Bulletin of Mathematical Analysis and Applications ISSN: 1821-1291, URL: http://www.bmathaa.org Volume 3 Issue 3(2010), Pages 109-121. A NEW SUBCLASS OF MEROMORPHIC FUNCTION WITH POSITIVE COEFFICIENTS S. KAVITHA, S. SIVASUBRAMANIAN, K. MUTHUNAGAI Abstract. In the present investigation, the authors deο¬ne a new class of meromorphic functions deο¬ned in the punctured unit disk Δ∗ := {π§ ∈ β : 0 < β£π§β£ < 1}. Coeο¬cient inequalities, growth and distortion inequalities, as well as closure results are obtained. We also prove a Property using an integral operator and its inverse deο¬ned on the new class. 1. Introduction Let Σ denote the class of normalized meromorphic functions π of the form ∞ π (π§) = 1 ∑ + ππ π§ π , π§ π=1 (1.1) deο¬ned on the punctured unit disk Δ∗ := {π§ ∈ β : 0 < β£π§β£ < 1}. The Hadamard product or convolution of two functions π (π§) given by (1.1) and ∞ π(π§) = 1 ∑ + ππ π§ π π§ π=1 (1.2) is deο¬ned by ∞ (π ∗ π)(π§) := 1 ∑ ππ ππ π§ π . + π§ π=1 A function π ∈ Σ is meromorphic starlike of order πΌ (0 ≤ πΌ < 1) if ( ′ ) π§π (π§) −ℜ > πΌ (π§ ∈ Δ := Δ∗ ∪ {0}). π (π§) The class of all such functions is denoted by Σ∗ (πΌ). Similarly the class of convex functions of order πΌ is deο¬ned. Let Σπ be the class of functions π ∈ Σ with ππ ≥ 0. The subclass of Σπ consisting of starlike functions of order πΌ is denoted by Σ∗π (πΌ). Now, we deο¬ne a new class of functions in Deο¬nition 1. 2000 Mathematics Subject Classiο¬cation. 30C50. Key words and phrases. Meromorphic functions, starlike function, convolution, positive coefο¬cients, coeο¬cient inequalities, integral operator. c β2010 Universiteti i PrishtineΜs, PrishtineΜ, KosoveΜ. Submitted September 14, 2009. Published September 3, 2010 . 109 110 KAVITHA ET AL. Deο¬nition 1. Let 0 ≤ πΌ < 1. Further, let π (π§) ∈ Σπ be given by (1.1), 0 ≤ π < 1 The class ππ (πΌ, π) is deο¬ned by { ( ) } π§π ′ (π§) ππ (πΌ, π) = π ∈ Σπ : ℜ >πΌ . (π − 1)π (π§) + ππ§π ′ (π§) Clearly, ππ (πΌ, 0) reduces to the class Σ∗π (πΌ). The class Σ∗π (πΌ) and various other subclasses of Σ have been studied rather extensively by Clunie [4], Nehari and Netanyahu [8], Pommerenke ([9], [10]), Royster [11], and others (cf., e.g., Bajpai [2] , Mogra et al. [7], Uralegaddi and Ganigi [16], Cho et al. [3], Aouf [3], and Uralegaddi and Somanatha [15]; see also Duren [[5], pages 29 and 137], and Srivastava and Owa [[13], pages 86 and 429]) (see also [1]). In this paper, we obtain the coeο¬cient inequalities, growth and distortion inequalities, as well as closure results for the class ππ (πΌ, π). Properties of a certain integral operator and its inverse deο¬ned on the new class ππ (πΌ, π) are also discussed. 2. Coefficients Inequalities Our ο¬rst theorem gives a necessary and suο¬cient condition for a function π to be in the class ππ (πΌ, π). Theorem 2.1. Let π (π§) ∈ Σπ be given by (1.1). Then π ∈ ππ (πΌ, π) if and only if ∞ ∑ {π + πΌ − πΌπ(1 + π)} ππ ≤ 1 − πΌ. (2.1) π=1 Proof. If π ∈ ππ (πΌ, π), then ∑∞ ) { } ( −1 + π=1 πππ π§ π+1 π§π ′ (π§) ∑∞ =ℜ > πΌ. ℜ (π − 1)π (π§) + ππ§π ′ (π§) −1 + π=1 (π − 1 + ππ)ππ π§ π+1 By letting π§ → 1− , we have { −1 + −1 + ∑∞ ∑∞ πππ (π − 1 + ππ)ππ π=1 π=1 } > πΌ. This shows that (2.1) holds. Conversely assume that (2.1) holds. It is suο¬cient to show that π§π ′ (π§) − {(π − 1)π (π§) + ππ§π ′ (π§)} π§π ′ (π§) + (1 − 2πΌ) {(π − 1)π (π§) + ππ§π ′ (π§)} < 1 (π§ ∈ Δ). we see that Using (2.1), π§π ′ (π§) − {(π − 1)π (π§) + ππ§π ′ (π§)} π§π ′ (π§) + (1 − 2πΌ) {(π − 1)π (π§) + ππ§π ′ (π§)} ∑∞ (1 − π)(π + 1)ππ π§ π+1 π=1 ∑∞ = −2(1 − πΌ) + π=1 [{1 + (1 − 2πΌ)π}π + (1 − 2πΌ)(π − 1)] ππ π§ π+1 ∑∞ π=1 (1 − π)(π + 1)ππ ∑∞ ≤ ≤ 1. 2(1 − πΌ) − π=1 [{1 + (1 − 2πΌ)π}π + (1 − 2πΌ)(π − 1)] ππ Thus we have π ∈ ππ (πΌ, π). For the choice of π = 0, we get the following. β‘ A NEW SUBCLASS OF MEROMORPHIC FUNCTION... 111 Remark 2.2. Let π (π§) ∈ Σπ be given by (1.1). Then π ∈ Σ∗π (πΌ) if and only if ∞ ∑ (π + πΌ)ππ ≤ 1 − πΌ. π=1 Our next result gives the coeο¬cient estimates for functions in ππ (πΌ, π). Theorem 2.3. If π ∈ ππ (πΌ, π), then 1−πΌ ππ ≤ , {π + πΌ − πΌπ(1 + π)} π = 1, 2, 3, . . . . The result is sharp for the functions πΉπ (π§) given by 1 1−πΌ πΉπ (π§) = + π§π, π§ {π + πΌ − πΌπ(1 + π)} π = 1, 2, 3, . . . . Proof. If π ∈ ππ (πΌ, π), then we have, for each π, {π + πΌ − πΌπ(1 + π)} ππ ≤ ∞ ∑ {π + πΌ − πΌπ(1 + π)} ππ ≤ 1 − πΌ. π=1 Therefore we have ππ ≤ 1−πΌ . {π + πΌ − πΌπ(1 + π)} Since 1−πΌ 1 + π§π π§ {π + πΌ − πΌπ(1 + π)} satisο¬es the conditions of Theorem 2.1, πΉπ (π§) ∈ ππ (πΌ, π) and the equality is attained for this