FUNCTIONS WITH NEGATIVE COEFFICIENTS

A NEW SUBCLASS OF k-uniformly CONVEX FUNCTIONS WITH NEGATIVE COEFFICIENTS H. M. SRIVASTAVA T. N. SHANMUGAM Department of Mathematics and Statistics University of Victoria British Columbia 1V8W 3P4, Canada EMail: harimsri@math.uvic.ca Department of Information Technology Salalah College of Technology PO Box 608, Salalah PC211, Sultanate of Oman EMail: drtns2001@yahoo.com C. RAMACHANDRAN S. SIVASUBRAMANIAN Department of Mathematics Department of Mathematics College of Engineering, Easwari Engineering College, Anna University Ramapuram, Chennai 600089, Chennai 600025, Tamilnadu, India Tamilnadu, India EMail: crjsp2004@yahoo.com EMail: sivasaisastha@rediffmail.com Received: 31 May, 2007 Accepted: 15 June, 2007 Communicated by: Th.M. Rassias 2000 AMS Sub. Class.: Primary 30C45. Key words: Analytic functions; Univalent functions; Coefficient inequalities and coefficient estimates; Starlike functions; Convex functions; -to-convex functions; k- Uniformly convex functions; k-uniformly starlike functions; Uniformly starlike functions; Hadamard product (or convolution); Extreme points; Radii of close-toconvexity, Starlikeness and convexity; Integral operators. Page 1 of 29

Abstract: Acknowledgements: The main object of this paper is to introduce and investigate a subclass U(λ, α, β, k) of normalized analytic functions in the open unit disk, which generalizes the familiar class of uniformly convex functions. The various properties and characteristics for functions belonging to the class U(λ, α, β, k) derived here include (for example) a characterization theorem, coefficient inequalities and coefficient estimates, a distortion theorem and a covering theorem, extreme points, and the radii of close-to-convexity, starlikeness and convexity. Relevant connections of the results, which are presented in this paper, with various known results are also considered. The present investigation was supported, in part, by the Natural Sciences and Engineering Research Council of Canada under Grant OGP0007353. Page 2 of 29

1 Introduction and Motivation 4 2 A Characterization Theorem and Resulting Coefficient Estimates 6 3 Distortion and Covering Theorems for the Function Class U(λ, α, β, k) 11 4 Extreme Points of the Function Class U(λ, α, β, k) 16 5 Modified Hadamard Products (or Convolution) 20 6 Radii of -to-convexity, Starlikeness and Convexity 23 7 Hadamard Products and Integral Operators 27 Page 3 of 29

1. Introduction and Motivation Let A denote the class of functions f normalized by (1.1) f(z) = z + a n z n, which are analytic in the open unit disk = {z : z C and z < 1}. As usual, we denote by S the subclass of A consisting of functions which are univalent in. Suppose also that, for 0 α < 1, S (α) and K(α) denote the classes of functions in A which are, respectively, starlike of order α in and convex of order α in (see, for example, [11]). Finally, let T denote the subclass of S consisting of functions f given by (1.2) f(z) = z a n z n (a n 0) with negative coefficients. Silverman [9] introduced and investigated the following subclasses of the function class T : (1.3) T (α) := S (α) T and C(α) := K(α) T (0 α < 1). Definition 1. A function f T is said to be in the class U(λ, α, β, k) if it satisfies the following inequality: ( ) zf (z) R > k zf (1.4) (z) F (z) F (z) 1 + β (0 α λ 1; 0 β < 1; k 0), Page 4 of 29

where (1.5) F (z) := λαz 2 f (z) + (λ α)zf (z) + (1 λ + α)f(z). The above-defined function class U(λ, α, β, k) is of special interest and it contains many well-known as well as new classes of analytic univalent functions. In particular, U(λ, α, β, 0) is the class of functions with negative coefficients, which was introduced and studied recently by Kamali and Kadıoğlu [3], and U(λ, 0, β, 0) is the function class which was introduced and studied by Srivastava et al. [12] (see also Aqlan et al. [1]). We note that the class of k-uniformly convex functions was introduced and studied recently by Kanas and Wiśniowska [4]. Subsequently, Kanas and Wiśniowska [5] introduced and studied the class of k-uniformly starlike functions. The various properties of the above two function classes were extensively investigated by Kanas and Srivastava [6]. Furthermore, we have [cf. Equation (1.3)] (1.6) U(0, 0, β, 0) T (α) and U(1, 0, β, 0) C(α). We remark here that the classes of k-uniformly starlike functions and k-uniformly convex functions are an extension of the relatively more familiar classes of uniformly starlike functions and uniformly convex functions investigated earlier by (for example) Goodman [2], Rønning [8], and Ma and Minda [7] (see also the more recent contributions on these function classes by Srivastava and Mishra [10]). In our present investigation of the function class U(λ, α, β, k), we obtain a characterization theorem, coefficient inequalities and coefficient estimates, a distortion theorem and a covering theorem, extreme points, and the radii of close-to-convexity, starlikeness and convexity for functions belonging to the class U(λ, α, β, k). Page 5 of 29

