On Quasi-Hadamard Product of Certain Classes of Analytic Functions

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1 Bulletin of Mathematical analysis Applications ISSN: , URL: Volume 1, Issue 2, (2009), Pages On Quasi-Hadamard Product of Certain Classes of Analytic Functions Wei-Ping Kuang, Yong Sun, & Zhi-Gang Wang Abstract The main purpose of this paper is to derive some quasi-hadamard product properties for certain classes of analytic functions which are defined by means of the Salagean operator. Relevant connections of the results presented here with those obtained in earlier works are also pointed out. 1 Introduction Let T denote the class of functions of the form f(z) = z a k z k (a k 0), (1.1) which are analytic in the open unit disk U := {z : z C z < 1}. For 0 α < 1, we denote by ST (α) CT (α) the usual subclasses of T consisting of functions with negative coefficients which are, respectively, starlike of order α convex of order α in U. A function f T is said to be in the class ST (α, β) if it satisfies the inequality ( zf ) (z) R > β zf (z) f(z) f(z) 1 + α (0 α < 1; β 0; z U). (1.2) Also, a function f T is said to be in the class UCT (α, β) if it satisfies the inequality: ( R 1 + zf ) (z) f > β zf (z) (z) f (z) + α (0 α < 1; β 0; z U). (1.3) Mathematics Subject Classifications: 30C45. Key words: Analytic functions, Quasi-Hadamard product, Salagean operator. c 2009 Universiteti i Prishtinës, Prishtinë, Kosovë. Submitted June, Published August,

2 Wei-Ping Kuang, Yong Sun, & Zhi-Gang Wang 37 Salagean [9] once introduced the following operator which called the Salagean operator D 0 f(z) = f(z), D 1 f(z) = Df(z) = zf (z), We note that D n f(z) = D(D n 1 f(z)) (n N := {1, 2,...}). D n f(z) = z k n a k z k (n N 0 := N {0}). (1.4) A function f T is said to be in the class ST n (α, β), if it satisfies the following inequality: ( D n+1 ) f(z) R D n > β D n+1 f(z) f(z) D n 1 f(z) +α (0 α < 1; β 0; n N 0; z U). (1.5) Making use of the similar arguments as Bharati et al. [2], we get the following necessary sufficient condition of the class ST n (α, β). A function f ST n (α, β) if only if k n [k(1 + β) (α + β)]a k. (1.6) The result is sharp for the function f given by f(z) = z k n [k(1 + β) (α + β)] zk (z U). (1.7) In view of (1.6), we have the following inclusion relationships for any positive integer n: ST n (α, β) ST n 1 (α, β) ST 2 (α, β) ST 1 (α, β) ST 0 (α, β). Indeed, by specializing the parameters n, α β, we obtain the following subclasses studied by various authors. (1) ST 0 (α, β) ST (α, β) ST 1 (α, β) UCT (α, β) (see [2]). (2) ST 0 (0, β) β ST ST 1 (0, β) β UCT (see [6]). (3) ST 0 (α, 0) ST (α) ST 1 (α, 0) CT (α) (see [10]). Let f j T (j = 1, 2,..., s) be given by f j (z) = z a k,j z k (a k,j 0; j = 1, 2,..., s). Then the quasi-hadamard product (or convolution) of these functions is defined by s (f 1 f s )(z) = z z k. (1.8) a k,j

3 38 On Quasi-Hadamard Product We note that, several interesting properties characteristics of the quasi- Hadamard product of two or more functions can be found in the recent investigations by (for example) Raina Bansal [7], Raina Prajapat [8], Aouf [1], Darwish Aouf [3], El-Ashwah Aouf [4], the references cited therein. In this paper, we aim at proving some new quasi-hadamard product properties for the function classes ST n (α, β). Relevant connections of the results presented here with those obtained in earlier works are also pointed out. 2 Main Results Throughout this paper, we assume that f j (z) = z a k,j z k ST n (α, β) (j = 1, 2,..., s), g j (z) = z b k,j z k ST m (α, β) (j = 1, 2,..., t). We begin by presenting the following quasi-hadamard product results for the class ST n (α, β). Theorem 2.1 Let f j ST n (α j, β) (j = 1, 2,..., s). Then δ = 1 (f 1 f s )(z) ST n (δ, β), (1 + β) s () 2 (s 1)n s (2 + β α j) s (). (2.1) The result is sharp for the functions f j (j = 1, 2,..., s) given by f j = z 2 n (2 + β α j ) z2 (j = 1, 2,..., s). (2.2) Proof. We use the principle of mathematical induction in the proof of Theorem 2.1. Let f j ST n (α j, β) (j = 1, 2), we need to find the largest δ such that k n [k(1 + β) (δ + β)] a k,1 a k, δ Since f j ST n (α j, β) (j = 1, 2), in view of (1.6), we have k n [k(1 + β) (α j + β)] a k,j 1 (j = 1, 2).

