Abstract. The objective of this paper is to introduce certain subclasses of an- alytic functions and seek an upper bound to the second Hankel determinant

COEFFICIENT INEQUALITY FOR CERTAIN SUBCLASSES OF ANALYTIC FUNCTIONS

D. VAMSHEE KRISHNA ¹ and T. RAMREDDY

(Received 12 December, 2012)

Abstract. The objective of this paper is to introduce certain subclasses of an- alytic functions and seek an upper bound to the second Hankel determinant

|a

a

− a

| for the function f belonging to these classes, using Toeplitz deter- minants.

1. Introduction

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

∞

X

n=2

a n z ⁿ (1)

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

H q (n) =

a n a n+1 · · · a n+q−1

a n+1 a n+2 · · · a n+q

.. . .. . .. . .. . a n+q−1 a n+q · · · a n+2q−2

. (2)

This determinant has been considered by several authors in the literature. For example, Noor [16] determined the rate of growth of H q (n) as n → ∞ for the func- tions 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 H 2 (1). Fekete-Szeg¨ o then further generalized the esti- mate |a 3 − µa ² ₂ | with µ real and f ∈ S. Ali [3] found sharp bounds on the first four coefficients and sharp estimate for the Fekete-Szeg¨ o functional |γ ₃ − tγ ² ₂ |, where t is real, for the inverse function of f defined as f ⁻¹ (w) = w + P ∞

n=2 γ _n w ⁿ to the class of strongly starlike functions of order α(0 < α ≤ 1) denoted by f ST (α). For our

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 sugges-

tions in the preparation of this paper.

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

a ₂ a ₃

a ₃ a ₄ = a 2 a 4 − a ² ₃ . (3)

Janteng, Halim and Darus [10] have considered the functional |a 2 a 4 − a ² ₃ | 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 |a 2 a 4 − a ² ₃ | ≤ ⁴ ₉ .

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 |a 2 a 4 −a ² ₃ | ≤ 1 and |a 2 a 4 −a ² ₃ | ≤ ¹ ₈ respectively. Mishra and Gochhayat [13] have obtained the sharp bound to the non- linear functional

|a 2 a 4 −a ² ₃ | for the class of analytic functions denoted by R λ (α, ρ)(0 ≤ ρ ≤ 1, 0 ≤ λ <

1, |α| < ^π ₂ ), defined as Re n

e ^{iα Ω}

^{f (z)} _z o

> ρ cos α, using the fractional differential operator denoted by Ω ^λ _z , defined by Owa and Srivastava [17]. These authors have shown that, if f ∈ R _λ (α, ρ) then |a ₂ a ₄ − a ² ₃ | ≤ n _(1−ρ)

(2−λ)

(3−λ)

cos

α

9

o . 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 |a 2 a 4 − a ² ₃ | 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 RT _α {α > 0}, if it satisfies the condition

{f ⁰ (z)} ^α > 0, ∀z ∈ E. (4)

It is observed that for α = 1, we obtain RT ₁ = RT .

Definition 1.2. A function f (z) ∈ A is said to be in the class ST _α {α > 0}, if it satisfies the condition

 zf ⁰ (z) f (z)

 ^α

> 0, ∀z ∈ E. (5)

It is observed that for α = 1, we obtain ST ₁ = ST .

Definition 1.3. A function f (z) ∈ A is said to be in the class CV _α {α > 0}, if it satisfies the condition



1 + zf ⁰⁰ (z) f ⁰ (z)

 ^α

> 0, ∀z ∈ E. (6)

It is observed that for α = 1, we obtain CV 1 = CV . We first state some preliminary

Lemmas required for proving our results.

