On a Subclass of k-uniformly Convex Functions with Negative Coefficients

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1 International Mathematical Forum, 1, 2006, no. 34, On a Subclass of k-uniformly Convex Functions with Negative Coefficients T. N. SHANMUGAM Department of Mathematics Anna University, Chennai Tamilnadu, India S. SIVASUBRAMANIAN Department of Mathematics Easwari Engineering College, Chennai Tamilnadu, India M. DARUS School of Mathematical Sciences Faculty of Science and Technology, UKM, Bangi Malaysia Abstract Certain classes of analytic functions are defined which will generate new, as well as k-uniformly convex functions. We provide necessary and sufficient conditions, distortion bounds, extreme points and radii of convexity for this class Mathematics Subject Classification: Primary 30C45 Keywords: Analytic functions, Hadamard product, k-uniformly convex functions. 1 Introduction and Definitions Let A be the class of all analytic functions f(z) =z + a n z n defined on the open unit disk Δ = {z : z < 1}. Let S denote the subclass of A consisting of functions that are univalent in Δ. Let S (α) and K(α) be the classes of

2 1678 T.N.Shanmugam, S.Sivasubramanian, and M.Darus functions respectively starlike of order α and convex of order α, Let T be the subclass of S, consisting of functions 0 α<1. f(z) =z a n z n, a n 0. (1) A function f(z) T is called a function with negative coefficients. The class T was introduced and studied by Silverman [4]. In [4], Silverman investigated the subclasses of T denoted by T (α), and C(α), 0 α<1, that are respectively starlike of order α and convex of order α. In this present paper, we study the following class of function: Definition 1.1 For 0 λ 1, 0 β<1and k 0, we let U(k, λ, β), consist of functions f T satisfying the condition { λz 3 f (z)+(1+2λ)z 2 f (z)+zf } (z) R λz 2 f (z)+zf (z) λz 3 f (z)+(1+2λ)z 2 f (z)+zf (z) >k λz 2 f (z)+zf (z) 1 + β. (2) The family U(k, λ, β) is of special interest for it contains many well known, as well as new, classes of analytic univalent functions. In particular U(k, 0, 0) is the class of k uniformly convex function introduced and studied by Kanas and Wisniowska [3] and U(0, 0,β):= C(β) is the class of uniformly convex functions of order β studied by Silverman [4]. We remark that the class of k uniformly convex functions is an extension of uniformly convex functions introduced by Goodman [2]. In this paper we provide necessary and sufficient conditions, coefficient bounds, extreme points, radius of starlikeness and convexity, closure theorem for functions in U(k, λ, β). 2 Characterization We employ the technique adopted by Aqlan et al. [1] to find the coefficient estimates for our class. Theorem 2.1 f U(k, λ, β) if and only if n [(n β)(λn λ +1)+k(λ + n 1)] a n (3) where 0 λ 1, 0 β<1, k 0.

3 On a Subclass of k-uniformly convex functions 1679 Proof. We have f U(k, λ, β) if and only if the condition (2) is satisfied. Upon the fact that R(ω) >k ω 1 + β R { ω(1 + ke iθ ) ke iθ} >β, π θ<π. Equation (2) may be written as { λz 3 f (z)+(1+2λ)z 2 f (z)+zf } (z) R (1 + ke iθ ) ke iθ >β λz 2 f (z)+zf (z) or equivalently { λz 3 f (z)+(1+2λ)z 2 f (z)+zf (z)(1 + ke iθ ) ke iθ (λz 2 f (z)+zf } (z)) R >β. λz 2 f (z)+zf (z) (4) Now, we let A(z) =λz 3 f (z)+(1+2λ)z 2 f (z)+zf (z)(1+ke iθ ) ke ( iθ λz 2 f (z)+zf (z) ) and let B(z) =λz 2 f (z)+zf (z). Then (4) is equivalent to A(z) +(1 β)b(z) > A(z) (1 + β)b(z) for 0 β<1. For A(z) and B(z) as above, we have A(z)+(1 β)b(z) = (2 β)z n(n +1 β)[λ(n 1) + 1] a n z n and similarly ke iθ (2 β) z n(n +1 β)[λ(n 1) + 1] a n z n k n(λ + n 1)a n z n n(λ + n 1)a n z n A(z) (1 + β)b(z) <β z + n(n )[λ(n 1) + 1] a n z n +k n(λ + n 1)a n z n. Therefore, A(z) +(1 β)b(z) A(z) (1 + β)b(z)

