COEFFICIENT INEQUALITY FOR CERTAIN SUBCLASSES OF ...

NEW ZEALAND JOURNAL OF MATHEMATICS Volume 42 (2012), 217-228

COEFFICIENT INEQUALITY FOR CERTAIN SUBCLASSES OF ANALYTIC FUNCTIONS D. VAMSHEE KRISHNA1 and T. RAMREDDY (Received 12 December, 2012)

Abstract. The objective of this paper is to introduce certain subclasses of analytic functions and seek an upper bound to the second Hankel determinant |a2 a4 − a23 | for the function f belonging to these classes, using Toeplitz determinants.

1. Introduction Let A denote the class of functions f of the form f (z) = z +

∞ X

an z n

(1)

n=2

in the open unit disc E = {z : |z| < 1}. Let S be the subclass of A, consisting of univalent functions. In 1976, Noonan and Thomas [15] defined the q th Hankel determinant of f for q ≥ 1 and n ≥ 1 as an an+1 · · · an+q−1 an+1 an+2 · · · an+q Hq (n) = . (2) .. .. .. .. . . . . an+q−1

an+q

···

an+2q−2

This determinant has been considered by several authors in the literature. For example, Noor [16] determined the rate of growth of Hq (n) as n → ∞ for the functions in S with bounded boundary. Ehrenborg [5] studied the Hankel determinant of exponential polynomials. The Hankel transform of an integer sequence and some of its properties were discussed by Layman in [11]. One can easily observe that the Fekete-Szeg¨ o functional is H2 (1). Fekete-Szeg¨o then further generalized the estimate |a3 − µa22 | with µ real and f ∈ S. Ali [3] found sharp bounds on the first four 2 coefficients and sharp estimate for the Fekete-Szeg¨o functional P∞|γ3 − tγn2 |, where t is −1 real, for the inverse function of f defined as f (w) = w + n=2 γn w to the class f (α). For our of strongly starlike functions of order α(0 < α ≤ 1) denoted by ST 2010 Mathematics Subject Classification 30C45; 30C50. Key words and phrases: Analytic function, upper bound, second Hankel determinant, positive real function, Toeplitz determinants. 1: Corresponding author. The authors would like to thank the esteemed referee for his/her valuable comments and suggestions in the preparation of this paper.

218

D. VAMSHEE KRISHNA and T. RAMREDDY

discussion in this paper, we consider the Hankel determinant in the case of q = 2 and n = 2, known as the second Hankel determinant a2 a3

a3 = a2 a4 − a23 . a4

(3)

Janteng, Halim and Darus [10] have considered the functional |a2 a4 − a23 | and found a sharp bound for the function f in the subclass RT of S, consisting of functions whose derivative has a positive real part studied by Mac Gregor [12]. In their work, they have shown that if f ∈ RT then |a2 a4 − a23 | ≤ 94 . These authors [9] also obtained the second Hankel determinant and sharp bounds for the familiar subclasses of S, namely, starlike and convex functions denoted by ST and CV and shown that |a2 a4 −a23 | ≤ 1 and |a2 a4 −a23 | ≤ 81 respectively. Mishra and Gochhayat [13] have obtained the sharp bound to the non- linear functional |a2 a4 −a23 | for the class of analytic functions denoted by Rλ (α, ρ)(0 ≤ ρ ≤ 1, 0 ≤ λ < n o 1, |α|
ρ cos α, using the fractional differential

Ωλz ,

operator denoted by defined by Owa and nSrivastava [17]. These authors have o 2 2 2 2 shown that, if f ∈ Rλ (α, ρ) then |a2 a4 − a23 | ≤ (1−ρ) (2−λ)9(3−λ) cos α . Similarly, the same coefficient inequality was calculated for certain subclasses of analytic functions by many authors ([1], [4], [14]). Motivated by the above mentioned results obtained by different authors in this direction, in this paper, we introduce certain subclasses of analytic functions and obtain an upper bound to the functional |a2 a4 − a23 | for the function f belonging to these classes, defined as follows. Definition 1.1. A function f (z) ∈ A is said to be in the class RTα {α > 0}, if it satisfies the condition α

{f 0 (z)} > 0,

∀z ∈ E.