2. Preliminary Results

Let P denote the class of functions p analytic in E, for which Re{p(z)} > 0, p(z) = (1 + c 1 z + c 2 z ² + c 3 z ³ + ...) =

"

1 +

∞

X

n=1

c n z ⁿ

#

, ∀z ∈ E. (7)

Lemma 2.1. ([18], [19]) If p ∈ P, then |c k | ≤ 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

D n =

2 c 1 c 2 · · · c n

c ₋₁ 2 c 1 · · · c n−1

.. . .. . .. . .. . .. . c −n c −n+1 c −n+2 · · · 2

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

and c _−k = c _k , are all non-negative. These are strictly positive except for p(z) = P m

k=1 ρ k p 0 (exp(it k )z), ρ k > 0, t k real and t k 6= t j , for k 6= j; in this case D n > 0 for n < (m − 1) and D n .

= 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 c 1 > 0. On using Lemma 2.2, for n = 2 and n = 3 respectively, we get

D ₂ =

2 c ₁ c ₂ c 1 2 c 1

c 2 c 1 2

= [8 + 2Re{c ² ₁ c ₂ } − 2 | c ₂ | ² − 4c ² ₁ ] ≥ 0,

which is equivalent to

2c ₂ = {c ² ₁ + x(4 − c ² ₁ )}, for some x, |x| ≤ 1. (8)

D 3 =

2 c 1 c 2 c 3

c 1 2 c 1 c 2

c 2 c 1 2 c 1

c 3 c 2 c 1 2 .

Then D 3 ≥ 0 is equivalent to

|(4c 3 − 4c 1 c 2 + c ³ ₁ )(4 − c ² ₁ ) + c 1 (2c 2 − c ² ₁ ) ² | ≤ 2(4 − c ² ₁ ) ² − 2|(2c 2 − c ² ₁ )| ² . (9) From the relations (8) and (9), after simplifying, we get

4c 3 = {c ³ ₁ + 2c 1 (4 − c ² ₁ )x − c 1 (4 − c ² ₁ )x ² + 2(4 − c ² ₁ )(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 a n z ⁿ ∈ RT α (α ≥ 1) then

|a ₂ a ₄ − a ² ₃ | ≤ 4

9α ² .

Proof. Since f (z) = z + P ∞

n=2 a n 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

[f ⁰ (z)] ^α = [p(z)]. (11)

Replacing f ⁰ (z) and p(z) with their equivalent series expressions in (11), we have (

1 +

∞

X

n=2

na n z ⁿ⁻¹ ) ^α

= (

1 +

∞

X

n=1

c n z ⁿ )

Using the Binomial expansion in the left hand side of the above expression, upon simplification , we obtain

1 + 2αa 2 z + 3αa 3 + 2α(α − 1)a ² ₂ z ² +



4αa 4 + 6α(α − 1)a 2 a 3 + 4

3 α(α − 1)(α − 2)a ³ ₂

 z ³ + ...



= [1 + c 1 z + c 2 z ² + c 3 z ³ + ...]. (12) Equating the coefficients of like powers of z, z ² and z ³ respectively in (12), after simplifying, we get

[a ₂ = c 1

2α ; a ₃ = 1

6α ² 2αc 2 − (α − 1)c ² ₁ ; a ₄ = 1

24α ³ 6α ² c ₃ − 6α(α − 1)c ₁ c ₂ + (α − 1)(2α − 1)c ³ ₁ ]. (13) Substituting the values of a ₂ , a ₃ and a ₄ from (13) in the second Hankel determinant

|a 2 a 4 − a ² ₃ | for the function f ∈ RT α , upon simplification, we obtain

|a 2 a 4 − a ² ₃ | = 1 144α ⁴ ×

 18α ² c 1 c 3 + 2α(1 − α)c ² ₁ c 2 − 16α ² c ² ₂ − (1 − α)(1 + 2α)c ⁴ ₁   . (14) Substituting the values of c 2 and c 3 from (8) and (10) respectively from Lemma 2.2 in the right hand side of (14), we have

