On the Class of Functions Starlike with Respect to a Boundary Point

Transcription

1 Journal of Mathematical Analysis and Applications 261, (2001) doi: /jmaa , available online at on On the Class of Functions Starlike with Respect to a Boundary Point Adam Lecko Department of Mathematics, Technical University of Rzeszów, ul. W. Pola 2, Rzeszów, Poland alecko@prz.rzeszow.pl Submitted by Stephan Ruscheweyh Received July 11, 2000 The aim of this paper is to present a new analytic characterization of the class of functions starlike with respect to a boundary point introduced by M. S. Robertson (1981, J. Math. Anal. Appl. 81, ) and characterized by A. Lyzzaik (1984, Proc. Amer. Math. Soc. 91, ). The method of proof, using Julia s lemma, seems to be new Academic Press Key Words: functions starlike w.r.t. a boundary point; geometric characterization; Julia lemma. 1. INTRODUCTION 1.1. Let =. For each z 0 and r>0 let z 0 r = z z z 0 <r denote the open disk with center at z 0 and radius r. Let = 0 1 and =. For each k>0 consider the oricycle Ɔ k = z 1 z 2 1 z 2 <k The oricycle Ɔ k is a disk in whose boundary circle Ɔ k is tangent to at z = 1. For ζ the set } Ɣ ζ b = z arg 1 ζz <b z ζ <ρ ( b 0 π ) 2 ρ < 2 cos b X/01 $35.00 Copyright 2001 by Academic Press All rights of reproduction in any form reserved.

2 650 adam lecko is called a Stolz angle at ζ. The value 2b 0 π of the opening angle of Ɣ ζ b will not be relevant. Therefore we will write Ɣ ζ for short. For simplicity of notation we will also write Ɣ instead of Ɣ 1. The set of all analytic functions ω in such that ω z < 1 for z will be denoted by B The Julia lemma ([3]; see also [10, pp ]) recalled below is the basis for our considerations. Lemma 1.1 (Julia). Let ω B. Assume that there exists a sequence z n of points in such that (1.1) and (1.2) Then (1.3) z n = 1 n and hence, for every k>0, ω z n =1 n 1 ω z n = α< n 1 z n 1 ω z 2 z 2 α 1 z 1 ω z 2 1 z 2 ω Ɔ k Ɔ αk Remark 1 1 Since 1 ω z 1 ω 0 z 1 z 1 + ω 0 for every ω B, the constant α defined in (1.2) is positive (see [10, pp ]). Definition 1.1. For α>0 let B α be the subclass of all functions ω B satisfying (1.1) and (1.2), and let α denote the class of all functions p of the form p z =4 1 ω z z 1 + ω z where ω B α. The next lemma is a converse of the Julia Lemma (see [10, Vol. II, p. 72]). Lemma 1.2. Let ω B. If (1.3) holds for some α>0, then ω B α 0 for some α 0 α; i.e., there exists a sequence z n of points in satisfying (1.1) and such that 1 ω z n (1.4) = α n 1 z n 0 α

3 Remark 1 2 starlike functions 651 Under the assumptions of Lemma 1.2 let 1 ω z 2 β = sup 1 ω z 1 } z 2 z 2 1 z 2 Since (1.3) holds, we have β α. Then it can be proved that 1 ω z (1.5) = β Ɣ z 1 1 z for every Stolz angle Ɣ (see [1, pp ]). On the other hand 1 ω x 1 x 1 ω x 1 x so (1.5) implies at once (1.4) with some α 0 β. x STARLIKENESS W.R.T. A FINITE BOUNDARY POINT 2.1. Let us start with the following definitions. Definition 2.1. A simply connected domain with 0 is called starlike with respect to the boundary point at the origin (written 0 ) if the segment 0 w = tw t 0 1 is contained in for every w. Definition 2.2. Let 0 denote the class of all univalent analytic functions f in such that 0 f and f is in 0. Functions belonging to will be called starlike w.r.t. the boundary point at the origin. 0 Construction of a Prime End for a Domain Starlike w.r.t. the Boundary Point at the Origin Now for an arbitrary domain in 0 we introduce special null-chain C n. Let us recall that a crosscut C of a domain G is an open Jordan arc in G such that C = C a b, where a b G. Let 0. Fix w 1. Then 0 w 1. For each t 0 1 consider the circle C t = w w =t w 1. It is clear that C t. By Proposition 2.13 [7, p. 28], for each t 0 1 there are countably many crosscuts C k t C t k, of, each of which is a subarc of C t. We denote by Q t the crosscut C k t of C t containing the point tw 1, and by 0 t the connected component of \Q t containing the segment 0 tw 1. So Q t 0 t. Let now t n be a strictly decreasing sequence of points in 0 1 such that n t n = 0 and let Q t n be the corresponding