(4)

It is observed that for α = 1, we obtain RT1 = RT . Definition 1.2. A function f (z) ∈ A is said to be in the class STα {α > 0}, if it satisfies the condition  0 α zf (z) > 0, ∀z ∈ E. (5) f (z) It is observed that for α = 1, we obtain ST1 = ST . Definition 1.3. A function f (z) ∈ A is said to be in the class CVα {α > 0}, if it satisfies the condition  α zf 00 (z) 1+ 0 > 0, ∀z ∈ E. (6) f (z) It is observed that for α = 1, we obtain CV1 = CV . We first state some preliminary Lemmas required for proving our results.

COEFFICIENT INEQUALITY FOR CERTAIN SUBCLASSES

219

2. Preliminary Results Let P denote the class of functions p analytic in E, for which Re{p(z)} > 0, " # ∞ X 2 3 n p(z) = (1 + c1 z + c2 z + c3 z + ...) = 1 + cn z , ∀z ∈ E. (7) n=1

Lemma 2.1. ([18], [19]) If p ∈ P, then |ck | ≤ 2, for each k ≥ 1. Lemma 2.2. ([7]) The power series for p given in (7) converges in the unit disc E to a function in P if and only if the Toeplitz determinants 2 c−1 Dn = . .. c−n

c1 2 .. .

c2 c1 .. .

··· ··· .. .

cn cn−1 .. .

c−n+1

c−n+2

···

2

, n = 1, 2, 3....

and Pm c−k = ck , are all non-negative. These are strictly positive except for p(z) = k=1 ρk p0 (exp(itk )z), ρk . > 0, tk real and tk 6= tj , for k 6= j; in this case Dn > 0 for n < (m − 1) and Dn = 0 for n ≥ m. This necessary and sufficient condition is due to Caratheodory and Toeplitz can be found in [7]. We may assume without restriction that c1 > 0. On using Lemma 2.2, for n = 2 and n = 3 respectively, we get 2 D2 = c1 c2

c1 2 c1

c2 c1 = [8 + 2Re{c21 c2 } − 2 | c2 |2 − 4c21 ] ≥ 0, 2

which is equivalent to 2c2 = {c21 + x(4 − c21 )}, 2 c1 D3 = c2 c3

for some c1 2 c1 c2

c2 c1 2 c1

x,

|x| ≤ 1.

(8)

c3 c2 . c1 2

Then D3 ≥ 0 is equivalent to |(4c3 − 4c1 c2 + c31 )(4 − c21 ) + c1 (2c2 − c21 )2 | ≤ 2(4 − c21 )2 − 2|(2c2 − c21 )|2 .

(9)

From the relations (8) and (9), after simplifying, we get 4c3 = {c31 + 2c1 (4 − c21 )x − c1 (4 − c21 )x2 + 2(4 − c21 )(1 − |x|2 )z} for some real value of z,

with |z| ≤ 1. (10)

3. Main Results Theorem 3.1. If f (z) = z +

P∞

n=2

an z n ∈ RT α (α ≥ 1) then

|a2 a4 − a23 | ≤

4 . 9α2

220

D. VAMSHEE KRISHNA and T. RAMREDDY

P∞ Proof. Since f (z) = z + n=2 an z n ∈ RT α , from the Definition 1.1, there exists an analytic function p ∈ P in the unit disc E with p(0) = 1 and Re{p(z)} > 0 such that α [f 0 (z)] = [p(z)]. (11) Replacing f 0 (z) and p(z) with their equivalent series expressions in (11), we have )α ( ) ( ∞ ∞ X X = 1+ 1+ nan z n−1 cn z n n=2

n=1

Using the Binomial expansion in the left hand side of the above expression, upon simplification , we obtain   1 + 2αa2 z + 3αa3 + 2α(α − 1)a22 z 2 +    4 3 3 4αa4 + 6α(α − 1)a2 a3 + α(α − 1)(α − 2)a2 z + ... 3 = [1 + c1 z + c2 z 2 + c3 z 3 + ...]. (12) Equating the coefficients of like powers of z, z 2 and z 3 respectively in (12), after simplifying, we get [a2 =

