Bounds on the Third Order Hankel Determinant for Certain ... - EMIS

DOI: 10.1515/auom-2017-0045 An. S ¸ t. Univ. Ovidius Constant ¸a

Vol. 25(3),2017, 199–214

Bounds on the Third Order Hankel Determinant for Certain Subclasses of Analytic Functions S.P. Vijayalakshmi, T.V. Sudharsan, Daniel Breaz and K.G.Subramanian

Abstract Let A be the class of analytic functions f (z) in the unit disc ∆ = {z ∈ C : |z| < P 1} with the Taylor series expansion about the origin given n by f (z) = z + ∞ n=2 an z , z ∈ ∆. The focus of this paper is on deriving upper bounds for the third order Hankel determinant H3 (1) for two new subclasses of A.

1

Introduction

Let A be the class of functions f (z) = z +

∞ X

an z n

(1)

n=2

which are analytic in ∆ = {z ∈ C : |z| < 1}. A function f ∈ A is respectively said to be with bounded turning, starlike or convex if and only if for z ∈ ∆, zf 0 (z) zf 00 (z) 0 Ref (z) > 0, Re f (z) > 0 or Re 1 + f 0 (z) > 0. The classes of these functions are respectively denoted by R, S ∗ and C. For n ≥ 0 and q ≥ 1, Key Words: Analytic functions, Hankel determinant, Functions with positive real part. 2010 Mathematics Subject Classification: Primary 30C45; Secondary 30C50. Received: 09.09.2016 Accepted: 24.10.2016

199

BOUNDS ON THE THIRD ORDER HANKEL DETERMINANT FOR CERTAIN SUBCLASSES OF ANALYTIC FUNCTIONS

the q th Hankel determinant is defined as follows: an an+1 . . . an+q−1 an+1 ... . . . Hq (n) = . . .. .. an+q−1 . . . an+2(q−1)

200

(2)

This determinant has been considered by several authors (see, for example, [1, 2, 3, 10, 17, 18, 22]). In fact Noor [18] determined the rate of growth of Hq (n) as n → ∞ for functions f given by (1) with bounded boundary. In particular, upper bounds for the second Hankel determinant were obtained by several authors [6, 9, 12, 20, 21] for different classes of analytic functions. Upper bound on the third Hankel determinant for different classes of functions has been studied recently [1, 2, 3, 10, 16, 22]. In the present investigation, the β and focus is on the third order Hankel determinant H3 (1) for the classes Rα β Sα in ∆ defined as follows: β Definition 1.1. Let f be given by (1). Then f ∈ Rα if and only if for any z ∈ ∆, 0 ≤ β < 1, 0 ≤ α ≤ 1,

Re{f 0 (z) + αzf 00 (z)} > β.

(3)

The choice α = 0, β = 0 yields Re f 0 (z) > 0, z ∈ ∆, defining the class R of bounded turning [15] while the choice α = 0, yields Re f 0 (z) > β [5]. Definition 1.2. Let f be given by (1). Then f ∈ Sαβ if and only if for any z ∈ ∆, 0 ≤ β < 1, 0 ≤ α ≤ 1, 0 zf (z) zf 00 (z) Re +α 0 > β. f (z) f (z) The choice α = 0, β = 0 yields Re

zf 0 (z) f (z)

> 0, z ∈ ∆, defining the class S ∗ 0

(z) of starlike functions [19] and the choice of α = 0 yields Re zff (z) > β, z ∈ ∆, ∗ defining the class S (β) starlike functions of order β [19]. Setting n = 1 in (2), H3 (1) is given by a1 a2 a3 H3 (1) = a2 a3 a4 a3 a4 a5

and for f ∈ A, H3 (1) = a3 (a2 a4 − a23 ) − a4 (a4 − a2 a3 ) + a5 (a3 − a22 ).


201

Using the triangle inequality, we have |H3 (1)| ≤ |a3 ||a2 a4 − a23 | + |a4 ||a2 a3 − a4 | + |a5 ||a3 − a22 |.

