The Noor Integral and Strongly Starlike Functions

Transcription

1 Journal of Mathematical Analysis and Applications 61, (001) doi: /jmaa , available online at on The Noor Integral and Strongly Starlike Functions Jinlin Liu Department of Mathematics, Yangzhou University, Yangzhou 500, Jiangsu, People s Republic of China Submitted by M. A. Noor Received November 13, 000 Let A denote the class of analytic functions f z defined in the unit disc satisfying the condition f 0 =f 0 1 = 0. We introduce some new subclasses of strongly starlike functions defined by the Noor integral and study their properties. 001 Academic Press Key Words: strongly starlike function; strongly convex function; Noor integral. 1. INTRODUCTION Let A denote the class of analytic functions defined in the unit disc E = z z < 1 and normalized by f 0 =0, f 0 =1. A function f z belonging to A is said to be starlike of order γ if it satisfies zf z Re >γ z E (1.1) f z for some γ 0 γ<1. We denote by S γ the subclass of A consisting of functions which are starlike of order γ in E. Also, a function f z in A is said to be convex of order γ if it satisfies zf z S γ, or Re 1 + zf z f z >γ z E (1.) for some γ 0 γ<1. We denote by C γ the subclass of A consisting of all functions which are convex order γ in E. If f z A satisfies ( zf arg z ) f z γ < π 441 β z E (1.3) 00-47X/01 $35.00 Copyright 001 by Academic Press All rights of reproduction in any form reserved.

2 44 jinlin liu for some γ 0 γ<1 and β 0 <β 1, then f z is said to be strongly starlike of order β and type γ in E. We denote this by f z S β γ. If f z A satisfies ( arg 1 + zf z ) f z γ < π β z E (1.4) for some γ 0 γ<1 and β 0 <β 1, then we say that f z is strongly convex of order β and type γ in E. We denote by C β γ the class of all such functions. It is obvious that f z A belongs to C β γ if and only if zf z S β γ. Also, we note that S 1 γ =S γ and C 1 γ =C γ. Let f A. Denote by D α A A the operator defined by which implies that D α f z = z f z 1 z α α+1 > 1 D n f z = z zn 1 f z n n! for n N 0 = 0 1 and the operator ( ) stands for the Hadamard product or convolution. We note that D 0 f z =f z and D 1 f z =zf z. The operator D n f z is called the Ruscheweyh derivative of nth order of f (see [13]). Several classes of analytic functions, defined by using this operator, have been studied; see [1,, 6, 14]. Recently, Noor [7] and Noor and Noor [8] defined and studied an integral operator I n A A analogous to D n f as follows. Let f n z =z/ 1 z n+1, n N 0, and let fn 1 be defined such that f n z fn 1 z z = (1.5) 1 z Then [ ] I n f = fn 1 z 1 f = f (1.6) 1 z n+1 We note that I 0 f = zf and I 1 f = f. The operator I n defined by (1.6) is called the Noor integral of nth order of f. For the properties and applications of the Noor integral, see Noor [7] and Noor and Noor [8]. From (1.5), (1.6), and a well-known identity for D n f, it follows z I n+1 f = n + 1 I n f ni n+1 f (1.7) The relation (1.7) plays an important and significant role in obtaining our results.

3 strongly starlike functions 443 Using the Noor integral, we introduce and study the properties of some new classes of the analytic functions: Sn β γ = f z A I n f z S β γ z I nf z γ for all z E I n f z and C n β γ = f z A I n f z C β γ 1+ z I nf z I n f z γ for all z E Clearly, f C n β γ if and only if zf Sn β γ. For n = 1, it is easy to see that S1 β γ is the class of strongly starlike functions of order β and type γ whereas C 1 β γ is the class of strongly convex functions of order β and type γ. Also, for n = 1 and β = 1, we note that S1 1 γ =S γ and C 1 1 γ =C γ.. MAIN RESULTS In order to give our results, we need the following result, which is due to Nunokawa [10]. Lemma.1 (see [10]). Let a function p z =1 + c 1 z + c z + be analytic in E and p z 0 z E. If there exists a point z 0 E such that arg p z < π β z < z 0 and arg p z 0 = π β 0 <β 1 then we have z 0 p z 0 /p z 0 =ikβ, where k 1 ( a + 1 a k 1 and p z 0 1/β =±ia a >0. ) (when arg p z 0 = π β ) ( a + 1 ) (when arg p z a 0 = π ) β Theorem.1. S n β γ S n+1 β γ for each n N 0. Let f z Sn β γ. Then we set z I n+1 f z I n+1 f z = γ + 1 γ p z (.1) where p z =1 + c 1 z + c z + is analytic in E and p z 0 for all z E. Using (1.7) and (.1), we have n + 1 I nf z = n + γ + 1 γ p z (.) I n+1 f z