1  c1 ; a3 = 2αc2 − (α − 1)c21 ; 2 2α 6α 1  2 a4 = 6α c3 − 6α(α − 1)c1 c2 + (α − 1)(2α − 1)c31 ]. (13) 3 24α

Substituting the values of a2 , a3 and a4 from (13) in the second Hankel determinant |a2 a4 − a23 | for the function f ∈ RT α , upon simplification, we obtain |a2 a4 − a23 | =

1 × 4 144α 18α2 c1 c3 + 2α(1 − α)c21 c2 − 16α2 c22 − (1 − α)(1 + 2α)c41 . (14)

Substituting the values of c2 and c3 from (8) and (10) respectively from Lemma 2.2 in the right hand side of (14), we have 18α2 c1 c3 + 2α(1 − α)c21 c2 − 16α2 c22 − (1 − α)(1 + 2α)c41 = 18α2 c1 × 1 {c31 + 2c1 (4 − c21 )x − c1 (4 − c21 )x2 + 2(4 − c21 )(1 − |x|2 )z} 4 1 +2α(1 − α)c21 × {c21 + x(4 − c21 )} − 16α2 × {c21 + x(4 − c21 )}2 4 −(1 − α)(1 + 2α)c41 . Using the facts that |z| < 1 and |xa + yb| ≤ |x||a| + |y||b|, where x, y, a and b are real numbers, after simplifying, we get 2 18α2 c1 c3 + 2α(1 − α)c21 c2 − 16α2 c22 − (1 − α)(1 + 2α)c41 ≤ |(3α2 − 2)c41 + 18α2 c1 (4 − c21 ) + 2αc21 (4 − c21 )|x| − α2 (c1 + 2)(c1 + 16)(4 − c21 )|x|2 |. (15)

COEFFICIENT INEQUALITY FOR CERTAIN SUBCLASSES

221

Since c1 ∈ [0, 2], using the result (c1 + a)(c1 + b) ≥ (c1 − a)(c1 − b), where a, b ≥ 0 in the right hand side of (15), upon simplification, we obtain 2 18α2 c1 c3 + 2α(1 − α)c21 c2 − 16α2 c22 − (1 − α)(1 + 2α)c41 ≤ |(3α2 − 2)c41 + 18α2 c1 (4 − c21 ) + 2αc21 (4 − c21 )|x| − α2 (c1 − 2)(c1 − 16)(4 − c21 )|x|2 |. (16) Choosing c1 = c ∈ [0, 2], applying Triangle inequality and replacing | x | by µ in the right hand side of (16), we get 2 18α2 c1 c3 + 2α(1 − α)c21 c2 − 16α2 c22 − (1 − α)(1 + 2α)c41 ≤ [(3α2 − 2)c4 + 18α2 c(4 − c2 ) + 2αc2 (4 − c2 )µ + α2 (c − 2)(c − 16)(4 − c2 )µ2 ] = F (c, µ)(say),

with

0 ≤ µ = |x| ≤ 1, (17)

where F (c, µ) = [(3α2 − 2)c4 + 18α2 c(4 − c2 )+ 2αc2 (4 − c2 )µ + α2 (c − 2)(c − 16)(4 − c2 )µ2 ]. (18) We next maximize the function F (c, µ) on the closed square [0, 2] × [0, 1]. Differentiating F (c, µ) in (18) partially with respect to µ, we get   ∂F = 2α c2 + α(c − 2)(c − 16)µ × (4 − c2 ). (19) ∂µ For 0 < µ < 1, for fixed c ∈ (0, 2) and for fixed α with (α ≥ 1), from (19), we observe that ∂F ∂µ > 0. Consequently, F (c, µ) is an increasing function of µ and hence it cannot have a maximum value at any point in the interior of the closed square [0, 2] × [0, 1]. Moreover, for fixed c ∈ [0, 2], we have max F (c, µ) = F (c, 1) = G(c)(say).

