Home Mathematics Hankel determinant of order three for familiar subsets of analytic functions related with sine function
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Hankel determinant of order three for familiar subsets of analytic functions related with sine function

  • Muhammad Arif , Mohsan Raza , Huo Tang EMAIL logo , Shehzad Hussain and Hassan Khan
Published/Copyright: December 31, 2019

Abstract

In this paper we define and consider some familiar subsets of analytic functions associated with sine functions in the region of unit disk on the complex plane. For these classes our aim is to find the Hankel determinant of order three.

MSC 2010: 30C45; 30C50

1 Introduction and definitions

Let 𝒜 denote the class of functions f of the form

f(z)=z+n=2anznzD, (1.1)

which are analytic in the region 𝔻 = {z ∈ 𝔻 : |z| < 1} . Let 𝒮 denote a subclass of 𝒜 which contains univalent functions in 𝔻 . An analytic function f is subordinate to an analytic function g (written as fg) if there exists an analytic function w with w(0) = 0 and |w(z)| < 1 for z ∈ 𝔻 such that f(z) = g(w(z)) . In particular if g is univalent in 𝔻, then f(0) = g(0) and f(𝔻) ⊂ g(𝔻) . The familiar classes of starlike, convex and bounded turning functions are defined respectively in terms of subordination as

S=fS:zf(z)fzψ(z),zD,C=fS:1+zf(z)f(z)ψ(z),zD,R=fS:f(z)ψ(z),zD, (1.2)

where ψ(z)=1+z1z. By choosing suitable function ψ in (1.2), we obtain several subfamilies of 𝒜 which have interesting geometric interpretation, see [1, 2, 3, 4, 5,6, 7, 8, 9, 10, 11]. From these subfamilies, we recall here which are connected with trigonometric functions and are defined as follows:

Csin=fS:1+zfzfz1+sinz, (1.3)
Ssin=fS:zfzfz1+sinz, (1.4)
Rsin=fS:fz1+sinz. (1.5)

The class Ssin of analytic function defined in (1.4) was introduced by Cho et al. [6] and they also study the radii problems for this class of functions.

The familiar coefficient conjecture for the functions f ∈ 𝒮 having the series form (1.1), was given by Bieberbach [12] in 1916 and it was later proved by de-Branges [13] in 1985. It was one of the most celebrated conjecture in classical analysis, one that has stood as a challenge to mathematician for a very long time. During this period, many mathematicians worked hard to prove this conjecture and as result they established coefficient bounds for some subfamilies of the class 𝒮 of univalent functions. They also develop some new inequalities related with coefficient bounds of some subclasses of univalent functions. Fekete-Szegő inequality is one of the inequality for the coefficients of univalent analytic functions found by Fekete and Szegő (1933), related to the Bieberbach conjecture. An other coefficient problem which is closely related with Fekete and Szegő is Hankel determinant. Hankel determinants are very useful in the investigations of the singularities and power series with integral coefficients.

For given parameters q, n ∈ ℕ = {1, 2, …}, the Hankel determinant Hq, n(f) of a function f ∈ 𝓢 of the form (1.1) was introduced by Pommerenke [14, 15] as:

Hq,nf=anan+1an+q1an+1an+2an+qan+q1an+qan+2q2. (1.6)

The growth of Hq, n(f) has been investigated for different subfamilies of univalent functions. In particular, the absolute sharp bound of the functional H2,2(f) = a2a4 a32 for each of the sets 𝓒, 𝓢* and 𝓡 were found by Janteng et al. [16, 17] while the exact estimate of this determinant for the family of close-to-convex functions is still unknown (see [18]. On the other hand, the best estimate of |H2,2(f)| for the set of Bazilevič functions was proved by Krishna et al. [19]. For more work on H2,2(f), see [20, 21, 22 23, 24] and reference therein.

The H3,1(f) determinant for a function f of the from (1.1) is defined as follows:

H3,1f=a1a2a3a2a3a4a3a4a5.

The sharp estimation of |H3,1(f)| is tedious as compared to |H2,2(f)|. Babalola [25] obtained the upper bound of |H3,1(f)| for the subfamilies of 𝓢*, 𝓒 and 𝓡 in 2010. Later on, the upper bound of |H3,1(f)| for various subfamilies of analytic and univalent functions were studied, see [26, 27, 28, 29, 30, 31, 32, 33]. In 2017, Zaprawa [34] improved the results of Babalola and showed that

H3,1f1,forfS,49540,forfC,4160,forfR.