 18α ² c ₁ c ₃ + 2α(1 − α)c ² ₁ c ₂ − 16α ² c ² ₂ − (1 − α)(1 + 2α)c ⁴ ₁   = 

18α ² c 1 × 1

4 {c ³ ₁ + 2c 1 (4 − c ² ₁ )x − c 1 (4 − c ² ₁ )x ² + 2(4 − c ² ₁ )(1 − |x| ² )z}

+2α(1 − α)c ² ₁ × {c ² ₁ + x(4 − c ² ₁ )} − 16α ² × 1

4 {c ² ₁ + x(4 − c ² ₁ )} ²

−(1 − α)(1 + 2α)c ⁴ ₁   . 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 

18α ² c ₁ c ₃ + 2α(1 − α)c ² ₁ c ₂ − 16α ² c ² ₂ − (1 − α)(1 + 2α)c ⁴ ₁   ≤

|(3α ² − 2)c ⁴ ₁ + 18α ² c ₁ (4 − c ² ₁ ) + 2αc ² ₁ (4 − c ² ₁ )|x| − α ² (c ₁ + 2)(c ₁ + 16)(4 − c ² ₁ )|x| ² |.

(15)

Since c 1 ∈ [0, 2], using the result (c 1 + a)(c 1 + b) ≥ (c 1 − a)(c 1 − b), where a, b ≥ 0 in the right hand side of (15), upon simplification, we obtain

2 

18α ² c 1 c 3 + 2α(1 − α)c ² ₁ c 2 − 16α ² c ² ₂ − (1 − α)(1 + 2α)c ⁴ ₁   ≤

|(3α ² − 2)c ⁴ ₁ + 18α ² c 1 (4 − c ² ₁ ) + 2αc ² ₁ (4 − c ² ₁ )|x| − α ² (c 1 − 2)(c 1 − 16)(4 − c ² ₁ )|x| ² |.

(16) Choosing c 1 = c ∈ [0, 2], applying Triangle inequality and replacing | x | by µ in the right hand side of (16), we get

2 

18α ² c 1 c 3 + 2α(1 − α)c ² ₁ c 2 − 16α ² c ² ₂ − (1 − α)(1 + 2α)c ⁴ ₁   ≤

[(3α ² − 2)c ⁴ + 18α ² c(4 − c ² ) + 2αc ² (4 − c ² )µ + α ² (c − 2)(c − 16)(4 − c ² )µ ² ]

= F (c, µ)(say), with 0 ≤ µ = |x| ≤ 1, (17) where

F (c, µ) = [(3α ² − 2)c ⁴ + 18α ² c(4 − c ² )+

2αc ² (4 − c ² )µ + α ² (c − 2)(c − 16)(4 − c ² )µ ² ]. (18) We next maximize the function F (c, µ) on the closed square [0, 2] × [0, 1]. Differen- tiating F (c, µ) in (18) partially with respect to µ, we get

∂F

∂µ = 2α c ² + α(c − 2)(c − 16)µ × (4 − c ² ). (19) 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

0≤µ≤1 max F (c, µ) = F (c, 1) = G(c)(say). (20) From the relations (18) and (20), upon simplification, we obtain

G(c) = 2 (α ² − α − 1)c ⁴ + 2α(2 − 7α)c ² + 64α ² . (21) G ⁰ (c) = 8 (α ² − α − 1)c ³ + α(2 − 7α)c . (22) From the expression (22), we observe that G ⁰ (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

0≤c≤2 G(c) = G max = G(c) = 128α ² . (23) From the expressions (17) and (23), after simplifying, we get

 18α ² c 1 c 3 + 2α(1 − α)c ² ₁ c 2 − 16α ² c ² ₂ − (1 − α)(1 + 2α)c ⁴ ₁ 

 ≤ 64α ² . (24) From the expressions (14) and (24), upon simplification, we obtain

|a ₂ a ₄ − a ² ₃ | ≤ 4

9α ² . (25)

This completes the proof of our Theorem 3.1.

