COEFFICIENT INEQUALITY FOR CERTAIN SUBCLASSES OF ANALYTIC FUNCTIONS
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1 NEW ZEALAND JOURNAL OF MATHEMATICS Volume 42 (2012), 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 analytic functions and seek an upper bound to the second Hankel determinant a 2 a 4 a 2 3 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 + 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 a n a n+1 a n+q 1 a n+1 a n+2 a n+q H q (n) =. (2).... a n+q 1 a n+q a n+2q 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 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-Szegö functional is H 2 (1). Fekete-Szegö then further generalized the estimate 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-Szegö functional γ 3 tγ2, 2 where t is real, for the inverse function of f defined as f 1 (w) = w + n=2 γ nw n to the class of strongly starlike functions of order α(0 < α 1) denoted by ST (α). For our n= 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.
2 218 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 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 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 and a 2 a 4 a 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 e iα Ω λ z f(z) > ρ cos α, using the fractional differential z 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 (1 ρ) 2 (2 λ) 2 (3 λ) 2 cos 2 α 9. 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 (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 α (z) > 0, z E. (5) f(z) 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 α (z) f > 0, z E. (6) (z) It is observed that for α = 1, we obtain CV 1 = CV. We first state some preliminary Lemmas required for proving our results.
3 COEFFICIENT INEQUALITY FOR CERTAIN SUBCLASSES Preliminary Results Let P denote the class of functions p analytic in E, for which Rep(z) > 0, [ ] p(z) = (1 + c 1 z + c 2 z 2 + c 3 z ) = 1 + c n z n, z E. (7) n=1 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 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) = m k=1 ρ kp 0 (exp(it k )z), ρ k > 0, t k real and t k t j, for k 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 = which is equivalent to 2 c 1 c 2 c 1 2 c 1 c 2 c 1 2 Then D 3 0 is equivalent to = [8 + 2Rec 2 1c 2 2 c 2 2 4c 2 1] 0, 2c 2 = c 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 (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 c 1 (4 c 2 1)x c 1 (4 c 2 1)x 2 + 2(4 c 2 1)(1 x 2 )z 3. Main Results. for some real value of z, with z 1. (10) Theorem 3.1. If f(z) = z + n=2 a nz n RT α (α 1) then a 2 a 4 a α 2.
4 220 D. VAMSHEE KRISHNA and T. RAMREDDY Proof. Since f(z) = z + n=2 a nz 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 Rep(z) > 0 such that [f (z)] α = [p(z)]. (11) Replacing f (z) and p(z) with their equivalent series expressions in (11), we have α 1 + na n z n 1 = 1 + c n z n n=2 Using the Binomial expansion in the left hand side of the above expression, upon simplification, we obtain [ 1 + 2αa2 z + 3αa 3 + 2α(α 1)a 2 2 z 2 + 4αa 4 + 6α(α 1)a 2 a ] 3 α(α 1)(α 2)a3 2 z n=1 = [1 + c 1 z + c 2 z 2 + c 3 z ]. (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 2αc2 6α 2 (α 1)c 2 1 ; a 4 = 1 6α 2 24α 3 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 1c 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 1c 2 16α 2 c 2 2 (1 α)(1 + 2α)c 4 1 = 18α2 c c c 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 x(4 c 2 1) 16α c2 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 1c 2 16α 2 c 2 2 (1 α)(1 + 2α)c 4 1 (3α 2 2)c α 2 c 1 (4 c 2 1) + 2αc 2 1(4 c 2 1) x α 2 (c 1 + 2)(c )(4 c 2 1) x 2. (15)
