Page 1

arXiv:0910.3779v1 [math.CV] 20 Oct 2009

Inequality Theory and Applications 6(2007),

Pages ???–???

On H3(1) Hankel determinant for some classes of univalent functions

K. O. BABALOLA1,2

Abstract. Focus in this paper is on the Hankel determinant, H3(1), for the

well-known classes of bounded-turning, starlike and convex functions in the open

unit disk E = {z ∈ C: |z| < 1}. The results obtained complete the series of

research works in the search for sharp upper bounds on H3(1) for each of these

classes.

1. Introduction

Let A be the class of functions

f(z) = z + a2z2+ ··· (1.1)

which are analytic in E. A function f ∈ A is said to be of bounded turning,

starlike and convex respectively if and only if, for z ∈ E, Re f′(z) > 0, Re

zf′(z)/f(z) > 0 and Re (1 + zf′′(z)/f′(z)) > 0. By usual notations we denote

these classes of functions respectively by R, S∗and C. Let n ≥ 0 and q ≥ 1, the

q-th Hankel determinant is defined as:

Hq(n) =

??????????

an

an+1

···an+q−1

...

...

an+2(q−1)

an+1

...

an+q−1

···

...

···

···

...

···

??????????

2000 Mathematics Subject Classification. 30C45; 30C50.

Key words and phrases. Hankel determinant, functions of bounded turning, starlike and

convex univalent functions.

1Current Address: Centre for Advanced Studies in Mathematics, Lahore University of Man-

agement Sciences, Lahore, Pakistan. E-mail: kobabalola@lums.edu.pk

2Permanent Address: Department of Mathematics, University of Ilorin, Ilorin, Nigeria. E-

mail: kobabalola@gmail.com; babalola.ko@unilorin.edu.ng

1

Page 2

2K. O. BABALOLA

(see [11] for example). The determinant has been investigated by several authors

with the subject of inquiry ranging from rate of growth of Hq(n) as n → ∞ [11, 12]

to the determination of precise bounds on Hq(n) for specific q and n for some

favored classes of functions [4, 5, 10]. In particlar, sharp upper bounds on H2(2)

were obtained by the authors of articles [4, 5, 10] for various classes of functions.

In the present investigation, our focus is on the Hankel determinant, H3(1), for

the well-known classes of bounded-turning, starlike and convex functions in E.

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− a2

3) − a4(a4− a2a3) + a5(a3− a2

2)

and by triangle inequality, we have

|H3(1)| ≤ |a3||a2a4− a2

3| + |a4||a2a3− a4| + |a5||a3− a2

2|.(1.2)

Incidentally, all of the functionals on the right side of the inequality (1.2) have

known (and sharp) upper bounds in the classes of functions which are of interest

in this paper, except |a2a3− a4|. The last one is the well-known Fekete-Szego

functional. For R, sharp bound 2/3 was reported in [1] (with R corresponding to

n = α = 1, β = 0 in the classes Tα

n(β) studied there) while for S∗and C, sharp

bounds 1 and 1/3 respectively were given in [6]. Janteng et-al [4, 5] obtained for

the functional |H2(2)| ≡ |a2a4− a2

3| sharp bounds 4/9, 1 and 1/8 repectively for

R, S∗and C. Furthermore, it is known that for k = 2,3,···, |ak| ≤ 2/k, |ak| ≤ k

and |ak| ≤ 1 also respectively for R, S∗and C (see [2, 9]). Thus finding the

best possible bounds on |a2a3−a4| for each of the classes and using those known

inequalities, then the sharp upper bounds on H3(1) follow as simple corollaries.

Our investigation follows a method of classical analysis devised by Libera and

Zlotkiewicz [7, 8]. The same has been employed by many authors in similar works

(see also [4, 5, 10]). In the next section we state the necessary lemmas while in

Section 3 we present our main results.

2. Preliminary Lemmas

Let P denote the class of functions p(z) = 1+c1z+c2z2+··· which are regular

in E and satisfy Re p(z) > 0, z ∈ E. To prove the main results in the next section

we shall require the following two lemmas.

Lemma 2.1. ([2]) Let p ∈ P, then |ck| ≤ 2, k = 1,2,···, and the inequality is

sharp.

Page 3

On H3(1) Hankel determinant for some classes of univalent functions3

Lemma 2.2. ([7, 8]) Let p ∈ P, then

2c2= c2

1+ x(4 − c2

1)(2.1)

and

4c3= c3

1+ 2xc1(4 − c2

1) − x2c1(4 − c2

1) + 2z(1 − |x|2)(4 − c2

1) (2.2)

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

3. Main Results

Theorem 3.1. Let f ∈ R. Then

|a2a3− a4| ≤1

2.

The inequality is sharp. Equality is attained by

f(z) =

?z

0

1 + t3

1 − t3dt.

Proof. Let f ∈ R. Then there exists a p ∈ P such that f′(z) = p(z), wherefrom

equating coefficients we find that 2a2= c1, 3a3= c2and 4a4= c3. Thus we have

|a2a3− a4| =

???c1c2

6

−c3

4

???. (3.1)

Substituting for c2and c3using Lemma 2, we obtain

c3

1

48−c1(4 − c2

|a2a3− a4| =

????

1)x

24

+c1(4 − c2

1)x2

16

−(4 − c2

1)(1 − |x|2)z

8

????. (3.2)

By Lemma 1, |c1| ≤ 2. Then letting c1= c, we may assume without restriction

that c ∈ [−2,0]. Thus applying the triangle inequality on (3.2), with ρ = |x|, we

obtain

|a2a3− a4| ≤c3

8

= F(ρ).