Remark. For the choice of α = 1, we get RT 1 = RT , for which, from (25), we get

|a 2 a ₄ − a ² ₃ | ≤ ⁴ ₉ . This inequality is sharp and coincides with that of Janteng, Halim

and Darus [10].

Theorem 3.2. If f (z) ∈ ST α (α ≥ 1), then |a 2 a 4 − a ² ₃ | ≤ _α ¹

. Proof. Since f (z) = z + P ∞

n=2 a n z ⁿ ∈ ST α , 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

 zf ⁰ (z) f (z)

 ^α

= [p(z)]. (26)

Replacing f (z), f ⁰ (z) and p(z) with their equivalent series expressions in (26), we have

"

z 1 + P ^∞ _n=2 na _n z ⁿ⁻¹ {z + P ∞

n=2 a n z ⁿ }

# α

=

"(

1 +

∞

X

n=1

c _n z ⁿ )#

. (27)

Using the Binomial expansion in the left hand side of the relation (27), upon sim- plification , we obtain



1 + a ₂ αz +



2a ₃ α +  α ² − 3α 2

 a ² ₂

 z ² +



3a 4 α + α(2α − 5)a 2 a 3 +  α ³ − 9α ² + 14α 6

 a ³ ₂



z ³ + ...]

= 1 + c 1 z + c 2 z ² + c 3 z ³ + ... . (28) Equating the coefficients of like powers of z, z ² and z ³ respectively in (28), after simplifying, we get

[a ₂ = c ₁

α ; a ₃ = 1

4α ² 2αc 2 + (3 − α)c ² ₁ ; a ₄ = 1

36α ³ 12α ² c ₃ + 6α(5 − 2α)c ₁ c ₂ + (4α ² − 15α + 17)c ³ ₁ ]. (29) Substituting the values of a 2 , a 3 and a 4 from (29) in the second Hankel deter- minant |a 2 a 4 − a ² ₃ | for the function f ∈ ST α , we have

|a 2 a 4 − a ² ₃ | =    

c 1

α × 1

36α ³ 12α ² c 3 + 6α(5 − 2α)c 1 c 2 + (4α ² − 15α + 17)c ³ ₁

− 1

16α ⁴ 2αc 2 + (3 − α)c ² ₁ 2     . Upon simplification, we obtain

|a 2 a 4 − a ² ₃ | = 1 144α ⁴ ×

 48α ² c 1 c 3 + 12α(1 − α)c ² ₁ c 2 − 36α ² c ² ₂ + (7α ² − 6α − 13)c ⁴ ₁   . The above expression is equivalent to

|a 2 a ₄ − a ² ₃ | = 1 144α ⁴ × 

d ₁ c ₁ c ₃ + d ₂ c ² ₁ c ₂ + d ₃ c ² ₂ + d ₄ c ⁴ ₁ 

 , (30)

where

{d ₁ = 48α ² ; d ₂ = 12α(1 − α); d ₃ = −36α ² ; d ₄ = (7α ² − 6α − 13))}. (31)

Substituting the values of c 2 and c 3 from (8) and (10) respectively from Lemma 2.2 in the right hand side of (30), we have

|d 1 c ₁ c ₃ + d ₂ c ² ₁ c ₂ + d ₃ c ² ₂ + d ₄ c ⁴ ₁ |

= |d ₁ c ₁ × 1

4 {c ³ ₁ + 2c ₁ (4 − c ² ₁ )x − c ₁ (4 − c ² ₁ )x ² + 2(4 − c ² ₁ )(1 − |x| ² )z}+

d ₂ c ² ₁ × 1

2 {c ² ₁ + x(4 − c ² ₁ )} + d ₃ × 1

4 {c ² ₁ + x(4 − c ² ₁ )} ² + d ₄ c ⁴ ₁ |.

