A Note on the Powers of Bazilevič Functions

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International Journal of Mathematical Analysis Vol. 9, 015, no. 4, 061-067 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.1988/ijma.015.56169 A Note on the Powers of Bailevič Functions Marjono Faculty of Mathematics and Natural Sciences Brawijaya University Malang, Jawa Timur 65145, Indonesia D. K. Thomas Department of Mathematics Swansea University, Singleton Park Swansea, SA 8PP, UK Copyright c 015 Marjono and D. K. Thomas. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract For α 0, let B 1 (α) be the set of Bailevič functions f, analytic in D = { : < 1}, given by f() = + n= a n n, and satisfying Re 1 α f () f( α > 0. For ( f() > 0, let = 1 + a n() n. We consider the problem of determining whether a n () b n (), where b n () are the coefficients of the extreme functions ( f() of for f B 1 (α). Mathematics Subject Classification: Primary 30C45; Secondary 30C50 Keywords: Univalent functions, powers, coefficients, starlike functions, Bailevič functions, Fekete-Segö 1. Introduction Let S be the class of analytic normalised univalent functions f, defined in D = { : < 1} and given by

06 Marjono and D. K. Thomas For > 0, suppose that ( f() f() = + = 1 + a n n. (1.1) n= a n () n, and ( k() = 1 + k n () n, where k() is the Koebe function (1 ). Hayman and Hummel [3] posed the question of whether a n () k n () for n 1. De Brange s proof of the Bieberbach conjecture [] easily extends to show that a n () k n () is true when 1, whereas Hayman and Hummel [loc.cit] showed that this is false when > 1. It is known [5] that for starlike functions, a n () k n () is true for n 1, and > 0, and false for close-to-convex functions [4]. The class B 1 (α), of Bailevič functions with logarithmic growth, defined as follows, has been extensively studied see e.g. [6,7, 8].. Definition Suppose that f is analytic in D = { : < 1} and be given by (1.1). Then for α 0, f B 1 (α), if and only if, Re f () > 0. (.1) f( α α Since f B 1 (α) is a subset of the Bailevič functions [1], B 1 (α) S. For f B 1 (α), the coefficients a n exhibit irregular behaviour, and sharp bounds are known only when n =, 3, 4 [6]. In this paper we consider the validity of a n () b n () in the cases n = 1,, 3 for functions in B 1 (α), where

A note on the powers of Bailevič functions 063 ( f() b n () is the relevant extreme function of. We also give a Fekete-Segö theorem. We first note that taking powers in the extreme functions in [6], the relevant coefficients b 1 (), b (, 1), b (, ), b 3 (, 1) and b 3 (, ) of the extreme function ( f() for are as follows. b 1 () := if α 0, (1 + α) ( + α + ) b (, 1) := if 0 α 1, (1 + α) ( + α) b (, ) := if α 1, ( + α) b 3 (, 1) := (3 + α) + 4( + 3 α α 3 α( 1 + 3 ) if 0 α 1, 3(1 + α) 3 ( + α) 3 b 3 (, ) := if α 1. (3 + α) From (.1) we can write f () = f( α α p(), so that p P, the set of functions with positive real part in D. Let p() = 1 + p n n. We will use the following well-known lemma. Lemma If p P with coefficients p n as above, then p n for n 1, and p 1 p 1 p 1. Without loss in generality, we can assume that α 0 and 1, since α = 0 corresponds to the starlike functions, and when = 1 there is nothing to prove.

064 Marjono and D. K. Thomas 3. Theorem 1 a 1 () b 1 () if α > 0 and > 0. If 0 < α 1. a () b (, 1) if 0 < < 1, and if > 1 provided α < 1 b (, ) if > 1 provided α > 1. If α 1. a () b (, ) if > 1, and if 0 < < 1 provided α > 1 b (, 1) if 0 < < 1 provided α < 1. All these inequalities for a () are sharp, and false on all complimentary intervals. The inequalities a 3 () b 3 (, 1) and a 3 () b 3 (, ) are false for all α > 0 and > 0. Proof. Using (.1) and equating coefficients we obtain a 1 () = a () = a 3 () = p 1 (1 + α), p ( + α) + (1 α)p 1 (1 + α), p 3 (3 + α) + 3 + α + + ( 3α α )p 1 p (1 + α)( + α)(3 + α) + (6 + α + 9 + + (1 + α) 3 ( + α) 3(1 + α) (3 + α + ) + α(5 + 3 ))p 3 1 6(1 + α) 3 ( + α)(3 + α) 3. (3.1) The first inequality in Theorem 1 is trivial, since p 1 from the lemma. From (3.1), we have

