# A Study of Fourth-Order Hankel Determinants for Starlike Functions Connected with the Sine Function

Authors: Hai-Yan Zhang; Huo Tang
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, upper bounds for the fourth-order Hankel determinantH41 for the function class Ss∗ associated with the sine function are given.


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## Body

## 1. Introduction

LetA denote the class of functions f which are analytic in the open unit disk D=z:∣z∣<1 of the form
(1)fz=z+a2z2+a3z3+⋯z∈D,and let S denote the subclass of A consisting of univalent functions.Suppose thatP is the class of analytic functions p normalized by
(2)pz=1+c1z+c2z2+c3z3+⋯and satisfying the condition
(3)Rpz>0z∈D.Assume thatf and g are two analytic functions in D. Then, we say that the function g is subordinate to the function f, and we write
(4)gz≺fzz∈D,if there exists a Schwarz function ωz with ω0=0 and ∣ωz∣<1, such that (see [1])
(5)gz=fωzz∈D.In 2018, Cho et al. [2] introduced the following function class Ss∗:
(6)Ss∗≔f∈A:zf′zfz≺1+sinzz∈D,which implies that the quantity zf′z/fz lies in an eight-shaped region in the right-half plane.In 1976, Noonan and Thomas [3] stated the qth Hankel determinant for q≥1 and n≥1 of functions f as follows:
(7)Hqn=anan+1⋯an+q−1an+1an⋯an+q−2⋮⋮⋮an+q−1an+q−2⋯ana1=1.In particular, we have(8)H21=a1a2a2a3=a3−a22a1=1,n=1,q=2,H22=a2a3a3a4=a2a4−a32n=2,q=2,H31=a1a2a3a2a3a4a3a4a5n=1,q=3,H41=a1a2a3a4a2a3a4a4a3a4a5a6a4a5a6a7n=1,q=4.Sincef∈S,a1=1, thus
(9)H41=a7a3a2a4−a32−a4a4−a2a3+a5a3−a22−a6a3a2a5−a3a4−a4a5−a2a4+a6a3−a22+a5a3a3a5−a42−a5a5−a2a4+a6a4−a2a3+a5a3a3a5−a42−a5a5−a2a4+a6a4−a2a3−a4a4a3a5−a42−a5a2a5−a3a4+a6a4−a2a3.We note thatH21 is the well-known Fekete-Szegö functional (see [4–6]).In recent years, many papers have been devoted to finding upper bounds for the second-order Hankel determinantH22 and the third-order Hankel determinant H31, whose elements are various classes of analytic functions; it is worth mentioning that [7–20]. For instance, Murugusundaramoorthy and Bulboacă [21] defined a new subclass of analytic functions MLcaλ,ϕ and got upper bounds for the Fekete-Szegö functional and the Hankel determinant of order two for f∈MLcaλ,ϕ. Islam et al. [22] examined the q-analog of starlike functions connected with a trigonometric sine function and discussed some interesting geometric properties, such as the well-known problems of Fekete-Szegö, the necessary and sufficient condition, the growth and distortion bound, closure theorem, and convolution results with partial sums for this class. Zaprawa et al. [23] obtained the bound of the third Hankel determinant for the univalent starlike functions. Very recently, Arif et al. [24] studied the problem of fourth Hankel determinant H41 for the first time for the class of bounded turning functions and successfully obtained the bound of H41. Recently, Khan et al. [25] discussed some classes of functions with bounded turning which are connected to the sine functions and obtained upper bounds for the third- and fourth-order Hankel determinants related to such classes. Inspired by the aforementioned works, in this paper, we mainly investigate upper bounds for the fourth-order Hankel determinant H41 for the function class Ss∗ associated with the sine function.