4 Wei-Ping Kuang, Yong Sun, & Zhi-Gang Wang 39 Furthermore, by the Cauchy-Schwartz inequality, we get k n [k(1 + β) (α 1 + β)][k(1 + β) (α 2 + β)] ak,1 a k,2 1. (1 )( 2 ) Thus, it suffices to show that k n [k(1 + β) (δ + β)] a k,1 a k,2 1 δ kn [k(1 + β) (α 1 + β)][k(1 + β) (α 2 + β)] ak,1 a k,2 (k N := N {1}), (1 )( 2 ) or equivalently, ak,1 a k,2 (1 δ) [k(1 + β) (α 1 + β)][k(1 + β) (α 2 + β)] [k(1 + β) (δ + β)] ( 1 )( 2 ) (k N ). By noting that (1 )( 2 ) ak,1 a k,2 k n [k(1 + β) (α 1 + β)][k(1 + β) (α 2 + β)] (k N ), consequently, we need only to prove that ( 1 )( 2 ) k n [k(1 + β) (α 1 + β)][k(1 + β) (α 2 + β)] (1 δ) [k(1 + β) (δ + β)] (k N ), which is equivalent to (k 1)(1 + β)( 1 )( 2 ) δ 1 k n [k(1 + β) (α 1 + β)][k(1 + β) (α 2 + β)] ( 1 )( 2 ) := A(k) (k N ). Since A(k) is an increasing function of k (k N ), letting k = 2, we obtain δ = 1 (1 + β)( 1 )( 2 ) 2 n (2 + β α 1 )(2 + β α 2 ) ( 1 )( 2 ), which implies that (2.1) holds for s = 2. We now suppose that (2.1) holds for s = 2, 3,..., t. Then γ = 1 (f 1 f t )(z) ST n (γ, β), (1 + β) t () 2 (t 1)n t (2 + β α j) t (). Then, by means of the above technique, we can show that (f 1 f t+1 )(z) ST n (δ, β),

5 40 On Quasi-Hadamard Product Since thus, we have δ = 1 1 γ = 2 + β γ = (1 + β)( t+1 )(1 γ) 2 n (2 + β α t+1 )(2 + β γ) ( t+1 )(1 γ). δ = 1 (1 + β) t () 2 (t 1)n t (2 + β α j) t () (1 + β)2 (t 1)n t (2 + β α j) 2 (t 1)n t (2 + β α j) t (), (1 + β) t+1 () 2 tn t+1 (2 + β α j) t+1 (), which shows that (2.1) holds for s = t + 1. Finally, for the functions f j (j = 1, 2,..., s) given by (2.2), we have s (f 1 f s )(z) = z 2 n z 2 := z A 2 z 2. (2 + β α j ) It follows that k n [k(1 + β) (δ + β)] A k = 1. 1 δ This evidently completes the proof of Theorem 2.1. By setting α j = α (j = 1, 2,..., s) in Theorem 2.1, we get Corollary 2.1 Let f j ST n (α, β) (j = 1, 2,..., s). Then δ = 1 (f 1 f s )(z) ST n (δ, β), (1 + β)() s. (2.3) 2 (s 1)n (2 + β α) s () s The result is sharp for the functions f j (j = 1, 2,..., s) given by f j = z 2 n (2 + β α) z2 (j = 1, 2,..., s). (2.4) Next, by similarly applying the method of proof of Theorem 2.1, we easily get the following result. Theorem 2.2 Let f j ST n (α, β j ) (j = 1, 2,..., s). Then (f 1 f s )(z) ST n (α, η),

