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Third Hankel determinant for a subclass of analytic univalent functions

T.V. Sudharsan

a,∗

, S.P. Vijayalakshmi

b

and B. Adolf Stephen

c

aDepartment of Mathematics, SIVET College, Chennai - 600 073, Tamil Nadu, India.

bDepartment of Mathematics, Ethiraj College, Chennai - 600 008, Tamil Nadu, India.

cDepartment of Mathematics, Madras Christian College, Chennai - 600 059, Tamil Nadu, India.

Abstract

This paper focuses on attaining the upper bounds onH3(1)for a classCαβ(0 ≤β < 1,α≥ 0) in the unit disk∆={z∈C:|z|<1}.

Keywords: Bounded turning, coefficient bounds, convex functions, Hankel determinant, starlike functions.

2010 MSC:30C45, 30C50. c2012 MJM. All rights reserved.

1

Introduction

LetAbe the class of functions

f(z) =z+a2z2+. . . (1.1)

which are analytic in∆={z∈C:|z|<1}.

A function f ∈Ais said to be of bounded turning, starlike and convex respectively if and only if forz∈∆,

Re f0(z)>0,Re z ff(z)0(z) >0 andRe1+z ff000(z)(z)

>0. By usual notations, we denote these classes of functions

respectively byR,S∗andC. Letn≥0 andq≥1. TheqthHankel determinant is defined as:

Hq(n) =

an an+1 . . . an+q−1

an+1 . . . .

..

. ...

an+q−1 . . . an+2(q−1)

.

This determinant has been considered by several authors. For example, Noor in [11] determined the rate of growth ofHq(n)asn→∞for functionsf given by (1.1) with bounded boundary. In particular, sharp upper bounds onH2(2)were obtained by authors of articles [5, 6, 7, 13, 14] for different classes of functions.

The classCαβis defined as follows.

Definition 1.1. Let f be given by (1.1). Then f ∈Cαβif and only if

Re

(z f0(z) +

αz2f00(z))0

f0(z)

>β, z∈∆, 0≤β<1, 0≤α≤1.

The choice ofα=0,β=0yields Re

n

1+z ff000(z)(z)

o

>0, z∈∆, the class of convex functions C [12].

The choice ofα=0, yields Re

n 1+z f

00(z)

f0(z)

o

>β, z∈∆, the class of convex functions of orderβdenoted by C(β)[12].

Corresponding author.

E-mail address:tvsudharsan@rediffmail.com(T.V. Sudharsan),vijishreekanth@gmail.com(S.P. Vijayalakshmi),adolfmcc2003@yahoo.co.in

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In the present investigation, our focus is on the Hankel determinant, H3(1) for the classCαβ in ∆. By definition,H3(1)is given by

H3(1) =

a1 a2 a3

a2 a3 a4

a3 a4 a5

for f ∈A,a1=1, so that

H3(1) =a3(a2a4−a23)−a4(a4−a2a3) +a5(a3−a22)

and by triangle inequality, we have

|H3(1)| ≤ |a3||a2a4−a23|+|a4||a2a3−a4|+|a5||a3−a22|. (1.2)

In this paper, we find the sharp upper bound for the functional|a2a3−a4|,|a2a4−a23|and|a3−a22|

respec-tively for the functions belonging to the classCαβ. Our proofs are based on the techniques employed by [8, 9] which has been widely used by many authors (see for example [5, 6, 7, 14]).

2

Preliminary Results

LetPdenote the class of functions

p(z) =1+c1z+c2z2+· · · (2.3)

which are regular in∆and satisfyRe p(z)>0,z∈∆. Throughout this paper, we assume thatp(z)is given by (2.3) and f(z)is given by (1.1). To prove the main results we shall require the following lemmas.

Lemma 2.1. [3] Let p∈P. Then|ck| ≤2, k=1, 2, . . . and the inequality is sharp.