function. β‘ πΉπ (π§) = For π = 0, we have the following corollary. Remark 2.4. If π ∈ Σ∗π (πΌ), then 1−πΌ ππ ≤ , π+πΌ π = 1, 2, 3, . . . . Theorem 2.5. If π ∈ ππ (πΌ, π), then 1 1−πΌ 1 1−πΌ − π ≤ β£π (π§)β£ ≤ + π π 1 + πΌ − 2πΌπ π 1 + πΌ − 2πΌπ The result is sharp for 1 1−πΌ π (π§) = + π§. π§ {1 + πΌ − 2πΌπ} ∑∞ Proof. Since π (π§) = π§1 + π=1 ππ π§ π , we have β£π (π§)β£ ≤ ∞ ∞ ∑ 1 ∑ 1 + ππ π π ≤ + π ππ . π π=1 π π=1 Since, ∞ ∑ ππ ≤ π=1 1−πΌ . 1 + πΌ − 2πΌπ Using this, we have β£π (π§)β£ ≤ 1−πΌ 1 + π. π 1 + πΌ − 2πΌπ (β£π§β£ = π). (2.2) 112 KAVITHA ET AL. Similarly 1 1−πΌ − π. π 1 + πΌ − 2πΌπ β£π (π§)β£ ≥ The result is sharp for π (π§) = 1 π§ + 1−πΌ 1+πΌ−2πΌπ π§. β‘ Similarly we have the following: Theorem 2.6. If π ∈ ππ (πΌ, π), then 1 1−πΌ 1−πΌ 1 − ≤ β£π ′ (π§)β£ ≤ 2 + π2 1 + πΌ − 2πΌπ π 1 + πΌ − 2πΌπ (β£π§β£ = π). The result is sharp for the function given by (2.2). (πΎ) 3. Neighborhoods for the class ππ (πΌ, π) (πΎ) In this section, we determine the neighborhood for the class ππ (πΌ, π), which we deο¬ne as follows: (πΎ) Deο¬nition 2. A function π ∈ Σπ is said to be in the class ππ (πΌ, π) if there exists a function π ∈ ππ (πΌ, π) such that π (π§) (π§ ∈ Δ, 0 ≤ πΎ < 1). (3.1) π(π§) − 1 < 1 − πΎ, Following the earlier works on neighborhoods of analytic functions by Goodman [6] and Ruscheweyh [14], we deο¬ne the πΏ-neighborhood of a function π ∈ Σπ by { } ∞ ∞ ∑ 1 ∑ π ππΏ (π ) := π ∈ Σπ : π(π§) = + ππ π§ and πβ£ππ − ππ β£ ≤ πΏ . (3.2) π§ π=1 π=1 Theorem 3.1. If π ∈ ππ (πΌ, π) and πΎ =1− πΏ(1 + πΌ − 2πΌπ) , 2πΌ − 2πΌπ (3.3) then ππΏ (π) ⊂ ππ(πΎ) (πΌ, π). Proof. Let π ∈ ππΏ (π). Then we ο¬nd from (3.2) that ∞ ∑ πβ£ππ − ππ β£ ≤ πΏ, (3.4) π=1 which implies the coeο¬cient inequality ∞ ∑ β£ππ − ππ β£ ≤ πΏ, (π ∈ β). (3.5) π=1 Since π ∈ ππ (πΌ, π), we have [cf. equation (2.1)] ∞ ∑ π=1 ππ ≤ 1−πΌ , 1 + πΌ − 2πΌπ (3.6) A NEW SUBCLASS OF MEROMORPHIC FUNCTION... 113 so that ∑∞ π (π§) < − 1 π(π§) β£ππ − ππ β£ π=1∑ ∞ 1 − π=1 ππ πΏ(1 + πΌ − 2πΌπ) 2πΌ − 2πΌπ 1 − πΎ, = = (πΎ) provided πΎ is given by (3.3). Hence, by deο¬nition, π ∈ ππ (πΌ, π) for πΎ given by (3.3), which completes the proof. β‘ 4. Closure Theorems Let the functions πΉπ (π§) be given by ∞ πΉπ (π§) = 1 ∑ ππ,π π§ π , + π§ π=1 π = 1, 2, ..., π. (4.1) We shall prove the following