Hacettepe Journal of Mathematics and Statistics Volume 34 (005), 9 15 CERTAIN SUBCLASSES OF STARLIKE AND CONVEX FUNCTIONS OF COMPLEX ORDER V Ravichandran, Yasar Polatoglu, Metin Bolcal, and Aru Sen Received 4 : 1 : 004 : Accepted 05 : 1 : 005 Abstract In the present investigation, we consider certain subclasses of starlike and convex functions of complex order, giving necessary and sufficient conditions for functions to belong to these classes Keywords: Starlike functions, Convex functions, Starlike functions of complex order, Convex functions of complex order 000 AMS Classification: 30 C 45 1 Introduction Let A be the class of all analytic functions (1) f() = + a + a 3 3 + in the open unit disk = { C; < 1} A function f A is subordinate to an univalent function g A, written f() g(), if f(0) = g(0) and f( ) g( ) Let Ω be the family of analytic functions ω() in the unit disc satisfying the conditions ω(0) = 0, ω() < 1 for Note that f() g() if there is a function w() Ω such that f() = g(ω()) Let S be the subclass of A consisting of univalent functions The class S (φ), introduced and studied by Ma and Minda [5], consists of functions in f S for which f() φ(), ( ) Department of Computer Applications, Sri Venkateswara College of Engineering, Sriperumbudur 60 105, India E-mail: vravi@svceacin Department of Mathematics, Faculty of Sciences and Arts, Kültür University, İstanbul E-mail: (Y Polatoglu) ypolatoglu@ikuedutr (M Bolcal) mbolcal@ikuedutr (A Sen) asen@ikuedutr
10 V Ravichandran, Y Polatoglu, M Bolcal, A Sen The functions h φn (n =, 3, ) are defined by h φn() h φn () = φ(n 1 ), h φn (0) = 0 = h φn(0) 1 The functions h φn are all functions in S (φ) We write h φ simply as h φ Clearly, ( ) φ(x) 1 () h φ () = exp dx x 0 Following Ma and Minda [5], we define a more general class related to the class of starlike functions of complex order as follows 11 Definition Let b 0 be a complex number Let φ() be an analytic function with positive real part on, which satisfies φ(0) = 1, φ (0) > 0, and which maps the unit disk onto a region starlike with respect to 1 and symmetric with respect to the real axis Then the class Sb (φ) consists of all analytic functions f A satisfying 1 + 1 b f() 1 φ() The class C b (φ) consists of the functions f A satisfying 1 + 1 b f () f () φ() Moreover, we let S (A, B, b) and C(A, B, b) (b 0, complex) denote the classes S b (φ) and C b (φ) respectively, where φ() = 1 + A, ( 1 B < A 1) 1 + B The class S (A, B, b), and therefore the class S b (φ), specialie to several well-known classes of univalent functions for suitable choices of A, B and b The class S (A, B, 1) is denoted by S (A, B) Some of these classes are listed below: (1) S (1, 1, 1) is the class S of starlike functions [1,, 7] () S (1, 1, b) is the class of starlike functions of complex order introduced by Wiatrowski [1] (3) S (1, 1, 1 β), 0 β < 1, is the class S (β) of starlike functions of order β This class was introduced by Robertson [8] (4) S (1, 1, e iλ cos λ), λ < π is the class of λ-spirallike functions introduced by Spacek [11] (5) S (1, 1, (1 β)e iλ cos λ), 0 β < 1, λ < π, is the class of λ-spirallike functions of order β This class was introduced by Libera [4] Let ST (b) denote 1 + 1 f () 1 Then we have the following: b f() (6) S (1, 0, b) is the set defined by ST (b) 1 < 1 (7) S (β, 0, b) is the set defined by ST (b) 1 < β, 0 β < 1 (8) S ST (b) 1 (β, β, b) is the set defined by < β, 0 β < 1 (ST (b)+1) (9) S (1, ( 1 + 1 ), b) is the set defined by ST (b) M < M M (10) S (1 β, 1, b) is the set defined by ReST (b) > β, 0 β < 1 To prove our main result, we need the following Lemma due to Miller and Mocanu:
Starlike and Convex Functions of Complex Order 11 1 Lemma [6, Corollary 34h1, p135] Let be univalent in and let ϕ() be analytic in a domain containing q( ) If q ()/ϕ() is starlike, then p ()ϕ(p()) q ()ϕ() implies that p(), and is the best dominant Let C be the class of convex analytic functions in We will also need the following result: 13 Lemma [10, Theorem 36, p 86] For f, h C and g h, we have f g f h A necessary and Sufficient Condition We begin with the following: 1 Lemma Let φ be a convex ( function defined on and satisfying φ(0) = 1 As in ) φ(x) 1 Equation (1) let h φ () = exp dx, and let = 1 + c 0 x 1 + be analytic in Then (3) 1 + q () φ() if and only if for all s 1 and t 1, we have (4) q(t) q(s) sh φ(t) th φ (s) Proof Our result and its proof are motivated by a similar result of Ruscheweyh [rus] for functions in the class S (φ) Also see Ruscheweyh [10, Theorem 37, pages 86-88] Let satisfy (3) Since the function ( s p() = 1 sx t ) dx 1 tx 0 is convex and univalent in for s, t := { C : = 1}, s t, by Lemma 1 we have: q () (5) p() (φ() 1) p() For an analytic function h() with h(0) = 0, we have (6) (h p)() = t s h(x) dx x, and using (6), we see that (5) is equivalent to t q (x) t φ(x) 1 dx dx, q(x) x s s which gives the desired assertion (4) upon exponentiation (7) To prove the converse, let us assume that (4) holds By taking t = 1 in (4), we have q(s) sh φ() h φ (s), and therefore we have (8) q(s) = sh φ(φ s()) h φ (sφ, s())
1 V Ravichandran, Y Polatoglu, M Bolcal, A Sen where φ s() are analytic in and satisfy φ s() Thus we can find a sequence s k 1 such that φ sk φ locally uniformly in, where φ () ( ) Therefore, by making use of (8), we have for any fixed, [ ] sk q(s k ) 1 + q () This shows that 1 + q () = lim k = lim k (s k 1) φ sk () h φ (φ sk ()) = φ ()h φ(φ ()) h φ (φ ()) ( h φ h φ [ hφ (s k φ sk ()) h φ (φ sk ()) s k φ sk () φ sk () ) ( ) = φ( ), ( ), which completes the proof of our Lemma 1 By making use of Lemma 1, we now have the following: Theorem ( Let φ be a convex function defined on which satisfies φ(0) = 1, and ) φ(x) 1 h φ () = exp dx be as in Equation (1) The the function f belongs to S 0 x b (φ) if and only if for all s 1 and t 1, we have ] (9) 1 sf(t) b sh φ (t) tf(s) th φ (s) Proof Define the function by 1/b f() (10) := Then a computation show that 1 + q () = 1 + 1 b f() 1 The result now follows from Lemma 1 As an immediate consequence of Theorem, we have: 3 Corollary Let φ() and h φ () be as in Theorem If f S b (φ), then we have (11) ( f() ) 1 b h φ () 3 Another Subordination Result In this section, we prove the following without the assumption that the function φ is convex We only require that the function φ be starlike with respect to the origin 31 Corollary If f Sb (φ), then we have 1 f() b h φ () (1), where h φ () is given by ()
Starlike and Convex Functions of Complex Order 13 Proof Define the functions p() and by 1/b f() p() :=, := h φ() Then a computation yields and 1 + p () p() q () = 1 + 1 b = h φ() h φ () Since f S b (φ), we have p () p() = 1 b f() 1 1 = φ() 1 f() 1 φ() 1 = q () The result now follows by an application of Lemma 11 4 The Fekete-Segö inequality In this section, we obtain the Fekete-Segö inequality for functions in the class S b (φ) 41 Theorem Let φ() = 1 + B 1 + B + B 3 3 + If f() given by Equation (1) belongs to Sb (φ), then { } a 3 µa max 1; B + (1 µ)bb 1 B 1 The result is sharp Proof If f() Sb (φ), then there is a Schwar function w(), analytic in, with w(0) = 0 and w() < 1 in and such that (13) 1 + 1 b f() 1 = φ(w()) Define the function p 1() by (14) p 1() := 1 + w() 1 w() = 1 + c1 + c + Since w() is a Schwar function, we see that Rp 1() > 0 and p 1(0) = 1 Define the function p() by (15) p() := 1 + 1 b f() 1 = 1 + b 1 + b + In view of the equations (13), (14) and (15), we have p1() 1 (16) p() = φ p 1() + 1 Since p 1() 1 p 1() + 1 = 1 [ ] c 1 + (c c 1 ) + (c 3 + c3 1 4 c1c)3 + and therefore p1() 1 φ = 1 + 1 [ 1 p 1() + 1 B1c1 + B1(c 1 c 1) + 1 ] 4 Bc 1 +,
14 V Ravichandran, Y Polatoglu, M Bolcal, A Sen from this equation and (16), we obtain and Since b 1 = 1 B1c1 b = 1 B1(c 1 c 1) + 1 4 Bc 1 f() = 1 + a + (a 3 a ) + (3a 4 + a 3 3a 3a ) 3 +, from Equation (15), we see that (17) (18) bb 1 = a, bb = a 3 a, or equivalently we have a = bb 1 = bb1c1, a 3 = 1 { bb + b b 1} = b 4 B1c1 + c { 1 b B1 b(b 1 B } ) 8 Therefore we have (19) a 3 µa = bb1 4 where v := 1 { c vc } 1, [ ] 1 B + (µ 1)bB 1 B 1 We recall from [5] that if p() = 1 + c 1 + c + is a function with positive real part, then c µc 1 max{1, µ 1 }, the result being sharp for the functions given by p() = 1 + 1, p() = 1 + 1 Our result now follows from an application of the above inequality, and we see that he result is sharp for the functions defined by 1 + 1 b f() 1 = φ( ) and 1 + 1 b f() 1 = φ() This completes the proof of the theorem
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