5 COEFFICIENT INEQUALITY FOR CERTAIN SUBCLASSES 221 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 1c 2 16α 2 c 2 2 (1 α)(1 + 2α)c 4 1 (3α 2 2)c α 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 1c 2 16α 2 c 2 2 (1 α)(1 + 2α)c 4 1 where [(3α 2 2)c α 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) = [(3α 2 2)c α 2 c(4 c 2 )+ = F (c, µ)(say), with 0 µ = x 1, (17) 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]. Differentiating 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 max F (c, µ) = F (c, 1) = G(c)(say). (20) 0 µ 1 From the relations (18) and (20), upon simplification, we obtain G(c) = 2 (α 2 α 1)c 4 + 2α(2 7α)c α 2. (21) G (c) = 8 (α 2 α 1)c 3 + α(2 7α)c. (22) From the expression (22), we observe that G (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) = G max = G(c) = 128α 2. (23) 0 c 2 From the expressions (17) and (23), after simplifying, we get 18α 2 c 1 c 3 + 2α(1 α)c 2 1c 2 16α 2 c 2 2 (1 α)(1 + 2α)c α 2. (24) From the expressions (14) and (24), upon simplification, we obtain a 2 a 4 a α 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 This inequality is sharp and coincides with that of Janteng, Halim
6 222 D. VAMSHEE KRISHNA and T. RAMREDDY and Darus [10]. Theorem 3.2. If f(z) ST α (α 1), then a 2 a 4 a α 2. Proof. Since f(z) = z + n=2 a nz 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 Rep(z) > 0 such that zf α (z) = [p(z)]. (26) f(z) Replacing f(z), f (z) and p(z) with their equivalent series expressions in (26), we have [ 1 + n=2 na nz n 1 ] α [ = z z + n=2 a nz n n=1 Using the Binomial expansion in the left hand side of the relation (27), upon simplification, we obtain [ 1 + a 2 αz + 2a 3 α + ( α 2 3α 2 3a 4 α + α(2α 5)a 2 a 3 + ) a 2 2 z ( α 3 9α α 6 c n z n ] ) a 3 2 z ]. (27) = [ 1 + c 1 z + c 2 z 2 + c 3 z ]. (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 2αc2 4α 2 + (3 α)c 2 1 ; a 4 = 1 12α 2 36α 3 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 determinant a 2 a 4 a 2 3 for the function f ST α, we have a 2 a 4 a 2 3 = Upon simplification, we obtain c 1 α 1 12α 2 36α 3 c 3 + 6α(5 2α)c 1 c 2 + (4α 2 15α + 17)c αc2 16α 4 + (3 α)c a 2 a 4 a 2 3 = 1 144α 4 48α 2 c 1 c α(1 α)c 2 1c 2 36α 2 c (7α 2 6α 13)c 4 1. The above expression is equivalent to a 2 a 4 a 2 3 = 1 144α 4 d1 c 1 c 3 + d 2 c 2 1c 2 + d 3 c 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)
7 COEFFICIENT INEQUALITY FOR CERTAIN SUBCLASSES 223 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 1c 2 + d 3 c d 4 c 4 1 = d 1 c c c 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 c2 1 + x(4 c 2 1) + d c2 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 1c 2 + d 3 c d 4 c 4 1 (d 1 + 2d 2 + d 3 + 4d 4 )c d 1 c 1 (4 c 2 1) + 2(d 1 + d 2 + d 3 )c 2 1(4 c 2 1) x (d1 + d 3 )c d 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); (d1 + d 3 )c d 1 c 1 4d 3 d 1 = 48α 2 ; (d 1 + d 2 + d 3 ) = 12α. (33) = 12α 2 ( c c ) = 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 d 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 1c 2 + d 3 c d 4 c 4 1 (4α 2 13)c α 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 1c 2 + d 3 c d 4 c 4 1 [(13 4α 2 )c α 2 c(4 c 2 )+ where 6αc 2 (4 c 2 )µ + 3α 2 (c 2)(c 6)(4 c 2 )µ 2 ]. F (c, µ) = [(13 4α 2 )c α 2 c(4 c 2 )+ = F (c, µ)(say), with 0 µ = x 1 (37) 6αc 2 (4 c 2 )µ + 3α 2 (c 2)(c 6)(4 c 2 )µ 2 ]. (38)
8 224 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 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 F (c, µ) = F (c, 1) = G(c)(say). (40) 0 µ 1 From the relations (38) and (40), upon simplification, we obtain G(c) = (1 α)(7α + 13)c α(1 α)c α 2. (41) G (c) = 4(1 α)c (7α + 13)c α. (42) 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) = 0 c 2 144α2. (43) From the expressions (37) and (43), after simplifying, we get d 1 c 1 c 3 + d 2 c 2 1c 2 + d 3 c d 4 c α 2. (44) From the expressions (30) and (44), upon simplification, we obtain This completes the proof of our Theorem 3.2. a 2 a 4 a α 2. (45) Remark. Choosing α = 1, we get ST 1 = ST, for which, from (45), we get a 2 a 4 a 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 + n=2 a nz 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 Rep(z) > 0 such that [ 1 + zf ] α (z) f = [p(z)] (46) (z) Replacing f (z), f (z) and p(z) with their equivalent series expressions in (46), we have [ 1 + z n=2 n(n 1)a nz n 2 ] α [ ] 1 + n=2 na = 1 + c nz n 1 n z n. (47) ]. n=1
