Surveys in Mathematics and its Applications ISSN 1842-6298 (electronic), 1843-7265 (print) Volume 4 (2009), 77 – 88 APPLICATIONS OF GENERALIZED RUSCHEWEYH DERIVATIVE TO UNIVALENT FUNCTIONS WITH FINITELY MANY COEFFICIENTS Abdul Rahman S. Juma and S. R. Kulkarni Abstract. By making use of the generalized Ruscheweyh derivative, the authors investigate several interesting properties of certain subclasses of univalent functions having the form X m f (z) = z − n=2 1 X ∞ (1 − β)en zn − ak z k . (1 − nβ + α(1 − n))B1λ,µ (n) k=m+1 Introduction Let A(n, p) denote the class of functions f normalized by f (z) = z p − ∞ P ak z k , (p, n ∈ k=n+p IN = {1, 2, 3, 4, ....}), which are analytic and multivalent in the unit disk U = {z : |z| < 1}. The function f (z) is said to be starlike of order δ(0 ≤ δ < p) if and only zf 0 (z) if Re f (z) > δ, (z ∈ U). On the other hand f (z) is said to be convex of order 00 (z) δ(0 ≤ δ < p) if and only if Re 1 + zff 0 (z) > δ, (z ∈ U). Now for α ≥ 0, 0 ≤ β < 1 and λ > −1, µ, ν ∈ IR, the following classes ( ( ) ) J λ,µ f (z) Jpλ,µ f (z) p λ,µ Mp (α, β) = f (z) ∈ A(n, p) : Re > α − p + β z(Jpλ,µ f (z))0 z(Jpλ,µ f (z))0 (1.1) ( Mλ,µ 1,1 (α, β) = f (z) ∈ A(1, 1) : Re ( J1λ,µ f (z) z(J1λ,µ f (z))0 ) ) J λ,µ f (z) 1 > α − 1 + β z(J λ,µ f (z))0 1 (1.2) 2000 Mathematics Subject Classification: 30C45 Keywords: Ruscheweyh derivative; Integral operator; Univalent function. ****************************************************************************** http://www.utgjiu.ro/math/sma 78 A. R. S. Juma and S. R. Kulkarni where J1λ,µ f (z) is the generalized Ruscheweyh derivative of f defined by J1λ,µ f (z) = Γ(µ − λ + ν + 2) λ,µ,ν µ−1 zJ (z f (z)) Γ(ν + 2)Γ(µ + 1) 0,z (1.3) λ,µ,ν where J0,z f (z) generalized fractional derivative operator of order λ (see eg.[1], [5], [7], [8]). We can write (1.3) as J1λ,µ f (z) =z− ∞ X ak B1λ,µ (k)z k (1.4) k=n+1 where B1λ,µ (k) = Γ(k + µ)Γ(ν + 2 + µ − λ)Γ(k + ν + 1) . Γ(k)Γ(k + ν + 1 + µ − λ)Γ(ν + 2)Γ(1 + µ) (1.5) λ,µ Consider the class Mλ,µ 1,1 (α, β, em ) is subclass of M1,1 (α, β) consisting of functions of the form f (z) = z − m X (1 − β)en λ,µ n=2 (1 − nβ + α(1 − n))B1 (n) zn − ∞ X ak z k , (1.6) k=m+1 where en = [1 − nβ + α(1 − n)]B1λ,µ (n) an . 1−β Many authors have studied the different cases (e.g. [2], [4], [6]), and for λ = µ, ν = 1, we get the case was studied by [3]. 