(4)

In obtaining an upper bound for |H3 (1)|, the approach used is to first determine upper bounds for the functionals |a2 a3 − a4 |,|a2 a4 − a23 | and |a3 − a22 |. Furthermore techniques employed in [13, 14] are useful in establishing the results (see, for example [6, 9, 12, 21]).

2

Preliminary Results

Some preliminary results required in the following sections are now listed. Let P denote the class of functions p(z) = 1 + c1 z + c2 z 2 + · · ·

(5)

which are regular in ∆ and satisfy Re p(z) > 0, z ∈ ∆. Throughout this paper, we assume that p(z) is given by (5) and f (z) is given by (1). To prove the main results, the following known Lemmas are required. Lemma 2.1. [4] Let p ∈ P . Then |ck | ≤ 2, k = 1, 2, . . . and the inequality is sharp. Lemma 2.2. [13, 14] Let p ∈ P . Then 2c2 = c21 + x(4 − c21 )

(6)

4c3 = c31 + 2xc1 (4 − c21 ) − x2 c1 (4 − c21 ) + 2y(1 − |x|2 )(4 − c21 )

(7)

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

3

Main Results

β , Lemma 3.1- Lemma 3.3 give the upper bounds for the For functions f ∈ Rα three functionals mentioned earlier while Theorem 3.1 presents an estimate for |H3 (1)|. β Lemma 3.1. Let f ∈ Rα . Then

|a2 a3 − a4 | ≤

(1 − β) 2(1 + 3α)

β Proof. Let f ∈ Rα . Then there exists a p such that

(8)


202

f 0 (z) + αzf 00 (z) = (1 − β)p(z) + β, p(0) = 1, Re p(z) > 0. Equating the coefficients, we find that a2 =

c1 (1 − β) c2 (1 − β) c3 (1 − β) c4 (1 − β) , a3 = , a4 = , a5 = . 2(1 + α) 3(1 + 2α) 4(1 + 3α) 5(1 + 4α)

The functional |a2 a3 − a4 | is given by c1 c2 (1 − β)2 c3 (1 − β) . |a2 a3 − a4 | = − 6(1 + α)(1 + 2α) 4(1 + 3α)

(9)

Substituting for c2 and c3 from (6) and (7) of Lemma 2.2, we obtain |a2 a3 − a4 | = (1 − β) 4(1 + 3α)(1 − β)c1 (c21 + x(4 − c21 )) 48(1 + α)(1 + 2α)(1 + 3α) −3(1 + α)(1 + 2α)(c31 + 2xc1 (4 − c21 ) − x2 c1 (4 − c21 ) + 2y(1 − |x|2 )(4 − c21 )) =A(α, β) c31 (2a − 3b) − 2c1 x(4 − c21 )(3b − a) − 3bx2 c1 (4 − c21 ) −2y × 3b(1 − |x|2 )(4 − c21 ) (1−β) where A(α, β) = 48(1+α)(1+2α)(1+3α) , a = 2(1 + 3α)(1 − β), b = (1 + α)(1 + 2α). Suppose now that c1 = c. Since |c| = |c1 | ≤ 2, using the Lemma 2.1, we may assume without restriction that c ∈ [0, 2] and on applying the triangle inequality with ρ = |x| ≤ 1, we get |a2 a3 − a4 | ≤A(α, β) c3 |2a − 3b| + 2cρ(4 − c2 )(3b − a) + 3bρ2 c(4 − c2 ) +2 × 3b(1 − ρ2 )(4 − c2 )

=A(α, β) c3 |2a − 3b| + 2cρ(4 − c2 )(3b − a) + 3bρ2 (4 − c2 )(c − 2) +6b(4 − c2 ) = F (ρ). Next we maximize the function F (ρ). F 0 (ρ) =A(α, β) 2c(4 − c2 )(3b − a) + 6bρ(4 − c2 )(c − 2)

(10)

c(3b−a) ∗ ∗ F 0 (ρ) = 0 implies ρ = c(3b−a) 3b(2−c) . Set ρ = 3b(2−c) . Now 0 ≤ ρ ≤ 1. Also we 00 2 have F (ρ) = A(α, β){6b(4 − c )(c − 2)} < 0, for c < 2. Thus ρ∗ is the only