4 652 adam lecko sequence of crosscuts of. It is easy to observe that (a) (b) (c) Q t n Q t n+1 = for every n. 0 t n+1 0 t n for every n. diam Q t n 0asn. Therefore C n = Q t n forms a null chain of [7, p. 29]. Notice also that the null chain C n is independent of the choices of w 1 and the sequence t n. The equivalence class of the null chain C n [7, p. 30] defines the prime end denoted by p 0. We can also derive that w 0 = 0 is a unique principal point of the prime end p 0 ; i.e., p 0 = 0, where p 0 denotes the set of all principal points of p 0 [7, p. 33]. Therefore, the following proposition follows. Proposition 2.1. of the second kind. Example. Let For every 0 the prime end p 0 is of the first or = z Re z < 1 Im z < 1 \ A n where A 1 = 1 0 A n = 1 i/n 1/n i/n 2 for n \ 1. Then 0, the impression I p 0 = 1 0, and p 0 = 0 I p 0. Hence, p 0 is a prime end of the second kind. Let f be a conformal mapping of onto ; i.e., let f 0.Bythe prime end theorem there exists a bijective mapping fˆ of onto the set of all prime ends of. Therefore, there exists a unique ζ 0 such that p 0 = fˆ ζ 0 [7, p. 30]. By using Proposition 2.1 and Corollary 2.17 [7, p. 35], we conclude the following Proposition 2.2. Proposition 2.2. Every function f 0 has an angular it at the point ζ 0 such that p 0 f = fˆ ζ 0 ; i.e., f z =0 Ɣ ζ 0 z ζ In the proof of the main theorems of this paper we will need two lemmas. Lemma 2.1. n Every sequence a n of positive numbers with (2.1) a 1a 2 a n =0 n has a convergent subsequence a nk and 0 a nk = a 1 k

5 The proof of the lemma is immediate. starlike functions 653 Remark 2 1 The condition (2.1) does not imply the convergence of the sequence a n itself: take for example the sequence 1 1/2 1 1/3 1 1/ Let f be a conformal mapping in. Assume that the radial it r 1 f r =v exists. Let (2.2) Q z v = z 1 f z z f z v denote the Visser Ostrowski quotient [7, p. 251]. Let Q z =Q z 0. Lemma 2.2. Let f be a conformal mappingin havinga radial it r 1 f r =v. Then there exist α 0 1 and a sequence r n with 0 < r n < 1 and n r n = 1 such that Q r n v = 2α n Proof. Since f is conformal in, f z dist f z f 4 1 r 2 for z =r<1 [7, p. 92]. As v f, f z f z v 4 1 r 2 for z =r<1. Hence Q r v = r 1 f r (2.3) f r v r for every r 0 1. The rest of the proof is immediate. 3. ANALYTIC CHARACTERIZATION OF FUNCTIONS STARLIKE W.R.T. A FINITE BOUNDARY POINT In this section, we present an analytic characterization of the class 0. Theorem 3.1. Let f be a univalent function in. Then f 0 and p 0 f = fˆ 1 if and only if f Ɔ k 0 for every k>0.