2. A Characterization Theorem and Resulting Coefficient Estimates We employ the technique adopted by Aqlan et al. [1] to find the coefficient estimates for the function class U(λ, α, β, k). Our main characterization theorem for this function class is stated as Theorem 1 below. Theorem 1. A function f T given by (1.2) is in the class U(λ, α, β, k) if and only if (2.1) {n(k + 1) (k + β)} {(n 1)(nλα + λ α) + 1} a n (0 α λ 1; 0 β < 1; k 0). The result is sharp for the function f(z) given by (2.2) f(z) = z {n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1} zn (n 2). Proof. By Definition 1, f U(λ, α, β, k) if and only if the condition (1.4) is satisfied. Since it is easily verified that R(ω) > k ω 1 + β R ( ω(1 + ke iθ ) ke iθ) > β ( π θ < π; 0 β < 1; k 0), Page 6 of 29 the inequality (1.4) may be rewritten as follows: ( ) zf (z) (2.3) R F (z) (1 + keiθ ) ke iθ > β

or, equivalently, ( ) zf (z)(1 + ke iθ ) F (z)ke iθ (2.4) R > β. F (z) Now, by setting (2.5) G(z) = zf (z)(1 + ke iθ ) F (z)ke iθ, the inequality (2.4) becomes equivalent to G(z) + ()F (z) > G(z) (1 + β)f (z) (0 β < 1), where F (z) and G(z) are defined by (1.5) and (2.5), respectively. We thus observe that G(z) + ()F (z) (2 β)z (n β + 1){(n 1)(nλα + λ α) + 1}a n z n (n 1){(n 1)(nλα + λ α) + 1}a n z n keiθ (2 β) z k (n β + 1){(n 1)(nλα + λ α) + 1}a n z n (n 1){(n 1)(nλα + λ α) + 1}a n z n (2 β) z {(n(k + 1) (k + β) + 1}{(n 1)(nλα + λ α) + 1}a n z n. Page 7 of 29

Similarly, we obtain G(z) (1 + β)f (z) β z + {(n(k + 1) (k + β) 1}{(n 1)(nλα + λ α) + 1}a n z n. Therefore, we have G(z) + ()F (z) G(z) (1 + β)f (z) 2() z 2 {(n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1}a n z n 0, which implies the inequality (2.1) asserted by Theorem 1. Conversely, by setting 0 z = r < 1, and choosing the values of z on the positive real axis, the inequality (2.3) reduces to the following form: (1 β) {(n β) ke iθ (n 1)}{(n 1)(nλα+λ α)+1}a n r n 1 (2.6) R 1 0, (n 1){(n 1)(nλα+λ α)+1}a n r n 1 which, in view of the elementary identity: Page 8 of 29 R( e iθ ) e iθ = 1,

becomes (1 β) {(n β) k(n 1)}{(n 1)(nλα+λ α)+1}a n r n 1 (2.7) R 1 0. (n 1){(n 1)(nλα+λ α)+1}a n r n 1 Finally, upon letting r 1 in (2.7), we get the desired result. By taking α = 0 and k = 0 in Theorem 1, we can deduce the following corollary. Corollary 1. Let f T be given by (1.2). Then f U(λ, 0, β, 0) if and only if (n β){(n 1)λ + 1}a n. By setting α = 0, λ = 1 and k = 0 in Theorem 1, we get the following corollary. Corollary 2 (Silverman [9]). Let f T be given by (1.2). Then f C(β) if and only if n(n β)a n. The following coefficient estimates for f U(λ, α, β, k) is an immediate consequence of Theorem 1. Theorem 2. If f U(λ, α, β, k) is given by (1.2), then (2.8) a n {n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1} (0 α λ 1; 0 β < 1; k 0). Equality in (2.8) holds true for the function f(z) given by (2.2). (n 2) Page 9 of 29