0≤µ≤1

(20)

From the relations (18) and (20), upon simplification, we obtain  G(c) = 2 (α2 − α − 1)c4 + 2α(2 − 7α)c2 + 64α2 . (21)  2 0 3 G (c) = 8 (α − α − 1)c + α(2 − 7α)c . (22) From the expression (22), we observe that G0 (c) ≤ 0 for all values of c ∈ [0, 2] and α with with (α ≥ 1). Therefore, G(c) is a monotonically decreasing function of c in the interval [0, 2] so that its maximum value occurs at c = 0. From (21), we obtain max G(c) = Gmax = G(c) = 128α2 .

0≤c≤2

From the expressions (17) and (23), after simplifying, we get 18α2 c1 c3 + 2α(1 − α)c21 c2 − 16α2 c22 − (1 − α)(1 + 2α)c41 ≤ 64α2 . From the expressions (14) and (24), upon simplification, we obtain 4 . |a2 a4 − a23 | ≤ 9α2 This completes the proof of our Theorem 3.1.

(23)

(24)

(25)

Remark. For the choice of α = 1, we get RT 1 = RT , for which, from (25), we get |a2 a4 − a23 | ≤ 49 . This inequality is sharp and coincides with that of Janteng, Halim

222

D. VAMSHEE KRISHNA and T. RAMREDDY

and Darus [10]. Theorem 3.2. If f (z) ∈ STα (α ≥ 1), then |a2 a4 − a23 | ≤

1 α2 .

P∞ Proof. Since f (z) = z + n=2 an z n ∈ STα , from the Definition 1.2, there exists an analytic function p ∈ P in the unit disc E with p(0) = 1 and Re{p(z)} > 0 such that  0 α zf (z) = [p(z)]. (26) f (z) Replacing f (z), f 0 (z) and p(z) with their equivalent series expressions in (26), we have )# "  #α "( P∞ ∞ X 1 + n=2 nan z n−1 n P∞ z = 1+ cn z . (27) {z + n=2 an z n } n=1 Using the Binomial expansion in the left hand side of the relation (27), upon simplification , we obtain    2   α − 3α 1 + a2 αz + 2a3 α + a22 z 2 + 2   3   α − 9α2 + 14α 3a4 α + α(2α − 5)a2 a3 + a32 z 3 + ...] 6   = 1 + c1 z + c2 z 2 + c3 z 3 + ... . (28) Equating the coefficients of like powers of z, z 2 and z 3 respectively in (28), after simplifying, we get c1 1  ; a3 = 2αc2 + (3 − α)c21 ; α 4α2 1  a4 = 12α2 c3 + 6α(5 − 2α)c1 c2 + (4α2 − 15α + 17)c31 ]. (29) 36α3 Substituting the values of a2 , a3 and a4 from (29) in the second Hankel determinant |a2 a4 − a23 | for the function f ∈ STα , we have c1 1  2 12α2 c3 + 6α(5 − 2α)c1 c2 + (4α2 − 15α + 17)c31 |a2 a4 − a3 | = × 3 α 36α 1  2 2 − 2αc2 + (3 − α)c1 . 16α4 [a2 =

Upon simplification, we obtain |a2 a4 − a23 | =

1 × 144α 4 48α2 c1 c3 + 12α(1 − α)c21 c2 − 36α2 c22 + (7α2 − 6α − 13)c41 .

The above expression is equivalent to 1 |a2 a4 − a23 | = × d1 c1 c3 + d2 c21 c2 + d3 c22 + d4 c41 , 4 144α where

(30)

{d1 = 48α2 ; d2 = 12α(1 − α); d3 = −36α2 ; d4 = (7α2 − 6α − 13))}. (31)