6 Wei-Ping Kuang, Yong Sun, & Zhi-Gang Wang 41 η = 2(s 1)n s (2 + β j α) () s 1 + α 2. (2.5) The result is sharp for the functions f j (j = 1, 2,..., s) given by f j = z 2 n (2 + β j α) z2 (j = 1, 2,..., s). (2.6) Putting β j = β (j = 1, 2,..., s) in Theorem 2.2, we get Corollary 2.2 Let f j ST n (α, β) (j = 1, 2,..., s). Then (f 1 f s )(z) ST n (α, η), η = 2(s 1)n (2 + β α) s () s 1 + α 2. The result is sharp for the functions f j (j = 1, 2,..., s) given by (2.4). Theorem 2.3 Let f j ST n (α j, β) (j = 1, 2,..., s) suppose that s F (z) = z z k (t > 1; z U). (2.7) a t k,j Then F ST n (δ s, β), δ s = 1 s(1 + β)() t 2 n(t 1) (2 + β α) t s() t ( ) α = min {α j } 1 j s, (2.8) 2 n(t 1) (2 + β α) t s(2 + β)() t. The result is sharp for the functions f j (j = 1, 2,..., s) given by (2.2). Proof. Since f j ST n (α j, β), in view of (1.6), we obtain ( k n ) [k(1 + β) (α j + β)] a k,j 1 (j = 1, 2,..., s). By virtue of the Cauchy-Schwarz inequality, we get ( k n ) ( t ) t [k(1 + β) (α j + β)] a t k n [k(1 + β) (α j + β)] k,j a k,j 1. (2.9) It follows from (2.9) that 1 s ( k n ) t [k(1 + β) (α j + β)] a t k,j 1. s

7 42 On Quasi-Hadamard Product By setting suppose also that δ s 1 α = min {α j }, 1 j s s(k 1)(1 + β)() t k n(t 1) [k(1 + β) (α + β)] t s() t (k N ). By virtue of (1.6), we get ( k n ) [k(1 + β) (δ s + β)] s a t k,j 1 δ s 1 ( k n ) t [k(1 + β) (α + β)] s s 1 s ( k n ) t [k(1 + β) (α j + β)] a t s Now let D(k) = 1 a t k,j k,j 1. s(k 1)(1 + β)() t k n(t 1) [k(1 + β) (α + β)] t s() t (k N ). Since D(k) is an increasing function of k (k N ), we readily have By noting that we can see that δ s = D(2) = 1 s(1 + β)() t 2 n(t 1) (2 + β α) t s() t. 2 n(t 1) (2 + β α) t s(2 + β)() t, 0 δ s < 1. The result is sharp for the functions f j (j = 1, 2,..., s) given by (2.2). The proof of Theorem 2.3 is thus completed. Taking t = 2 α j = α (j = 1, 2,..., s) in Theorem 2.3, we get the following result. Corollary 2.3 Let f j ST n (α, β) (j = 1, 2,..., s) suppose that s F (z) = z z k (z U). (2.10) Then F ST n (δ s, β), δ s = 1 a 2 k,j s(1 + β)() 2 2 n (2 + β α) 2 s() 2,

8 Wei-Ping Kuang, Yong Sun, & Zhi-Gang Wang 43 2 n (2 + β α) 2 s(2 + β)() 2. The result is sharp for the functions f j (j = 1, 2,..., s) given by (2.4). By similarly applying the method of proof of Theorem 2.3, we easily get the following result. Theorem 2.4 Let f j ST n (α, β j ) (j = 1, 2,..., s) function F be defined by (2.7). Then F ST n (α, η s ), η s = 2n(t 1) (2 + β α) t s() t 1 + α 2 ( ) β = min {β j }, 1 j s 2 n(t 1) (2 + β α) t s(2 α)() t 1. The result is sharp for the functions f j (j = 1, 2,..., s) given by (2.6). Taking t = 2 β j following result. = β (j = 1, 2,..., s) in Theorem 2.4, we get the Corollary 2.4 Let f j ST n (α, β) (j = 1, 2,..., s) function F be defined by (2.10). Then F ST n (α, η s ), η s = 2n (2 + β α) 2 s() + α 2, 2 n (2 + β α) 2 s(2 α)(). The result is sharp for the functions f j (j = 1, 2,..., s) given by (2.4). Finally, we derive some quasi-hadamard product results for ST n (α, β) ST m (α, β). Theorem 2.5 Let f j ST n (α, β) (j = 1, 2,..., s) g j ST m (α, β) (j = 1, 2,..., t). Then Proof. Suppose that (f 1 f s g 1 g t )(z) ST sn+tm+s+t 1 (α, β). h(z) := (f 1 f s g 1 g t )(z). Then we have s h(z) = z t a k,j b k,j z k. (2.11)