By taking α = k = 0 in Theorem 2, we obtain the following corollary. Corollary 3. Let f T be given by (1.2). Then f U(λ, 0, β, 0) if and only if (2.9) a n (n β){(n 1)λ + 1} (n 2). Equality in (2.9) holds true for the function f(z) given by (2.10) f(z) = z (n β){(n 1)λ + 1} zn (n 2). Lastly, if we set α = 0, λ = 1 and k = 0 in Theorem 1, we get the following familiar result. Corollary 4 (Silverman [9]). Let f T be given by (1.2). Then f C(β) if and only if (2.11) a n n(n β) (n 2). Equality in (2.11) holds true for the function f(z) given by (2.12) f(z) = z n(n β) zn (n 2). Page 10 of 29

3. Distortion and Covering Theorems for the Function Class U(λ, α, β, k) Theorem 3. If f U(λ, α, β, k), then (3.1) r (2 + k β)(2λα + λ α) r2 f(z) r + (2 + k β)(2λα + λ α) r2 ( z = r < 1). Equality in (3) holds true for the function f(z) given by (3.2) f(z) = z (2 + k β)(2λα + λ α) z2. Proof. We only prove the second part of the inequality in (3), since the first part can be derived by using similar arguments. Since f U(λ, α, β, k), by using Theorem 1, we find that (2 + k β)(2λα + λ α + 1) = a n (2 + k β)(2λα + λ α + 1)a n {n(k + 1) (k + β)} {(n 1)(nλα + λ α) + 1} a n, Page 11 of 29

which readily yields the following inequality: (3.3) a n Moreover, it follows from (1.2) and (3.3) that f(z) = z a n z n (2 + k β)(2λα + λ α + 1). z + z 2 r + r 2 r + a n a n (2 + k β)(2λα + λ α + 1) r2, which proves the second part of the inequality in (3). Theorem 4. If f U(λ, α, β, k), then (3.4) 1 2() (2 + k β)(2λα + λ α) r f (z) 2() 1 + r ( z = r < 1). (2 + k β)(2λα + λ α) Equality in (3.4) holds true for the function f(z) given by (3.2). Page 12 of 29

Proof. Our proof of Theorem 4 is much akin to that of Theorem 3. Indeed, since f U(λ, α, β, k), it is easily verified from (1.2) that (3.5) f (z) 1 + and (3.6) f (z) 1 na n z n 1 1 + r na n z n 1 1 + r na n na n. The assertion (3.4) of Theorem 4 would now follow from (3.5) and (3.6) by means of a rather simple consequence of (3.3) given by (3.7) na n This completes the proof of Theorem 4. 2() (2 + k β)(2λα + λ α + 1). Theorem 5. If f U(λ, α, β, k), then f T (δ), where δ := 1 (2 + k β)(2λα + λ α) (). The result is sharp with the extremal function f(z) given by (3.2). Page 13 of 29 Proof. It is sufficient to show that (2.1) implies that (3.8) (n δ)a n 1 δ,

that is, that (3.9) n δ 1 δ {n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1} Since (3.9) is equivalent to the following inequality: (n 1)() δ 1 {n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1} () =: Ψ(n), and since (3.9) holds true for Ψ(n) Ψ(2) (n 2), n 2, 0 λ 1, 0 α 1, 0 β < 1 and k 0. This completes the proof of Theorem 5. By setting α = k = 0 in Theorem 5, we can deduce the following result. Corollary 5. If f U(λ, α, β, k), then ( ) λ(2 β) + β f T. λ(2 β) + 1 This result is sharp for the extremal function f(z) given by f(z) = z (λ + 1)(2 β) z2. (n 2), (n 2) Page 14 of 29

For the choices α = 0, λ = 1 and k = 0 in Theorem 5, we obtain the following result of Silverman [9]. Corollary 6. If f C(β), then ( ) 2 f T. 3 β This result is sharp for the extremal function f(z) given by f(z) = z 2(2 β) z2. Page 15 of 29

