c36−c12β2β5−5c141152+c12c2192+c1c324+c2232−c48=c24β3β4c13144−16c3−c1c24−c12β2β5c14+c22+2c1c3−3c12c2−c432+19c12192c2−c122−c1c3−2/3c1348−3c432.
Using the result of Lemma7, we can write
(106)a2a5−a3a4≤12β3β4c3144+26+12β2β5c4+19c396−19c5384+c224=12β3β4Hc+12β2β5Mc,where
(107)Hc=13+c3144,Mc=c4+c224+19c396−19c5384.
Obviously,(108)M′c=14+c12+19c232+95c4384.
ForM′c=0, we can get c=1.71468508801 and consequently M′′1.71468508801=−2.8693. As M′′0<0, the maximum value at c=0 is
(109)M0≤0.8156.
Also,(110)H′c=c248,
whereHc attains its maximum value at c=2, so
(111)Hc=H2≤718.
Using the results of (109) and (111) in (106), we can get
(112)a2a5−a3a4≤736β3β4+0.81562β2β5.Theorem 13.
If the functionfz∈Ss∗ and is of the form (1), then we have
(113)a5−a2a4≤718β2β4+0.50β5.Proof.
From (52) and (53), we have
(114)a5−a2a4=12β2β4c14144+c12c224−c1c36−1β5−5c141152+c12c2192+c1c324+c2232−c48=12β2β4c14144−c16c3−c1c24−1β5c14+c22+2c1c3−3c12c2−c432+19c12192c2−c122−c1c3−2/3c1348−3c432.
Lettingx=t∈0,1 and c1=c∈0,2 and using the results of Lemmas 6 and 7, we get
(115)a5−a2a4≤12β2β4c4144+2c6+1β514+c24+19c296−19c4384.
Suppose that(116)Fc=12β2β4Hc+1β5Mc,where
(117)Hc=c4144+c3,Mc=14+c24+19c296−19c4384.
We see thatH′c≥0 and the maximum value of Hc can be attained at c=2, which is H2≤7/9. Also,
(118)M′c=124+19c48−19c396.
If we setM′c=0, then we get c=1.46416723. Consequently, M′′1.46416723=−0.86. As M′′0<0, the maximum value at c=0 is given by M0≤0.50. Hence,
(119)a5−a2a4≤718β2β4+0.50β5.Theorem 14.
If the functionfz∈Ss∗ and is of the form (1), then we have
(120)a5a3−a42≤85324β42+0.5079002β2β5.Proof.
From (52) and (53), we have
(121)a5a3−a42=−1β42c1620736+c14c21728−c13c3432+c12c22576−c1c2c372+c3236−c24β3β5−5c141152+c12c2192+c1c324+c2232−c48=−1β42c1620736+c3236−c13432c3−c1c24−c1c272c3−c1c28−c24β3β5c14+c22+2c1c3−3c12c2−c432+19c12192c2−c122−3c432−c1c3−2/3c1348.
Now, using the results of Lemmas6 and 7, we obtain
(122)a5a3−a42≤1β4219+c18+c3216+c620736+12β3β514+c24+19c296−19c4384,a5a3−a42≤1β42Hc+12β2β5Mc,where
(123)Hc=19+c18+c3216+c620736,Mc=14+c24+19c296−19c4384.
It is clear thatHc is an increasing function, so it attains its maximum value at c=2, which is
(124)H2≤85324β42.
Also, for allc∈0,2, we have
(125)M′c=124+19c48−19c396,M′′c=1948−19c232.
When we setM′c=0, then we get c=1.464167. Obviously,
(126)M′′c=M′′1.464167=−0.87703<0,and it attains its maximum value at c=1.464167, which is given by
(127)M1.464167≤0.507900686.
Hence,(128)a5a3−a42≤85324β42+0.5079002β2β5,which completes the proof of Theorem 14.Theorem 15.
If the functionfz∈Ss∗ and is of the form (1), then we have
(129)Δ4,1f≤2.9161×10−2β44+7.0313×10−2β53+448928800β3β62+0.006204β42β6+558786400β33β7+5.7491×10−2β42β7+4.7616×10−2β2β3β52+0.10567β2β4β52+2.8228×10−2β2β42β5+7.2391×10−2β2β42β6+0.0735β3β42β5+4698640β32β4β6+558757600β3β5β7+0.16287β4β5β6+558764800β2β3β4β7+0.11384β2β3β5β6.Proof.
We know that(130)Δ4,1f=a7a3a2a4−a32−a4a4−a2a3+a5a3−a22−a6a3a2a5−a3a4−a4a5−a2a4+a6a3−a22+a5a3a3a5−a42−a5a5−a2a4+a6a4−a2a3−a4a4a3a5−a42−a5a2a5−a3a4+a6a4−a2a3,which further implies that
(131)Δ4,1f=a7a3a2a4−a32−a4a4−a2a3+a5a3−a22−a6a3a2a5−a3a4−a4a5−a2a4+a6a3−a22+a5a3a3a5−a42−a5a5−a2a4+a6a4−a2a3−a4a4a3a5−a42−a5a2a5−a3a4+a6a4−a2a3.
Using the triangle inequality, we can write(132)Δ4,1f≤a7a3a2a4−a32+a4a7a4−a2a3+a5a7a3−a22+a6a3a2a5−a3a4+a4a6a5−a2a4+a62a3−a22+a5a3a3a5−a42+a52a5−a2a4+a5a6a4−a2a3+a42a3a5−a42+a4a5a2a5−a3a4+a4a6a4−a2a3.
By substituting the results of (41), (80), (89), (98), (104), (113), (120), and (128) in (132), we can get the desired result in (129).


## 3. Conclusion

In the present investigation, first, we have extended the well-known Bernardi theorem to a specific classSs∗ of univalent starlike functions in the open unit disk D. We have proven that if g is a starlike univalent function in the unit disk D and g∈Ss∗, then Jz∈Ss∗, where
(133)Jz=γ+1zγ∫0ztγ−1gtdt,γ>−1.Additionally, we have estimated the upper bounds of the fourth-order Hankel determinant for the functions classSs∗ associated with the sine function systematically.



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*Source: 4227493-2022-09-09.xml*