The results in [34] are not sharp for the families 𝓢*, 𝓒 and 𝓡. Moreover, Zaprawa found the sharp bounds of |H3,1(f)| for subfamilies of 𝓢*, 𝓒 and 𝓡 comprising of m -fold symmetric functions. Recently, Kowalczyk et al. [35] and Lecko et al. [36] obtained the sharp inequalities

H3,1f4/135,andH3,1f1/9,

for the sets 𝓒 and 𝓢*(1/2) respectively, where 𝓢*(1/2) is the family of starlike functions of order 1/2. More recently in 2019, Kwon et al.[37] got an improved bound |H3,1(f)| ≤ 8/9 for f ∈ 𝓢*. In this paper, our aim is to study |H3,1(f)| for the families of functions defined in the relations (1.3), (1.4), (1.5).

2 A set of lemmas

Let 𝓟 denote the family of all functions p which are analytic in 𝔻 with 𝔎𝔢(p(z)) > 0 and has the following series representation

pz=1+n=1cnznzD. (2.1)

Lemma 2.1

If p ∈ 𝓟 and has the form (2.1), then

cn2forn1, (2.2)
cn+kμcnck<2,for0μ1, (2.3)
cmcnckcl4form+n=k+l, (2.4)
cn+2kμcnck22(1+2μ);forμR, (2.5)
c2c1222c122, (2.6)

and for complex number λ, we have

c2λc12max2,2λ1. (2.7)

For the results in (2.2), (2.6), (2.3), (2.5), (2.4) see [38]. Also, see [39] for the inequality (2.7).

Lemma 2.2

Let p ∈ 𝓟 and has the form (2.1), then

Jc13Kc1c2+Lc32J+2K2J+2JK+L. (2.8)

Proof

Consider the left hand side of (2.8) and rearranging the terms, we have

Jc13Kc1c2+Lc3=Jc132c1c2+c3K2Jc1c2c3+JK+Lc3Jc132c1c2+c3+K2Jc1c2c3+JK+Lc32J+2K2J+2JK+L,

where we have used (2.2), (2.3) and the result c132c1c2+c32 due to [40]. □

3 Bound of |H3,1(f)| for the set 𝓒sin

Theorem 3.1

If f ∈ 𝓒sin and of the form (1.1)), then

a212, (3.1)
a316, (3.2)
a413144, (3.3)
a54097200. (3.4)

The first two bounds are best possible.

Proof

If f ∈ 𝓒sin, then the relation (1.1)) leads us to

1+zfzfz=1+sinwz, (3.5)

where w is a Schwarz function. Consider a function p such that

pz=1+wz1wz=1+c1z+c2z2+, (3.6)

then p ∈ 𝓟. This implies that

wz=pz1pz+1=c1z+c2z2+c3z3+2+c1z+c2z2+c3z3+.

From (1.1), we can write

1+zfzfz=1+2a2z+6a34a22z2+12a418a2a3+8a23z3+20a518a3232a2a4+48a22a316a24z4+ (3.7)

After some simple calculations, we obtain

1+sinwz=1wz+wz33!wz55!+wz77!=1+12c1z+c22c124z2+548c13+c32c1c22z3+c42+516c12c2c1432c1c32c224z4+. (3.8)

From (3.5), (3.7) and (3.8), it follows that

a2=c14, (3.9)
a3=c212, (3.10)
a4=112c32c1c28c1348, (3.11)
a5=1205288c14+c42c1c36c12c248c228. (3.12)

Applying relation (2.2) in (3.9) and (3.10), we obtain

a212 and a316.

By rearranging (3.11), it gives

a4=11214c3c1c22+14c3c1312.

Application of triangle inequality along with (2.3) and (2.5) leads us to

a411212+712=13144.

From (3.12), it follows that

a5=12014c4718c22+14c423c1c35144c12c2c122c236c2c122.

Using triangle inequality along with (2.2), (2.6) and (2.3), we get

a512012+12+5144c122c122+1182c122.

Suppose that |c1| = x and x ∈ [0, 2], therefore

a51201+5144x22x22+1182x22.

The above function attains its maximum value at x=65=1.0954, hence

a54097200.

Equality for the bounds given in (3.1) and (3.2), is obtained by taking

1+zfzfz=1+sinz. (3.13)

Theorem 3.2

Let f ∈ 𝓒sin. Then for a complex number γ

a3γa22max16,1123γ2. (3.14)

This result is sharp.