COEFFICIENT INEQUALITY FOR CERTAIN SUBCLASSES

223

Substituting the values of c2 and c3 from (8) and (10) respectively from Lemma 2.2 in the right hand side of (30), we have |d1 c1 c3 + d2 c21 c2 + d3 c22 + d4 c41 | 1 = |d1 c1 × {c31 + 2c1 (4 − c21 )x − c1 (4 − c21 )x2 + 2(4 − c21 )(1 − |x|2 )z}+ 4 1 1 d2 c21 × {c21 + x(4 − c21 )} + d3 × {c21 + x(4 − c21 )}2 + d4 c41 |. 2 4 Using the facts that |z| < 1 and |xa + yb| ≤ |x||a| + |y||b|, where x, y, a and b are real numbers, after simplifying, we get 4|d1 c1 c3 + d2 c21 c2 + d3 c22 + d4 c41 | ≤ |(d1 + 2d2 + d3 + 4d4 )c41 + 2d1 c1 (4 − c21 ) + 2(d1 + d2 + d3 )c21 (4 − c21 )|x|−  (d1 + d3 )c21 + 2d1 c1 − 4d3 (4 − c21 )|x|2 |. (32) Using the values of d1 , d2 , d3 and d4 from the relation (31), upon simplification, we obtain {(d1 + 2d2 + d3 + 4d4 ) = 4(4α2 − 13); d1 = 48α2 ; (d1 + d2 + d3 ) = 12α}. (33) 

(d1 + d3 )c21 + 2d1 c1 − 4d3


= 12α2 c21 + 8c1 + 12 = 12α2 (c1 + 2)(c1 + 6). (34)

Since c1 ∈ [0, 2], using the result (c1 + a)(c1 + b) ≥ (c1 − a)(c1 − b), where a, b ≥ 0 in the right hand side of (34), upon simplification, we obtain  − (d1 + d3 )c21 + 2d1 c1 − 4d3 ≤ −12α2 (c1 − 2)(c1 − 6). (35) Substituting the calculated values from (33) and (35) in the right hand side of (32), after simplifying, we obtain |d1 c1 c3 + d2 c21 c2 + d3 c22 + d4 c41 | ≤ |(4α2 − 13)c41 + 24α2 c1 (4 − c21 )+ 6αc21 (4 − c21 )|x| − 3α2 (c1 − 2)(c1 − 6)(4 − c21 )|x|2 |. (36) Choosing c1 = c ∈ [0, 2], applying Triangle inequality and replacing | x | by µ in the right hand side of (36), we get |d1 c1 c3 + d2 c21 c2 + d3 c22 + d4 c41 | ≤ [(13 − 4α2 )c4 + 24α2 c(4 − c2 )+ 6αc2 (4 − c2 )µ + 3α2 (c − 2)(c − 6)(4 − c2 )µ2 ]. = F (c, µ)(say),

with

0 ≤ µ = |x| ≤ 1

(37)

where F (c, µ) = [(13 − 4α2 )c4 + 24α2 c(4 − c2 )+ 6αc2 (4 − c2 )µ + 3α2 (c − 2)(c − 6)(4 − c2 )µ2 ]. (38)

224

D. VAMSHEE KRISHNA and T. RAMREDDY

We next maximize the function F (c, µ) on the closed square [0, 2] × [0, 1]. Differentiating F (c, µ) in (38) partially with respect to µ, we get   ∂F = 3α 2c2 + α(c − 2)(c − 6)µ × (4 − c2 ). (39) ∂µ For 0 < µ < 1, for fixed c with 0 < c < 2 and for fixed α with (α ≥ 1), from (39), we observe that ∂F ∂µ > 0. Consequently, F (c, µ) is an increasing function of µ and hence it cannot have a maximum value at any point in the interior of the closed square [0, 2] × [0, 1]. Moreover, for fixed c ∈ [0, 2], we have max F (c, µ) = F (c, 1) = G(c)(say).

0≤µ≤1

From the relations (38) and (40), upon simplification, we obtain  G(c) = (1 − α)(7α + 13)c4 + 24α(1 − α)c2 + 144α2 .  G0 (c) = 4(1 − α)c (7α + 13)c2 + 12α .