203

value in [0, 1] at which F (ρ) attains a maximum. Hence F (ρ) ≤ F (ρ∗ ).Thus c2 F (ρ) ≤ A(α, β) c3 |2a − 3b| + (3b − a)2 (2 + c) + 6b(4 − c2 ) 3b (3b − a)2 2(3b − a)2 = A(α, β) c3 {|2a − 3b| + } − c2 {[6b − ]} + 24b 3b 3b F (ρ) ≤ A(α, β) c3 γ − c2 δ + 24b = G(c), 2

2

, δ = [6b − 2(3b−a) ]. G0 (c) = 0 implies c = 0 and where γ = |2a − 3b| + (3b−a) 3b 3b at c = 0, G00 (c) < 0. Thus, the upper bound of F (ρ) corresponds to ρ = ρ∗ (1−β) . and c = 0. Hence |a2 a3 − a4 | ≤ 2(1+3α) Corollary 3.1. Choosing α = 0, β = 0 in (8), we get |a2 a3 − a4 | ≤ 21 . This result coincides with the corresponding result in [3]. β Lemma 3.2. Let f ∈ Rα . Then

|a2 a4 − a23 | ≤

4 (1 − β)2 9 (1 + 2α)2

(11)

β Proof. Let f ∈ Rα . In a manner similar to the proof of Lemma 3.1, we can derive c1 c3 (1 − β)2 c22 (1 − β)2 2 − (12) |a2 a4 − a3 | = 8(1 + α)(1 + 3α) 9(1 + 2α)2

Substituting for c2 and c3 from (6) and (7) of Lemma 2.2, we obtain c1 (1 − β)2 = [c3 + 2xc1 (4 − c21 ) − x2 c1 (4 − c21 ) 32(1 + α)(1 + 3α) 1 (1 − β)2 2 2 2 2 2 +2y(1 − |x| )(4 − c1 )] − [c1 + x(4 − c1 )] . 2 36(1 + 2α)

=

(1 − β)2 9c1 (1 + 2α)2 [c31 + 2xc1 (4 − c21 ) 2 288(1 + α)(1 + 2α) (1 + 3α)

−x2 c1 (4 − c21 ) + 2y(1 − |x|2 )(4 − c21 )] −8(1 + α)(1 + 3α)[c41 + x2 (4 − c21 )2 + 2xc21 (4 − c21 )] .


204

2

(1−β) 2 Let N = 288(1+α)(1+2α) 2 (1+3α) ,a = 9(1 + 2α) , b = 8(1 + α)(1 + 3α) 2 and a − b = 9(1 + 2α) − 8(1 + α)(1 + 3α) = 1 + 12α2 + 4α ≥ 0, since α ≥ 0. |a2 a4 − a23 | = N ac1 [c31 + 2xc1 (4 − c21 ) − x2 c1 (4 − c21 ) + 2y(1 − |x|2 )(4 − c21 )] −b[c41 + x2 (4 − c21 )2 + 2xc21 (4 − c21 )] = N c41 (a − b) + 2xc21 (4 − c21 )(a − b) − x2 (4 − c21 )[ac21 + b(4 − c21 )] +2yac1 (1 − |x|2 )(4 − c21 ) .

Suppose now that c1 = c. Since |c| = |c1 | ≤ 2, using the Lemma 2.1, we may assume without restriction that c ∈ [0, 2] and on applying the triangle inequality with ρ = |x| ≤ 1, we get |a2 a4 − a23 | ≤ N c4 (a − b) + 2ρc2 (4 − c2 )(a − b) +ρ2 (4 − c2 )[c2 (a − b) − 2ac + 4b] + 2ac(4 − c2 ) = N c4 (a − b) + 2ρc2 (4 − c2 )(a − b) 2b ) + 2ac(4 − c2 ) = F (ρ). +ρ2 (4 − c2 )(a − b)(c − 2)(c − (a − b) Differentiating F (ρ), we get F 0 (ρ) = N [2c2 (4 − c2 )(a − b) + 2ρ(4 − c2 )(a − b)(c − 2)(c −