2 Coefficient Bounds In order to prove our results, we need the following Lemma due to A.R.S.Juma and S. R. Kulkarni [2]. Lemma 1. Let f (z) = z p − ∞ P ak z k . Then f (z) ∈ Mλ,µ p (α, β) if and only if k=m+p ∞ X [(1 + α) − k(αp + β)] λ,µ B (k)ak < 1. [(1 + α) − p(αp + β)] p (2.1) k=m+p First we prove ****************************************************************************** Surveys in Mathematics and its Applications 4 (2009), 77 – 88 http://www.utgjiu.ro/math/sma 79 Applications of generalized Ruscheweyh derivative Theorem 2. Let the function f (z) defined by (1.6). Then f ∈ Mλ,µ 1,1 (α, β, em ) if and only if ∞ m X X (1 − kβ) + α(1 − k) λ,µ B1 (k)ak < 1 − en . (2.2) 1−β n=2 k=m+1 Proof. Let an = (1 − β)en (1 − nβ + α(1 − n))B1λ,µ (n) . (2.3) λ,µ We can say that f ∈ Mλ,µ 1,1 (α, β, em ) ⊂ M1,1 (α, β) if and only if m X [(1 − nβ) + α(1 − n)]B λ,µ (n) 1 1−β n=2 or an + ∞ X [1 − kβ + α(1 − k)]B1λ,µ (k) ak < 1 1−β k=m+1 ∞ m X X [1 − kβ + α(1 − k)]B1λ,µ (k) ak < 1 − en . 1−β n=2 k=m+1 Corollary 3. Let f (z) defined by (1.6) be in Mλ,µ 1,1 (α, β, em ). Then for k ≥ m + 1 we have ∞ P en ) (1 − β)(1 − n=2 ak ≤ , (2.4) (1 − kβ + α(1 − k))B1λ,µ (k) this result is sharp for h(z) = z − m X (1 − β)(1 − (1 − β)en n λ,µ n=2 [(1 − nβ) + α(1 − n)]B1 (n) z − ∞ P en ) n=2 zk . [(1 − kβ) + α(1 − k)]B1λ,µ (k) (2.5) Corollary 4. Let f defined by (1.6). Then f ∈ M0,0 1,1 (0, β, em ), that is, f (z) = z − m X en (1 − β) n=2 1 − nβ n z − ∞ X ak z k , k=m+1 if and only if ∞ m X X 1 − kβ ak < 1 − en . 1−β k=m+1 n=2 ****************************************************************************** Surveys in Mathematics and its Applications 4 (2009), 77 – 88 http://www.utgjiu.ro/math/sma 80 A. R. S. Juma and S. R. Kulkarni Corollary 5. Let f defined by (1.6). Then f ∈ M1,0 1,1 (0, β, em ), that is, m ∞ X X (1 − β)(ν + 1)en n f (z) = z − =z − ak z k (1 − nβ)Γ(n + ν) n=2 if and only if k=m+1 m ∞ X X (1 − kβ)Γ(k + ν) ak < 1 − en . (1 − β)(ν + 1) n=2 k=m+1 Corollary 6. Let f defined by (1.6). Then f ∈ M0,0 1,1 (α, 0, em ), that is, f (z) = z − m X n=2 if and only if ∞ X ∞ X en zn − ak z k 1 + α(1 − n) k=m+1 (1 + α(1 − k))ak < 1 − m X en . n=2 k=m+1 Corollary 7. Let f defined by (1.6). Then f ∈ M1,0 1,1 (α, 0, em ) , that is, f (z) = z − m X n=2 if and only if ∞ X en (ν + 1) n ak z k z − (1 + α(1 − n))Γ(n + ν) k=m+1 ∞ m X X [1 + α(1 − k)]Γ(k + ν) ak < 1 − en . (ν + 1) n=2 k=m+1 We claim that all these results are entirely new. 3 Extreme points and other results Theorem 8. Let f1 (z), f2 (z), ..., f` (z) defined by fi (z) = z − m X (1 − β)en n λ,µ n=2 [1 − nβ + α(1 − n)]B1 (n) z − ∞ X ak,i z k , (3.1) k=m+1 (i = 1, · · · , `) be in the class Mλ,µ 1,1 (α, β, em ). Then G(z) defined by G(z) = ` X i=1 λi fi and ` X i=1 λi = 1, 0 ≤ m X en ≤ 1, 0 ≤ en ≤ 1 n=2 is also in this class. ****************************************************************************** Surveys in Mathematics and its Applications 4 (2009), 77 – 88 http://www.utgjiu.ro/math/sma 81 Applications of generalized Ruscheweyh derivative Proof. In view of Theorem 2, we have ∞ m X X [1 − kβ − α(1 − k)]B1λ,µ (k) ak,i < 1 − en 1−β n=2 k=m+1 for every i = 1, 2, · · · , `. Here G(z) = ` X λi fi = z − i=1 m X (1 − β)en n=2 (1 − nβ + α(1 − n))B1λ,µ (n) ∞ X − ` X ( λi ak,i )z k . k=m+1 i=1 Thus, ` ∞ X (1 − kβ + α(1 − k))B1λ,µ (k) X ( λi ak,i ) 1−β i=1 k=m+1 ! ` ∞ X X (1 − kβ + α(1 − k))B1λ,µ (k) = ak,i λi 1−β i=1 k=m+1 < ` X i=1 (1 − m X en )λi = 1 − n=2 Remark 9. The function G(z) = m X en . n=2 1 2 [f1 (z) + f2 (z)] belongs to Mλ,µ 1,1 (α, β, em ) if f1 (z), f2 (z) are in the class Mλ,µ 1,1 (α, β, em ). Remark 10. The class Mλ,µ 1,1 (α, β, em ) is convex set. Theorem 11. Let fi (z), (i = 1, · · · , `) defined by (3.1) be in Mλ,µ 1,1 (α, β, en ). Then the function F (z) = z − m X en (1 − β) λ,µ n=2 (1 − nβ + α(1 − n))B1 (n) is also in the class Mλ,µ 1,1 (α, β, em ), where bk = ∞ X n 1 ` z − P̀ bk z k (bk ≥ 0) k=m+1 ak,i . i=1 Proof. It is clear that ∞ ∞ ` X X (1 − kβ + α(1 − k))B1λ,µ (k) (1 − kβ + α(1 − k))B1λ,µ (k) X bk = ( ak,i ) 1−β `(1 − β) i=1 k=m+1 k=m+1 ! ` ∞ λ,µ X (1 − kβ + α(1 − k))B (k) 1X 1 = ak,i , ` 1−β i=1 k=m+1 ****************************************************************************** Surveys in Mathematics and its Applications 4 (2009), 77 – 88 http://www.utgjiu.ro/math/sma 82 A. R. S. Juma and S. R. Kulkarni by Theorem 2, we have the last expression is less than m P 1 ` P̀ m P (1 − i=1 en ) = 1 − n=2 en . n=2 The next Theorem is very useful to obtain the extreme points of the class Mλ,µ 1,1 (α, β, em ). Theorem 12. Let fm (z) = z − m X en (1 − β) λ,µ n=2 (1 − nβ + α(1 − n))B1 (n) zn (3.2) and for k ≥ m + 1 fk (z) = z − m X n=2 (1 − β)(1 − en (1 − β) (1 − nβ + α(1 − n))B1λ,µ (n) zn − m P en ) n=2 (1 − kβ + α(1 − k))B1λ,µ (k) zk . (3.3) Then the function f (z) ∈ if and only if it can be expressed in the ∞ ∞ P P form f (z) = δk fk (z), where δk ≥ 0, (k ≥ m) and δk = 1. Mλ,µ 1,1 (α, β, em ) k=m k=m Proof. Let f (z) = ∞ X δk fk (z) + δm fm (z) k=m+1 = δm z − δm + ∞ X m X en (1 − β) λ,µ n=2 (1 − nβ + α(1 − n))B1 (n) δk z − ∞ X δk m X zn (1 − nβ + α(1 − (1 − en )(1 − β) ∞ X n=2 k − δk z (1 − kβ + α(1 − k))B λ,µ (k) 1 k=m+1 k=m+1 ∞ X δk )z − (δm + k=m+1 − ∞ X k=m+1 ∞ X k=m+1 (1 − n))B1λ,µ (n) zn m P = (δm + n=2 k=m+1 ! en (1 − β) m P δk ) m X en (1 − β) λ,µ n=2 (1 − nβ + α(1 − n))B1 (n) zn en )(1 − β) n=2 (1 − kβ + α(1 − δk z k))B1λ,µ (k) k ****************************************************************************** Surveys in Mathematics and its Applications 4 (2009), 77 – 88 http://www.utgjiu.ro/math/sma 83 Applications of generalized Ruscheweyh derivative = z− m X en (1 − β) n=2 (1 − nβ + α(1 − n))B1λ,µ (n) (1 − ∞ X zn − m P en )(1 − β) n=2 k=m+1 (1 − kβ + α(1 − P e )(1−β) δk z k))B1λ,µ (k) k , m therefore at the end we can write (1−kβ+α(1−k))(1− ∞ P (1−β)(1−kβ+α(1−k))B1λ,µ (k) k=m+1 = (1 − m X ∞ X en ) n=2 n n=2 δk = (1 − m X en )(1 − δm ) < 1 − n=2 k=m+1 δk B1λ,µ (k) m X en . n=2 Thus f ∈ Mλ,µ 1,1 (α, β, em ). Conversely, let f ∈ Mλ,µ 1,1 (α, β, em ), that is, f (z) = z − m X en (1 − β) λ,µ n=2 (1 − nβ + α(1 − n))B1 (n) zn − ∞ X ak z k , k=m+1 put δk = (1 − kβ + α(1 − k))B1λ,µ (k) ak (k ≥ m + 1) m P (1 − β)(1 − en ) n=2 ∞ P we have δk ≥ 0 and if we set δm = 1 − δk , then we have k=m+1 f (z) = z − m X en (1 − β) n λ,µ n=2 (1 − nβ + α(1 − n))B1 (n) = fm (z) − = fm (z) − = (1 − ∞ X k=m+1 ∞ X k=m+1 ∞ X (1 − m P en )(1 − β) n=2 λ,µ k=m+1 (1 − kβ + α(1 − k))B1 (k) ! m X en (1 − β) z− z n − fk (z) δk λ,µ (1 − β + α(1 − n))B (n) 1 n=2 (fm (z) − fk (z))δk δk )fm (z) + k=m+1 z − ∞ X ∞ X k=m+1 δk fk (z) = ∞ X δk fk (z). k=m+1 Corollary 13. The extreme points of the class Mλ,µ 1,1 (α, β, em ) are the functions fk (z), (k ≥ m), defined by (3.2), (3.3). ****************************************************************************** Surveys in Mathematics and its Applications 4 (2009), 77 – 88 http://www.utgjiu.ro/math/sma δk z k 84 A. R. S. Juma and S. R. Kulkarni Theorem 14. Let f (z) defined by (1.6) be in the class Mλ,µ 1,1 (α, β, em ). Then f (z) is starlike of order δ(0 ≤ δ < 1) in |z| < r, where r is the largest value such that m X n=2 < en (1 − nβ + α(1 − n))B1λ,µ (n) r n−1 (1 − ∞ X + k=m+1 m P en ) n=2 (1 − kβ + α(1 − k))B1λ,µ (k) 1 , (k ≥ m + 1). 1−β rk−1 (3.4) Proof. It is sufficient to show that 0 zf (z) f (z) − 1 < 1 − δ. (3.5) Therefore m ∞ P P (1−β)en n− k z n ka z z − 0 k λ,µ (1−nβ+α(1−n))B (n) zf (z) 1 n=2 k=m+1 − 1 m ∞ f (z) − 1 = P (1−β)en z− P zn − ak z k (1−nβ+α(1−n))B1λ,µ (n) n=2 k=m+1 ∞ P (n−1)(1−β)en |z|n−1 + (k − 1)ak |z|k−1 λ,µ (1−nβ+α(1−n))B (n) 1 n=2 k=m+1 m ∞ P P (1−β)en 1− |z|n−1 − ak |z|k−1 λ,µ n=2 (1−nβ+α(1−n))B1 (n) k=m+1 ∞ m P P (k−1)(1−β)(1−m (n−1)(1−β)en n=2 en ) rn−1 + rk−1 λ,µ λ,µ (1−nβ+α(1−n))B (1−kβ+α(1−k))B (n) (k) 1 1 n=2 k=m+1 m P ≤ < Pe ) m 1− m P (1−β)en rn−1 λ,µ n=2 (1−nβ+α(1−n))B1 (n) − ∞ P k=m+1 (1−β)(1− . n n=2 (1−kβ+α(1−k))B1λ,µ (k) rk−1 Thus (3.5) holds true if the last expression is less than 1 − δ or, m X n=2 + ∞ X k=m+1 (n − δ)(1 − β)en n))B1λ,µ (n) (1 − δ)(1 − nβ + α(1 − m P (k − δ)(1 − β)(1 − en ) n=2 (1 − δ)(1 − kβ + α(1 − k))B1λ,µ (k) rn−1 rk−1 < 1 at the end we find (3.4). Making use the following Theorem we obtain the next corollary. ****************************************************************************** Surveys in Mathematics and its Applications 4 (2009), 77 – 88 http://www.utgjiu.ro/math/sma 85 Applications of generalized Ruscheweyh derivative Theorem 15. [Alexander’s Theorem] Let f be analytic in U with f (0) = f 0 (0) − 1 = 0. Then f is convex function if and only if zf 0 is starlike function. Corollary 16. Let f ∈ Mλ,µ 1,1 (α, β, em ). Then f (z) is convex of order δ(0 ≤ δ < 1) in |z| < r where r is the largest value such that m X n=2 k(1 − nen (1 − nβ + α(1 − n))B1λ,µ (n) rn−1 + m P en ) n=2 (1 − kβ + α(1 − k))B1λ,µ (k) rk−1 < 1 . 