4 1680 T.N.Shanmugam, S.Sivasubramanian, and M.Darus or 2()z 2 n(n β)[λ(n 1) + 1] a n z n 2k n(λ + n 1)a n z n 0 () {n(n β)[λ(n 1)+1]+nk(λ + n 1)} a n which yields (3). On the other hand, we must have { λz 3 f (z)+(1+2λ)z 2 f (z)+zf } (z) R (1 + ke iθ ) ke iθ >β, π θ<π. λz 2 f (z)+zf (z) Upon choosing the values of z on the positive real axis where 0 z = r<1, the above inequality reduces to { () R n(n β)[λ(n 1) + 1] a nr n 1 ke iθ n(λ + n 1)a nr n 1 } 1 0 n [1 + λ(n 1)] rn 1 Since R( e iθ ) e iθ = 1, the above inequality reduces to { () R n(n β)[λ(n 1) + 1] a nr n 1 k n(λ + n 1)a nr n 1 } 1 0. n [1 + λ(n 1)] rn 1 Letting r 1, we get the desired result. By taking λ =0, k = 0 in Theorem 2.1, we get Corollary 2.2 Let f T. Then f C(β), if and only if n(n β)a n. The above result was obtained by Silverman [4]. Theorem 2.3 If f U(k, λ, β), then a n n [(n β)(λn λ +1)+k(λ + n 1)] (5) Equality holds for the function f(z) =z (0 λ 1, 0 β<1, k 0). n [(n β)(λn λ +1)+k(λ + n 1)] zn. (6)

5 On a Subclass of k-uniformly convex functions 1681 Proof. Since f U(k, λ, β), (3) holds. Since n [(n β)(λn λ +1)+k(λ + n 1)] a n we have, a n n [(n β)(λn λ +1)+k(λ + n 1)]. Clearly the function given by (6) satisfies (5) and therefore f(z) given by (6) is in U(k, λ, β). For this function, the result is clearly sharp. By taking k =0, and λ = 0, in Theorem 2.3, we get Corollary 2.4 If f C(β), then Equality holds for the function a n f(z) =z n(n β). n(n β) zn. 3 Distortion and Covering theorems for U(k, λ, β) Theorem 3.1 If f U(k, λ, β), then () () r 2(1 + λ)(2 + k β) r2 f(z) r + 2(1 + λ)(2 + k β) r2, z = r<1. (7) Equality holds for the function () f(z) =z 2(1 + λ)(2 + k β) z2. Proof. We only prove the right hand side inequality in (7), since the other inequality can be justified using similar arguments. Since f U(k, λ, β), by Theorem 2.1, n [(n β)(λn λ +1)+k(λ + n 1)] a n. Now, 2(1 + λ)(2 + k β) a n = 2(1 + λ)(2 + k β)a n n [(n β)(λn λ +1)+k(λ + n 1)] a n

6 1682 T.N.Shanmugam, S.Sivasubramanian, and M.Darus and therefore a n Since, f(z) =z a n z n, 2(1 + λ)(2 + k β). (8) f(z) = z a n z n z + z 2 a n z n 2 r + r 2 = r + a n () 2(1 + λ)(2 + k β) r2, which yields the right hand side inequality of (7). Theorem 3.2 If f U(k, λ, β), then 1 () (1 + λ)(2 + k β) r f () (z) 1+ r, z = r<1. (9) (1 + λ)(2 + k β) Equality holds for the function Proof. We have f(z) =z () (1 + λ)(2 + k β) z2. f (z) 1+ na n z n 1 1+r na n (10) Since, f U(k, λ, β), we have (2 β)(1 + λ)(1 + k) na n n [(n β)(λn λ +1)+k(λ + n 1)] a n. Hence, () na n (2 β)(1 + λ)(1 + k). (11) Substituting (11) in (10), we get f (z) 1+ () r. (12) (2 + k β)(1 + λ)