## 2. Main Results

By proving our desired results, we need the following lemmas.Lemma 1 (see [26]).
Ifpz∈P, then exists somex,zwith∣x∣≤1,∣z∣≤1, such that(10)2c2=c12+x4−c12,4c3=c13+2c1x4−c12−4−c12c1x2+24−c121−x2z.Lemma 2 (see [27]).
Letpz∈P, then(11)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,n=1,2,⋯.Lemma 3 (see [28]).
Letpz∈P, then we have(12)c2−c122≤2−c122,cn+k−μcnck<2,0≤μ≤1,∣cn+2k−μcnck2∣≤21+2μ.We now state and prove the main results of our present investigation.Theorem 4.
If the functionfz∈Ss∗and of the form ((1)), then(13)a2≤1,a3≤12,a4≤0.344,a5≤38,a6≤67120,∣a7∣≤558710800.Proof.
Sincefz∈Ss∗, according to subordination relationship, thus there exists a Schwarz function ωz with ω0=0 and ∣ωz∣<1, satisfying
(14)zf′zfz=1+sinωz.
Here,(15)zf′zfz=z+∑n=2∞nanznz+∑n=2∞anzn=1+∑n=2∞nanzn−11−a2z+a22−a3z2−a23−2a2a3+a4z3+a24−3a22a3+2a2a4−a5z4+⋯=1+a2z+2a3−a22z2+a23−3a2a3+3a4z3+4a5−a24+4a22a3−4a2a4−2a32z4+5a6−5a2a5+a25−5a3a4−5a23a3+5a22a4+5a2a32z5+6a7−6a2a6+6a22a5−6a3a5+12a2a3a4−a26−6a23a4−3a42+2a33−9a22a32+6a24a3z6+.
Now, we define a function(16)pz=1+ωz1−ωz=1+c1z+c2z2+⋯.
It is easy to see thatpz∈P and
(17)ωz=pz−11+pz=c1z+c2z2+c3z3+⋯2+c1z+c2z2+c3z3+⋯.
On the other hand,(18)1+sinωz=1+12c1z+c22−c124z2+5c1348+c3−c1c22z3+c4−c1c32+5c12c216−c224−c1432z4+c5−c1c4−c2c32+5c12c3+c1c2216−c13c28+c153840z5+c6−c1c5−c2c42+5c1c2c38+5c2348−c324+5c16512+c14c2768−3c12c2216+5c12c416−c13c38z6+⋯.
Comparing the coefficients ofz,z2,z3,z4,z5,z6 between equations (15) and (18), we obtain
(19)a2=c12,a3=c24,a4=c36−c1c224−c13144,a5=c48−c1c324+5c141152−c12c2192−c2232,(20)a6=−3c1c480−7c2c3120−11c154800−43c1c22960+71c13c25760+c510,(21)a7=c12c4480+c1c2c3480+833c16691200−41c12c223840−109c14c211520−c1c530−5c2c496+5c231152+c612+c13c3144.
Applying Lemma2, we easily get
(22)∣a2∣≤1,∣a3∣≤12,∣a4∣=c36−c1c224−c13144=16c3−c1c23+c172c2−c122.
Letc1=c,c∈0,2; by using Lemma 3, we show
(23)a4=16c3−c1c23+c172c2−c122≤13+c2−c2/272;also, let
(24)Fc=13+c2−c2/272;obviously, we find
(25)F′c=136−c248.
SettingF′c=0, we have c=23/3, and so, Fc has a maximum value attained at c=23/3, also which is
(26)a4≤F233=13+3162≈0.344,∣a5∣=c48−c1c324+5c141152−c12c2192−c2232=18c4−c1c33−c12576c2−c122−c232c2−c122−7c12c2576.
Letc1=c,c∈0,2, according to Lemma 3, we obtain
(27)∣a5∣≤14+5c22−c2/2576+1162−c22+7c2288.
Putting(28)Fc=14+5c22−c2/2576+1162−c22+7c2288,we get
(29)F′c=−7c144−5c3288≤0.
Therefore, the functionFc has a maximum value attained at c=0, also which is
(30)a5≤F0=38,∣a6∣=−3c1c480−7c2c3120−11c154800−43c1c22960+71c13c25760+c510=124c5−9c1c410+7120c5−c2c3+11c132400c2−c122−43c1c2960c2−c122−211c13c214400.
Letc1=c,c∈0,2, in view of Lemma 3, we have that
(31)∣a6∣≤760+112+11c32−c2/22400+432402−c22+211c37200.
Taking(32)Fc=760+112+11c32−c2/22400+432402−c22+211c37200,we obtain
(33)F′c=277c22400−55c44800−c240.
Thus,c=0 is the root of the function F′c=0 and F′′0<0; we are easy to see that the function Fc has a maximum value attained at c=0, also which is