closure theorems for the class ππ (πΌ, π). Theorem 4.1. Let the function πΉπ (π§) deο¬ned by (4.1) be in the class ππ (πΌ, π) for every π = 1, 2, ..., π. Then the function π (π§) deο¬ned by ∞ π (π§) = 1 ∑ ππ π§ π + π§ π=1 belongs to the class ππ (πΌ, π), where ππ = 1 π (ππ ≥ 0) ∑π π=1 ππ,π (π = 1, 2, ..) Proof. Since πΉπ (π§) ∈ ππ (πΌ, π), it follows from Theorem 2.1 that ∞ ∑ {π + πΌ − πΌπ(1 + π)} ππ,π ≤ 1 − πΌ (4.2) π=1 for every π = 1, 2, .., π. Hence ∞ ∑ {π + πΌ − πΌπ(1 + π)} ππ = π=1 ∞ ∑ ( {π + πΌ − πΌπ(1 + π)} π=1 = π 1 ∑ ππ,π π ) π=1 (∞ ) π 1 ∑ ∑ {π + πΌ − πΌπ(1 + π)} ππ,π π π=1 π=1 ≤ 1 − πΌ. By Theorem 2.1, it follows that π (π§) ∈ ππ (πΌ, π). β‘ Theorem 4.2. The class ππ (πΌ, π) is closed under convex linear combination. Proof. Let the function πΉπ (π§) given by (4.1) be in the class ππ (πΌ, π). Then it is enough to show that the function π»(π§) = ππΉ1 (π§) + (1 − π)πΉ2 (π§) (0 ≤ π ≤ 1) is also in the class ππ (πΌ, π). Since for 0 ≤ π ≤ 1, ∞ π»(π§) = 1 ∑ + [πππ,1 + (1 − π)ππ,2 ]π§ π , π§ π=1 114 KAVITHA ET AL. we observe that ∞ ∑ {π + πΌ − πΌπ(1 + π)} [πππ,1 + (1 − π)ππ,2 ] π=1 = π ∞ ∑ {π + πΌ − πΌπ(1 + π)} ππ,1 + (1 − π) π=1 ∞ ∑ {π + πΌ − πΌπ(1 + π)} ππ,2 π=1 ≤ 1 − πΌ. By Theorem 2.1, we have π»(π§) ∈ ππ (πΌ, π). β‘ 1−πΌ π§ π for π = 1, 2, . . .. Theorem 4.3. Let πΉ0 (π§) = π§1 and πΉπ (π§) = π§1 + {π+πΌ−πΌπ(1+π)} Then if π (π§) can be expressed in the form π (π§) = ∑∞ π (π§) ∈ ππ (πΌ, π) if and only ∑∞ π πΉ (π§) where π ≥ 0 and π π=0 π π π=0 ππ = 1. Proof. Let π (π§) = ∞ ∑ ππ πΉπ (π§) π=0 ∞ = 1 ∑ ππ (1 − πΌ) + π§π. π§ π=1 {π + πΌ − πΌπ(1 + π)} Then ∞ ∑ π=1 ππ ∞ 1−πΌ {π + πΌ − πΌπ(1 + π)} ∑ = ππ = 1 − π0 ≤ 1. {π + πΌ − πΌπ(1 + π)} (1 − πΌ) π=1 By Theorem 2.1, we have π (π§) ∈ ππ (πΌ, π). Conversely, let π (π§) ∈ ππ (πΌ, π). From Theorem 2.3, we have ππ ≤ 1−πΌ {π + πΌ − πΌπ(1 + π)} for π = 1, 2, .. we may take ππ = {π + πΌ − πΌπ(1 + π)} ππ 1−πΌ for π = 1, 2, ... and π0 = 1 − ∞ ∑ ππ . π=1 Then π (π§) = ∞ ∑ ππ πΉπ (π§). π=0 β‘ A NEW SUBCLASS OF MEROMORPHIC FUNCTION... 