6 654 adam lecko Proof. Assume that f 0 and ζ 0 = 1 corresponds to the prime end p 0 f. For each t 0 1 define ω t z =f 1 tf z z Since f is starlike w.r.t. the origin, we have tf z f for every t 0 1 and z, and the univalence of f shows that ω t is well defined for each t 0 1. Now fix t 0 1 and w 1 f. Hence 0 w 1 f. Forn let w n = t n 1 w 1 and z n = f 1 w n. Using the same notations as before let C t n = w w = w n and let Q t n C t n denote the crosscut of f containing w n. Therefore Q t n is a null-chain representing the prime end p 0 f. By the prime end theorem, f 1 Q t n is a nullchain in that separates the origin from ζ 0 = 1 for large n. Since z n = f 1 w n f 1 Q t n and diam f 1 Q t n 0asn, we conclude that n z n = 1. Observe that Now let Hence for all n. Consequently, a 1a 2 a n = n n ω t z n =f 1 tw n =f 1 t n w 1 =z n+1 a n = 1 ω t z n 1 z n a n = 1 z n+1 1 z n n ( 1 z2 1 z 2 1 z 1 1 z 3 1 z n 1 z n 1 1 z = n+1 = 0 n 1 z 1 Lemma 2.1 yields a convergent subsequence a nk such that which means that 0 a nk = α t 1 k 1 ω t z nk = α t 1 k 1 z nk ) 1 z n+1 1 z n for every fixed t 0 1. In fact, in view of Remark 1.1, α t > 0 for every t 0 1. Hence each ω t satisfies the assumptions of the Julia lemma, and since α t 1 for every t 0 1, we derive that ω t Ɔ k Ɔ α t k Ɔ k

7 starlike functions 655 for every k>0. This yields tf Ɔ k f Ɔ k for every t 0 1 and hence f Ɔ k 0. Conversely, assume that f Ɔ k 0 for every k > 0. Since 0 k>0 f Ɔ k and f = f Ɔ k k>0 it follows that 0 f and the domain f is starlike w.r.t. boundary point at the origin. Observe also that there exists a prime end p 0 f corresponding to a point ζ 0. We need to show that ζ 0 = 1. To this end, fix k>0 and suppose that ζ 0 1. Let w 1 f. Then 0 w 1 f. Let C t = w w =t w 1 t 0 1. Repeating the construction of a null-chain for the domain f and using the same notation let Q t t 0 1, denote the crosscut of f containing the point tw 1. Choosing a strictly decreasing sequence t n of points in 0 1 such that n t n = 0 let Q t n be the corresponding sequence of crosscuts of f which represents the prime end p 0 f corresponding in a unique way to ζ 0. By the prime end theorem, f 1 Q t n is a null-chain that separates in the points 0 and ζ 0 for large n. Since ζ 0 1 and diam f 1 Q t n 0asn we see that (3.1) f 1 Q t n Ɔ k = for large n. On the other hand, f Ɔ k is in 0, which implies that Q t n f Ɔ k for large n. This contradicts (3.1) and finally proves that ζ 0 = 1, so p 0 f = fˆ 1. The proof of the theorem is finished. Using Theorem 3.1 we characterize the functions of 0 as follows. Theorem 3.2. If f 0 and p 0 f = fˆ 1, then there exists α 0 1 such that Proof. 1 z 2 f z f z α z First, we show that Re 1 z 2 f } z < 0 z f z Fix k>0, and define the open circular arc γ k by (3.2) γ k θ = 1 + keiθ θ 0 2π 1 + k

8 656 adam lecko Then γ k is the positively oriented Ɔ k \ 1. By Theorem 3.1, d/dθ arg f γ k θ 0. Since 1 γ k θ 2 k 2 = 1 + k 1 2 eiθ 2 = 4k ( ) sin2 θ/2 k k + 1 i k + 1 eiθ i = 4k sin2 θ/2 γ k k + 1 θ i = 2Re 1 γ k θ γ k θ i θ 0 2π d dθ arg f γ k θ = d dθ Im log f γ k θ = Im γ k } θ f γ k θ f γ k θ k + 1 = Im 4k sin 2 θ/2 i 1 γ k θ 2 f } γ k θ f γ k θ 1 = 2Re 1 γ k θ Re 1 γ k θ 2 f } γ k θ 0 f γ k θ for θ 0 2π. Since any z belongs to Ɔ k \ 1 for some k, Re 1 z 2 f } z 0 f z for z. Further, by the maximum principle, the strict inequality holds in. Let p z = 1 z 2 f z /f z, and let (3.3) ω z = 4 p z z 4 + p z Then ω maps into. We now prove that ω B α for some α 0 1. From the assumption that p 0 f = fˆ 1 and Proposition 2.1 it follows that r 1 f r =0. Recall the Visser Ostrowski quotient (2.2). Consequently, we can write p z = 1 z Q z z By (2.3) in the proof of Lemma 2.2 and by the above we have (3.4) 1 r Q r = p r =0 r 1 r 1 Now (3.3) yields r 1 ω r =1 so the condition (1.1) of the Julia lemma is satisfied. By Lemma 2.2, there exist α and a sequence r n in (0, 1) with n r n = 1 such that n Q r n = 2α 0

