Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2013, Article ID 847287, 6 pages http://dx.doi.org/10.1155/2013/847287 Research Article Certain Properties of a Class of Close-to-Convex Functions Related to Conic Domains Wasim Ul-Haq and Shahid Mahmood Department of Mathematics, Abdul Wali Khan University Mardan, Mardan 23200, Pakistan Correspondence should be addressed to Wasim Ul-Haq; wasim474@hotmail.com Received 30 September 2012; Revised 3 March 2013; Accepted 17 March 2013 Academic Editor: Nikolaos Papageorgiou Copyright © 2013 W. Ul-Haq and S. Mahmood. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We aim to define a new class of close-to-convex functions which is related to conic domains. Many interesting properties such as sufficiency criteria, inclusion results, and integral preserving properties are investigated here. Some interesting consequences of our results are also observed. 1. Introduction Let A be the class of functions π ∞ π (π§) = π§ + ∑ ππ π§π , (1) was considered by Acu [4]; for study details on these classes, we refer to [5–7]. All these above mentioned classes were generalized to the classes SD(π, πΏ), CD(π, πΏ), and KD(π, π½, πΏ) by Shams et al. [8] and Srivastava et al. [9], respectively. The classes SD(π, πΏ) and CD(π, πΏ) are defined as π=2 which are analytic in the open unit disc πΈ = {π§ ∈ C : |π§| < 1}. Let π and π be analytic in πΈ, and we say that π is subordinate to π, written as π(π§) βΊ π(π§) if there exists a Schwarz function π€, which is analytic in πΈ with π€(0) = 0 and |π€(π§)| < 1 (π§ ∈ πΈ), such that π(π§) = π(π€(π§)). In particular, when π is univalent, then the above subordination is equivalent to π(0) = π(0) and π(πΈ) ⊆ π(πΈ); see [1]. Kanas and Wisniowska [2, 3] introduced and studied the classes of π-uniformly convex functions denoted by π-UCV and the corresponding class π-ST related by the Alexandertype relation. Later, the class π-uniformly close-to-convex functions denoted by π-UK defined as σ΅¨σ΅¨ σΈ σ΅¨σ΅¨σ΅¨ π§πσΈ (π§) σ΅¨ π§π (π§) ) > π σ΅¨σ΅¨σ΅¨σ΅¨ − 1σ΅¨σ΅¨σ΅¨σ΅¨ , π (π§) σ΅¨σ΅¨ π (π§) σ΅¨σ΅¨ π (π§) ∈ π-ST, π§ ∈ πΈ} Re ( σ΅¨σ΅¨ σΈ σ΅¨σ΅¨ π§πσΈ (π§) σ΅¨ π§π (π§) σ΅¨ ) > π σ΅¨σ΅¨σ΅¨σ΅¨ − 1σ΅¨σ΅¨σ΅¨σ΅¨ π (π§) σ΅¨σ΅¨ π (π§) σ΅¨σ΅¨ + πΏ, π§ ∈ πΈ} , (3) CD (π, πΏ) = {π ∈ A : Re (1 + π-UK = {π ∈ A : Re ( SD (π, πΏ) = {π ∈ A : (2) σ΅¨σ΅¨ σΈ σΈ σ΅¨ π§πσΈ σΈ (π§) σ΅¨σ΅¨ π§π (π§) σ΅¨σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨ ) > π σ΅¨σ΅¨ σΈ πσΈ (π§) σ΅¨σ΅¨ π (π§) σ΅¨σ΅¨σ΅¨ + πΏ, π§ ∈ πΈ} , where π ≥ 0, 0 ≤ π½, πΏ < 1. The class KD(π, π½, πΏ) known as π-uniformly close-to-convex functions of order π½ type πΏ 2 Abstract and Applied Analysis is the class of all those functions π ∈ A which satisfies the following condition: Re ( σ΅¨σ΅¨ σΈ σ΅¨σ΅¨ πσΈ (π§) σ΅¨σ΅¨ π (π§) σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨ + π½, ) ≥ π − 1 σ΅¨σ΅¨ σΈ σ΅¨σ΅¨ πσΈ (π§) σ΅¨σ΅¨ π (π§) σ΅¨ (4) (π ≥ 0, 0 ≤ π½, πΏ < 1; π§ ∈ πΈ) , for some π ∈ CD(π, πΏ). Motivated by the work of Noor et al. [10–13], we define the following. Definition 1. Let π ∈ A. Then, π is in the class TD(π, πΌ, π½, πΎ, πΏ) if and only if, for πΌ ≥ 0, 0 ≤ π½, πΎ, πΏ < 1, σ΅¨ σ΅¨ Re (π» (πΌ, π½, πΎ; π, π)) ≥ π σ΅¨σ΅¨σ΅¨π» (πΌ, π½, πΎ; π, π) − 1σ΅¨σ΅¨σ΅¨ , (π ≥ 0; π§ ∈ πΈ) , (5) 1 − πΌ πσΈ (π§) − π½] [ 1 − π½ πσΈ (π§) + πΌ [ 1−πΎ πσΈ (π§) [ (6) σΈ (π§πσΈ (π§)) 1+π§ , π = 0, { { { 1−π§ { 2 { { 1 + √π§ 2 { { { π = 1, ), 1 + 2 (log { { π 1 − √π§ { { { { { 2 2 { { 1+ sinh2 [( arccos π) arctan β√π§] , { 2 { 1 − π π { { ππ (π§) = { 0 < π < 1, { { { { { { { { 1 1 π π’(π§)/√π‘ { { 1+ ππ₯ ) sin( ∫ { { 2 −1 { π 2π (π‘) 0 { √1−π₯2 √1−(π‘π₯)2 { { { { { { 1 { + , π > 1, { π2 − 1 (8) where π’(π§) = (π§ − √π‘)/(1 − √π‘π§), π‘ ∈ (0, 1), π§ ∈ πΈ, and π§ is chosen such that π = cosh(ππ σΈ (π‘)/4π (π‘)), where π (π‘) is Legendre’s complete elliptic integral of the first kind and π σΈ (π‘) is complementary integral of π (π‘); see [2, 3]. for some π ∈ CD(π, πΏ), where π» (πΌ, π½, πΎ; π, π) = Extremal functions for these conic regions are denoted by ππ (π§), which are analytic in πΈ and map πΈ onto Ωπ such that ππ (0) = 1 and ππσΈ (0) > 1. These functions are given as: − πΎ] . ] We require the following results which are essential in our investigations. Lemma 2 (see [17, page 70]). Let β be convex function in πΈ and π : πΈ σ³¨→ πΆ with Re π(π§) > 0, π§ ∈ πΈ. If π is analytic in πΈ with π(0) = β(0), then Special Cases (i) (π, 0, π½, πΎ, πΏ) = KD(π, π½, πΏ); see [9]. (ii) TD(1, 0, π½, πΎ, 0) = UK(π½) and TD(1, 