(40)

(41) (42)

0

From the expression (42), we observe that G (c) ≤ 0 for all values of 0 ≤ c ≤ 2 and α with (α ≥ 1). Therefore, G(c) is a monotonically decreasing function of c in the interval [0, 2] so that its maximum value occurs at c = 0. From (41), we obtain max G(0) = 144α2 .

0≤c≤2

(43)

From the expressions (37) and (43), after simplifying, we get |d1 c1 c3 + d2 c21 c2 + d3 c22 + d4 c41 | ≤ 144α2 .

(44)

From the expressions (30) and (44), upon simplification, we obtain |a2 a4 − a23 | ≤

1 . α2

(45)

This completes the proof of our Theorem 3.2. Remark. Choosing α = 1, we get ST1 = ST , for which, from (45), we get |a2 a4 − a23 | ≤ 1. This inequality is sharp and coincides with that of Janteng, Halim and Darus [9]. Theorem 3.3. If f (z) ∈ CV α (1 ≤ α < 3), then   (1 + α)(17 + α) 2 |a2 a4 − a3 | ≤ . 72α2 (1 + 3α) P∞ Proof. Since f (z) = z + n=2 an z n ∈ CV α , from the Definition 1.3, there exists an analytic function p ∈ P in the unit disc E with p(0) = 1 and Re{p(z)} > 0 such that  α zf 00 (z) 1+ 0 = [p(z)] (46) f (z) Replacing f 0 (z), f 00 (z) and p(z) with their equivalent series expressions in (46), we have " )# P∞ #α "( ∞ n−2 X z n=2 n(n − 1)an z n P∞ 1+ = 1+ cn z . (47) {1 + n=2 nan z n−1 } n=1

COEFFICIENT INEQUALITY FOR CERTAIN SUBCLASSES

225

Applying the Binomial expansion in the left hand side of the above expression (47), upon simplification, we obtain   1 + 2αa2 z + 2α(3a3 − 2a22 ) + 2α(α − 1)a22 z 2 +   4α(α − 1)(α − 2) 3 3 3 2 a2 z + ...] 2α(6a4 − 9a2 a3 + 4a2 ) + 4α(α − 1)a2 (3a3 − 2a2 ) + 3   = 1 + c1 z + c2 z 2 + c3 z 3 + ... . (48) Equating the coefficients of like powers of z, z 2 and z 3 respectively in (48), after simplifying, we get [a2 =

1  c1 ; a3 = 2αc2 + (3 − α)c21 ; 2 2α 12α 1  a4 = 12α2 c3 + 6α(5 − 2α)c1 c2 + (4α2 − 15α + 17)c31 ] 3 144α

(49)

Substituting the values of a2 , a3 and a4 from (49) in the second Hankel functional |a2 a4 − a23 | for the function f ∈ CV α , upon simplification, we obtain |a2 a4 − a23 | =

1 × 4 288α 12α2 c1 c3 + 2α(3 − 2α)c21 c2 − 8α2 c22 + (2α2 − 3α − 1)c41 . (50)

The above expression is equivalent to 1 |a2 a4 − a23 | = × d1 c1 c3 + d2 c21 c2 + d3 c22 + d4 c41 , 288α4 where

(51)

{d1 = 12α2 ; d2 = 2α(3 − 2α); d3 = −8α2 ; d4 = (2α2 − 3α − 1)}. (52) Applying the same procedure as described in Theorem 3.2, we get 4|d1 c1 c3 + d2 c21 c2 + d3 c22 + d4 c41 | ≤ |(d1 + 2d2 + d3 + 4d4 )c41 + 2d1 c1 (4 − c21 ) + 2(d1 + d2 + d3 )c21 (4 − c21 )|x|−  (d1 + d3 )c21 + 2d1 c1 − 4d3 (4 − c21 )|x|2 |. (53) Using the values of d1 , d2 , d3 and d4 from the relation (52), upon simplification, we obtain {(d1 + 2d2 + d3 + 4d4 ) = 4(α2 − 1); d1 = 12α2 ; (d1 + d2 + d3 ) = 6α}. (54) 

(d1 + d3 )c21 + 2d1 c1 − 4d3


= 4α2 c21 + 6c1 + 8 = 4α2 (c1 + 2)(c1 + 4). (55)

Applying the same procedure as described in Theorem 3.2, we get  − (d1 + d3 )c21 + 2d1 c1 − 4d3 ≤ −4α2 (c1 − 2)(c1 − 4).