1−β Theorem 17. Let f ∈ Mλ,µ 1,1 (α, β, em ) and λ > 0, if dn = (1 − β)e2n (2 ≤ n ≤ m), (1 − nβ + α(1 − n))B1λ,µ (n) (3.6) then the function H(z) = z − m X (1 − β)dn λ,µ n=2 ((1 − nβ) + α(1 − n))B1 (n) zn − ∞ X ak z k k=m+1 is also in the class Mλ,µ 1,1 (α, β, em ). Proof. By assumption we have (1 − nβ + α(1 − n))B1λ,µ (n) > 1 therefore, dn = So, 0 ≤ m P n=2 dn < m P (1 − β)e2n (1 − nβ + α(1 − n))B1λ,µ (n) < en ≤ 1. en ≤ 1, thus n=2 ∞ ∞ X X (1 − kβ + α(1 − k))B1λ,µ (k) (1 − kβ + α(1 − k))B1λ,µ (k) a < < 1. k m m P P k=m+1 k=m+1 (1 − β)(1 − dn ) (1 − β)(1 − en ) n=2 n=2 Theorem 18. Let f, g ∈ Mλ,µ 1,1 (α, β, em ) and λ > 0. Then (f ∗ g)(z) = z − m X (1 − β)2 e2n zn − λ,µ 2 2 n=2 (1 − nβ + α(1 − n)) (B1 (n)) is also in the class Mλ,µ 1,1 (α, β, dm ) if λ1 ≤ inf k ∞ X ak bk z k k=m+1 (B1λ,µ (k))2 P d ) − 1, where dn(2 ≤ n ≤ m) m (1− n n=2 are defined by (3.6). ****************************************************************************** Surveys in Mathematics and its Applications 4 (2009), 77 – 88 http://www.utgjiu.ro/math/sma 86 A. R. S. Juma and S. R. Kulkarni Proof. By making use of (3.6), we have (f ∗ g)(z) = z − m X (1 − β)dn λ,µ n=2 (1 − nβ + α(1 − n))B1 (n) ∞ X zn − ak bk z k . k=m+1 By Theorem 17, we have ∞ X (1 − kβ + α(1 − k))B1λ,µ (k) ak < 1 m P k=m+1 (1 − β)(1 − dn ) n=2 and ∞ X (1 − kβ + α(1 − k))B1λ,µ (k) bk < 1. ∞ P k=m+1 (1 − β)(1 − dn ) n=2 Therefore, by Cauchy-Schwarz inequality we obtain ∞ X (1 − kβ + α(1 − k))B1λ,µ (k) p ak bk < 1. m P k=m+1 dn ) (1 − β)(1 − (3.7) n=2 Now, we want to prove that ∞ X (1 − kβ + α(1 − k))B1λ,µ (k) ak bk < 1. m P k=m+1 (1 − β)(1 − dn ) (3.8) n=2 In view of (3.7) the inequality (3.8) holds true if p ak bk B1λ1 ,µ (k) B1λ,µ (k) < 1. (3.9) But we have B1λ,µ (k) p (1 − kβ + α(1 − k))B1λ,µ (k) p a b < ak bk < 1. k k m m P P 1− dn (1 − β)(1 − dn ) n=2 n=2 Therefore (3.9) holds true if 1− m P dn n=2 B1λ,µ (k) < B1λ,µ (k) B1λ1 ,µ (k) , or B1λ1 ,µ (k) < (B1λ,µ (k))2 . m P 1− dn n=2 Then λ1 < (B1λ,µ (k))2 m 1− dn n=2 P − 1. ****************************************************************************** Surveys in Mathematics and its Applications 4 (2009), 77 – 88 http://www.utgjiu.ro/math/sma Applications of generalized Ruscheweyh derivative 87 Theorem 19. Let f (z) ∈ Mλ,µ 1,1 (α, β, em ). Then Z z E(t) − α λ,µ J1 f (z) = exp dt , |E(z)| < 1, z ∈ U. 0 (βE(t) − α)t Proof. The case α = 0 is obvious. Therefore, suppose that α 6= 0. Then for f ∈ J λ,µ f (z) 1 Mλ,µ 1,1 (α, β, em ) and let w = z(J λ,µ f (z))0 we have