Lemma 2.2. [8, 9] Let p∈P. Then

2c2=c21+x(4−c21) (2.4)

and

4c3=c31+2xc1(4−c21)−x2c1(4−c21) +2y(1− |x|2)(4−c21) (2.5)

for some x,y such that|x| ≤1and|y| ≤1.

Lemma 2.3. [2] Let p∈P. Then

c2−σc

2 1

2

=

   

  

2(1−σ) ifσ≤0,

2 if0≤σ≤2,

2(σ−1) ifσ≥2.

3

Main Results

Lemma 3.1. Let f ∈Cαβ. Then, we have the best possible bound for

|a2a3−a4| ≤   

 

4

9√3 α=0,β=0

(1−β)

MA2 q

A1

A2[B1+ (4A2−A1)(B2+B3)] 0<α≤1, 0<β<1, where,

A1=4(4+23α+48α2+36α3−β−2αβ),

A2=3(4+20α+64α2+48α3+2β+20αβ−2β2−12αβ2),

B1=−3α+3β+22αβ−2β2−12αβ2+16α2+12α3,

B2=3+16α+32α2+24α3,

B3=1+7α+16α2+12α3,

(3)

Proof. For f ∈Cαβ, there exists ap∈Psuch that

f0(z) +z f00(z) +αz2f000(z) +2αz f00(z) = [(1−β)p(z) +β]f0(z). Equating the coefficients,

a2= c1(1−β)

2(1+2α), a3=

c2(1−β)

6(1+3α)+

c2

1(1−β)2

6(1+2α)(1+3α),

a4=

c3(1−β)

12(1+4α)+

c1c2(3+8α)(1−β)2 24(1+2α)(1+3α)(1+4α)+

c3

1(1−β)3

24(1+2α)(1+3α)(1+4α),

a5=

1 20(1+5α)

(

c1c3(1−β)2(4+4α) 3(1+2α)(1+4α) +

c41(1−β)4

6(1+2α)(1+3α)(1+4α)+

c22(1−β)2 2(1+3α)

+ c

2

1c2(1−β)3(6+20α)

6(1+2α)(1+3α)(1+4α)+c4(1−β)

) .

Thus, we have

|a2a3−a4|=

c1c2(1−β)2(−1)

24(1+2α)(1+3α)(1+4α)+

c31(1−β)3(1+6α) 24(1+2α)2(1+)(1+)

− c3(1−β) 12(1+4α)

(3.6)

Suppose now thatc1 = c. Since |c| = |c1| ≤ 2 ,using the Lemma 2.1, we may assume without restriction

c ∈ [0, 2]. Substituting forc2andc3, from Lemma 2.2 and applying the triangle inequality withρ =|x|, we obtain

|a2a3−a4| ≤

c3(1β)[−3α++22αβ212αβ2+16α2+12α3]

48(1+2α)2(1+)(1+)

+ρc(1−β)(4−c

2)(3+10α+12α2β)

48(1+2α)(1+3α)(1+4α)

+ρ

2(4c2)(1

β)(c−2) 48(1+4α) +

2(1−β)(4−c2) 48(1+4α) =F(ρ).

DifferentiatingF(ρ), we get

F0(ρ) = c(1−β)(4−c

2)(3+10α+12α2β)

48(1+2α)(1+3α)(1+4α) +

ρ(4−c2)(1−β)(c−2) 24(1+4α) .

Note also thatF0(ρ)≥F0(1)≥0. Then there exist ac∗∈[0, 2]such thatF0(ρ)>0 forc∈(c∗, 2]andF0(ρ)≤0 otherwise.

Then, for ac∈(c∗, 2],F(ρ)≤F(1), that is:

|a2a3−a4| ≤ c

3(1β)[++22αβ212αβ2+16α2+12α3]

48(1+2α)2(1+)(1+)

+c(1−β)(4−c

2)(3+10α+12α2

β) 48(1+2α)(1+3α)(1+4α)

+(4−c

2)(1β)(c2)

48(1+4α) +

2(1−β)(4−c2) 48(1+4α) =G(c).