9 44 On Quasi-Hadamard Product We need to show that s k sn+tm+s+t 1 [k(1 + β) (α + β)] a k,j b k,j t. (2.12) Since f j ST n (α, β), in view of (1.6), we have k n [k(1 + β) (α + β)]a k,j (j = 1, 2,..., s). (2.13) Therefore, we get a k,j k n [k(1 + β) (α + β)] (j = 1, 2,..., s; k N ). Since which implies that k(1 + β) (α + β) 1 k (0 α < 1; β 0; k N ), a k,j k (1+n) (j = 1, 2,..., s; k N ). (2.14) Similarly, for g j ST m (α, β), we have k m [k(1 + β) (α + β)]b k,j (j = 1, 2,..., t). (2.15) Therefore, we get b k,j k m [k(1 + β) (α + β)] (j = 1, 2,..., t; k N ), which implies that b k,j k (1+m) (j = 1, 2,..., t; k N ). (2.16) By using (2.14) for j = 1, 2,..., s, (2.16) for j = 1, 2,..., t 1, (2.15) for j = t, we obtain s t ksn+tm+s+t 1 [k(1 + β) (α + β)] a k,j b k,j = { k sn+tm+s+t 1 [k(1 + β) (α + β)] k s(1+n) k (t 1)(1+m)} b k,t k m [k(1 + β) (α + β)]b k,t.

10 Wei-Ping Kuang, Yong Sun, & Zhi-Gang Wang 45 Therefore, we know that h ST sn+tm+s+t 1 (α, β). The proof of Theorem 2.5 is evidently completed. Setting β = 0 in Theorem 2.5, we get the following result. Corollary 2.5 Let f j ST n (α, 0) (j = 1, 2,..., s) g j ST m (α, 0) (j = 1, 2,..., t). Then (f 1 f s g 1 g t )(z) ST (α). Taking into account the quasi-hadamard product of functions f j (j = 1, 2,..., s) only, in the proof of the above theorem, using (2.14) for j = 1, 2,..., s 1, (2.13) for j = s, we obtain Corollary 2.6 Let f j ST n (α, β) (j = 1, 2,..., s). Then (f 1 f s )(z) ST s(n+1) 1 (α, β). Remark 2.1 By specializing the parameters m = 0 n = 1 in Theorem 2.5, we get the corresponding results obtained by Frasin [5]. Acknowledgements The authors would like to thank Prof. R. K. Raina the referee for their careful reading valuable comments which have essentially improved the presentation of this paper. The present investigation was supported by the Scientific Research Fund of Huaihua University under Grant HHUQ the Scientific Research Fund of Hunan Provincial Education Department under Grant 08C118 of the People s Republic of China. References [1] M. K. Aouf, The quasi-hadamard product of certain analytic functions, Appl. Math. Lett. 21 (2008), [2] R. Bharati, R. Parvatham A. Swaminathan, On subclasses of uniformly convex functions corresponding class of starlike functions, Tamkang J. Math. 28 (1997), [3] H. E. Darwish M. K. Aouf, Generalizations of modified-hadamard products of p-valent functions with negative coefficients, Math. Comput. Modelling. 49 (2009), [4] R. M. El-Ashwah M. K. Aouf, Hadamard product of certain meromorphic starlike convex functions, Comput. Math. Appl. 57 (2009),

11 46 On Quasi-Hadamard Product [5] B. A. Frasin, Quasi-Hadamard product of certain classes of uniformly analytic functions, Gen. Math. 16 (2007), [6] S. Kanas H. M. Srivastava, Linear operators associated with kuniformly convex functions, Integral Transforms Spec. Funct. 9 (2000), [7] R. K. Raina D. Bansal, Some properties of a new class of analytic functions defined in terms of a hadamard product, J. Inequal. Pure Appl. Math. 9 (2008), Art. 22, 1 9. (electronic) [8] R. K. Raina J. K. Prajapat, On a certain new subclass of multivalently analytic functions, Math. Balkanica (N.S.) 23 (2009), [9] G. S. Salagean, Subclasses of univalent functions, Lecture Notes in Math. Springer-Verlag, 1013 (1983), [10] H. Silverman, Univalent functions with negative coefficients, Proc. Amer. Math. Soc. 51 (1975), Wei-Ping Kuang Department of Mathematics, Huaihua University, Huaihua , Hunan, P. R. China sy785153@126.com Yong Sun Department of Mathematics, Huaihua University, Huaihua , Hunan, P. R. China yongsun2008@foxmail.com Zhi-Gang Wang School of Mathematics Computing Science, Changsha University of Science Technology, Yuntang Campus, Changsha , Hunan, P. R. China zhigangwang@foxmail.com