Re(w) > α|w − 1| + β, therefore, 1 w−1 1 w−1 = E(z) , where |E(z)| < 1, z ∈ U. This yields w−β < α or w−β α Thus J1λ,µ f (z) z(J1λ,µ f (z))0 −1 J1λ,µ f (z) z(J1λ,µ f (z))0 −β (J1λ,µ f (z))0 (J1λ,µ f (z)) = = E(z) E(z) J1λ,µ f (z) − z(J1λ,µ f (z))0 = or . λ,µ λ,µ 0 α α J1 f (z) − βz(J1 f (z)) E(z) − α . (βE(z) − α)z By the integration we get the result. References [1] S. P. Goyal and Ritu Goya, On a class of multivalent functions defined by generalized Ruscheweyh derivatives involving a general fractional derivative operator, Journal of Indian Acad. Math., 27 (2) (2005), 439-456. MR2259538(2007d:30005). Zbl 1128.30008. [2] A. R. S. Juma and S. R. Kulkarni, Application of generalized Ruscheweyh derivative to multivalent functions, J.Rajasthan Acad. Phy. Sci. 6 (2007), no.3, 261-272. MR2355701. [3] Sh. Najafzadeh and S. R. Kulkarni, Application of Ruscheweyh derivative on univalent functions with negative and fixed many coefficients, J. Rajasthan Acad. Phy. Sci., 5 (1) (2006), 39-51. MR2213990. Zbl 1137.30005. [4] S. Owa, M. Nunokawa and H. M. Srivastava, A certain class of multivalent functions, Appl. Math. Lett., 10 (2) (1997), no.2, 7-10. MR1436490. [5] R. K. Raina and T. S. Nahar, Characterization properties for starlikeness and convexity of some subclasses of analytic functions involving a class of fractional derivative operators, Acta Math. Univ. Comenian, 69 (N.S.) (2000), 1-8. MR1796782(2001h:30014). [6] S. Shams and S. R. Kulkarni, class of univalent functions with negative and fixed finitely many coefficients, Acta Cienica Indica Math. 29 (3) (2003), no. 3, 587-594. MR2024372. ****************************************************************************** Surveys in Mathematics and its Applications 4 (2009), 77 – 88 http://www.utgjiu.ro/math/sma 88 A. R. S. Juma and S. R. Kulkarni [7] H. M. Srivastava, Distortion inequalities for analytic and univalent functions associated with certain fractional calculus and other linear operators (In analytic and Geometric Inequalities and Applications, eds. T. M. Rassias and H.M. Srivastava), Kluwar Academic Publishers, 478 (1999), 349-374. MR1785879(2001h:30016). [8] H. M. Srivastava and R. K. Saxena, Operators of fractional integration and their applications, Applied Math. and Computation, 118 (2001), no. 1, 1-52. MRMR1805158(2001m:26016). Zbl 1022.26012. Abdul Rahman S. Juma S. R. Kulkarni Department of Mathematics Department of Mathematics Al-Anbar University, Fergusson College, Ramadi, Iraq. Pune 411004, India e-mail: absa662004@yahoo.com e-mail: kulkarni− ferg@yahoo.com ****************************************************************************** Surveys in Mathematics and its Applications 4 (2009), 77 – 88 http://www.utgjiu.ro/math/sma