Ifα=0,β=0, we haveG(c) = c(4−c

2)

12 . By elementary calculus, we haveG0(c) = 4−3c2

12 ,G00(c) =− c

2 <0. Since c ∈[0, 2]by our assumption, it follows thatG(c)is maximum atc =2/√3. Otherwise, again by elementary calculusG(c)is maximum atc=qA1

A2 and is given by

G(c)≤ (1−β)

MA2 s

A1

(4)

Now supposec∈[0,c∗], thenF(ρ)≤F(0), that is:

F(ρ)≤ c

3(1

β)[−3α+3β+22αβ−2β2−12αβ2+16α2+12α3] 48(1+2α)2(1+)(1+) +

2(1−β)(4−c2) 48(1+4α) =G(c),

which implies thatG(c)turns atc=0 andc= 4(1+2α)2(1+3α)

[−3α+3β+22αβ−2β2−12αβ2+16α2+12α3]

with its maximum atc=0. That isG(c)≤ (1−β) 6(1+4α).

Thus for all admissiblec∈[0, 2], the maximum of the functional|a2a3−a4|are given by the inequalities of the

theorem.

If p(z)∈ P, withc1=2/√3,c2=−2/3 andc3=−10/3√3, then we obtainp(z) = 1+√2 3z−

2

3z2−310√3z3+

· · · ∈Pwhich shows that the result is sharp.

Lemma 3.2. Let f ∈Cαβ. Then ,we have the best possible bound for

|a2a4−a23| ≤   

  1

8 α=0,β=0

(1−β)2

N [M1V1V2+ (4V2−V1){M2V1+V1P1+P2}] 0<α≤1, 0<β≤1, where,

M1= [22α3+31α2+11α−2β2−5β−3αβ−8α2β],

M2=3+118α2−45α+44α3−β−3αβ−8α2β,

P1= (1+27α2−10α)(1+2α),

P2= (8+48α+64α2)(1+2α),

V1=2M1+8M2+8P1−2P2,

V2=4M2+4P1,

N=288(1+2α)2(1+3α)2(1+4α).

Proof. Let f ∈Cαβ. Then proceeding as in Lemma 3.1, we have

|a2a4−a23|=

c1c3(1−β)2 24(1+2α)(1+4α)+

c2

1c2(3+8α)(1−β)3 48(1+2α)2(1+)(1+)+

c4

1(1−β)4

48(1+2α)2(1+)(1+)

− c

2

2(1−β)2

36(1+3α)2−

c41(1−β)4

36(1+2α)2(1+)2−

2c21c2(1−β)3 36(1+2α)(1+3α)2

. (3.7)

Suppose now thatc1=c. Since|c|=|c1| ≤2, Using Lemma 2.1, we may assume without restrictionc∈(0, 2].

Substituting forc2andc3, from Lemma 2.2 and applying triangle inequality withρ=|x|, we obtain

|a2a4−a23| ≤ 1 144

(

1−β)2c4[22α3+23α2+8α2+11α−2β2−5β−3αβ−8α2β 2(1+2α)2(1+)2(1+)

+ρc

2(4c2)(1

β)2[3+118α2−45α+44α3−β−3αβ−8α2β] 2(1+2α)2(1+)2(1+)

+ρ

2(4c2)(1β)2[8+c2+48α+64α2+27c2α210c2α]

2(1+2α)2(1+)2(1+)

+3(4−c

2)(1

ρ)(1−β)2

(1+2α)(1+4α)

=F(ρ).

DifferentiatingF(ρ), we get,

F0(ρ) = 1

144

c2(4c2)(1β)2[3+118α245α+44α2β3αβ2β]

2(1+2α)2(1+)2(1+)

+2ρ(4−c

2)(1β)2[8+c2+48α+64α2+27c2α10c2α]

2(1+2α)2(1+)2(1+)

+3ρ(4−c

2)(1

β)2

(1+2α)(1+4α)

(5)

Note also thatF0(ρ)≥F0(1)≥0. Then there exist ac∗∈[0, 2]such thatF0(ρ)>0 forc∈(c∗, 2]andF0(ρ)≤0 otherwise.