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|d 1 c 1 c 3 + d 2 c ² ₁ c 2 + d 3 c ² ₂ + d 4 c ⁴ ₁ | ≤ |(d 1 + 2d 2 + d 3 + 4d 4 )c ⁴ ₁ + 2d 1 c 1 (4 − c ² ₁ ) + 2(d 1 + d 2 + d 3 )c ² ₁ (4 − c ² ₁ )|x|−

(d 1 + d 3 )c ² ₁ + 2d 1 c 1 − 4d 3 (4 − c ² ₁ )|x| ² |. (32) Using the values of d 1 , d 2 , d 3 and d 4 from the relation (31), upon simplification, we obtain

{(d 1 + 2d 2 + d 3 + 4d 4 ) = 4(4α ² − 13);

d 1 = 48α ² ; (d 1 + d 2 + d 3 ) = 12α}. (33)

(d 1 + d 3 )c ² ₁ + 2d 1 c 1 − 4d 3

= 12α ² c ² ₁ + 8c 1 + 12
= 12α ² (c 1 + 2)(c 1 + 6). (34) Since c ₁ ∈ [0, 2], using the result (c ₁ + a)(c ₁ + b) ≥ (c ₁ − a)(c ₁ − b), where a, b ≥ 0 in the right hand side of (34), upon simplification, we obtain

− (d 1 + d 3 )c ² ₁ + 2d 1 c 1 − 4d 3 ≤ −12α ² (c 1 − 2)(c 1 − 6). (35) Substituting the calculated values from (33) and (35) in the right hand side of (32), after simplifying, we obtain

|d 1 c 1 c 3 + d 2 c ² ₁ c 2 + d 3 c ² ₂ + d 4 c ⁴ ₁ | ≤ |(4α ² − 13)c ⁴ ₁ + 24α ² c 1 (4 − c ² ₁ )+

6αc ² ₁ (4 − c ² ₁ )|x| − 3α ² (c 1 − 2)(c 1 − 6)(4 − c ² ₁ )|x| ² |. (36) Choosing c 1 = c ∈ [0, 2], applying Triangle inequality and replacing | x | by µ in the right hand side of (36), we get

|d 1 c 1 c 3 + d 2 c ² ₁ c 2 + d 3 c ² ₂ + d 4 c ⁴ ₁ | ≤ [(13 − 4α ² )c ⁴ + 24α ² c(4 − c ² )+

6αc ² (4 − c ² )µ + 3α ² (c − 2)(c − 6)(4 − c ² )µ ² ].

= F (c, µ)(say), with 0 ≤ µ = |x| ≤ 1 (37) where

F (c, µ) = [(13 − 4α ² )c ⁴ + 24α ² c(4 − c ² )+

6αc ² (4 − c ² )µ + 3α ² (c − 2)(c − 6)(4 − c ² )µ ² ]. (38)

We next maximize the function F (c, µ) on the closed square [0, 2] × [0, 1]. Differen- tiating F (c, µ) in (38) partially with respect to µ, we get

∂F

∂µ = 3α 2c ² + α(c − 2)(c − 6)µ × (4 − c ² ). (39) 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

0≤µ≤1 F (c, µ) = F (c, 1) = G(c)(say). (40) From the relations (38) and (40), upon simplification, we obtain

G(c) = (1 − α)(7α + 13)c ⁴ + 24α(1 − α)c ² + 144α ² . (41) G ⁰ (c) = 4(1 − α)c (7α + 13)c ² + 12α . (42) From the expression (42), we observe that G ⁰ (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

0≤c≤2 G(0) = 144α ² . (43)

From the expressions (37) and (43), after simplifying, we get

|d 1 c ₁ c ₃ + d ₂ c ² ₁ c ₂ + d ₃ c ² ₂ + d ₄ c ⁴ ₁ | ≤ 144α ² . (44) From the expressions (30) and (44), upon simplification, we obtain

|a 2 a 4 − a ² ₃ | ≤ 1

α ² . (45)

This completes the proof of our Theorem 3.2.

Remark. Choosing α = 1, we get ST ₁ = ST , for which, from (45), we get

|a ₂ a ₄ − a ² ₃ | ≤ 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

|a 2 a 4 − a ² ₃ | ≤  (1 + α)(17 + α) 72α ² (1 + 3α)

 . Proof. Since f (z) = z + P ∞

n=2 a n 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



1 + zf ⁰⁰ (z) f ⁰ (z)

 ^α

= [p(z)] (46)

Replacing f ⁰ (z), f ⁰⁰ (z) and p(z) with their equivalent series expressions in (46), we have

"

1 + z P ∞

n=2 n(n − 1)a _n z ⁿ⁻² {1 + P ∞

n=2 na n z ⁿ⁻¹ }

# ^α

=

"(

1 +

∞

X

n=1

c _n z ⁿ )#

. (47)

Applying the Binomial expansion in the left hand side of the above expression (47), upon simplification, we obtain

1 + 2αa 2 z + 2α(3a 3 − 2a ² ₂ ) + 2α(α − 1)a ² ₂ z ² +



2α(6a 4 − 9a 2 a 3 + 4a ³ ₂ ) + 4α(α − 1)a 2 (3a 3 − 2a ² ₂ ) + 4α(α − 1)(α − 2)

3 a ³ ₂

 z ³ + ...]

= 1 + c 1 z + c 2 z ² + c 3 z ³ + ... . (48) Equating the coefficients of like powers of z, z ² and z ³ respectively in (48), after simplifying, we get

[a ₂ = c 1

2α ; a ₃ = 1

12α ² 2αc 2 + (3 − α)c ² ₁ ; a 4 = 1

144α ³ 12α ² c 3 + 6α(5 − 2α)c 1 c 2 + (4α ² − 15α + 17)c ³ ₁ ] (49) Substituting the values of a 2 , a 3 and a 4 from (49) in the second Hankel functional

|a 2 a 4 − a ² ₃ | for the function f ∈ CV α , upon simplification, we obtain

|a 2 a 4 − a ² ₃ | = 1 288α ⁴ ×

 12α ² c 1 c 3 + 2α(3 − 2α)c ² ₁ c 2 − 8α ² c ² ₂ + (2α ² − 3α − 1)c ⁴ ₁   . (50) The above expression is equivalent to

|a 2 a 4 − a ² ₃ | = 1 288α ⁴ × 

d 1 c 1 c 3 + d 2 c ² ₁ c 2 + d 3 c ² ₂ + d 4 c ⁴ ₁ 

 , (51)

where

{d 1 = 12α ² ; d 2 = 2α(3 − 2α); d 3 = −8α ² ; d 4 = (2α ² − 3α − 1)}. (52) Applying the same procedure as described in Theorem 3.2, we get

4|d 1 c 1 c 3 + d 2 c ² ₁ c 2 + d 3 c ² ₂ + d 4 c ⁴ ₁ | ≤ |(d 1 + 2d 2 + d 3 + 4d 4 )c ⁴ ₁ + 2d 1 c 1 (4 − c ² ₁ ) + 2(d 1 + d 2 + d 3 )c ² ₁ (4 − c ² ₁ )|x|−

(d 1 + d 3 )c ² ₁ + 2d 1 c 1 − 4d 3 (4 − c ² ₁ )|x| ² |. (53) Using the values of d ₁ , d ₂ , d ₃ and d ₄ from the relation (52), upon simplification, we obtain

{(d ₁ + 2d ₂ + d ₃ + 4d ₄ ) = 4(α ² − 1);

d ₁ = 12α ² ; (d ₁ + d ₂ + d ₃ ) = 6α}. (54)

(d 1 + d ₃ )c ² ₁ + 2d ₁ c ₁ − 4d 3

= 4α ² c ² ₁ + 6c ₁ + 8
= 4α ² (c ₁ + 2)(c ₁ + 4). (55) Applying the same procedure as described in Theorem 3.2, we get