2b )] ≥ 0, (a − b)

since a − b > 0, 2b/(a − b) > 2 so that c − 2b/(a − b) < c − 2 < 0 and 2b (c − 2)(c − (a−b) ) > 0 for all c ∈ [0, 2]. This implies that F (ρ) is an increasing function of ρ on the closed interval [0,1]. Hence F (ρ) ≤ F (1) for all ρ ∈ [0, 1]. That is, |a2 a4 − a23 | ≤ N c4 (a − b) + 2c2 (4 − c2 )(a − b) 2b ) + 2ac(4 − c2 ) +(4 − c2 )(a − b)(c − 2)(c − (a − b) 4 2 = N −2c (a − b) − 4c (4b − 3a) + 16b = G(c). G0 (c) = 0 implies c = 0 so that at c = 0,G00 (c) < 0. Therefore c = 0 is a point of maximum for G(c). Thus, the upper bound of F (ρ) corresponds to ρ = 1 (1−β)2 and c = 0. Hence, |a2 a4 − a23 | ≤ 94 (1+2α) 2. Corollary 3.2. Choosing α = 0, β = 0 in (11), we get |a2 a4 − a23 | ≤ 94 . This result coincides with [7].


205

Corollary 3.3. Choosing α = 0 in (11), we get |a2 a4 − a23 | ≤ 94 (1 − β)2 . This result coincides with [11]. β Lemma 3.3. Let f ∈ Rα . Then for 1/3 ≤ β < 1,

|a3 − a22 | ≤

2(1 − β) 3(1 + 2α)

β Proof. Let f ∈ Rα . Then by proceeding as in Lemma 3.1, we have c2 (1 − β) c2 (1 − β)2 |(a3 − a22 )| = − 1 3(1 + 2α) 4(1 + α)2

|(a3 − a22 )| =

(13)

(14)

(1 − β) 4(1 + α)2 c2 − 3(1 + 2α)(1 − β)c21 . 2 12(1 + 2α)(1 + α)

Substituting for c2 from (6) of Lemma 2.2, we obtain |(a3 − a22 )| =

(1 − β) 2(1 + α)2 [c21 + x(4 − c21 )] − 3(1 + 2α)(1 − β)c21 2 12(1 + 2α)(1 + α)

= M |k1 c21 + k1 x(4 − c21 ) − k2 c21 |, (1−β) 2 where M = 12(1+2α)(1+α) 2 , k1 = 2(1 + α) , k2 = 3(1 + 2α)(1 − β). Set c1 = c. Since |c| = |c1 | ≤ 2, using the Lemma 2.1, we may assume without restriction that c ∈ [0, 2] and on applying the triangle inequality with ρ = |x| ≤ 1, we get

|a3 − a22 | ≤ M [c2 |k1 − k2 | + k1 ρ(4 − c2 )] = F (ρ). Differentiating F (ρ), we get F 0 (ρ) = M [k1 (4 − c2 )] ≥ 0, implying that F (ρ) is an increasing function of ρ on a closed interval [0,1]. Hence F (ρ) ≤ F (1) for all ρ ∈ [0, 1]. That is, |(a3 − a22 )| ≤ M [c2 |k1 − k2 | + k1 (4 − c2 )] = G(c). By hypothesis, β ≥ 1/3 and hence k1 − k2 = 2α2 − 2α − 1 + 3β(1 + 2α) ≥ 2α2 . Hence G(c) = M [4k1 − c2 k2 ], G0 (c) = −2M k2 c and G00 (c) = −2M k2 . Since c ∈ [0, 2], it follows that G(c) attains the maximum at c = 0. Thus, the upper bound of F (ρ) corresponds to ρ = 1 and c = 0. Hence |a3 − a22 | ≤ M [4k1 ] = 2(1−β) 3(1+2α) .