115 5. Partial Sums Silverman [12] determined sharp lower bounds on the real part of the quotients between the normalized starlike or convex functions and their sequences of partial sums. As a natural extension, one is interested to search results analogous to those of Silverman for meromorphic univalent functions. In this section, motivated essentially by the work of silverman [12] and Cho and Owa [3] we will investigate the ratio of a function of the form ∞ 1 ∑ π (π§) = + ππ π§ π , (5.1) π§ π=1 to its sequence of partial sums π 1 ∑ 1 and ππ (π§) = + ππ π§ π π1 (π§) = π§ π§ π=1 (5.2) when the coeο¬cients are suο¬ciently small to satisfy the condition analogous to ∞ ∑ {π + πΌ − πΌπ(1 + π)} ππ ≤ 1 − πΌ. π=1 For the sake of brevity we rewrite it as ∞ ∑ ππ β£ππ β£ ≤ 1 − πΌ, (5.3) π=1 where ππ := π + πΌ − πΌπ(1 + π) (5.4) More precisely we will determine sharp lower bounds for ℜ{π (π§)/π } ℜ{ππ (π§)/π (π§)}. { π (π§)} and 1+π€(π§) > 0 (π§ ∈ In this connection we make use of the well known results thatℜ 1−π€(π§) ∞ ∑ Δ) if and only if π(π§) = ππ π§ π satisο¬es the inequality β£π(π§)β£ ≤ β£π§β£. Unless otherπ=1 wise stated, we will assume that π is of the form (1.1) and its sequence of partial ∑π sums is denoted by ππ (π§) = π§1 + π=1 ππ π§ π . Theorem 5.1. Let π (π§) ∈ ππ (πΌ, π) be given by (5.1)satisο¬es condition, (2.1) } { ππ+1 (π, πΌ) − 1 + πΌ π (π§) ≥ (π§ ∈ π ) (5.5) Re ππ (π§) ππ+1 (π, πΌ) where { ππ (π, πΌ) ≥ 1 − πΌ, ππ+1 (π, πΌ), ππ π = 1, 2, 3, . . . , π ππ π = π + 1, π + 2, . . . . (5.6) The result (5.5) is sharp with the function given by π (π§) = 1 1−πΌ + π§ π+1 . π§ ππ+1 (π, πΌ) Proof. Deο¬ne the function π€(π§) by [ ] 1 + π€(π§) ππ+1 (π, πΌ) π (π§) ππ+1 (π, πΌ) − 1 + πΌ = − 1 − π€(π§) 1−πΌ ππ (π§) ππ+1 (π, πΌ) (5.7) 116 KAVITHA ET AL. 1+ π ∑ ππ π§ π+1 + π=1 = 1+ ( ππ+1 (π,πΌ) 1−πΌ π ∑ ππ ) ∑ ∞ ππ π§ π+1 π=π+1 (5.8) π§ π+1 π=1 It suο¬ces to show that β£π€(π§)β£ ≤ 1. Now, from (5.8) we can write ( π€(π§) = 2+2 π ∑ ππ+1 (π,πΌ) 1−πΌ ππ π§ π+1 + ) ∑ ∞ ππ π§ π+1 π=π+1 ( π=1 ππ+1 (π,πΌ) 1−πΌ . ) ∑ ∞ ππ π§ π+1 π=π+1 Hence we obtain ( β£π€(π§)β£ ≤ 2−2 π ∑ ππ+1 (π,πΌ) 1−πΌ β£ππ β£ − ( π=1 ) ∑ ∞ β£ππ β£ π=π+1 ππ+1 (π,πΌ) 1−πΌ ) ∑ ∞ β£ππ β£ π=π+1 Now β£π€(π§)β£ ≤ 1 if ( 2 ππ+1 (π, πΌ) 1−πΌ ) ∑ ∞ β£ππ β£ ≤ 2 − 2 π ∑ β£ππ β£ π=1 π=π+1 or, equivalently, π ∑ β£ππ β£ + π=1 ∞ ππ+1 (π, πΌ) ∑ β£ππ β£ ≤ 1. 