(34)a6≤F0=67120,∣a7∣=c12c4480+c1c2c3480+833c16691200−41c12c223840−109c14c211520−c1c530−5c2c496+5c231152+c612+c13c3144=−37c16691200−25c12c225760−c1c530+c12c4−c22480+c1c2c3−c1c2480+c13c3−c1c2144−29c14c2−c12/211520+5c22c2−c12/21152+c6−5/8c2c412.
Letc1=c,c∈0,2, by virtue of Lemma 3, we have that
(35)∣a7∣≤16+c2240+9c120+29c42−c2/211520+37c6691200+c372+25c21440+52−c2/2288.
Letting(36)Fc=16+c2240+9c120+29c42−c2/211520+37c6691200+c372+25c21440+52−c2/2288,so we get
(37)F′c≥0.
Thus, the functionFc has a maximum value attained at c=2, also which is
(38)∣a7∣≤F2=558710800.
Hence, the proof is complete.Theorem 5.
If the functionfz∈Ss∗and of the form ((1)), then we have(39)∣a3−a22∣≤12.Proof.
Applying equation (21), we have
(40)∣a3−a22∣=c24−c124.
Then, by applying Lemma1, we get
(41)∣a3−a22∣=x4−c128−c128.
Suppose thatx=t,t∈0,1,c1=c,c∈0,2. Then, using the triangle inequality, we obtain
(42)∣a3−a22∣≤t4−c28+c28.
Suppose(43)Fc,t=t4−c28+c28,then for any t∈0,1 and c∈0,2, we get
(44)∂F∂t=4−c28>0,which means that Fc,t is an increasing function on the closed interval [0,1] about t. Therefore, the function Fc,t can get the maximum value at t=1, that is,
(45)maxFc,t=Fc,1=4−c28+c28=12.
So, obviously,(46)∣a3−a22∣≤12.
Hence, the proof is complete.Theorem 6.
If the functionfz∈Ss∗and of the form ((1)), then we have(47)∣a2a3−a4∣≤13.Proof.
From (21), we have
(48)∣a2a3−a4∣=∣c1c28+c13144−c36+c1c224∣=c1c26−c36+c13144.
Now, in view of Lemma1, we get
(49)∣a2a3−a4∣=7c13144+4−c12c1x224−4−c121−x2z12.
Letx=t,t∈0,1,c1=c,c∈0,2. Then, using the triangle inequality, we deduce that
(50)∣a2a3−a4∣≤7c3144+4−c2ct224+4−c21−t212.
Assume that(51)Fc,t=7c3144+4−c2ct224+4−c21−t212.
Therefore, for anyt∈0,1 and c∈0,2, we have
(52)∂F∂t=4−c2tc−212<0,that is, Fc,t is an decreasing function on the closed interval [0,1] about t. This implies that the maximum value of Fc,t occurs at t=0, which is
(53)maxFc,t=Fc,0=7c3144+4−c212.
Define(54)Gc=4−c212+7c3144;we clearly see that the function Gc has a maximum value attained at c=0, also which is
(55)∣a2a3−a4∣≤G0=13.
Hence, the proof is complete.Theorem 7.
If the functionfz∈Ss∗and of the form ((1)), then we have(56)∣a2a4−a32∣≤14.Proof.
Letfz∈Ss∗, then by equation (21), we get
(57)∣a2a4−a32∣=c1c312−c12c248−c14288−c2216.
Now, in terms of Lemma1, we obtain
(58)∣a2a4−a32∣=c1c312−c12c248−c14288−c2216=−5c14576−x2c124−c1248−x24−c12264+c14−c121−x2z24.
Letx=t,t∈0,1,c1=c,c∈0,2. Then, using the triangle inequality, we get
(59)∣a2a4−a32∣≤t2c24−c248+1−t2c4−c224+t24−c2264+5c4576.
Setting(60)Fc,t=t2c24−c248+1−t2c4−c224+t24−c2264+5c4576,then, for any t∈0,1 and c∈0,2, we have
(61)∂F∂t=tc2−8c+124−c296>0,which implies that Fc,t increases on the closed interval [0,1] about t. That is, that Fc,t has a maximum value at t=1, which is
(62)maxFc,t=Fc,1=c24−c248+4−c2264+5c4576.
Putting(63)Gc=c24−c248+4−c2264+5c4576,then we have
(64)G′c=c4−c224−c324−c4−c216+5c3144.
IfG′c=0, then the root is c=0. Also, since G′′0=−1/12<0, so the function Gc can take the maximum value at c=0, which is