Proof

Using (3.9) and (3.10), one may write

a3γa22=c212γ16c12=112c21232γc12.

Application of relation (2.7), gives

a3γa22max16,1123γ2.

Thus we have the required result. □

Taking γ = 1 in last Theorem, we obtain

Corollary 3.3

If f ∈ 𝓒sin with the series form (1.1), then

a3a2216. (3.15)

This inequality is sharp.

Theorem 3.4

Let f ∈ 𝓒sin. Then

a2a3a413144. (3.16)

Proof

From (3.9), (3.10) and (3.11), we have

a2a3a4=c1c232+c13576c324.

Application of Lemma 2.2, Leads us to

a2a3a413144.

This completes the proof. □

Theorem 3.5

Let f ∈ 𝓒sin. Then

a2a4a327144. (3.17)

Proof

From (3.9), (3.10) and (3.11), we may write

a2a4a32=c1c396c142304c12c2384c22144=c1288c334c1c2+c1c3c22144c142304.

Application of triangle inequality as well as (2.2), (2.3) and (2.4)), we obtain

a2a4a324288+4144+162304=7144.

This completes the result. □

Theorem 3.6

Let f ∈ 𝓒sin. Then

H3,1f13333518400.

Proof

From (1.6), it is easy to see that

H3,1f=a3a2a4a32a4a4a2a3+a5a3a22,

where a1 = 1. This implies that

H3,1fa3a2a4a32+a4a4a2a3+a5a3a22.

By using (3.1), (3.2), (3.3), (3.4), (3.15), (3.16) and (3.17), we obtain the required result. □

4 Bound of |H3,1(f)| for the set Ssin

Theorem 4.1

Let f Ssin. Then

a21, (4.1)
a312, (4.2)
a41336, (4.3)
a54091440. (4.4)

The first two coefficients bounds are best possible.

Proof

Let f Ssin. Then we can write (1.4), in terms of Schwarz function as

zfzfz=1+sinwz,zD.

Since,

zfzfz=1+a2z+2a3a22z2+3a43a2a3+a23z3++4a52a324a2a4+4a22a3a24z4+. (4.5)

By comparing (4.5) and (3.8), we may write

a2=c12, (4.6)
a3=c24, (4.7)
a4=13c32c1c28c1348, (4.8)
a5=145288c14+c42c1c36c12c248c228. (4.9)

Using coefficient bounds for class 𝓟 given in (2.2) in (4.6) and (4.7), we get

a21 and a312.

From (4.7), we write

a4=11214c3c1c22+14c3c1312.

Application of triangle inequality along with (2.3) and (2.5) lead us to

a41312+712=1336.

Now

a5=1414c423c1c3+14c4718c225144c12c2c122c236c2c122.

By using (2.2), (2.6) and (2.3), we have

a51412+12+5144c122c123+1182c123.

Let |c1| = x ∈ [0, 2] . Then

a5141+5144x22x23+1182x23.

Suppose that

Φx=141+5144x22x23+1182x23.

The function Φ has maximum value at x=65, therefore

a54091440.

Equality is attained for the first two bounds for the function

fz=zexp0zsint1tdt=z+z2+z32+z49z572+. (4.10)

Theorem 4.2

Let f Ssin. Then for a complex number γ

a3γa22max12,12γ1. (4.11)

Equality is acheived for the function defined in (4.10).

Proof

Using (4.6) and (4.7), we may write

a3γa22=c24γ4c12=14c2γc12.

Application of (2.7), leads us to

a3γa22max12,γ1.

Hence the result is completed. □

If we put γ = 1 in the last result, we get the following result.

Corollary 4.3

If f Ssin, then

a3a2212. (4.12)

This result is sharp.

Theorem 4.4

If f Ssin, then

a2a3a4718. (4.13)

Proof

From (4.6), (4.7) and (4.8), we have

a2a3a4=c1c26+c13144c36.

Application of triangle inequality and Lemma 2.2, leads us to

a2a3a4718.

Thus the proof is completed. □

Theorem 4.5

If f Ssin, then

a2a4a321736. (4.14)

Proof

From (4.6), (4.7) and (4.8), we have

a2a4a32=c1c312c14288c12c248c2216=116c1c3c22+c148c3c1c2c14288.

Using triangle inequality as well as (2.2), (2.3) and (2.4)), we get

a2a4a3214+112+118=1736.

Theorem 4.6

If f Ssin, then

H3,1f1344125920.

Proof

From (1.6), we may write

H3,1fa3a2a4a32+a4a4a2a3+a5a3a22.