1−πΌ π=π+1 From the condition (2.1), it is suο¬cient to show that ∞ ∞ ∑ ππ+1 (π, πΌ) ∑ ππ (π, πΌ) β£ππ β£ + β£ππ β£ ≤ β£ππ β£ 1 − πΌ 1−πΌ π=1 π=1 π ∑ π=π+1 which is equivalent to ) π ( ∑ ππ (π, πΌ) − 1 + πΌ π=1 + 1−πΌ β£ππ β£ ) ∞ ( ∑ ππ (π, πΌ) − ππ+1 (π, πΌ) β£ππ β£ 1−πΌ π=π+1 ≥ 0 (5.9) To see that the function given by (5.7) gives the sharp result, we observe that for π§ = ππππ/π π (π§) 1−πΌ 1−πΌ = 1+ π§π → 1 − ππ (π§) ππ+1 (π, πΌ) ππ+1 (π, πΌ) ππ+1 (π, πΌ) − 1 + πΌ when π → 1− . = ππ+1 (π, πΌ) A NEW SUBCLASS OF MEROMORPHIC FUNCTION... which shows the bound (5.5) is the best possible for each π ∈ β. 117 β‘ The proof of the next theorem is much akin to that of the earlier theorem and hence we state the theorem without proof. Theorem 5.2. Let π (π§) ∈ ππ (πΌ, π) be given by (5.1)satisο¬es condition, (2.1) { } ππ (π§) ππ+1 (π, πΌ) Re ≥ (π§ ∈ π ) (5.10) π (π§) ππ+1 (π, πΌ) + 1 − πΌ where ππ+1 (π, πΌ) ≥ 1 − πΌ { 1 − πΌ, ππ π = 1, 2, 3, . . . , π ππ (π, πΌ) ≥ (5.11) ππ+1 (π, πΌ), ππ π = π + 1, π + 2, . . . . The result (5.10) is sharp with the function given by π (π§) = 1−πΌ 1 + π§ π+1 . π§ ππ+1 (π, πΌ) (5.12) 6. Radius of meromorphic starlikeness and meromorphic convexity The radii of starlikeness and convexity for the class are given by the following theorems for the class ππ (πΌ, π). Theorem 6.1. Let the function π be in the class ππ (πΌ, π). Then π is meromorphically starlike of order π(0 ≤ π < 1), in β£π§β£ < π1 (πΌ, π, π), where 1 [ ] π+1 (1 − π)(1 − πΌ) π1 (πΌ, π, π) = inf , (6.1) π≥1 (π + 2 − π) {π + πΌ − πΌπ(1 + π)} Proof. Since, ∞ π (π§) = 1 ∑ + ππ π§ π , π§ π=1 we get ∞ π ′ (π§) = − ∑ 1 πππ π§ π−1 . + 2 π§ π=1 It is suο¬cient to show that π§π ′ (π§) − π (π§) − 1 ≤ 1 − π or equivalently or ∞ ∑ π (π + 1)ππ π§ ′ π§π (π§) π=1 ≤1−π ∞ π (π§) + 1 = 1 ∑ − ππ π§ π π§ π=1 ) ∞ ( ∑ π+2−π π=1 1−π ππ β£π§β£π+1 ≤ 1, for 0 ≤ π < 1, and β£π§β£ < π1 (πΌ, π, π). By Theorem 2.1, (6.2) will be true if ( ) π+2−π 1−πΌ β£π§β£π+1 ≤ 1−π {π + πΌ − πΌπ(1 + π)} (6.2) 118 KAVITHA ET AL. or, if [ (1 − π)(1 − πΌ) (π + 2 − π) {π + πΌ − πΌπ(1 + π)} This completes the proof of Theorem 6.1. 1 ] π+1 β£π§β£ ≤ , π ≥ 1. (6.3) β‘ Theorem 6.2. Let the function π in the class ππ (πΌ, π). Then π is meromorphically convex of order π, (0 ≤ π < 1), in β£π§β£ < π2 (πΌ, π, π), where 1 [ ] π+1 (1 − π)(1 − πΌ) , π β§ 1, (6.4) π2 (πΌ, π, π) = inf π≥1 π (π + 2 − π) {π + πΌ − πΌπ(1 + π)} Proof. Since, ∞ π (π§) = we get 1 ∑ − ππ π§ π , π§ π=1 ∞ π ′ (π§) = − ∑ 1 − πππ π§ π−1 . 