Third Hankel determinant for a subclass of analytic univalent functions
T.V. Sudharsan
a,∗, S.P. Vijayalakshmi
band B. Adolf Stephen
caDepartment 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
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,
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+3α)(1+4α)
− 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(1−β)[−3α+3β+22αβ−2β2−12αβ2+16α2+12α3]
48(1+2α)2(1+3α)(1+4α)
+ρc(1−β)(4−c
2)(3+10α+12α2−β)
48(1+2α)(1+3α)(1+4α)
+ρ
2(4−c2)(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+12α2−β)
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(1−β)[−3α+3β+22αβ−2β2−12αβ2+16α2+12α3]
48(1+2α)2(1+3α)(1+4α)
+c(1−β)(4−c
2)(3+10α+12α2−
β) 48(1+2α)(1+3α)(1+4α)
+(4−c
2)(1−β)(c−2)
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
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+3α)(1+4α) +
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+3α)(1+4α)+
c4
1(1−β)4
48(1+2α)2(1+3α)(1+4α)
− c
2
2(1−β)2
36(1+3α)2−
c41(1−β)4
36(1+2α)2(1+3α)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+3α)2(1+4α)
+ρc
2(4−c2)(1−
β)2[3+118α2−45α+44α3−β−3αβ−8α2β] 2(1+2α)2(1+3α)2(1+4α)
+ρ
2(4−c2)(1−β)2[8+c2+48α+64α2+27c2α2−10c2α]
2(1+2α)2(1+3α)2(1+4α)
+3(4−c
2)(1−
ρ)(1−β)2
(1+2α)(1+4α)
=F(ρ).
DifferentiatingF(ρ), we get,
F0(ρ) = 1
144
c2(4−c2)(1−β)2[3+118α2−45α+44α2−β−3αβ−8α2β]
2(1+2α)2(1+3α)2(1+4α)
+2ρ(4−c
2)(1−β)2[8+c2+48α+64α2+27c2α−10c2α]
2(1+2α)2(1+3α)2(1+4α)
+3ρ(4−c
2)(1−
β)2
(1+2α)(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:
|a2a4−a23| ≤
1 144
(
1−β)2[22α3+31α2+11α−2β2−5β−3αβ−8α2β]c4 2(1+2α)2(1+3α)2(1+4α)
+(1−β)
2[3+118α2−45α+44α3−β−3αβ−8α2β]c2(4−c2)
2(1+2α)2(1+3α)2(1+4α)
+(1−β)
2(4−c2)(8+c2+48α+64α2+27c2α2−10c2α)
2(1+2α)(1+3α)2(1+4α)
=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+3α)2(1+4α)
+3(4−c
2)(1−β)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+3α)
(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).
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+2β2+13α−7β−8αβ)
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+500α−576αβ−432α2
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+2α−2β)
3(1+2α)(1+3α)
(1−β)2
N [M1V1V2+ (4V2−V1){M2V1+V1P1+P2}]
+(1−β)(6+6α
2+2β3+13α−7β−8αβ)
6(1+2α)(1+3α)(1+4α)
(
(1−β)
MA2
s A1
A2
[B1+ (4A2−A1)(B2+B3)]
)
+(1−β)(120+408α
2+500α−576αβ−432α2β+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.
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Received: March 05, 2014;Accepted: July 05, 2014
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