Estimates for derivatives of holomorphic functions in a hyperbolic domain

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1 J. Math. Anal. Appl. 9 (007) Estimates for derivatives of holomorphic functions in a hyperbolic domain Jian-Lin Li College of Mathematics and Information Science, Shaanxi Normal University, Xi an 71006, PR China Received 10 May 005 Available online 4 August 006 Submitted by J. Noguchi Abstract Let f(z) be a holomorphic function in a hyperbolic domain Ω. For n 8, the sharp estimate of f (n) (z)/f (z) associated with the Poincaré density λ Ω (z) and the radius of convexity ρ Ωc (z) at z Ω is established for f(z) univalent or convex in each Δ c (z) and z Ω. The detailed equality condition of the estimate is given. Further application of the results to the Avkhadiev Wirths conjecture is also discussed. 006 Elsevier Inc. All rights reserved. Keywords: Univalent function; Convex function; Hyperbolic domain; Poincaré density; Radii of univalency and convexity 1. Introduction Suppose that φ (w 0 ) 0 at a point w 0 D := {w C: w < 1} for φ holomorphic in D. Then there exists ρ(w 0,φ)>0, the greatest r such that 0 <r 1 and φ is univalent in { w C: w w 0 1 w 0 w }, <r (1.1) which is the non-euclidean disk of center w 0 and the non-euclidean radius arctanh r, and also is the disc of center C(w 0,r)and radius R(w 0,r), where C(w 0,r) w 0(1 r ) 1 r w 0 D and R(w 0,r) r(1 w 0 ) 1 r w 0 1 C(w0,r). address: jllimath@yahoo.com.cn X/$ see front matter 006 Elsevier Inc. All rights reserved. doi: /j.jmaa

2 58 J.-L. Li / J. Math. Anal. Appl. 9 (007) We call ρ(w 0,φ) the radius of univalency of φ at w 0. Similarly, the greatest r, 0 <r 1, such that φ is univalent and convex in (1.1) is denoted by ρ c (w 0,φ) and is called the radius of convexity of φ at w 0. With the aid of a well-known theorem in the theory of univalent functions [7], we have ( )ρ(w 0,φ) ρ c (w 0,φ) ρ(w 0,φ). (1.) A domain Ω in the complex plane C is called hyperbolic if C \ Ω contains at least two points. Let φ be a universal covering projection from D onto a hyperbolic domain Ω in C: φ is holomorphic and φ is zero-free in D.ThePoincaré density λ Ω is then the function defined in Ω by 1 λ Ω (z) = (1 w ) φ, z Ω, (1.) (w) where z = φ(w). The choice of φ and w is immaterial as far as z = φ(w) is satisfied. We next set ρ Ω (z) = ρ(w,φ) for z = φ(w) Ω.Againρ Ω (z) is independent of the particular choice of φ and w as far as z = φ(w) is satisfied. Such ρ Ω (z) is called the radius of univalency of Ω at z. Similarly, for z of a hyperbolic domain Ω,wesetρ Ωc (z) = ρ c (w, φ), where z = φ(w) is a universal covering projection. Then ρ Ωc is a well-defined function in Ω and ρ Ωc (z) is called the radius of convexity of Ω at z. Finally set ({ }) Δ c (z) φ ζ C: ζ w 1 wζ <ρ Ωc(z), z= φ(w) Ω. (1.4) Then Δ c (z) is a simply connected domain and is independent of the particular choice of φ and w as far as z = φ(w) is satisfied. In (1.4), if ρ Ωc (z) is replaced by ρ Ω (z), the corresponding symbol is denoted by Δ(z). In geometric function theory, many problems are devoted to the estimate f (n) (z)/f (z) for f univalent in a hyperbolic domain Ω with Poincaré density λ Ω (z) and the radius of univalency ρ Ω (z) or the radius of convexity ρ Ωc (z) at z Ω. In [17], Yamashita obtained the sharp estimate ρω (z) n 1 f (n) (z) λ Ω (z) f (z) n!4n 1, (1.5) for each n and z Ω, where f(z)is holomorphic in the hyperbolic domain Ω and univalent in each Δ(z), z Ω. The related estimate has also been obtained by Chua [5], Avkhadiev and Wirths [1]. One can see the reference material in these papers for the earlier partial results obtained before de Branges celebrated proof of the Bieberbach conjecture. In the case of convexity, by applying a result of Chua [5], Yamashita [17] only obtained the sharp estimate ρωc (z) n 1 f (n) (z) λ Ω (z) f (z) (n + 1)!n, (1.6) for n 4 and z Ω, where f(z)is holomorphic in the hyperbolic domain Ω and univalent in each Δ c (z), z Ω. The case n 5 is open. Chua s conjecture made after Theorem in [5] as well as the conjecture made by Avkhadiev and Wirths in [1] support the assertion that the inequality (1.6) should be true for each n and at each z Ω. In the present paper, we shall show that (1.6) also holds for 5 n 8 and z Ω. The equality condition is given in detail. Further application of the results to the conjecture of Avkhadiev and Wirths formulated in [1] is discussed in the final section.