2 π§ π=1 It is suο¬cient to show that ′′ −1 − π§π (π§) − 1 ≤ 1 − π or equivalently ′ π (π§) ∞ ∑ π−1 π(π + 1)ππ π§ ′′ π§π (π§) π=1 ≤ 1 − π or ∞ π ′ (π§) + 2 = ∑ 1 π−1 πππ π§ − 2 − π§ π=1 ( ) ∞ ∑ π(π + 2 − π) ππ β£π§β£π+1 ≤ 1, 1 − π π=1 for 0 ≤ π < 1, and β£π§β£ < π2 (πΌ, π, π). By Theorem 2.1, (6.5) will be true if ( ) π(π + 2 − π) (1 − πΌ) β£π§β£π+1 ≤ 1−π {π + πΌ − πΌπ(1 + π)} or, if 1 [ ] π+1 (1 − π)(1 − πΌ) β£π§β£ ≤ , π ≥ 1. π(π + 2 − π) {π + πΌ − πΌπ(1 + π)} This completes the proof of Theorem 6.2. (6.5) (6.6) β‘ 7. Integral Operators In this section, we consider integral transforms of functions in the class ππ (πΌ, π). Theorem 7.1. Let the function π (π§) given by (1) be in ππ (πΌ, π). Then the integral operator ∫ 1 πΉ (π§) = π π’π π (π’π§)ππ’ (0 < π’ ≤ 1, 0 < π < ∞) 0 is in ππ (πΏ, π), where πΏ= (π + 2) {1 + πΌ − 2πΌπ} − π(1 − πΌ) . π(1 − πΌ) {1 − 2π} + (1 + πΌ) {1 − 2π} (π + 2) A NEW SUBCLASS OF MEROMORPHIC FUNCTION... The result is sharp for the function π (π§) = 1 π§ + 119 1−πΌ {1+πΌ−2πΌπ} π§. Proof. Let π (π§) ∈ ππ (πΌ, π). Then ∫ 1 π’π π (π’π§)ππ’ πΉ (π§) = π 0 ) ∫ 1 ( π−1 ∑ ∞ π’ = π + ππ π’π+π π§ π ππ’ π§ 0 π=1 ∞ = 1 ∑ π + ππ π§ π . π§ π=1 π + π + 1 It is suο¬cient to show that ∞ ∑ π {π + πΏ − πΏπ(1 + π)} ππ ≤ 1. (π + π + 1)(1 − πΏ) π=1 (7.1) Since π ∈ ππ (πΌ, π), we have ∞ ∑ {π + πΌ − πΌπ(1 + π)} ππ ≤ 1. (1 − πΌ) π=1 Note that (7.1) is satisο¬ed if π {π + πΏ − πΏπ(1 + π)} {π + πΌ − πΌπ(1 + π)} ≤ . (π + π + 1)(1 − πΏ) (1 − πΌ) Rewriting the inequality, we have π {π + πΏ − πΏπ(1 + π)} (1 − πΌ) ≤ (π + π + 1)(1 − πΏ) {π + πΌ − πΌπ(1 + π)} . Solving for πΏ, we have πΏ≤ (π + π + 1) {π + πΌ − πΌπ(1 + π)} − ππ(1 − πΌ) = πΉ (π). π(1 − πΌ) {1 − π(1 + π)} + {(π + πΌ − πΌπ(1 + π)} (π + π + 1) A simple computation will show that πΉ (π) is increasing and πΉ (π) ≥ πΉ (1). Using this, the results follows. β‘ For the choice of π = 0, we have the following result of Uralegaddi and Ganigi [15]. Remark 7.2. Let the function π (π§) deο¬ned by (1) be in Σ∗π (πΌ). Then the integral operator ∫ 1 πΉ (π§) = π π’π π (π’π§)ππ’ (0 < π’ ≤ 1, 0 < π < ∞) 0 is in Σ∗π (πΏ), where πΏ = 1+πΌ+ππΌ 1+πΌ+π . The result is sharp for the function π (π§) = Also we have the following: 1 1−πΌ + π§. π§ 1+πΌ 120 KAVITHA ET AL. Theorem 7.3. Let