(56)

226

D. VAMSHEE KRISHNA and T. RAMREDDY

Substituting the calculated values from (54) and (56) in the right hand side of (53), after simplifying, we obtain |d1 c1 c3 + d2 c21 c2 + d3 c22 + d4 c41 | ≤ |(α2 − 1)c41 + 6α2 c1 (4 − c21 )+ 3αc21 (4 − c21 )|x| − α2 (c1 − 2)(c1 − 4)(4 − c21 )|x|2 |. (57) Choosing c1 = c ∈ [0, 2], applying Triangle inequality and replacing | x | by µ in the right hand side of (57), we get |d1 c1 c3 + d2 c21 c2 + d3 c22 + d4 c41 | ≤ [(α2 − 1)c4 + 6α2 c(4 − c2 )+ 3αc2 (4 − c2 )µ + α2 (c − 2)(c − 4)(4 − c2 )µ2 ]. = F (c, µ)(say),

with

0 ≤ µ = |x| ≤ 1, (58)

where F (c, µ) = [(α2 − 1)c4 + 6α2 c(4 − c2 ) + 3αc2 (4 − c2 )µ + α2 (c − 2)(c − 4)(4 − c2 )µ2 ]. (59) We next maximize the function F (c, µ) on the closed square [0, 2] × [0, 1]. Differentiating F (c, µ) in (59) partially with respect to µ, we get   ∂F = 3αc2 + 2α2 (c − 2)(c − 4)µ × (4 − c2 ). (60) ∂µ For 0 < µ < 1, for fixed c with 0 < c < 2 and for fixed α with (1 ≤ α < 3), from (60), we observe that ∂F ∂µ > 0. Consequently, F (c, µ) is an increasing function of µ and hence it cannot have a maximum value at any point in the interior of the closed square [0, 2] × [0, 1]. Further, for fixed c ∈ [0, 2], we have max F (c, µ) = F (c, 1) = G(c)(say).

0≤µ≤1

(61)

In view of the expression (61), replacing µ by 1 in (59), after simplifying, we get  G(c) = −4(1 + 3α)c4 + 16α(3 − α)c2 + 128α2 . (62)  0 3 G (c) = −16(1 + 3α)c + 32α(3 − α)c . (63)  G00 (c) = −48(1 + 3α)c2 + 32α(3 − α) . (64) For Optimum value of G(c), consider G0 (c) = 0. From (63), we get  −16c (1 + 3α)c2 − 2α(3 − α) = 0. (65) We now discuss the following cases. Case 1) If c = 0, then, from (64), we obtain G00 (c) = {32α(3 − α)} > 0,

for

1 ≤ α < 3.

From the second derivative test, G(c) has minimum value at c = 0. Case 2) If c 6= 0, then, from (65), we get     20 2α 20 2α(3 − α) 2 = − − . c = (1 + 3α) 9 3 9(1 + 3α) Using the value of c2 given in (66) in (64), after simplifying, we obtain G00 (c) = − {64α(3 − α)} < 0,

for

1 ≤ α < 3.

(66)

COEFFICIENT INEQUALITY FOR CERTAIN SUBCLASSES

227

By the second derivative test, G(c) has maximum value at c, where c2 given in (66). Using the value of c2 given by (66) in (62), upon simplification, we obtain  2  4α (1 + α)(17 + α) max G(c) = Gmax = . (67) 0≤c≤2 (1 + 3α) Considering, the maximum value of G(c) at c, where c2 is given by (66) , from (58) and (67), we get  2  d1 c1 c3 + d2 c21 c2 + d3 c22 + d4 c41 ≤ 4α (1 + α)(17 + α) . (68) (1 + 3α) From the expressions (51) and (68), upon simplification, we obtain   (1 + α)(17 + α) |a2 a4 − a23 | ≤ . 72α2 (1 + 3α)