3 J.-L. Li / J. Math. Anal. Appl. 9 (007) Extension of Yamashita s estimates Theorem 1. Let f(z) be holomorphic in a hyperbolic domain Ω and univalent in each Δ c (z), z Ω. Then (1.6) holds for n 8 and z Ω. If the equality holds in (1.6) at a point z Ω and for n with n 8, then the following items (I) and (II) hold. (I) There exist complex constants Q 0 and R such that Ω = { z C: Re(Qz + R) > 1/ } ; (.1) in particular, ρ Ωc (z) 1. (II) The function f(z)is of the form S(R + Qz)(1 + R + Qz) f(z)= Q(1 + R + Qz) + T, (.) where S 0 and T are complex constants. Conversely, suppose that f of (.) is given in Ω of (.1). Then the equality holds in (1.6) at each point of the half line { } (1 + R)t R L = : 1 <t<1 (.) Q(1 t) and for each n, whereas the inequality (1.6) is strict at each point of Ω \ L and for each n. Proof. We first suppose that 0 Ω and φ(0) = φ (0) 1 = 0 for a projection φ : D Ω. Then λ Ω (0) = 1. Supposing further that f(0) = f (0) 1 = 0, we shall prove that ρc n 1 f (n) (0) (n + 1)! n, (.4) for n 8, where ρ c = ρ Ωc (0). Observe that the function Φ(w) = ρc 1 φ(ρ c w), w D, is a normalized convex function, and the function F(w)= ρc 1 f ( φ(ρ c w) ) = ρc 1 f ( ρ c Φ(w) ), w D, (.5) is a normalized univalent function. Let ( Φ 1 (ξ) ) k = B n,k ξ n, k N, ξ= Φ(w) Φ(D). (.6) n=k Then B k,k = 1 (k N) and F Φ 1 F (k) (0) (ξ) = B n,k ξ n k! n=k { F (k) (0) = B n,k }ξ n. (.7) k! n=1

4 584 J.-L. Li / J. Math. Anal. Appl. 9 (007) From (.5), ρ 1 c ρ n 1 c f (n) (0) = n! f(ρ c ξ)= F Φ 1 (ξ), and (.7) gives F (k) (0) B n,k. (.8) k! Combining with the de Branges coefficient theorem [6], we have ρc n 1 f (n) (0) n! k B n,k. (.9) Let B n,1 = B n (n N). Then from (.6) ( Φ 1 (ξ) ) k ( k = ξ k k ) j B n ξ n 1 = B n,k ξ n, j j=0 n= n=k which yields k k k B k+1,k = B, B k+,k = B + B 1 1, k k k B k+,k = B 4 + B B + B 1, k k (B B k+4,k = B B ) k k B 4 + B B + B 4 4, k k k (B B k+5,k = B 6 + (B B 5 + B B 4 ) + 1 B 4 + B B ) k k + 4 B 4 B + B 5 5, k k (B B k+6,k = B B ) k (B B 6 + B B B B ) 5 + 6B B B 4 k (B + 4 B 4 + B ) k k B + 5 B 4 5 B + B 6 6, k k B k+7,k = B 8 + (B B 7 + B B 6 + B 4 B 5 ) 1 k (B + B 6 + B B B 5 + B B4 + B B ) 4 ( k (B B 5 + B B B 4 + B B ) k (B ) B 4 + B ) B k k + 6 B 5 6 B + B 7 7, (.10) where ( k j) are the binomial coefficients. For the convex function Φ(w),wehave B n 1 (n =,,...,8), (.11) and this bound is sharp. See Libera and Zlotkiewicz [1,1] and Campschroer [4]. Hence we deduce from the above expression (.10) that