(65)∣a2a4−a32∣≤G0=14.
Hence, the proof is complete.Theorem 8.
If the functionfz∈Ss∗and of the form ((1)), then we have(66)∣a2a5−a3a4∣≤1136.Proof.
Letfz∈Ss∗, then by using (21), we have
(67)∣a2a5−a3a4∣=5c152304+c1c416−c1c22192−c12c348−c13c21152−c2c324=−c13c2−c12/21152−c3c2−c12/224+c1c4−c1c324+c15576+c1c4−1/4c2248.
Letc1=c,c∈0,2, according to Lemma 3, we obtain
(68)a5≤14+5c22−c2/2576+1162−c22+7c2288.
Taking(69)Fc=c32−c2/21152+2−c2/212+c8+c5576.
Then,∀c∈0,2, we have
(70)F′c=c2192+c4128−c12+18>0,which implies that Fc increases on the closed interval [0,2] about c. Namely, the maximum value of Fc attains at c=2, also which is
(71)∣a2a5−a3a4∣≤F2=1136.
The proof of Theorem8 is completed.Theorem 9.
If the functionfz∈Ss∗and of the form ((1)), then we have(72)∣a5−a2a4∣≤1332.Proof.
Assume thatfz∈Ss∗, then from (21), we obtain
(73)∣a5−a2a4∣=c14128−c1c38+c12c264−c2232+c48=c14+c22+2c1c3−3c12c2−c432−5c12c2−c12/264−3c4−2/3c1c332.
Next, by virtue of Lemma3, we obtain
(74)∣a5−a2a4∣≤14+5c22−c2/264.
Setting(75)Fc=14+5c22−c2/264.
Then, we have(76)F′c=5c16−5c332.
LetF′c=0, we get c=0 or c=2 and F′2<0, which implies that the maximum value of Fc attains at c=2, also which is
(77)∣a5−a2a4∣≤F2=1332.
Hence, the proof is complete.Theorem 10.
If the functionfz∈Ss∗and of the form ((1)), then we have(78)∣a5a3−a42∣≤97324.Proof.
Assume thatfz∈Ss∗, then from (21), we obtain
(79)∣a5a3−a42∣=7c14c213824+c2c432+c1c2c3288−c23128+c13c3432−7c12c222304−c3236−c1620736=c2c4−c1c3/932−c3c3−c1c2/436−c22c2−c12/2128−c12c2c2−c12/2144+c13c3−31/32c1c2432−5c14c26912−c1620736.
Next, in terms of Lemma3, we obtain
(80)∣a5a3−a42∣≤18+19+2−c2/232+c22−c2/272+c3216+5c43456+c620736.
Putting(81)Fc=18+19+2−c2/232+c22−c2/272+c3216+5c43456+c620736.
Then, for anyc∈0,2, we have F′c>0, which means that the maximum value of Fc arrives at t=2, also which is
(82)∣a5a3−a42∣≤F2=97324.
Hence, the proof is complete.Theorem 11.
If the functionfz∈Ss∗and of the form ((1)), then we have(83)∣H41∣≤0.81945.Proof.
Because of(84)H41=a7a3a2a4−a32−a4a4−a2a3+a5a3−a22−a6a3a2a5−a3a4−a4a5−a2a4+a6a3−a22−a6a3a2a5−a3a4−a4a5−a2a4+a6a3−a22+a5a3a3a5−a42−a5a5−a2a4+a6a4−a2a3−a4a4a3a5−a42−a5a2a5−a3a4+a6a4−a2a3,then, by applying the triangle inequality, we get
(85)∣H41∣=a7a3∣a2a4−a32+a7a4a4−a2a3+a7a5a3−a22+a6a3a2a5−a3a4+a6a4a5−a2a4+a62a3−a22+a5a3a3a5−a42+a52a5−a2a4+a5a6a4−a2a3+a42a3a5−a42+a4a5a2a5−a3a4+a4a6a4−a2a3.
Next, substituting (13) and (39)–(78) into (85), we easily obtain the desired assertion (83).


## 3. Conclusion

In the present paper, we mainly get upper bounds of the fourth-order Hankel determinantH41 of starlike functions connected with the sine function. However, the results obtained in this paper are not sharp. In the future, we will consider the sharpness of the results. Also, we can discuss the related research of the fifth-order Hankel determinant and fifth-order Toeplitz determinant for this function class.



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