48+(4 − c2)

+c(4 − c2)ρ

24

+(c − 2)(4 − c2)ρ2

16

Now we have

F′(ρ) =c(4 − c2)

24

+(c − 2)(4 − c2)ρ

8

< 0.

Hence F(ρ) is a decreasing function of ρ on the closed interval [0,1], so that

F(ρ) ≤ F(0). That is

F(ρ) ≤c3

48+4 − c2

= G(c).

8

Obviously G(c) is increasing on [−2,0]. Hence we have G(c) ≤ G(0) = 1/2.

Page 4

4 K. O. BABALOLA

By setting c1= c = 0 and selecting x = 0 and z = 1 in (2.1) and (2.2) we find

that c2= 0 and c3= 2. Thus equality is attained by f(z) defined in theorem and

the proof is complete.

?

Let f ∈ R. Then using the above result in (1.2) together with the known

inequalities |a3− a2

[9], we have the sharp inequality:

2| ≤ 2/3 [1], |a2a4− a2

3| ≤ 4/9 [4] and |ak| ≤ 2/k, k = 2,3,···

Corollary 3.2. Let f ∈ R. Then

|H3(1)| ≤

993

1620.

Theorem 3.3. Let f ∈ S∗. Then

|a2a3− a4| ≤ 2.

The inequality is sharp. Equality is attained by the Koebe function k(z) = z/(1−

z)2.

Proof. Let f ∈ S∗. Then there exists a p ∈ P such that zf′(z) = f(z)p(z).

Equating coefficients we find that a2= c1, 2a3= c2+c2

Thus we have

|a2a3− a4| =1

1and 6a4= 2c3+3c1c2+c3

1.

3|c3

1− c3|. (3.3)

Substituting for c3from Lemma 2, we obtain

1

12|3c3

Since |c1| ≤ 2 by Lemma 1, let c1 = c and assume without restriction that

c ∈ [0,2]. Applying the triangle inequality on (3.4), with ρ = |x|, we obtain

|a2a3− a4| =

1− 2c1(4 − c2

1)x + c1(4 − c2

1)x2− 2(4 − c2

1)(1 − |x|2)z|. (3.4)

|a2a3− a4| ≤

1

12[3c3+ 2(4 − c2) + 2c(4 − c2)ρ + (c − 2)(4 − c2)ρ2]

= F(ρ).

Differentiating F(ρ), we have

F′(ρ) =

1

12[2c(4 − c2) + 2(c − 2)(4 − c2)] > 0.

This implies that F(ρ) is an increasing function of ρ on [0,1] if c ∈ [1,2]. In this

case F(ρ) ≤ F(1) = c ≤ 2 for all ρ ∈ [0,1]. It follows therefore that F(ρ) ≤ 2.

On the other hand suppose c ∈ [0,1), then F(ρ) is decreasing on [0,1] so that

F(ρ) ≤ F(0). That is

F(ρ) ≤3c3− 2c2+ 8

12

= G(c).

Page 5

On H3(1) Hankel determinant for some classes of univalent functions5

Hence we have G(c) ≤ G(0) = 2/3, c ∈ [0,1). This is less than 2, which is the

case when c ∈ [1,2]. Thus the maximum of the functional |a2a3−a4| corresponds

to ρ = 1 and c = 2.

If c1 = c = 2 in (2.1) and (2.2), then we have c2 = c3 = 2. Using these

in (3.3) we see that equality is attained which shows that our result is sharp.

Furthermore, it is easily seen that the extremal function in this case is the well

known Koebe function k(z) = z/(1 − z)2.

?

For f ∈ S∗, using the known inequalities |ak| ≤ k, k = 2,3,··· [2], |a2a4−a2

1 [5] and |a3− a2

2| ≤ 1 [6] together with Theorem 2 we have the next corollary.

Corollary 3.4. Let f ∈ S∗. Then

3| ≤

|H3(1)| ≤ 16.

The inequality is sharp. Equality is attained by a rotation, k1(z) = z/(1+z)2, of

the Koebe function.

Theorem 3.5. Let f ∈ C. Then

|a2a3− a4| ≤1

6.

The inequality is sharp. Equality is attained by

f(z) =

?z

0

?

s.exp

??s

0

2t3

1 − t3dt

??

ds.

Proof. For f ∈ C given by (1.1), there exists a p ∈ P such that (zf′(z))′=

f′(z)p(z). Then equating coefficients we find that 2a2= c1, 6a3= c2+ c2

24a4= 2c3+ 3c1c2+ c3

1. Thus we have

1

24|c3

Substituting for c2and c3using Lemma 2, we obtain

1

48| − 3c1(4 − c2

With |c1| ≤ 2 from Lemma 1, we let c1= c and assume also without restriction

that c ∈ [−2,0]. Thus applying the triangle inequality on (3.6), with ρ = |x|, we

obtain

|a2a3− a4| ≤(4 − c2)

24

= F(ρ).

Differentiating F(ρ), we get

F′(ρ) =c(4 − c2)

16

1and

|a2a3− a4| =

1− c1c2− 2c3|. (3.5)

|a2a3− a4| =

1)x + c1(4 − c2

1)x2− 2(4 − c2

1)(1 − |x|2)z|. (3.6)

+c(4 − c2)ρ

16

+(c − 2)(4 − c2)ρ2

48

+(c − 2)(4 − c2)ρ

24

< 0.