9 starlike functions 657 From (3.4) it follows that } 1 ω r n n 1 r = 2 p r n n n 4 + p r n 1 r n } 2 = n 4 + p r n Q r n = α But 1 ω r n 1 r n 1 ω r n 1 r n so we can find a subsequence r nk such that 1 ω r nk = α α k 1 r 0 nk By Remark 1.1, we see that α 0 1. Hence ω satisfies the assumptions of the Julia lemma. This ends the proof of the theorem. Remark 3 1 In the above proof, it is shown that d/dθ arg f γ k θ 0 if and only if Re 1 z 2 f z /f z < 0in. Corollary 3.1. If f 0 and p 0 f = fˆ 1, then there exists β 0 1 such that Q z =2β Ɣ z 1 for every Stolz angle Ɣ. Proof. In view of Remark 1.2, with ω and α as in the proof of Theorem 3.2, 1 ω z = β Ɣ z 1 1 z where Ɣ is any Stolz angle and 0 β α. Since (3.5) Q z = ω z 1 ω z 1 z z the proof follows at once. Theorem 3.3. Let f be an analytic function in with an angular it Ɣ z 1 f z =0. Assume that there exist α 0 1 and a function ω B such that z 1 ω z =1 and 1 ω z (3.6) = α z 1 1 z

10 658 adam lecko If (3.7) 1 z 2 f z f z ω z = 41 z 1 + ω z then f 0 and p 0 f = fˆ 1. Proof. Using Theorem 3.1, it is enough to show that f is univalent in and f Ɔ k 0 for every k>0. It is immediate from (3.7) that f is an exponential function of an analytic function in, and that f is never vanishing there. Since Ɣ z 1 f z = 0 0 f. Also, by Remark 3.1, d/dθ arg f γ k θ 0, θ 0 2π, where γ k is given by (3.2). It remains prove that f is univalent in. By monotonicity the right-hand and left-hand its ϕ k = arg f γ k θ ψ k = arg f γ k θ θ 0 + θ 2π (possibly infinite) both exist and ψ k ϕ k > 0 (see Theorem 1 in [5, p. 17]). We will prove that ψ k ϕ k 2π for every k>0. First we introduce some notation. For every k>0 and ε 0 2k/ k + 1 consider two arcs in : γ k ε = γ k \ 1 ε σ k ε = Ɔ k 1 ε Let z 1 and z 2 be the end points of the closed arc γ k ε. The same points are the end points of the open arc σ k ε so γ k ε σ k ε is a closed curve in. We will use the following parametrization: σ k ε z = σ t =1 + εe it t t k ε 2π t k ε, where t k ε π/2 π is such that ( ) 2k 2 tan t k ε = 1 ε 1 + k Hence z 1 = σ t k ε and z 2 = σ 2π t k ε The arc γ k ε will be parametrized by (3.2), where θ θ k ε 2π θ k ε, with θ k ε 0 π/2 being such that cos θ k ε =1 1 ( ε 1 + k ) 2 2 k Moreover z 1 = γ k θ k ε and z 2 = γ k 2π θ k ε. In view of (3.5) and (3.6), ( (3.8) Q z = z 1 z ω z ) 1 ω z = 2α z