206

Corollary 3.4. Choosing α = 0 in (13), we get |a3 − a22 | ≤ 32 (1 − β). This result coincides with [10], for 1/3 ≤ β < 1. β Remark 3.1. Let f ∈ Rα . By Lemma 2.1, we have c2 (1 − β) ≤ 2(1 − β) , |a3 | = 3(1 + 2α) 3(1 + 2α) c3 (1 − β) ≤ (1 − β) , |a4 | = 4(1 + 3α) 2(1 + 3α) c4 (1 − β) ≤ 2(1 − β) . |a5 | = 5(1 + 4α) 5(1 + 4α) β Using the above results, the upper bound for |H3 (1)| , f ∈ Rα is immediately obtained. β . Then for 1/3 ≤ β < 1, Theorem 3.1. Let f ∈ Rα

|H3 (1)| ≤

(1 − β)2 4(1 − β)2 8(1 − β)3 + + . 3 2 27(1 + 2α) 4(1 + 3α) 15(1 + 2α)(1 + 4α)

In the following results, with similar approach and technique, an upper bound for |H3 (1)| is attained for f ∈ Sαβ . As before, we first derive estimates for the functionals |a2 a3 − a4 |, |a2 a4 − a23 | and |a3 − a22 |. Their estimates are given in Lemmas 3.4, 3.5, and 3.6. Lemma 3.4. Let f ∈ Sαβ . Then |a2 a3 − a4 | ≤

2(1 − β) 3(1 + 4α)

(15)

Proof. Let f ∈ Sαβ . Then there exists a p ∈ P such that zf 0 (z) + αz 2 f 00 (z) = [(1 − β)p(z) + β]f (z), for some z ∈ ∆. Equating the coefficients, we have a2 =

a4 =

c1 (1 − β) , 1 + 2α

a3 =

c2 (1 − β) c21 (1 − β)2 + , 2(1 + 3α) 2(1 + 2α)(1 + 3α)

c3 (1 − β) c1 c2 (3 + 8α)(1 − β)2 c31 (1 − β)3 + + , 3(1 + 4α) 6(1 + 2α)(1 + 3α)(1 + 4α) 6(1 + 2α)(1 + 3α)(1 + 4α)


207

and c41 (1 − β)4 c22 (1 − β)2 + 24(1 + 2α)(1 + 3α)(1 + 4α)(1 + 5α) 8(1 + 3α)(1 + 5α) c21 c2 (1 − β)3 (20α + 6) c1 c3 (1 − β)2 (4 + 14α) + + 24(1 + 2α)(1 + 3α)(1 + 4α)(1 + 5α) 12(1 + 2α)(1 + 4α)(1 + 5α) c4 (1 − β) + . 4(1 + 5α)

a5 =

Thus, we have (1 − β) |c1 c2 (1 − β)4α(1 + 2α) 6(1 + + 3α)(1 + 4α) +2c31 (1 − β)2 (1 + 5α) − 2c3 (1 + 2α)2 (1 + 3α)

|a2 a3 − a4 | =

2α)2 (1

(16)

Substituting for c2 and c3 from (6) and (7) of Lemma 2.2, we have |a2 a3 − a4 | 4α(1 + 2α)(1 − β)c1 2 (c1 + x(4 − c21 )) = B(α, β) 2 2(1 + 2α)2 (1 + 3α) 3 +c31 2(1 − β)2 (1 + 5α) − [c1 + 2xc1 (4 − c21 ) 4 −x2 c1 (4 − c21 ) + 2y(1 − |x|2 )(4 − c21 )] r3 r3 = B(α, β) r1 c31 + r2 c1 x(4 − c21 ) + x2 c1 (4 − c21 ) − y(1 − |x|2 )(4 − c21 ) , 4 2 where B(α, β) =

(1 − β) , 6(1 + 2α)2 (1 + 3α)(1 + 4α)

(1 + 2α)2 (1 + 3α) , 2 r2 = 2α(1 + 2α)(1 − β) − (1 + 2α)2 (1 + 3α), r3 = (1 + 2α)2 (1 + 3α). r1 = 2α(1 + 2α)(1 − β) + 2(1 − β)2 (1 + 5α) −

Suppose now that c1 = c. Since |c| = |c1 | ≤ 2, using the Lemma 2.1, we may assume without restriction that c ∈ [0, 2] and on applying the triangle inequality with ρ = |x| ≤ 1, we get, |a2 a3 − a4 | r3 2 ρ c(4 − c2 ) + r3 (4 − c2 ) − r3 ρ2 (4 − c2 )} 2 r3 = β(α, β){|r1 |c3 + |r2 |ρc(4 − c2 ) + ρ2 (c − 2)(4 − c2 ) + r3 (4 − c2 )} = F (ρ). 2