π (π§), given by (1), be in ππ (πΌ, π), ∞ πΉ (π§) = 1 1 ∑ π+π+1 [(π + 1)π (π§) + π§π ′ (π§)] = + ππ π§ π , π π§ π=1 π π > 0. (7.2) Then πΉ (π§) is in ππ (πΌ, π) for β£π§β£ ≤ π(πΌ, π, π½) where ( π(πΌ, π, π½) = inf π π(1 − π½) {π + πΌ − πΌπ(1 + π)} (1 − πΌ)(π + π + 1) {π + π½ − π½π(1 + π)} The result is sharp for the function ππ (π§) = Proof. Let π€ = π§π ′ (π§) (π−1)π (π§)+ππ§π ′ (π§) . 1 π§ )1/(π+1) , 1−πΌ + {π+πΌ−πΌπ(1+π)} π§π, π = 1, 2, 3, . . . . π = 1, 2, 3, . . . . Then it is suο¬cient to show that π€−1 π€ + 1 − 2π½ < 1. A computation shows that this is satisο¬ed if ∞ ∑ {π + π½ − π½π(1 + π)} (π + π + 1) ππ β£π§β£π+1 ≤ 1. (1 − π½)π π=1 (7.3) Since π ∈ ππ (πΌ, π), by Theorem 2.1, we have ∞ ∑ {π + πΌ − πΌπ(1 + π)} ππ ≤ 1 − πΌ. π=1 The equation (7.3) is satisο¬ed if {π + π½ − π½π(1 + π)} (π + π + 1) {π + πΌ − πΌπ(1 + π)} ππ ππ β£π§β£π+1 ≤ . (1 − π½)π 1−πΌ Solving for β£π§β£, we get the result. β‘ For the choice of π = 0, we have the following result of Uralegaddi and Ganigi [15]. Remark 7.4. Let the function π (π§) deο¬ned by (1) be in Σ∗π (πΌ) and πΉ (π§) given by (7.2). Then πΉ (π§) is in Σ∗π (πΌ) for β£π§β£ ≤ π(πΌ, π½) where ( π(πΌ, π½) = inf π π(1 − π½)(π + πΌ) (1 − πΌ)(π + π + 1)(π + π½) The result is sharp for the function ππ (π§) = 1 π§ )1/(π+1) + , 1−πΌ π π+πΌ π§ , π = 1, 2, 3, . . . . π = 1, 2, 3, . . . . Acknowledgments. The authors thank Prof.G. Murugusundaramoorthy for his suggestions and comments on the original manuscript( in particular on the partial sums chapter). The authors also would like to thank the anonymous referee for his/her comments. A NEW SUBCLASS OF MEROMORPHIC FUNCTION... 121 References [1] M. K. Aouf, On a certain class of meromorphic univalent functions with positive coeο¬cients, Rend. Mat. Appl. 7 (11), (1991), no. 2, 209-219. [2] S. K. Bajpai, A note on a class of meromorphic univalent functions, Rev. Roumaine Math. Pures Appl. 22 (1977), no. 3, 295-297. [3] N. E. Cho, S. H. Lee, and S. Owa, A class of meromorphic univalent functions with positive coeο¬cients, Kobe J. Math. 4 (1987), no. 1, 43-50. [4] J. 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Muthunagai Department of Mathematics, Sri Sairam Engineering College Tambaram, Chennai-44, India E-mail address: muthunagaik@yahoo.com