Then for ac∈(c∗, 2],F(ρ)≤F(1), that is:

|a2a4−a23| ≤

1 144

(

1−β)2[22α3+31α2+11α−2β2−5β−3αβ−8α2β]c4 2(1+2α)2(1+)2(1+)

+(1−β)

2[3+118α245α+44α3β3αβ2β]c2(4c2)

2(1+2α)2(1+)2(1+)

+(1−β)

2(4c2)(8+c2+48α+64α2+27c2α210c2α)

2(1+2α)(1+3α)2(1+)

=G(c).

Ifα =0, β =0, we haveG(c) = 3c

2(4−c2)

2 +

(4−c2)(c2+8)

2 . By elementary calculus we have,G0(c) = 8c−8c3, G00(c) =8−24c2<0. Sincec∈(0, 2], by our assumption it follows thatG(c)is maximum atc=1. Otherwise, again by elementary calculusG(c)is maximum atc=qV1

V2 and is given by

G(c)≤ (1−β)

2

N [M1V1V2+ (4V2−V1){M2V1+V1P1+P2}].

Now supposec∈[0,c∗], thenF(ρ)≤F(0), that is:

F(ρ)≤ 1

144

(1−β)2[22α3+31α2+11α−2β2−5β−3αβ−8α2β] 2(1+2α)2(1+)2(1+)

+3(4−c

2)(1β)2

(1+2α)(1+4α)

=G(c),

which implies thatG(c)turns atc=0 andc=

r

(22α3+31α2+11α−2β2−5β−3αβ−8α2β)

3(1+2α)(1+3α)2 ,

with its maximum atc=0. That is,G(c)≤ 12(1−β)2

(1+2α)(1+4α).

Thus for all admissiblec∈[0, 2], the maximum of the functional|a2a4−a23|are given by the inequalities of the

theorem.

Ifp(z)∈P, withc1=1,c2=−1,c3=−2, thenp(z) = 1−z2

1−z+z2 =1+z−z2−2z3+· · · ∈Pwhich shows that

the result is sharp.

Lemma 3.3. Let f ∈Cαβ. Then we have the best possible bound for

|a3−a22| ≤ (1

3 α=0,β=0 1−β

3(1+3α) 0<α≤1, 0<β≤1.

Proof. Let f ∈Cαβ. Then proceeding as in Lemma 3.1, we have

|a3−a22|=

c2(1−β) 6(1+3α)−

c21(1−β)2(1+5α) 12(1+2α)2(1+)

(3.8)

and

|a3−a22|= (1−β)

6(1+3α)

c2−c

2

1(1−β)(1+5α)

2(1+2α)2 .

Settingσ= (1−(1+2β)(1+5α)

α)2 , using Lemma 2.3, we have|a3−a22| ≤ (1−β) 3(1+3α).

(6)

Remark 3.1. Let f ∈Cαβ. By Lemma 2.1, we have

|a3|=

c2(1−β)

6(1+3α)+

c2

1(1−β)2

6(1+2α)(1+3α)

,

≤ (1−β)(3+2α−2β) 3(1+2α)(1+3α) ,

|a4|=

c3(1−β) 12(1+4α)+

c1c2(3+8α)(1−β)2 24(1+2α)(1+3α)(1+4α)+

c3

1(1−β)3

24(1+2α)(1+3α)(1+4α)

,

≤ (1−β)(6+6α

2+2+13α8αβ)

6(1+2α)(1+3α)(1+4α) ,

|a5|=

1 20(1+5α)

(

c1c3(1−β)2(4+4α) 3(1+2α)(1+4α) +

c41(1−β)4

6(1+2α)(1+3α)(1+4α)+

c22(1−β)2 2(1+3α)

+ c

2

1c2(1−β)3(6+20α)

6(1+2α)(1+3α)(1+4α)+c4(1−β)

)

≤ (1−β)(120+408α

2+500α576αβ432α2

β+160αβ2+188β+96β2) 120(1+2α)(1+3α)(1+4α)(1+5α) .