≤ β(α, β){|r1 |c3 + |r2 |ρc(4 − c2 ) +


208

Next we maximize the function F (ρ). Differentiating F (ρ), we get F 0 (ρ) = B(α, β)[|r2 |c(4 − c2 ) + r3 ρ(4 − c2 )(c − 2)]. 2 |c 2 |c . Set ρ∗ = r3|r(2−c) . Now, 0 ≤ ρ∗ ≤ 1. Also we F 0 (ρ) = 0 implies ρ = r3|r(2−c) 00 2 have F (ρ) = B(α, β)r3 (4 − c )(c − 2) ≤ 0. Thus ρ∗ is the only value in [0, 1] at which F (ρ) attains maximum. Hence F (ρ) ≤ F (ρ∗ ).Thus

r22 c2 (2 + c) + 4r3 − r3 c2 ] 2r3 = B(α, β)[c3 γ − c2 δ + 4r3 ] = G(c),

F (ρ) ≤ B(α, β)[|r1 |c3 +

r2

r2

where γ = |r1 | + 2r23 , δ = r3 − r23 ≥ 0, G0 (c) = 0 implies c = 0 and at c = 0, G00 (c) < 0. Therefore c = 0 is a point of maximum of G(c). Thus the upper bound of F (ρ) corresponds to ρ = ρ∗ and c = 0. Hence |a2 a3 − a4 | ≤ 2(1−β) 3(1+4α) . Corollary 3.5. Choosing α = 0, β = 0 in (15), we get |a2 a3 − a4 | ≤ 32 . Corollary 3.6. Choosing α = 0, in (15), we get |a2 a3 − a4 | ≤

2(1 − β) . 3

Lemma 3.5. Let f ∈ Sαβ . Then |a2 a4 − a23 | ≤

(1 − β)2 (1 + 3α)2

(17)

Proof. Let f ∈ Sαβ . Then by proceeding as in Lemma 3.4, we have |a2 a4 − a23 | = c1 c3 (1 − β)2 c22 (1 − β)2 c41 (1 − β)4 (1 + 6α) − − 3(1 + 2α)(1 + 4α) 4(1 + 3α)2 12(1 + 2α)2 (1 + 3α)2 (1 + 4α) 2 3 c1 c2 (1 − β) (2α) − (18) 2 2 12(1 + 2α) (1 + 3α) (1 + 4α)

=

(1 − β)2 16(1 + 2α)(1 + 3α)2 c1 c3 2 2 48(1 + 2α) (1 + 3α) (1 + 4α)

−12c22 (1 + 2α)2 (1 + 4α) − 4c41 (1 − β)2 (1 + 6α) − 4(1 − β)2αc21 c2 .


209

Substituting for c2 and c3 from (6) and (7) of Lemma 2.2, we obtain |a2 a4 − a23 | = M k1 c1 [c31 + 2xc1 (4 − c21 ) − x2 c1 (4 − c21 ) + 2y(1 − |x|2 )(4 − c21 )] −k2 [c41 + x2 (4 − c21 )2 + 2xc21 (4 − c21 )] − k3 c41 − k4 c21 [c21 + x(4 − c21 ) , 2