(69)

This completes the proof of our Theorem 3.3. Remark. Choosing α = 1, we have CV1 = CV , for which, from (69), we get |a2 a4 − a23 | ≤ 18 . This inequality is sharp and coincides with that of Janteng, Halim and Darus [9]. References [1] A. Abubaker and M. Darus, Hankel determinant for a class of analytic functions involving a generalized linear differential operator, Int. J. Pure Appl.Math., 69 (4) (2011), 429-435. [2] J. W. Alexander, Functions which map the interior of the unit circle upon simple regions, Annl. of Math., 17 (1915), 12-22. [3] R.M Ali, Coefficients of the inverse of strongly starlike functions, Bull. Malays. Math. Sci. Soc., (second series) 26 (1) (2003), 63-71. [4] O. Al-Refai and M. Darus, Second Hankel determinant for a class of analytic functions defined by a fractional operator, European J. Sci. Res., 28 (2) (2009), 234-241. [5] R. Ehrenborg, The Hankel determinant of exponential polynomials, Amer. Math. Monthly, 107 (6) (2000), 557-560. [6] A.W. Goodman, Univalent Functions Vol.I and Vol.II, Mariner publishing Comp. Inc., Tampa, Florida, 1983. [7] U. Grenander and G. Szeg¨ o, Toeplitz Forms and their Application, Second edition: Chelsea publishing Co., New York, 1984. [8] A. Janteng, S. A. Halim and M. Darus, Estimate on the second Hankel functional for functions whose derivative has a positive real part, J. Qual. Meas. Anal. (JQMA), 4 (1) (2008), 189-195. [9] A. Janteng, S. A. Halim and M. Darus, Hankel determinant for starlike and convex functions, Int. J. Math. Anal., 1 (13) (2007), 619-625. [10] A. Janteng, S. A. Halim and M. Darus, Coefficient inequality for a function whose derivative has a positive real part, J. Inequal. Pure Appl. Math., 7 (2) (2006), 1-5. [11] J. W. Layman, The Hankel transform and some of its properties, J. Integer Seq., 4 (1) (2001), 1-11.

228

D. VAMSHEE KRISHNA and T. RAMREDDY

[12] T. H. Mac Gregor, Functions whose derivative have a positive real part, Trans. Amer. Math. Soc., 104 (3) (1962), 532-537. [13] A. K. Mishra and P. Gochhayat, Second Hankel determinant for a class of analytic functions defined by fractional derivative, Int. J. Math. Math. Sci., 2008 (2008), Article ID 153280, 1-10. [14] G. Murugusundaramoorthy and N. Magesh, Coefficient inequalities for certain classes of analytic functions associated with Hankel determinant, Bull Math Anal. Appl., 1 (3) (2009), 85-89. [15] J. W. Noonan and D. K. Thomas, On the second Hankel determinant of a really mean p-Valent functions, Trans. Amer. Math. Soc., 223 (2) (1976), 337-346. [16] K. I. Noor, Hankel determinant problem for the class of functions with bounded boundary rotation, Rev. Roum. Math. Pures Et Appl., 28 (8) (1983), 731-739. [17] S. Owa and H. M. Srivastava, Univalent and starlike generalised hypergeometric functions, Canad. J. Math., 39 (5) (1987), 1057-1077. [18] C. Pommerenke, Univalent Functions, Vandenhoeck and Ruprecht, Gottingen (1975). [19] B. Simon, Orthogonal polynomials on the unit circle, part 1. classical theory, American Mathematical Society Colloquium Publications, 54, part 1, American Mathematical Society, Providence, RI, 2005.

D. Vamshee Krishna, Assistant Professor, Department of Engineering Mathematics, Gitam Institute of Technology, Rushikonda, GITAM University, Visakhapatnam-530 045, Andhra Pradesh, India. [email protected]

T. Ramreddy, Professor in Mathematics, Department of Mathematics, Kakatiya University, Vidyaranyapuri- 506 009, Warangal Dt., Andhra Pradesh, India. [email protected]