4 444 jinlin liu Differentiating both sides of (.) logarithmically, it follows that z I n f z = z I n+1f z 1 γ zp z + I n f z I n+1 f z n + γ + 1 γ p z 1 γ zp z = γ + 1 γ p z + n + γ + 1 γ p z or z I n f z 1 γ zp z γ = 1 γ p z + (.3) I n f z n + γ + 1 γ p z Suppose that there exists a point z 0 E such that arg p z < π β z < z 0 and arg p z 0 = π β Then, applying Lemma.1, we can write that z 0 p z 0 /p z 0 =ikβ and p z 0 1/β =±ia a >0. Therefore, if arg p z 0 = π β, then z 0 I n f z 0 [ γ = 1 γ p z I n f z = 1 γ a β e iπβ/ [1 + This implies that z0 I arg n f z 0 γ I n f z 0 = π 1 β + arg ikβ + n + γ + 1 γ a β e iπβ/ = π β + tan 1 z 0 p ] z 0 /p z 0 n + γ + 1 γ p z 0 ikβ n + γ + 1 γ a β e iπβ/ kβ n + γ + 1 γ a β cos πβ/ n + γ + n + γ 1 γ a β cos πβ/ + 1 γ a β kβ 1 γ a β sin πβ/ π ( β where k 1 ( a + 1 ) ) 1 a which contradicts the condition f z Sn β γ. Similarly, if arg p z 0 = π β, then we obtain that z0 I arg n f z 0 γ π I n f z 0 β which also contradicts the hypothesis that f z Sn β γ. Thus the function p z has to satisfy arg p z < π β z E. This shows that arg z In+1 f z γ < π β z E I n+1 f z or f z Sn+1 β γ. ]

5 strongly starlike functions 445 Theorem.. C n β γ C n+1 β γ for each n N 0. f z C n β γ I n f z C β γ z I n f z S β γ I n zf z S β γ zf z Sn β γ zf z S n+1 β γ I n+1 zf z S β γ z I n+1 f z S β γ I n+1 f z C β γ f z C n+1 β γ For c> 1 and f z A, we define the integral operator L c f as L c f = c + 1 z c z 0 t c 1 f t dt (.4) The operator L c f when c N = 1 3 was studied by Bernardi [3]. For c = 1 L 1 f was introduced by Libera [5]. Theorem.3. Let c> γ and 0 γ<1. If f z Sn β γ with z I n L c f z /I n L c f z γ for all z E, then we have L c f Sn β γ. Set z I n L c f z I n L c f z = γ + 1 γ p z (.5) where p z is analytic in E, p 0 =1, and p z 0 z E. From (.4), we have z I n L c f z = c + 1 I n f z ci n L c f z (.6) Using (.5) and (.6), we get c + 1 I nf z = c + γ + 1 γ p z (.7) I n L c f z Differentiating both sides of (.7) logarithmically, we obtain z I n f z 1 γ zp z γ = 1 γ p z + I n f z c + γ + 1 γ p z Suppose that there exists a point z 0 E such that arg p z < π β z < z 0 and arg p z 0 = π β Then, applying Lemma.1, we can write that z 0 p z 0 /p z 0 =ikβ and p z 0 1/β =±ia a >0.

6 446 jinlin liu If arg p z 0 = π β, then z 0 I n f z 0 I n f z 0 [ γ = 1 γ p z = 1 γ a β e iπβ/ [1 + This shows that z0 I arg n f z 0 γ I n f z 0 = π β + arg ikβ 1 + c + γ + 1 γ a β e iπβ/ = π β + tan 1 z 0 p ] z 0 /p z 0 c + γ + 1 γ p z 0 ikβ c + γ + 1 γ a β e iπβ/ kβ c + γ + 1 γ a β cos πβ/ c + γ + c + γ 1 γ a β cos πβ/ + 1 γ a β + kβ 1 γ a β sin πβ/ π ( β where k 1 ( a + 1 ) ) 1 a which contradicts the condition f z Sn β γ. Similarly, we can prove the case arg p z 0 = π β. Thus we conclude that the function p z has to satisfy arg p z < π β for all z E. This gives that arg z In L c f z γ < π β z E I n L c f z or L c f z Sn β γ. Theorem.4. Let c> γ and 0 γ<1. If f z C n β γ and 1 + z I n L c f z / I n L c f z γ for all z E, then we have L c f z C n β γ. f z C n β γ zf z Sn β γ L c zf z Sn β γ z L c f z Sn β γ L c f z C n β γ ] ACKNOWLEDGMENT The author expresses his grateful thanks to the referee for his/her useful suggestions.

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