4. Extreme Points of the Function Class U(λ, α, β, k) Theorem 6. Let f 1 (z) = z and f n (z) = z {n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1} zn (n 2). Then f U(λ, α, β, k) if and only if it can be represented in the form: ( ) (4.1) f(z) = µ n f n (z) µ n 0; µ n = 1. n=1 Proof. Suppose that the function f(z) can be written as in (4.1). Then ) f(z) = µ n (z {n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1} zn n=1 ( ) = z µ n z n. {n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1} Now ( ) {n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1}() µ n (){n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1} = µ n n=1 Page 16 of 29 = 1 µ 1 1,

which implies that f U(λ, α, β, k). Conversely, we suppose that f U(λ, α, β, k). Then, by Theorem 2, we have a n {n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1} (n 2). Therefore, we may write and Then µ n = {n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1} µ 1 = 1 f(z) = µ n. µ n f n (z), with f n (z) given as in (4.1). This completes the proof of Theorem 6. n=1 a n (n 2) Corollary 7. The extreme points of the function class f U(λ, α, β, k) are the functions f 1 (z) = z and f n (z) = z {n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1} zn (n 2). Page 17 of 29 For α = k = 0 in Corollary 7, we have the following result.

Corollary 8. The extreme points of f U(λ, 0, β, 0) are the functions f 1 (z) = z and f n (z) = z {n β}{(n 1)λ + 1} zn (n 2). By setting α = 0, λ = 1 and k = 0 in Corollary 7, we obtain the following result obtained by Silverman [9]. Corollary 9. The extreme points of the class C(β) are the functions f 1 (z) = z and f n (z) = z n(n β) zn (n 2). Theorem 7. The class U(λ, α, β, k) is a convex set. Proof. Suppose that each of the functions f j (z) (j = 1, 2) given by (4.2) f j (z) = z a n,j z n (a n,j 0; j = 1, 2) is in the class U(λ, α, β, k). It is sufficient to show that the function g(z) defined by g(z) = µf 1 (z) + (1 µ)f 2 (z) (0 µ 1) is also in the class U(λ, α, β, k). Since g(z) = z [µa n,1 + (1 µ)a n,2 ]z n, Page 18 of 29

with the aid of Theorem 1, we have (4.3) {n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1} [µa n,1 + (1 µ)a n,2 ] µ {n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1}a n,1 + (1 µ) {n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1}a n,2 µ() + (1 µ)(), which implies that g U(λ, α, β, k). Hence U(λ, α, β, k) is indeed a convex set. Page 19 of 29

5. Modified Hadamard Products (or Convolution) For functions f(z) = a n z n and g(z) = n=0 b n z n, the Hadamard product (or convolution) (f g)(z) is defined, as usual, by (5.1) (f g)(z) := a n b n z n =: (g f)(z). On the other hand, for functions f j (z) = z n=0 n=0 a n,j z n (j = 1, 2) in the class T, we define the modified Hadamard product (or convolution) as follows: (5.2) (f 1 f 2 )(z) := z Then we have the following result. a n,1 a n,2 z n =: (f 2 f 1 )(z). Theorem 8. If f j (z) U(λ, α, β, k) (j = 1, 2), then (f 1 f 2 )(z) U(λ, α, β, k, ξ), where (2 β){2 + k β}{2λα + λ α + 1} 2()2 ξ := (2 β){2 + k β}{2λα + λ α + 1} (). 2 The result is sharp for the functions f j (z) (j = 1, 2) given as in (3.2). Page 20 of 29

Proof. Since f j (z) U(λ, α, β, k) (j = 1, 2), we have (5.3) {n(k+1) (k+β)}{(n 1)(nλα+λ α)+1}a n,j 1 β (j = 1, 2), which, in view of the Cauchy-Schwarz inequality, yields {n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1} (5.4) an,1 a n,2 1. We need to find the largest ξ such that {n(k + 1) (k + ξ)}{(n 1)(nλα + λ α) + 1} (5.5) a n,1 a n,2 1. 1 ξ Thus, in light of (5.4) and (5.5), whenever the following inequality: n ξ an,1 a n,2 n β 1 ξ (n 2) holds true, the inequality (5.5) is satisfied. We find from (5.4) that (5.6) an,1 a n,2 {n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1} Thus, if ( ) ( ) n ξ 1 ξ {n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1} or, if n β (n 2). (n 2), Page 21 of 29 ξ (n β){n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1} n()2 (n β){n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1} () 2 (n 2),

then (5.4) is satisfied. Setting Φ(n) := (n β){n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1} n()2 (n β){n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1} () 2 (n 2), we see that Φ(n) is an increasing function for n 2. This implies that ξ Φ(2) = (2 β){2 + k β}{2λα + λ α + 1} 2()2 (2 β){2 + k β}{2λα + λ α + 1} () 2. Finally, by taking each of the functions f j (z) (j = 1, 2) given as in (3.2), we see that the assertion of Theorem 8 is sharp. Page 22 of 29