A note on the powers of Bailevič functions 065 p a () = ( + α) + (1 α)p 1 (1 + α) 1 (1 α)( + α) = p + p ( + α) (1 + α) 1 1 = p 1 ( + α + ) ( + α) p 1 + (1 + α) p 1 1 ( 1 ( + α) p 1 ( + α + ) ) + (1 + α) p 1 := φ( p 1, α, ), where we have used the above lemma. Thus we need to maximise φ( p 1, α, ) over [0, ] for α > 0 and > 0, ignoring the case = 1. Simple consideration of the sign of p 1 in φ( p 1, α, ) gives the following. Case (i) 0 < α 1. (a) If 0 < < 1, then the maximum is a () b (, 1) is true. ( + a + ), and so (1 + α) ( + a) ( + α + ) (b) If > 1, then the maximum is again, provided α < 1, (1 + α) ( + α) and so a () b (, 1) is true provided α < 1. If > 1 and α > 1, then the maximum is is true in this case. Case (ii) α 1. (a) If 0 < < 1, the maximum is is true if α > 1. If 0 < < 1 and α < 1, the maximum is b (, 1). (b) If > 1, the maximum is ( + α), and so a () b (, ) ( + α) if α > 1 and so a () b (, ) ( + a + ) (1 + α) ( + a) and so a () ( + α), and so a () b (, ) is true.

066 Marjono and D. K. Thomas We note that the inequality for a 1 () is sharp when p 1 =, and the inequalities for a () are sharp on choosing either p 1 = p =, or p 1 = 0 and p = in (3.1). Choosing p 1 = p = in the expressions for a 1 () and a () in (3.1), it is a simple exercise to check that all the inequalities are false on all complementary intervals. To see that the inequalities for a 3 () are false, we choose the following values for p 1, p and p 3 in the expression for a 3 () in (3.1), noting that since the denominators are all positive, the algebraic calculations required to establish the inequalities are not difficult. Taking p 1 =, p = and p 3 =, shows that a 3 () > b 3 (, 1) when α < 1, and choosing p 1 =, p = 0 and p 3 =, that a 3 () > b 3 (, 1) when α > 1. Similarly taking p 1 = p = p 3 =, shows that a 3 () > b 3 (, ) when α < 1 and choosing p 1 = p 3 = and p = 0, that a 3 () > b 3 (, ) when α > 1. This completes the proof of Theorem 1. Finally we give a sharp Fekete-Segö result, noting that since we are considering the coefficients of, the relevant coefficients are a 1 () and a (). ( f() The simple proof is omitted. 4. Theorem For f B 1 (α) with a 1 () and a () defined as above, a () µa 1 () ( + α) ( + α) + (1 α µ) (1 + α) if µ (1 α), (1 α) if µ.

A note on the powers of Bailevič functions 067 References [1] I. E. Bailevič, On a case of integrability in quadratures of the Löwner-Kufarev equation, Mat. Sb., 37 (1955), no. 79, 471-476. (in Russian) MR 17, 356. [] L. De Branges, A proof of the Bieberbach conjecture, Acta Mathematica, 154 (1985), 137-15. [3] W. K. Hayman, & J. A. Hummel, Coefficients of powers of univalent functions, Complex Variables Theory Appl., 7 (1986), 51-70. http://dx.doi.org/10.1007/bf03981 [4] M. Jahangiri, On the coefficients of powers of a class of Bailevič functions, Indian J. Pure Appl. Math., 17 (1986), no. 9, 1140-1144. [5] M. Klein, Functions starlike of order Alpha, Trans. Amer. Math. Soc., 131 (1968), 99-106. http://dx.doi.org/10.1090/s000-9947-1968-019716-1 [6] R. Singh, On Bailevič Functions, Proc. Amer. Math. Soc., 38 (1973), no., 61-71. http://dx.doi.org/10.1090/s000-9939-1973-0311887-9 [7] D. K. Thomas, On a Subclass of Bailevič functions, Int. Journal. Math. & Math. Sci., 8 (1985), no. 4, 779-783. http://dx.doi.org/10.1155/s016117185000850 [8] D. K. Thomas, Proc. of an International Conference on New Trends in Geometric Function Theory and Applications, World Scientific (1991), 146-158. Received: July 4, 015; Published: August 17, 015