1, π½, πΎ, 0) = UQ(πΎ), the classes of uniformly close-to-convex and quasiconvex functions introduced and investigated in [14]. (iii) TD(0, πΌ, 0, 0 , 0) = QπΌ , the class of alpha quasiconvex functions, introduced and studied in [11]. (iv) TD(0, 0 π½, πΎ, πΏ) = K(π½, πΏ), the class of close-toconvex functions of order π½ type πΏ, [15]. (v) TD(0, 1, π½, πΎ, πΏ) = πΆ∗ (πΎ, πΏ), the class of quasiconvex functions of order πΎ type πΏ, [16]. The conditions π ≥ 0, πΌ ≥ 0, 0 ≤ π½, πΎ, πΏ < 1 on the parameters are assumed throughout the entire paper unless otherwise mentioned. Geometric Interpretation. A function π ∈ A is in the class TD(π, πΌ, π½, πΎ, πΏ) if and only if the functional π»(πΌ, π½, πΎ, πΏ; π, π) takes all the values in the conic domain Ωπ defined as follows: Ωπ = {π’ + πV; π’ > π√(π’ − 1)2 + V2 } . 2. Preliminaries Result (7) π (π§) + π (π§) π§πσΈ (π§) βΊ β (π§) implies π (π§) βΊ β (π§) . (9) Lemma 3 (see [17, page 195]). Let β be convex function in πΈ with β(0) = 0 and π΄ ≥ 0. Suppose that π ≥ 4/βσΈ (0) and that π΅(π§), πΆ(π§), and π·(π§) are analytic in πΈ and satisfy Re π΅ (π§) ≥ π΄ + |πΆ (π§) − 1| − Re (πΆ (π§) − 1) + ππ· (π§) , (10) for π§ ∈ πΈ. If π is analytic in πΈ with π(π§) = π1 π§+π2 π§2 +π3 π§3 +⋅ ⋅ ⋅ and the following subordination relation holds: π΄π§2 πσΈ σΈ (π§) + π΅ (π§) π§πσΈ (π§) + πΆ (π§) π (π§) + π· (π§) βΊ β (π§) , (11) then π (π§) βΊ β (π§) . (12) Lemma 4 (see [12]). If π(π§) βΊ β(π§) and π(π§) βΊ β(π§), then for πΌ ∈ [0, 1], (1 − πΌ) π (π§) + πΌπ (π§) βΊ β (π§) . (13) 3. Main Results First, we prove the following sufficiency criteria for the functions in the class TD(π, πΌ, π½, πΎ, πΏ). Abstract and Applied Analysis 3 Theorem 5. A function π ∈ A is said to be in the class TD(π, πΌ, π½, πΎ, πΏ), if ∞ ∑ Φπ (π; πΌ, π½, πΎ, πΏ) < (1 − π½) (1 − πΎ) , (14) π=2 π Using (1) and the series π(π§) = π§ + ∑∞ π=2 ππ π§ in (17), we have σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨π» (πΌ, π½, πΎ, πΏ; π, π) − 1σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ π−1 σ΅¨σ΅¨ (1 − πΌ) (1 − πΎ) (1 + ∑∞ ) π=2 πππ π§ = σ΅¨σ΅¨σ΅¨σ΅¨ ∞ π−1 σ΅¨σ΅¨ (1 − π½) (1 − πΎ) (1 + ∑π=2 πππ π§ ) σ΅¨ + where σ΅¨ σ΅¨ = (π + 1) [(1 − πΌ) (1 − πΎ) π +πΌ (1 − π½) π2 ] σ΅¨σ΅¨σ΅¨ππ σ΅¨σ΅¨σ΅¨ (15) + Proof. Let