This leads to the next theorem which gives a sharp result by using Lemmas 3.1,3.2 and 3.3 and Remark 3.1.

Theorem 3.1. Let f ∈Cβα. Then

|H3(1)| ≤ (1−β)

2(3+)

3(1+2α)(1+3α)

(1−β)2

N [M1V1V2+ (4V2−V1){M2V1+V1P1+P2}]

+(1−β)(6+6α

2+3+13α8αβ)

6(1+2α)(1+3α)(1+4α)

(

(1−β)

MA2

s A1

A2

[B1+ (4A2−A1)(B2+B3)]

)

+(1−β)(120+408α

2+500α576αβ432α2β+160αβ2+188β+96β2)

120(1+2α)(1+3α)(1+4α)(1+5α)

× (1−β) 3(1+3α).

Whenα=0,β=0, we have the following corollary due to [1].

Corollary 3.1. Ifα=0,β=0, then

|H3(1)| ≤

32+33√3

72√3 =0.714933452973167.

4

Acknowledgement

The authors thank the referees for their valuable comments and suggestions to improve the presentation of the paper.

References

[1] K.O. Babalola, OnH3(1)Hankel determinant for some classes of univalent functions,Inequality Theory

and Applications, S.S. Dragomir and J.Y. Cho, Editors, Nova Science Publishers, 6(2010), 1–7.

[2] K.O. Babalola and T.O. Opoola, On the coefficients of certain analytic and univalent functions,Advances

(7)

[3] P.L. Duren,Univalent Functions, Springer Verlag, New York Inc, 1983.

[4] R.M. Goel and B.S. Mehrok, A class of univalent functions,J. Austral. Math. Soc., (Series A), 35(1983), 1–17.

[5] T. Hayami and S. Owa, Generalized Hankel determinant for certain classes,Int. Journal of Math. Analysis, 4(52)(2010), 2473–2585.

[6] A. Janteng, S.A. Halim and M. Darus, Estimate on the second Hankel functional for functions whose derivative has a positive real part,Journal of Quality Measurement and Analysis, 4(1)(2008), 189–195.

[7] N. Kharudin, A. Akbarally, D. Mohamad and S.C. Soh, The second Hankel determinant for the class of close to convex functions,European Journal of Scientific Research, 66(3)(2011), 421–427.

[8] R.J. Libera and E.J. Zlotkiewiez, Early coefficients of the inverse of a regular convex function,Proc. Amer.

Math. Soc., 85(2)(1982), 225–230.

[9] R.J. Libera and E.J. Zlotkiewiez, Coefficient bounds for the inverse of a function with derivative in P,

Proc. Amer. Math. Soc.,87(2)(1983), 251–257.

[10] T.H. MacGregor, Functions whose derivative has a positive real part,Trans. Amer. Math. Soc., 104(1962), 532–537.

[11] K.I. Noor, Hankel determinant problem for the class of functions with bounded boundary rotation,Rev.

Roum. Math. Pures Et Appl., 28(c)(1983), 731–739.

[12] M.S. Robertson, On the theory of univalent functions,Annals of Math., 37(1936), 374–408.

[13] C. Selvaraj and N. Vasanthi, Coefficient bounds for certain subclasses of close-to-convex functions,Int.

Journal of Math. Analysis, 4(37)(2010), 1807–1814.

[14] G. Shanmugam, B. Adolf Stephen and K.G. Subramanian, Second Hankel determinant for certain classes of analytic functions,Bonfring International Journal of Data Mining, 2(2) (2012), 57–60.

Received: March 05, 2014;Accepted: July 05, 2014

UNIVERSITY PRESS

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