7 On a Subclass of k-uniformly convex functions 1683 Similarly, This completes the proof. f (z) 1 () (2 + k β)(1 + λ) r. Theorem 3.3 If f U(k, λ, β), then f T (δ) where δ =1 The result is sharp, with being the extremal. () (2 + 2k β)(λ +1)+(1 β). f(z) =z 2(1 + λ)(2 + k β) z2 Proof. It is sufficient to show that (3) implies that is, n δ 1 δ Since, (14) is equivalent to (n δ)a n 1 δ, (13) n(n β)(λn λ +1)+nk(n λ +1), n 2. (14) δ (n 1)() 1 n(n β)(λn λ +1)+nk(λ + n 1) () = Ψ(n) n 2 and Ψ(n) Ψ(2), (14) holds true for any n 2, 0 λ 1, 0 β< 1, k 0. This completes the proof of Theorem 3.3. If we take λ = 0 and k =0, in Theorem 3.3, then we have the following result of Silverman [4]. Corollary 3.4 If f C(β), then f T ( 2 ). The result is sharp, with 3 β being the extremal. f(z) =z () 2(2 β) z2

8 1684 T.N.Shanmugam, S.Sivasubramanian, and M.Darus 4 Extreme points of the class U(k, λ, β) Theorem 4.1 Let f 1 (z) =z and f n (z) =z n [(n β)(λn λ +1)+k(λ + n 1)] zn, n =2, 3, 4... Then f U(k, λ, β), if and only if it can be represented in the form f(z) = λ n f n (z), (15) n=1 where λ n 0 and n=1 λ n =1. Proof. Suppose f(z) can be expressed as in (15). Then f(z) = } λ n {z n=1 n [(n β)(λn λ +1)+k(λ + n 1)] zn, = z λ n n [(n β)(λn λ +1)+k(λ + n 1)] zn Now, λ n n [(n β)(λn λ +1)+k(λ + n 1)] Thus f U(k, λ, β). Conversely, suppose f U(k, λ, β). then a n and therefore we may set λ n = n [(n β)(λn λ +1)+k(λ + n 1)] = λ n =1 λ 1 1, n =2, 3,... n [(n β)(λn λ +1)+k(λ + n 1)] n [(n β)(λn λ +1)+k(λ + n 1)] a n, n =2, 3,... and λ 1 =1 λ n. Then f(z) = λ n f n (z), n=1 and hence the proof is complete.

9 On a Subclass of k-uniformly convex functions 1685 and Corollary 4.2 The extreme points of U(k, λ, β) are the functions f n (z) =z f 1 (z) =z n [(n β)(λn λ +1)+k(λ + n 1)] zn, n =2, 3, 4... If we take k = 0 and λ = 0 in Corollary 4.2, we have the following result by Silverman [4]. Corollary 4.3 The extreme points of f C(β) are the functions f 1 (z) =z and f n (z) =z n(n β) zn, n =2, 3, Radius of Convexity Theorem 5.1 If f U(k, λ, β), then f is convex of order γ in z <R, where R = inf n 2 { [(n β)(λn λ +1)+k(λ + n 1)] Proof. By a computation, we have zf f = Thus f is convex of order γ if Since f U(k, λ, β), we have n(n 1)a n z n 1 1 na n z n 1 n(n 1)a n z n 1 1. na n z n 1 } 1 n 1 1 γ. (16) (n γ) n(n γ) 1 γ a n z n 1 1. (17) [ ] (n β)(λn λ +1)+k(n 1+λ) n a n 1. (18)