11 starlike functions 659 (a) Assume first that α 0 1. Fix k>0. Then by (3.8) there exists δ 0 2k/ k + 1 such that Q z 2 for z 1 δ. Take ε 0 δ. Then 1 ε 1 δ. Letγ k ε and σ k ε be as above. Since γ k ε σ k ε is a closed curve lying in, we have (3.9) arg f γ k 2π θ k ε arg f γ k θ k ε f z = γk ε arg f z =Im γ k ε f z dz Q z = Im γ k ε 1 z dz = Im Q z σ k ε 1 z dz 2π t k ε Q σ t = Im iεe it dt = t k ε εe it t k ε If ε 0 +, then Therefore (3.10) 2π t k ε t k ε 0 <ψ k ϕ k 2π t k ε Q σ t dt 4 π t k ε < 2π t k ε π/2 + θ k ε 0 + Re Q σ t dt = ε 0 + arg f γ k 2π θ k ε arg f γ k θ k ε 2π Since 0 f Ɔ k, we deduce that f Ɔ k V k, where V k is the sector with vertex at the origin and with opening angle ψ k ϕ k. The inclusion f Ɔ k1 f Ɔ k2 for 0 <k 1 <k 2 yields V k1 V k2. Hence V = k>0 V k is a sector with vertex at the origin and with opening angle ψ ϕ = sup ψ k ϕ k k>0. Obviously, 0 <ψ ϕ 2π. As there exists ϕ 0 such that (3.11) f = f Ɔ k V k = V k>0 k>0 ϕ 0 < arg f z =Im log f z = Im g z <ϕ 0 + ψ ϕ ϕ 0 + 2π z where g z =log f z z, is well defined in. But (3.7) yields Re 1 z 2 f } z = Re 1 z 2 g z < 0 z f z Hence g is close-to-convex in w.r.t. the convex univalent function h z = z/ 1 z z. Therefore g is univalent in (see [8; 2, Vol. II, p. 2]). Since f z =exp g z z, using (3.11) we finally deduce that f is univalent in.

12 660 adam lecko (b) Now let α = 1. Fix k > 0. For n there exists δ n 0 2k/ k + 1 such that Q z 2 + 1/n for z 1 δ n. For each n take ε n 0 δ n in such a way that n ε n =0. Clearly, 1 ε n 1 δ n for every n. Letγ k ε n and σ k ε n be defined analogously as γ k ε and σ k ε. Since γ k ε n σ k ε n is a closed curve lying in, we have, as in (3.9), γk ε n arg f z 2π t k ε n t k ε n and we complete the proof of univalence as before. ( Q σ t dt ) π t k ε n n Remark 3 1 (1) The above proof of univalence is based on the notion of close-to-convexity. It is interesting, in the case when the prime end p 0 f is of the first kind, that the univalence of f can be proved using only geometrical properties of f. To this end, fix k>0. As was observed in the proof of Theorem 3.3 the function 0 2π θ arg f γ k θ is strictly increasing on γ k = Ɔ k \ 1. From (3.10) and its counterpart in case (b) we deduce that f is injective on γ k. But I p 0 f = 0 since the prime end p 0 f = fˆ 1 is of the first kind. In view of Corollary 2.17 of [7, p. 35] the function f has an unrestricted it at z = 1, which means that f is continuous in 1 with f 1 =0. Consequently, f is analytic in Ɔ k \ 1 and has a continuous extension to Ɔ k. In this way f is continuous and injective on γ k 1. Applying Corollary 9.5 of [6, p. 282] we deduce that f is univalent in Ɔ k. Since this is true for every k>0 we obtain the univalence of f in. If p 0 f is of the second kind, then f has no continuous extension at z = 1 and we are not able to use the cited Corollary 9.5. The author does not know how to prove then the univalence of f using only geometrical arguments. (2) Since 1 ω r 1 r (3.6) implies that ω is in the class B α. (3) In view of (3.3) we have 1 ω z 1 z = = 1 ω r 1 r 4 + p z 4 p z 4 + p z 1 z 1 + z 4 + p z 4 + p z + 4 p z 4 + p z 2 4 p z 2 1 z 2 z