us assume that relation (6) holds. Now, it is sufficient to show that σ΅¨ σ΅¨ π σ΅¨σ΅¨σ΅¨π» (πΌ, π½, πΎ, πΏ; π, π)− 1σ΅¨σ΅¨σ΅¨ − Re [π» (πΌ, π½, πΎ, πΏ; π, π)− 1] < 1. (16) First, we consider σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨π» (πΌ, π½, πΎ, πΏ; π, π) − 1σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨ 1 − πΌ πσΈ (π§) = σ΅¨σ΅¨σ΅¨σ΅¨ − π½] [ σ΅¨σ΅¨ 1 − π½ πσΈ (π§) σ΅¨σ΅¨ σΈ σΈ σ΅¨σ΅¨σ΅¨ πΌ [ (π§π (π§)) + − πΎ] − 1σ΅¨σ΅¨σ΅¨ σΈ σ΅¨σ΅¨ 1−πΎ π (π§) σ΅¨σ΅¨ ] [ σ΅¨σ΅¨ σΈ σΈ σ΅¨σ΅¨ 1 − πΌ πσΈ (π§) πΌ (π§π (π§)) σ΅¨σ΅¨ = σ΅¨σ΅¨ + σ΅¨σ΅¨ 1 − π½ πσΈ (π§) 1 − πΎ πσΈ (π§) σ΅¨σ΅¨ σ΅¨σ΅¨ πΌπΎ (1 − πΌ) π½ σ΅¨ − − − 1σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ 1−π½ 1−πΎ σ΅¨σ΅¨ (1 − πΌ) (1 − πΎ) πσΈ (π§) σ΅¨ = σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ (1 − π½) (1 − πΎ) πσΈ (π§) + − πΌ (1 − π½) (π§πσΈ (π§)) σΈ (1 − π½) (1 − πΎ) πσΈ (π§) [(1 − πΎ) + πΌ (πΎ − π½)] πσΈ (π§) σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ . (1 − π½) (1 − πΎ) πσΈ (π§) σ΅¨σ΅¨σ΅¨ π−1 ) (1 − π½) (1 − πΎ) (1 + ∑∞ π=2 πππ π§ π−1 σ΅¨σ΅¨ [(1 − πΎ) + πΌ (πΎ − π½)] (1 + ∑∞ ) σ΅¨σ΅¨σ΅¨ π=2 πππ π§ σ΅¨σ΅¨ − ∞ (1 − π½) (1 − πΎ) (1 + ∑π=2 πππ π§π−1 ) σ΅¨σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨ ∞ π−1 σ΅¨σ΅¨ (1 − πΌ) (1 − πΎ) (∑π=2 πππ π§ ) = σ΅¨σ΅¨σ΅¨σ΅¨ π−1 ) σ΅¨σ΅¨ (1 − π½) (1 − πΎ) (1 + ∑∞ π=2 πππ π§ σ΅¨ Φπ (π; πΌ, π½, πΎ, πΏ) σ΅¨ σ΅¨ + [(π + 2 − π½) (1 − πΎ) +πΌ (π + 1) (πΎ − π½) π σ΅¨σ΅¨σ΅¨ππ σ΅¨σ΅¨σ΅¨] . 2 π−1 πΌ (1 − π½) (1 + ∑∞ ) π=2 π ππ π§ (17) 2 π−1 πΌ (1 − π½) (∑∞ ) π=2 π ππ π§ π−1 ) (1 − π½) (1 − πΎ) (1 + ∑∞ π=2 πππ π§ π−1 σ΅¨σ΅¨ [(1 − πΎ) + πΌ (πΎ − π½)] (∑∞ ) σ΅¨σ΅¨ π=2 πππ π§ σ΅¨σ΅¨ − σ΅¨ ∞ (1 − π½) (1 − πΎ) (1 + ∑π=2 πππ π§π−1 ) σ΅¨σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨ σ΅¨ σ΅¨ ∞ π−1 σ΅¨σ΅¨ (1 − πΌ) (1 − πΎ) (∑π=2 π σ΅¨σ΅¨σ΅¨ππ σ΅¨σ΅¨σ΅¨ |π§| ) ≤ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ π−1 ) σ΅¨σ΅¨ (1 − π½) (1 − πΎ) (1 − ∑∞ π=2 π σ΅¨σ΅¨ππ σ΅¨σ΅¨ |π§| σ΅¨ π−1 2 σ΅¨σ΅¨ σ΅¨σ΅¨ πΌ (1 − π½) (∑∞ ) π=2 π σ΅¨σ΅¨ππ σ΅¨σ΅¨ |π§| + σ΅¨ σ΅¨σ΅¨ π−1 ∞ σ΅¨ (1 − π½) (1 − πΎ) (1 − ∑π=2 π σ΅¨σ΅¨ππ σ΅¨σ΅¨ |π§| ) σ΅¨σ΅¨ σ΅¨σ΅¨ π−1 σ΅¨σ΅¨σ΅¨ [(1 − πΎ) + πΌ (πΎ − π½)] (∑∞ ) σ΅¨σ΅¨ π=2 π σ΅¨σ΅¨ππ σ΅¨σ΅¨ |π§| + σ΅¨ σ΅¨σ΅¨ π−1 σ΅¨σ΅¨σ΅¨σ΅¨ ∞ σ΅¨ (1 − π½) (1 − πΎ) (1 − ∑π=2 π σ΅¨σ΅¨ππ σ΅¨σ΅¨ |π§| ) σ΅¨σ΅¨ σ΅¨σ΅¨ (1 − πΌ) (1 − πΎ) (∑∞ π σ΅¨σ΅¨π σ΅¨σ΅¨) σ΅¨ π=2 σ΅¨σ΅¨ π σ΅¨σ΅¨ ≤ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ (1 − π½) (1 − πΎ) (1 − ∑∞ π=2 π σ΅¨σ΅¨ππ σ΅¨σ΅¨) 2 σ΅¨σ΅¨ σ΅¨σ΅¨ πΌ (1 − π½) (∑∞ π=2 π σ΅¨σ΅¨ππ σ΅¨σ΅¨) + σ΅¨σ΅¨ σ΅¨σ΅¨ (1 − π½) (1 − πΎ) (1 − ∑∞ π=2 π σ΅¨σ΅¨ππ σ΅¨σ΅¨) σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ [(1 − πΎ) + πΌ (πΎ − π½)] (∑∞ π=2 π σ΅¨σ΅¨ππ σ΅¨σ΅¨) σ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ . (1 − π½) (1 − πΎ) (1 − ∑∞ π=2 π σ΅¨σ΅¨ππ σ΅¨σ΅¨) σ΅¨σ΅¨ (18) Now, σ΅¨ σ΅¨ π σ΅¨σ΅¨σ΅¨π» (πΌ, π½, πΎ, πΏ; π, π) − 1σ΅¨σ΅¨σ΅¨ − Re [π» (πΌ, π½, πΎ, πΏ; π, π) − 1] σ΅¨ σ΅¨ ≤ (π + 1) σ΅¨σ΅¨σ΅¨π» (πΌ, π½, πΎ, πΏ; π, π) − 1σ΅¨σ΅¨σ΅¨ σ΅¨ σ΅¨ ∞ (1 − πΌ) (1 − πΎ) (∑π=2 π σ΅¨σ΅¨σ΅¨ππ σ΅¨σ΅¨σ΅¨) ≤ (π + 1) [ σ΅¨σ΅¨ σ΅¨σ΅¨ (1 − π½) (1 − πΎ) (1 − ∑∞ π=2 π σ΅¨σ΅¨ππ σ΅¨σ΅¨) 2 σ΅¨σ΅¨ σ΅¨σ΅¨ πΌ (1 − π½) (∑∞ π=2 π σ΅¨σ΅¨ππ σ΅¨σ΅¨) + σ΅¨σ΅¨ σ΅¨σ΅¨ (1 − π½) (1 − πΎ) (1 − ∑∞ π=2 π σ΅¨σ΅¨ππ σ΅¨σ΅¨) σ΅¨σ΅¨ σ΅¨σ΅¨ [(1 − πΎ) + πΌ (πΎ − π½)] (∑∞ π=2 π σ΅¨σ΅¨ππ σ΅¨σ΅¨) + σ΅¨σ΅¨ σ΅¨σ΅¨ ] . (1 − π½) (1 − πΎ) (1 − ∑∞ π=2 π σ΅¨σ΅¨ππ σ΅¨σ΅¨) (19) 4 Abstract and Applied Analysis σΈ where π(π§) = (π§πσΈ (π§)) /πσΈ (π§). Using (24) and (25) in relation (6), we obtain The last inequality is bounded above by 1, if ∞ ∞ π=2 π=2 σ΅¨ σ΅¨ σ΅¨ σ΅¨ (π + 1) [(1 − πΌ) (1 − πΎ) ( ∑ π σ΅¨σ΅¨σ΅¨ππ σ΅¨σ΅¨σ΅¨) + πΌ (1 − π½) ( ∑ π2 σ΅¨σ΅¨σ΅¨ππ σ΅¨σ΅¨σ΅¨) π» (πΌ, π½, πΎ, πΏ; π (π§)) ∞ σ΅¨ σ΅¨ + [(1 − πΎ) + πΌ (πΎ − π½)] ( ∑ π σ΅¨σ΅¨σ΅¨ππ σ΅¨σ΅¨σ΅¨)] π=2 ∞ σ΅¨ σ΅¨ ≤ (1 − π½) (1 − πΎ) (1 − ∑ π σ΅¨σ΅¨σ΅¨ππ σ΅¨σ΅¨σ΅¨) . π=2 = πΌ 1−πΌ [π (π§) − π½] + [π§πσΈ (π§) + π (π§) π (π§) − πΎ] 1−π½ 1−πΎ = (1−πΌ) (1−πΎ)+πΌ (1−π½) π (π§) πΌ ] π (π§) π§πσΈ (π§)+[ 1−πΎ (1 − π½) (1 − πΎ) − (20) π½ (1 − πΌ) (1 − πΎ) + πΌπΎ (1 − π½) (1 − π½) (1 − πΎ) = π΅ (π§) π§πσΈ (π§) + πΆ (π§) π (π§) + π· (π§) , Hence, (26) ∞ ∑ Φπ (π; πΌ, π½, πΎ, πΏ) ≤ (1 − π½) (1 − πΎ) , (21) π=2 where Φπ (π; πΌ, π½, πΎ, πΏ) is given by (15). This completes the proof. When we take πΌ = 0, π = 1, and π(π§) = π§ in the above theorem, we obtain the following sufficient condition for the functions to be in the class UK(π½) which is proved in [14]. Corollary 6 (see [14]). A function π ∈ π΄ is said to be in the class UK(π½) if ∞ σ΅¨ σ΅¨ (1 − π½) ∑ π σ΅¨σ΅¨σ΅¨ππ σ΅¨σ΅¨σ΅¨ ≤ . 