9 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(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) a 3 2 z ] 3 = [ 1 + c 1 z + c 2 z 2 + c 3 z ]. (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 2αc2 12α 2 + (3 α)c 2 1 ; a 4 = 1 12α 2 144α 3 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 1c 2 8α 2 c (2α 2 3α 1)c 4 1. (50) The above expression is equivalent to where a 2 a 4 a 2 3 = 1 288α 4 d 1 c 1 c 3 + d 2 c 2 1c 2 + d 3 c d 4 c 4 1, (51) 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 1c 2 + d 3 c d 4 c 4 1 (d 1 + 2d 2 + d 3 + 4d 4 )c d 1 c 1 (4 c 2 1) + 2(d 1 + d 2 + d 3 )c 2 1(4 c 2 1) x (d1 + d 3 )c d 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) (d1 + d 3 )c d 1 c 1 4d 3 = 4α 2 ( c c ) = 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 d 1 c 1 4d 3 4α 2 (c 1 2)(c 1 4). (56)
10 226 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 d 1 c 1 c 3 + d 2 c 2 1c 2 + d 3 c d 4 c 4 1 (α 2 1)c α 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 1c 2 + d 3 c d 4 c 4 1 [(α 2 1)c 4 + 6α 2 c(4 c 2 )+ where 3αc 2 (4 c 2 )µ + α 2 (c 2)(c 4)(4 c 2 )µ 2 ]. F (c, µ) = [(α 2 1)c 4 + 6α 2 c(4 c 2 ) = F (c, µ)(say), with 0 µ = x 1, (58) + 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]. Differentiating 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 max F (c, µ) = F (c, 1) = G(c)(say). (61) 0 µ 1 In view of the expression (61), replacing µ by 1 in (59), after simplifying, we get G(c) = 4(1 + 3α)c α(3 α)c α 2. (62) G (c) = 16(1 + 3α)c α(3 α)c. (63) G (c) = 48(1 + 3α)c α(3 α). (64) For Optimum value of G(c), consider G (c) = 0. From (63), we get We now discuss the following cases. Case 1) If c = 0, then, from (64), we obtain 16c (1 + 3α)c 2 2α(3 α) = 0. (65) G (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 0, then, from (65), we get [ 2α(3 α) 20 c 2 = = (1 + 3α) 9 2α ] (66) 9(1 + 3α) Using the value of c 2 given in (66) in (64), after simplifying, we obtain G (c) = 64α(3 α) < 0, for 1 α < 3.
11 COEFFICIENT INEQUALITY FOR CERTAIN SUBCLASSES 227 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 [ 4α 2 ] max G(c) = G (1 + α)(17 + α) max =. (67) 0 c 2 (1 + 3α) 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 1c 2 + d 3 c d 4 c 4 4α 2 ] (1 + α)(17 + α) 1. (68) (1 + 3α) From the expressions (51) and (68), upon simplification, we obtain [ ] (1 + α)(17 + α) a 2 a 4 a α 2. (69) (1 + 3α) 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 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), [2] J. W. Alexander, Functions which map the interior of the unit circle upon simple regions, Annl. of Math., 17 (1915), [3] R.M Ali, Coefficients of the inverse of strongly starlike functions, Bull. Malays. Math. Sci. Soc., (second series) 26 (1) (2003), [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), [5] R. Ehrenborg, The Hankel determinant of exponential polynomials, Amer. Math. Monthly, 107 (6) (2000), [6] A.W. Goodman, Univalent Functions Vol.I and Vol.II, Mariner publishing Comp. Inc., Tampa, Florida, [7] U. Grenander and G. Szegö, Toeplitz Forms and their Application, Second edition: Chelsea publishing Co., New York, [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), [9] A. Janteng, S. A. Halim and M. Darus, Hankel determinant for starlike and convex functions, Int. J. Math. Anal., 1 (13) (2007), [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.
12 228 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), [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 , [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), [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), [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), [17] S. Owa and H. M. Srivastava, Univalent and starlike generalised hypergeometric functions, Canad. J. Math., 39 (5) (1987), [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, D. Vamshee Krishna, Assistant Professor, Department of Engineering Mathematics, Gitam Institute of Technology, Rushikonda, GITAM University, Visakhapatnam , Andhra Pradesh, India. vamsheekrishna1972@gmail.com T. Ramreddy, Professor in Mathematics, Department of Mathematics, Kakatiya University, Vidyaranyapuri , Warangal Dt., Andhra Pradesh, India. reddytr2@yahoo.com
COEFFICIENT INEQUALITY FOR CERTAIN SUBCLASS OF ANALYTIC FUNCTIONS
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