− (d 1 + d ₃ )c ² ₁ + 2d ₁ c ₁ − 4d 3 ≤ −4α ² (c ₁ − 2)(c 1 − 4). (56)

Substituting the calculated values from (54) and (56) in the right hand side of (53), after simplifying, we obtain

|d 1 c 1 c 3 + d 2 c ² ₁ c 2 + d 3 c ² ₂ + d 4 c ⁴ ₁ | ≤ |(α ² − 1)c ⁴ ₁ + 6α ² c 1 (4 − c ² ₁ )+

3αc ² ₁ (4 − c ² ₁ )|x| − α ² (c 1 − 2)(c 1 − 4)(4 − c ² ₁ )|x| ² |. (57) Choosing c ₁ = c ∈ [0, 2], applying Triangle inequality and replacing | x | by µ in the right hand side of (57), we get

|d ₁ c ₁ c ₃ + d ₂ c ² ₁ c ₂ + d ₃ c ² ₂ + d ₄ c ⁴ ₁ | ≤ [(α ² − 1)c ⁴ + 6α ² c(4 − c ² )+

3αc ² (4 − c ² )µ + α ² (c − 2)(c − 4)(4 − c ² )µ ² ].

= F (c, µ)(say), with 0 ≤ µ = |x| ≤ 1, (58) where

F (c, µ) = [(α ² − 1)c ⁴ + 6α ² c(4 − c ² )

+ 3αc ² (4 − c ² )µ + α ² (c − 2)(c − 4)(4 − c ² )µ ² ]. (59) We next maximize the function F (c, µ) on the closed square [0, 2] × [0, 1]. Differen- tiating F (c, µ) in (59) partially with respect to µ, we get

∂F

∂µ = 3αc ² + 2α ² (c − 2)(c − 4)µ × (4 − c ² ). (60) 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

0≤µ≤1 max F (c, µ) = F (c, 1) = G(c)(say). (61) In view of the expression (61), replacing µ by 1 in (59), after simplifying, we get

G(c) = −4(1 + 3α)c ⁴ + 16α(3 − α)c ² + 128α ² . (62) G ⁰ (c) = −16(1 + 3α)c ³ + 32α(3 − α)c . (63) G ⁰⁰ (c) = −48(1 + 3α)c ² + 32α(3 − α) . (64) For Optimum value of G(c), consider G ⁰ (c) = 0. From (63), we get

−16c (1 + 3α)c ² − 2α(3 − α) = 0. (65) We now discuss the following cases.

Case 1) If c = 0, then, from (64), we obtain

G ⁰⁰ (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 c ² =  2α(3 − α)

(1 + 3α)



=  20 9 − 2α

3 − 20

9(1 + 3α)



. (66)

Using the value of c ² given in (66) in (64), after simplifying, we obtain

G ⁰⁰ (c) = − {64α(3 − α)} < 0, for 1 ≤ α < 3.

By the second derivative test, G(c) has maximum value at c, where c ² given in (66).

Using the value of c ² given by (66) in (62), upon simplification, we obtain max

0≤c≤2 G(c) = G _max =  4α ² (1 + α)(17 + α) (1 + 3α)



. (67)

Considering, the maximum value of G(c) at c, where c ² is given by (66) , from (58) and (67), we get

 d 1 c 1 c 3 + d 2 c ² ₁ c 2 + d 3 c ² ₂ + d 4 c ⁴ ₁ 

 ≤  4α ² (1 + α)(17 + α) (1 + 3α)



. (68)

From the expressions (51) and (68), upon simplification, we obtain

|a 2 a 4 − a ² ₃ | ≤  (1 + α)(17 + α) 72α ² (1 + 3α)



. (69)

This completes the proof of our Theorem 3.3.

Remark. Choosing α = 1, we have CV 1 = CV , for which, from (69), we get

|a 2 a 4 − a ² ₃ | ≤ ¹ ₈ . This inequality is sharp and coincides with that of Janteng, Halim and Darus [9].

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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]