2 π=2 (22) Corollary 7 (see [14]). A function π ∈ π΄ is said to be in the class UQ(πΎ) if where π΅ (π§) = πΆ (π§) = π· (π§) = − π½ (1 − πΌ) (1 − πΎ) + πΌπΎ (1 − π½) . (1 − π½) (1 − πΎ) (27) Now, since π ∈ TD(π, πΌ, π½, πΎ, πΏ), we have π΅ (π§) π§πσΈ (π§) + πΆ (π§) π (π§) + π· (π§) βΊ ππ (π§) . (23) π΅ (π§) π§π∗σΈ (π§) + πΆ (π§) π∗ (π§) + π·∗ (π§) βΊ ππ∗ (π§) , Proof. Let σΈ π (π§) = π (π§) , πσΈ (π§) (24) where π(π§) is analytic and π(0) = 1. Now differentiating (24), we have σΈ (π§πσΈ (π§)) πσΈ (π§) = π§πσΈ (π§) + π (π§) π (π§) , (25) (29) where π·∗ (π§) = πΆ(π§) + π·(π§) − 1. Using Lemma 3 with π΄ = 0, we obtain π∗ (π§) βΊ ππ∗ (π§) . (30) πσΈ (π§) = π (π§) βΊ ππ (π§) . πσΈ (π§) (31) This implies that The above corollary is obtained when we take πΌ = 1, π = 1, and π(π§) = π§ in Theorem 5. Theorem 8. Let π ∈ TD(π, πΌ, π½, πΎ, πΏ) and πΌ ≥ 4(π½/(1 + 3π½)). Then, π ∈ KD(π, 0, πΏ). (28) Replacing π(π§) by π∗ (π§) = π(π§) − 1 and ππ (π§) by ππ∗ (π§) = ππ (π§) − 1, the above subordination is equivalent to ∞ σ΅¨ σ΅¨ (1 − πΎ) ∑ π2 σ΅¨σ΅¨σ΅¨ππ σ΅¨σ΅¨σ΅¨ ≤ . 2 π=2 (1−πΌ) (1−πΎ)+πΌ (1−π½) π (π§) , (1−π½) (1−πΎ) πΌ , 1−πΎ Hence, π ∈ KD(π, 0, πΏ). This completes the proof. Corollary 9. Let π ∈ TD(π, πΌ, 0, 0, 0) = QπΌ . Then, π ∈ KD(0, 0, 0) = K. That is, QπΌ ⊂ K, πΌ ≥ 0. The above result is well-known inclusion proved in [11]. For π ∈ π΄, consider the following integral operator defined by πΉ (π§) = πΌπ [π] = π + 1 π§ π−1 ∫ π‘ π (π‘) ππ‘, π§π 0 π = 1, 2, 3 . . . . (32) This operator was given by Bernardi [18] in 1969. In particular, the operator πΌ1 was considered by Libera [19]. Now let us prove the following. Abstract and Applied Analysis 5 Theorem 10. Let π ∈ TD(π, πΌ, π½, πΎ, πΏ). Then, πΌπ [π] ∈ KD(π, 0, πΏ). Now proceeding in the similar manner as in the proof of Theorem 5 and using Lemma 3 with π΄ = 0, we obtain Proof. Let the function π be such that (6) is satisfied. It can easily be seen that according to [4], the function πΊ = πΌπ [π] ∈ CD(π, πΏ), and from (32), we deduce πσΈ (π§) = β (π§) βΊ ππ (π§) . πσΈ (π§) (1 + π) πσΈ (π§) = (1 + π) πΉσΈ (π§) + π§πΉσΈ σΈ (π§) , σΈ σΈ σΈ σΈ (1 + π) π (π§) = (1 + π) πΊ (π§) + π§πΊ (π§) . σΈ σΈ If we let π(π§) = πΉ (π§)/πΊ (π§) and π(π§) = 1/(π + 1 + (π§πΊσΈ σΈ (π§)/πΊσΈ )), then simple computations yield us σΈ σΈ From (36), it implies that (33) (34) π (π§) + π§πσΈ (π§) π (π§) βΊ ππ (π§) . Theorem 11. For πΌ > πΌ1 ≥ 0, TD (π, πΌ, π½, πΎ, πΏ) ⊆ TD (π, πΌ1 , π½, πΎ, πΏ) . π§π (π§) πσΈ (π§) + π§π (π§) πσΈ σΈ (π§) + π§πσΈ (π§) πσΈ (π§) (35) = (1 + π) πΊσΈ (π§) + π§πΊσΈ σΈ (π§) π» (πΌ1 , π½, πΎ, πΏ; π, π) = σΈ (36) σΈ (37) σΈ where π(π§) = (π§π (π§)) /π (π§). Using (36) and (37) in (6), we have π» (πΌ, π½, πΎ, πΏ; π (π§)) = πΌ 1−πΌ [β (π§) − π½] + [π§βσΈ (π§) + β (π§) π (π§) − πΎ] 1−π½ 1−πΎ = πΌ π§βσΈ (π§) 1−πΎ +[ − (1 − πΌ) (1 − πΎ) + πΌ (1 − π½) π (π§) ] β (π§) (1 − π½) (1 − πΎ) = π΅ (π§) π§β (π§) + πΆ (π§) β (π§) + π· (π§) , (38) where πΌ , 1−πΎ = πσΈ (π§) πΌ 1 − π½] (1 − 1 ) [ σΈ 1−π½ πΌ π (π§) πΆ (π§) = (π§πσΈ (π§)) πΌ1 [1 − πΌ πσΈ (π§) πΌ + − π½) + − πΎ)] ( ( σΈ πΌ 1 − π½ πσΈ (π§) 1−πΎ π (π§) ] [ πΌ πΌ = (1 − 1) π» (0, π½, πΎ, πΏ; π, π)+ 1 π» (πΌ, π½, πΎ, πΏ; π, π) . πΌ πΌ (44) Now, since π ∈ TD(π, πΌ, π½, πΎ, πΏ), we have π»(πΌ, π½, πΎ, πΏ; π, π) βΊ ππ (π§). Also, Theorem 8 gives us that π»(0, π½, πΎ, πΏ; π, π) βΊ ππ (π§). The use of Lemma 4 leads us to the required relation; that is, π»(πΌ1 , π½, πΎ, πΏ; π, π) βΊ ππ (π§). This completes the proof. Acknowledgment π½ (1 − πΌ) (1 − πΎ) + πΌπΎ (1 − π½) (1 − π½) (1 − πΎ) σΈ π΅ (π§) = After some simple computations, we have σΈ σΈ σΈ = π (π§) β (π§) + π§βσΈ (π§) , (43) π» (πΌ1 , π½, πΎ, πΏ; π, π) where β(π§) is analytic and π(0) = 1. From (36), we have πσΈ (π§) 1 − πΌ1 πσΈ (π§) − π½] [ 1 − π½ πσΈ (π§) (π§πσΈ (π§)) πΌ + 1 [ − πΎ] . 1−πΎ πσΈ (π§) ] [ Let (π§πσΈ (π§)) (42) Proof. Let π ∈ TD(π, πΌ, π½, πΎ, πΏ). Then, consider = π (π§) + π (π§) π§πσΈ (π§) . πσΈ (π§) = π (π§) + π§πσΈ (π§) π (π§) = β (π§) , πσΈ (π§) (41) By employing Lemma 2, we immediately obtain the desired result. σΈ σΈ π (π§) (1 + π) πΉ (π§) + π§πΉ (π§) = πσΈ (π§) (1 + π) πΊσΈ (π§) + π§πΊσΈ σΈ (π§) (40) (1 − πΌ) (1 − πΎ) + πΌ (1 − π½) π (π§) , (1 − π½) (1 − πΎ) π½ (1 − πΌ) (1 − πΎ) + πΌπΎ (1 − π½) π· (π§) = − . (1 − π½) (1 − πΎ) (39) The authors would like to thank the reviewer of this paper for his/her valuable comments on the earlier version of this paper. 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