6. Radii of -to-convexity, Starlikeness and Convexity Theorem 9. Let the function f(z) defined by (1.2) be in the class U(λ, α, β, k). Then f(z) is close-to-convex of order ρ (0 ρ < 1) in z < r 1 (λ, α, β, ρ, k), where r 1 (λ, α, β, ρ, k) ( (1 ρ){n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1} := inf n n() The result is sharp for the function f(z) given by (2.2). Proof. It is sufficient to show that f (z) 1 1 ρ Since (6.1) f (z) 1 = na n z n 1 we have if (6.2) ) 1 n 1 ( 0 ρ < 1; z < r1 (λ, α, β, ρ, k) ). na n z n 1, f (z) 1 1 ρ (0 ρ < 1), ( ) n a n z n 1 1. 1 ρ Hence, by Theorem 1, (6.2) will hold true if ( ) n z n 1 {n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1}, 1 ρ (n 2). Page 23 of 29

that is, if ( (1 ρ){n(k+1) (k+β)}{(n 1)(nλα+λ α)+1} (6.3) z n(1 β) The assertion of Theorem 9 would now follow easily from (6.3). ) 1 n 1 (n 2). Theorem 10. Let the function f(z) defined by (1.2) be in the class U(λ, α, β, k). Then f(z) is starlike of order ρ (0 ρ < 1) in z < r 2 (λ, α, β, ρ, k), where r 2 (λ, α, β, ρ, k) ( (1 ρ){n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1} := inf n (n ρ)() The result is sharp for the function f(z) given by (2.2). Proof. It is sufficient to show that zf (z) f(z) 1 1 ρ ( 0 ρ < 1; z < r 2 (λ, α, β, ρ, k) ). Since (6.4) we have zf (z) f(z) 1 (n 1)a n z n 1 1, a n z n 1 zf (z) f(z) 1 1 ρ (0 ρ < 1), ) 1 n 1 (n 2). Page 24 of 29

if (6.5) ( ) n ρ a n z n 1 1. 1 ρ Hence, by Theorem 1, (6.5) will hold true if ( ) n ρ z n 1 {n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1}, 1 ρ that is, if ( (1 ρ){n(k+1) (k+β)}{(n 1)(nλα+λ α)+1} (6.6) z (n ρ)(1 β) The assertion of Theorem 10 would now follow easily from (6.6). ) 1 n 1 (n 2). Theorem 11. Let the function f(z) defined by (1.2) be in the class U(λ, α, β, k). Then f(z) is convex of order ρ (0 ρ < 1) in z < r 3 (λ, α, β, ρ, k), where r 3 (λ, α, β, ρ, k) ( (1 ρ){n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1} := inf n n(n ρ)() The result is sharp for the function f(z) given by (2.2). Proof. It is sufficient to show that zf (z) f (z) 1 ρ ( 0 ρ < 1; z < r3 (λ, α, β, ρ, k) ). ) 1 n 1 (n 2). Page 25 of 29

Since (6.7) we have if (6.8) zf (z) f (z) zf (z) f (z) n(n 1)a n z n 1 1, na n z n 1 1 ρ (0 ρ < 1), ( ) n(n ρ) a n z n 1 1. 1 ρ Hence, by Theorem 1, (6.8) will hold true if ( ) n(n ρ) z n 1 {n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1}, 1 ρ that is, if ( (1 ρ){n(k+1) (k+β)}{(n 1)(nλα+λ α)+1} (6.9) z n(n ρ)(1 β) Theorem 11 now follows easily from (6.9). ) 1 n 1 (n 2). Page 26 of 29

7. Hadamard Products and Integral Operators Theorem 12. Let f U(λ, α, β, k). Suppose also that (7.1) g(z) = z + g n z n (0 g n 1). Then f g U(λ, α, β, k). Proof. Since 0 g n 1 (n 2), {n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1}a n g n (7.2) {n(k + 1) (k + β)}{(n 1)(nλα + λ α) + 1}a n, which completes the proof of Theorem 12. Corollary 10. If f U(λ, α, β, k), then the function F(z) defined by (7.3) F(z) := 1 + c z t c 1 f(t)dt (c > 1) z c is also in the class U(λ, α, β, k). Proof. Since F(z) = z + 0 ( ) ( c + 1 z n 0 < c + 1 ) c + n c + n < 1, the result asserted by Corollary 10 follows from Theorem 12. Page 27 of 29

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