where M = 48(1+2α)2(1−β) (1+3α)2 (1+4α) , k1 = 4(1 + 2α)(1 + 3α)2 ,k2 = 3(1 + 2α)2 (1 + 4α) , k3 = 4(1 − β)2 (1 + 6α) and k4 = 8α(1 − β). |a2 a4 − a23 | = M c41 [k1 − k2 − k3 − k4 ] + xc21 (4 − c21 )[2k1 − 2k2 − k4 ] − x2 c21 (4 − c21 )k1 −x2 (4 − c21 )2 k2 + 2yc1 k1 (1 − |x|2 )(4 − c21 ) . Suppose now that c1 = c. Since |c| = |c1 | ≤ 2, using the Lemma 2.1, we may assume without restriction that c ∈ [0, 2] and on applying triangle inequality with ρ = |x| ≤ 1, we obtain |a2 a4 − a23 | ≤ M c4 |k1 − k2 − k3 − k4 | + ρc2 (4 − c2 )|2k1 − 2k2 − k4 | +ρ2 (4 − c2 )(c2 (k1 − k2 ) − 2ck1 + 4k2 ) + 2ck1 (4 − c2 ) = M c4 |k1 − k2 − k3 − k4 | + ρc2 (4 − c2 )|2k1 − 2k2 − k4 | 2k2 +ρ2 (4 − c2 )(k1 − k2 )(c − 2)(c − ) + 2ck1 (4 − c2 ) = F (ρ). k1 − k2 Differentiating F (ρ), we get F 0 (ρ) = M [c2 (4 − c2 )|2k1 − 2k2 − k4 | + 2ρ(4 − c2 )(k1 − k2 )(c − 2)(c −

2k2 ) ≥ 0, (k1 − k2 )

since 2k2 /(k1 − k2 ) > 2 so that c − 2k2 /(k1 − k2 ) < c − 2 < 0 and k1 − k2 = 2 (1 + 2α)(36α2 + 12α + 1) > 0 as α > 0, and so (c − 2)(c − (k12k −k2 ) > 0 for all c ∈ [0, 2]. This implies that F (ρ) is an increasing function of ρ on a closed interval [0,1]. Hence F (ρ) ≤ F (1) for all ρ ∈ [0, 1]. That is, F (ρ) ≤ M c4 |k1 − k2 − k3 − k4 | + (4 − c2 )[c2 |2k1 − 2k2 − k4 | + (c2 (k1 − k2 ) + 4k2 )] = M [c4 [|k1 − k2 − k3 − k4 | − (|2k1 − 2k2 − k4 | − (k1 − k2 ))] −c2 [4k2 − 4(|2k1 − 2k2 − k4 | − 4(k1 − k2 ))] + 16k2 = G(c). G0 (c) = 0 implies c = 0 so that at c = 0, G00 (c) < 0. Therefore c = 0 is a point of maximum for G(c). Thus the upper bound of F (ρ) corresponds to ρ = 1 (1−β)2 and c = 0. Hence |a2 a4 − a23 | ≤ (1+3α) 2.


210

Corollary 3.7. Choosing α = 0, β = 0 in (17), we get |a2 a4 − a23 | ≤ 1. This result coincides with [8]. Corollary 3.8. Choosing α = 0, in (17), we get |a2 a4 − a23 | ≤ (1 − β)2 . Lemma 3.6. Let f ∈ Sαβ . Then for 1/2 ≤ β < 1, |a3 − a22 | ≤

1−β 1 + 3α

Proof. Let f ∈ Sαβ . Then by proceeding as in Lemma 3.4, we have c2 (1 − β) c21 (1 − β)2 (1 + 4α) 2 |a3 − a2 | = − 2(1 + 3α) 2(1 + 2α)2 (1 + 3α)

(19)

(20)

Substituting for c2 from Lemma 2.2 we obtain (1 − β) 1 2 c21 (1 − β)2 (1 + 4α) 2 2 |a3 − a2 | = [c + x(4 − c1 )] − 2(1 + 3α) 2 1 2(1 + 3α)(1 + 2α)2 2 (1 − β) [c1 + x(4 − c21 )](1 + 2α)2 − 2c21 (1 − β)(1 + 4α) = 2 4(1 + 2α) (1 + 3α) = M b1 [c21 + x(4 − c21 )] − b2 c21 , where M =