By using(4.1), (4.2), (4.3), (4.4), (4.12), (4.13) and (4.14), we obtain the required result. □

5 Bound of |H3,1(f)| for the set 𝓡sin

Theorem 5.1

If f ∈ 𝓡sin, then

a212, (5.1)
a313, (5.2)
a414, (5.3)
a51740. (5.4)

Proof

Let f ∈ 𝓡sin. Then we can write

fz=1+sinwz,zD.

Since,

fz=1+2a2z+3a3z2+4a4z3+5a5z4+. (5.5)

By comparing (5.5) and (3.8), we may get

a2=c14, (5.6)
a3=13c22c124, (5.7)
a4=14548c13+c32c1c22, (5.8)
a5=15c42+516c12c2c1432c1c32c224. (5.9)

By using (2.2) in (5.6), we obtain

a212.

Now using (2.6), in (5.7), we get

a313.

Application of triangle inequality and Lemma 2.2 in (5.8) lead us to

a414.

From (5.9), we may write

a5=15c12c28+12c4c1c3+c1216c2c122c24c2c122.

By using (2.2), (2.6) and (2.3), we obtain

a515c124+1+c12162c122+122c122.

Let |c1| = x ∈ [0, 2]. Then

a515x24+1+x2162x22+122x22.

Consider the function

Φ1x=15x24+1+x2162x22+122x22.

The above function has its maximum at x=2, therefore we obtain

a51740.

Theorem 5.2

If f ∈ 𝓡sin, then

a3γa22max23,1123γ4, (5.10)

where γ is any complex number.

Proof

From (5.6) and (5.7), we may get

a3γa22=c23c1212γ16c12=13c2124+3γ8c12.

Application of (2.7), leads us to

a3γa22max23,1123γ4.

Thus the proof is completed. □

Making γ = 1 in the last result, we get the following result.

Corollary 5.3

If f ∈ 𝓡sin, then

a3a2223. (5.11)

Theorem 5.4

If f ∈ 𝓡sin, then

a2a3a40.39434. (5.12)

Proof

From (5.6), (5.7) and (5.8), we have

a2a3a4=c1c26364c13c38=332c1c2c12218c3712c1c2.

Application of triangle inequality along with (2.2), (2.6) and (2.3) give us

a2a3a4332c12c122+14.

If we replace |c1| = x ∈ [0, 2], then

a2a3a4332x2x22+14.

Consider the function

Φ2x=332x2x22+14.

The function Φ2 has its maximum value at x = 1.15472. Therefore

a2a3a40.39434

and this completes the proof. □

Theorem 5.5

If f ∈ 𝓡sin, then

a2a4a320.45324. (5.13)

Proof

From (5.6), (5.7), and (5.8), we have

a2a4a32=7288c12c2+c1c332c142304c229=c132c3c1c2c142304c29c2c122.

Using triangle inequality along with (2.2), (2.6) and (2.3), we get

a2a4a32c116+c142304+292c122.

If we put |c1| = x ∈ [0, 2], then we have

a2a4a32x16+x42304+292x22.

Now consider the function

Φ3x=x16+x42304+292x22.

Then the above function takes its maximum at x = 0.2814. Thus

a2a4a320.45324.

This completes the proof. □

Theorem 5.6

If f ∈ 𝓡sin, then

H3,1f0.53299.

Proof

From (1.6), we may write

H3,1fa3a2a4a32+a4a4a2a3+a5a3a22.

By using (5.1), (5.2), (5.3), (5.4), (5.11), (5.12) and (5.13), we get the required result. □

6 Bounds of |H3,1(f) | for %2-fold symmetric and 3-fold symmetric functions

Let m ∈ ℕ = {1, 2, …}. A domain Λ is said to be m-fold symmetric if a rotation of Λ about the origin through an angle 2π/m carries Λ on itself. It is easy to see that, an analytic function f is m-fold symmetric in 𝔻, if

fe2πi/mz=e2πi/mfz,zD.

By 𝓢(m), we mean the set of m-fold symmetric univalent functions having the following Taylor series form

fz=z+k=1amk+1zmk+1,zD. (6.1)

The subfamilies Rsin(m),Ssin(m) and Csin(m) of 𝓢(m) are the sets of m-fold symmetric bounded turning, starlike and convex functions respectively associated with sine functions. More intuitively, an analytic function f of the form (6.1), belongs to the families Rsin(m),Ssin(m) and Csin(m) , if and only if

f(z)=1+sinpz1pz+1,pP(m), (6.2)
zf(z)f(z)=1+sinpz1pz+1,pP(m), (6.3)
1+zfzfz=1+sinpz1pz+1,pP(m), (6.4)

where the set 𝓟(m) is defined by

P(m)=pP:pz=1+k=1cmkzmk,zD. (6.5)

Now we can prove the following theorem.