10 1686 T.N.Shanmugam, S.Sivasubramanian, and M.Darus Now, (17) holds if n(n γ) 1 γ a n z n 1 n [ ] (n β)(λn λ +1)+k(n 1+λ) or if z n 1 [(n β)(λn λ +1)+k(λ + n 1)] 1 γ (19) (n γ) which yields the desired result. As observed earlier for λ = 0, and k = 0 in Theorem 5.1, we get the following result. Corollary 5.2 If f C(β), then f is convex of order γ in z <R,where R = inf n 2 { (n β) 1 γ (n γ) 6 Modified Hadamard Product } 1 n 1. (20) For functions f j (z) =z a n,j z n (j =1, 2) in the class A, we define the modified Hadamard product f 1 f 2 (z) off 1 (z) and f 2 (z) by f 1 (z) f 2 (z) =z a n,1 a n,2 z n. Then, we prove Theorem 6.1 If f j (z) U(k, λ, β), j=1, 2, then f 1 f 2 (z) U(k, λ, ξ), where ξ = 2(2 β)2 (λ +1)+2k(λ + 1)(2 β) 2() 2 2(2 β) 2 (λ +1)+2k(λ + 1)(2 β) () 2. Proof. Since, f j (z) U(k, λ, β), j =1, 2, n [(n β)(λn λ +1)+k(λ + n 1)] a n,j (j =1, 2), (21) the Cauchy Schwartz inequality leads to [ ] (n β)(λn λ +1)+k(n 1+λ) n an,1 a n,2 1. (22)

11 On a Subclass of k-uniformly convex functions 1687 Note that we have to find the largest ξ such that [ ] (n ξ)(λn λ +1)+k(n 1+λ) n a n,1 a n,2 1 1 ξ (j =1, 2). (23) Therefore, in view of (22) and (23), if n ξ an,1 a n,2 n β 1 ξ (n 2), (24) then (23) is satisfied. We have, from (22) [ ] an,1 a n,2 (n β)(λn λ +1)+k(n 1+λ) (25) Thus, if [ n ξ ] 1 ξ (n β)(λn λ +1)+k(n 1+λ) n β (n 2), (26) or, if ξ n(n β)2 (λn λ +1) n() 2 + nk(λ + n 1)(n β), (n 2) n(n β) 2 (λn λ +1) () 2 + nk(λ + n 1)(n β) then (22) is satisfied. Letting Φ(n) = n(n β)2 (λn λ +1) n() 2 + nk(λ + n 1)(n β), (n 2), n(n β) 2 (λn λ +1) () 2 + nk(λ + n 1)(n β) we see that ϕ(n) is increasing on n. This implies that, ξ Φ(2) = 2(2 β)2 (λ +1)+2k(λ + 1)(2 β) 2() 2 2(2 β) 2 (λ +1)+2k(λ + 1)(2 β) () 2, Finally, by taking the function f(z) =z we can see that the result is sharp. 2(1 + λ)(2 + k β) z2,

12 1688 T.N.Shanmugam, S.Sivasubramanian, and M.Darus 7 Convolution and Integral Operators Let f(z) = a n z n, g(z) = b n z n. Then the Hadamard product or convolution of f(z) and g(z), written as (f g)(z) is defined n=0 n=0 by (f g)(z) = a n b n z n. (27) n=0 Theorem 7.1 Let f U(k, λ, β), and g(z) =z + g n z n, 0 g n 1. Then f g U(k, λ, β). Proof. Since, 0 g n 1, n [(n β)(λn λ +1)+k(λ + n 1)] a n g n which completes the proof. n [(n β)(λn λ +1)+k(λ + n 1)] a n, Corollary 7.2 If f U(k, λ, β), then F (z) = 1+c z c is also in U(k, λ, β). z 0 t c 1 f(t)dt (c > 1) Proof. Since, c +1 F (z) = c + n a nz n, 0 < c +1 c + n < 1, the result follows from Theorem 7.1. ACKNOWLEDGEMENT: The author Maslina Darus is supported by IRPA EAR. References [1] E.Aqlan,J.M.Jahangiri and S.R.Kulkarni, Classes of k-uniformly convex and starlike functions, Tamkang Journal of Mathematics 35(3), (2004), [2] A. W. Goodman, On uniformly convex functions, Ann. Polon. Math. 56 (1991), no. 1,

13 On a Subclass of k-uniformly convex functions 1689 [3] S. Kanas and A. Wisniowska, Conic regions and k-uniform convexity, J. Comput. Appl. Math. 105 (1999), no. 1-2, [4] H Silverman, Univalent Functions with Negative Coefficients, Proc. Amer. Math. Soc. 51(1975), Received: December 30, 2005