13 starlike functions 661 Therefore, conditions (1.1) and (1.2) can be translated as and (3.12) z n = 1 n p z n =0 n 4 + p z n 2 4 p z n 2 = 16α 0 16 n 1 z n 2 (4) Using again (3.3) we have 1 ω z 1 z 2 p z = z 4 + p z 1 z Therefore, conditions z 1 ω z = 1 and (3.6) can be translated as z 1 p z =0 and p z = 2α 0 2 z 1 1 z 4. EXAMPLES Now we present some examples of functions. (1) f z =exp z/ 1 z z. Hence p z =1and ω z =3/5, so ω/ B α for every α 0 1. Therefore p/ α for every α 0 1. In consequence, f / 0 by Theorem 3.2. (2) f z = 1 z / 1 + z β β 0 2 z. Then p z = 2β 1 z / 1 + z. Hence Re p z > 0 z, and z 1 p z =0. Moreover p z z 1 1 z = 2β = β 0 2 z z Consequently, f 0 by Theorem 3.3. (3) f z = 1 z 2 / 1 2z cos µ + z 2 µ 0 π z. Then p z =4 sin 2 µ/2 1 z 2 / 1 2z cos µ + z 2. It is easy to calculate that Re p e it =0for t 0 2π. Since p 0 > 0, we have Re p z > 0 for z. Moreover z 1 p z =0 and p z z 1 1 z = 2 Consequently, f 0 by Theorem 3.3 for every µ 0 π. Notice that for µ = π we obtain the function considered in Example 3 with β = 1.

14 662 adam lecko (4) f z = 1 z β e z β>0 z. Then p z = 1 z z + β 1 = z β z + β 1. Easy calculations show that for each β 0 2 we have p x < 0 for x 1 1 β so p/ α for every α 0 1 and f / 0 by Theorem 3.2. For β 2 we have Re p e iθ = 2 cos 2 θ β 2 cos θ + β 0 θ 0 2π which implies that Re p z > 0 for z. Moreover z 1 p z =0. It remains to check the condition (3.12). For β = 2 we see that p z z 1 1 z = 2 Consequently, f 0 by Theorem 3.3. Now let β>2. We have 4 + p z 2 4 p z 2 1 z 2 if x 2 + y 2 < 1. But = z2 + β 2 z β z 2 + β 2 z β z 2 = 8 z2 + z 2 + β 2 z + z 2 β 1 1 z 2 = 16 β 1 β 2 x x2 + y 2 β 1 β 2 x x x 2 y 2 1 x 2 β 1 β 2 x x 2 = β x 1 1 x 2 2 > 1 Consequently, f / 0 by (3.12) and Theorem STARLIKENESS W.R.T. THE BOUNDARY POINT AT THE INFINITY The previous results can be easily transformed to domains starlike with respect to an arbitrary finite boundary point (by a suitable translation) or with respect to the boundary point at infinity (by the transformation g z = 1/f z z ). The latter domains are defined as follows. Definition 5.1. For w, let l + w = sw s 1 + be the radial halfline with end point at w. A simply connected domain, with is called starlike with respect to the boundary point at infinity (written )ifl+ w for every w.

15 starlike functions 663 Remark 5 1 Observe first that the origin lies outside every domain in. On the contrary, assume that 0 for some. Since, there exists a finite point w 0 such that 0 w 0. Hence for no point w of the segment 0 w 0 l + w lies in. Definition 5.2. Let denote the class of all univalent analytic functions f in such that f and f is in. Functions belonging to the class will be called starlike w.r.t. the boundary point at infinity. To find an analytic characterization of functions in the class it is enough to observe that f if and only if the function g z =1/f z z, is in the class REMARKS Robertson [9] introduced two classes and of univalent functions in the following way: Definition 6.1. (1) is class of all functions f analytic in with f 0 =1 such that (6.1) Re 2z f z f z z } > 0 z 1 z (2) is the class of all functions f analytic and univalent in such that f 0 =1 r 1 f r =0 f is starlike w.r.t. the origin, i.e., f 0, and Re eiδ f z > 0 for δ and z. It is worth noticing that the analytic condition (6.1) was known for Styer [11], where also the concept of starlikeness with respect to the boundary point can be found. Robertson conjectured that the classes and are equal, which was proved by Lyzzaik [4]. Observe at this moment that f iff e iδ f 2 0 for some δ. Theorem 3.2 provides another characterization of class. It says that functions in the class have a very nice inner property, i.e., preserve the starlikeness w.r.t. the origin on each oricycle Ɔ k for k>0. ACKNOWLEDGMENT The author thanks the referee for his very helpful suggestions.

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