Theorem 6.1

Let f Rsin(2) , be of the form (6.1). Then

H3,1f115.

Proof

Since f Rsin(2) , therefore there exists a function p ∈ 𝓟(2), such that

f(z)=1+sinpz1pz+1.

For f Rsin(2) , using the series form (6.1) and (6.5), when m = 2 in the above relation, we can write

a3=16c2,a5=120c22+110c4.

It is clear that for f Rsin(2) ,

H3,1f:=a3a5a33.

Therefore,

H3,1f=160c2(c429c22).

Using (2.3) and triangle inequality, we get

H3,1f115.

Hence the proof is completed.□

Theorem 6.2

If f Rsin(3) , then

H3,1f116.

This result is sharp for the function

fz=0zsin(t3)dt=z+14z4154z10+.

Proof

Since f Rsin(3) , therefore there exists a function p ∈ 𝓟(3) such that

f(z)=1+sinpz1pz+1.

For f Rsin(3) , using the series form (6.1) and (6.5), when m = 2 in the above relation, we can write

a4=18c3.

It is easy to see that

H3,1f:=a42.

Therefore,

H3,1f=164c32.

Using coefficient estimates for class 𝓟 and triangle inequality, we get

H3,1f116.

Hence the proof is completed.□

Theorem 6.3

Let f Ssin(2) , be of the form (6.1). Then

H3,1f18.

Proof

Since f Ssin(2) , therefore there exists a function p ∈ 𝓟(2) such that

zf(z)f(z)=1+sinpz1pz+1.

Using the series form (6.1) and (6.5), when m = 2 in the above relation, we can write

a3=14c2,a5=132c22+18c4.

Now

H3,1f:=a3a5a33.

Therefore,

H3,1f=132c2(c434c22).

Using (2.3) and triangle inequality, we get

H3,1f18.

Thus the proof is completed.□

Theorem 6.4

If f Ssin(3) , then

H3,1f19.

This result is sharp for the function

fz=zexp0z(sin(t3)1t)dt=z+13z4+118z7+.

Proof

Since f Ssin(3) , therefore there exists a function p ∈ 𝓟(3) such that

zf(z)f(z)=1+sinpz1pz+1.

Using the series form (6.1) and (6.5), when m = 2 in the above relation, we can write

a4=16c3.

Then

H3,1f:=a42.

Therefore,

H3,1f=136c32.

Using coefficient estimates for class 𝓟 and triangle inequality, we get

H3,1f19.

Hence the proof is completed.□

Theorem 6.5

Let f Csin(2) , be of the form (6.1). Then

H3,1f1120.

Proof

Since f Csin(2) , therefore there exists a function p ∈ 𝓟(2) such that

1+zfzfz=1+sinpz1pz+1.

Using the series form (6.1) and (6.5), when m = 2 in the above relation, we can write

a3=112c2,a5=1160c22+140c4.

Therefore,

H3,1f=1480c2(c41936c22).

Using (2.3) and triangle inequality, we get

H3,1f1120.

Hence the proof is completed.□

Theorem 6.6

Let f Csin(3) , be of the form (6.1). Then

H3,1f1144.

This result is sharp for the function

1+zfzfz=1+sin(z3)=1+z316z9+.

Proof

Since f Csin(3) , therefore there exists a function p ∈ 𝓟(3) such that

f(z)=1+sinpz1pz+1.

Using the series form (6.1) and (6.5), when m = 2 in the above relation, we can write

a4=124c3.

Therefore,

H3,1f=1576c32.

Using coefficient estimates for class 𝓟 and triangle inequality, we get

H3,1f1144.

This completes the proof.□

  1. Funding Source: This work was supported by the Natural Science Foundation of the People’s Republic of China under Grants 11561001 and 11271045, the Program for Young Talents of Science and Technology in Universities of Inner Mongolia Autonomous Region under Grant NJYT-18-A14 and the Natural Science Foundation of Inner Mongolia of the People’s Republic of China under Grant 2018MS01026.

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Received: 2019-04-16
Accepted: 2019-10-30
Published Online: 2019-12-31

© 2019 Muhammad Arif et al., published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 Public License.

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