(1−β) 4(1+2α)2 (1+3α) , b1

= (1 + 2α)2 , b2 = 2(1 − β)(1 + 4α). Therefore

|a3 − a22 | = M b1 c21 + b1 x(4 − c21 ) − b2 c21 = M (b1 − b2 )c21 + b1 x(4 − c21 ) . Suppose now that c1 = c. Since |c| = |c1 | ≤ 2, using the Lemma 2.1, we may assume without restriction that c ∈ [0, 2] and on applying triangle inequality with ρ = |x| ≤ 1, we obtain |a3 − a22 | ≤ M [c2 |b1 − b2 | + b1 ρ(4 − c2 )] = F (ρ). Differentiating F (ρ), we get F 0 (ρ) = M b1 (4 − c2 ) > 0, implying that F (ρ) is an increasing function of ρ on a closed interval [0,1]. Hence F (ρ) ≤ F (1) for all ρ ∈ [0, 1]. That is |(a3 − a22 )| ≤ M [c2 |b1 − b2 | + b1 (4 − c2 )] = G(c). By hypothesis, β ≥ 1/2 and hence b1 − b2 = 4α2 − 4α − 1 + 2β(1 + 4α) ≥ 4α2 . Hence G(c) = M [4b1 − b2 c2 ], G0 (c) = −2b2 M c and G00 (c) = −2b2 M . Since c ∈ [0, 2], it follows that G(c) attains a maximum at c = 0. Thus the upper (1−β) bound of F (ρ) corresponds to ρ = 1 and c = 0. Hence |(a3 −a22 )| ≤ (1+3α) .


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Corollary 3.9. Choosing α = 0 in (19),we get |a3 − a22 | ≤ (1 − β). Using Lemma 2.1,the following estimates can be deduced. Remark 3.2. Let f ∈ Sαβ . By Lemma 2.1, we have c2 (1 − β) c21 (1 − β)2 , |a3 | = + 2(1 + 3α) 2(1 + 2α)(1 + 3α) (1 − β)(3 + 2α − 2β) ≤ , (1 + 2α)(1 + 3α) c3 (1 − β) c1 c2 (3 + 8α)(1 − β)2 c31 (1 − β)3 , |a4 | = + + 3(1 + 4α) 6(1 + 2α)(1 + 3α)(1 + 4α) 6(1 + 2α)(1 + 3α)(1 + 4α) (1 − β)[12 + 12α2 + 4β 2 − 16αβ − 14β + 26α] ≤ and 3(1 + 2α)(1 + 3α)(1 + 4α) c41 (1 − β)4 |a5 | = 24(1 + 2α)(1 + 3α)(1 + 4α)(1 + 5α) c22 (1 − β)2 c21 c2 (1 − β)3 (20α + 6) + + 8(1 + 3α)(1 + 5α) 24(1 + 2α)(1 + 3α)(1 + 4α)(1 + 5α) c1 c3 (1 − β)2 (4 + 14α) c4 (1 − β) + + 12(1 + 2α)(1 + 4α)(1 + 5α) 4(1 + 5α) 120 + 288α2 − 16β 3 + 744α2 + 548α − 188β (1 − β) +96β 2 − 600αβ + 144α2 β + 160αβ 2 ≤ . 24(1 + 2α)(1 + 3α)(1 + 4α)(1 + 5α) Finally, using the above results, an upper bound for |H3 (1)| , f ∈ Sαβ is immediately obtained. 3.2. Theorem 3.2. Let f ∈ Sαβ . Then for 1/2 ≤ β < 1, (1 − β)3 (3 + 2α − 2β) (1 + 2α)(1 + 3α)3 2(1 − β)2 [12 + 12α2 + 4β 2 − 16αβ − 14β + 26α] + 9(1 + 2α)(1 + 3α)(1 + 4α)2 120 + 288α2 − 16β 3 + 744α2 + 548α − 188β 2 (1 − β) +96β 2 − 600αβ + 144α2 β + 160αβ 2 + . 24(1 + 2α)(1 + 3α)2 (1 + 4α)(1 + 5α)

|H3 (1)| ≤

Remark 3.3. The determination of the sharp estimates for |H3 (1)| for funcβ tions belonging to the classes Rα and Sαβ remain to be explored.


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BOUNDS ON THE THIRD ORDER HANKEL DETERMINANT FOR CERTAIN SUBCLASSES OF ANALYTIC FUNCTIONS Daniel Breaz Department of Mathematics, ”1 Decembrie 1918” University of Alba-Iulia, Alba-Iulia, Romania email: [email protected] K.G.Subramanian Department of Mathematics, Madras Christian College, Chennai 600059 India Email:[email protected]

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