5 J.-L. Li / J. Math. Anal. Appl. 9 (007) k k k k + 1 B k+1,k, B k+,k + =, 1 1 k k k k + B k+,k + + =, 1 k k k k k + B k+4,k =, 1 4 k k k k k k + 4 B k+5,k =, k k k k k k k + 5 B k+6,k =, k k k k k k k k + 6 B k+7,k =, that is, n 1 B n,k (n = k,k + 1,...,k+ 7). (.1) It follows from (.9) and (.1) that, for n 8 ρc n 1 f (n) (0) n 1 n! k = (n + 1)! n, which is (.4). Suppose that the equality holds in (.4) for n with n 8. Then w F(w)= (1 αw) and Φ(w) = w 1 βw, (.1) where α, β C and α = β =1. If ρ c < 1, then from f(φ(ρ c w)) = ρ c F(w)= ρ c w/(1 αw), we see that f(z) has a pole φ(ρ c ᾱ) Ω. This contradiction shows that ρ c = 1, so that φ(w)= Φ(w) = w/(1 βw) and Ω = { z C: Re(βz) > 1/ }, (.14) which is (.1) with Q = β and R = 0. On the other hand, it follows from (.8) that n 1 f (n) (0) = n! kα k 1 ( β) n k (.15) with f (n) (0) =(n + 1)! n and n 8. Hence, for n 8, n 1 k ( α β) k = f (n) (0) n 1 = (n + 1) n = k. n! That is, the triangle inequality n z k n z k with z k = k ( n 1 k 1) ( α β) k takes equality, this shows that z k /z j 0 (k j, k,j = 1,,...,n), which gives α = β. Consequently, for z Ω,wehave f(z)= F Φ 1 (z) = z(1 + βz) (1 + βz), (.16)

6 586 J.-L. Li / J. Math. Anal. Appl. 9 (007) which is (.) with S = 1, R = 0, T = 0 and Q = β. To complete the proof of (1.6) at z = a Ω in the general case, we choose a projection φ with φ(0) = a, and let g(ζ) = f(a+ φ (0)ζ ) f(a) φ (0)f (.17) (a) for the variable ζ in the domain { } z a Σ = φ (0) : z Ω (.18) onto which ψ(w)= (φ(w) a)/φ (0) is a projection from D with ψ(0) = ψ (0) 1 = 0. Then g(ζ) is holomorphic in the hyperbolic domain Σ and univalent in each Δ c (ζ ) (ζ Σ) with g(0) = g (0) 1 = 0, where ({ }) Δ c (ζ ) ψ η C: η w 1 wη <ρ Σc(ζ ), ζ = ψ(w) Σ. Since g (n) (0) = f (n) (a)(φ (0)) n 1 f, ρ Σc (0) = ρ Ωc (a) and φ (0) 1 = (a) λ Ω (a), the above proved conclusion (.4) applied to g(ζ) at 0 with ρ c = ρ Σc (0) gives ρωc (a) n 1 f (n) (a) λ Ω (a) f (a) = ( ρ Σc (0) ) n 1 g (n) (0) (n + 1)! n. (.19) This is (1.6) for z = a and n 8. The equality in this general case yields (.1) and (.) with Q = β/φ (0), R = aβ/φ (0), S = βφ (0)f (a), and T = f(a), for β C and β =1. Conversely, given f of (.) in Ω of (.1) and n, we have f (n) (z) = Sn ( Q) n 1 (n + 1)! (1 + R + Qz) n+ (.0) and f (n) (z) f (z) = n ( Q) n 1 (n + 1)!, z Ω. (1 + R + Qz) n 1 Since (1 + R)w R z = (.1) Q(1 w) maps D univalently onto Ω, it follows that and λ Ω (z) = Q 1 w 1 w (.) 1 n 1 f (n) ( (z) 1 w ) n 1 λ Ω (z) f (z) = (n + 1)!n 1 w.

7 J.-L. Li / J. Math. Anal. Appl. 9 (007) Hence, for n, 1 n 1 f (n) (z) λ Ω (z) f (z) = (n + 1)!n if and only if 1 w = 1 w or w {w D: 1 <w<1} =( 1, 1), an open interval in the real axis. In conclusion, the equality holds in (1.6) at z Ω for n with n 8 if and only if z is on L, the image of the open interval ( 1, 1) by (.1). This completes the proof of Theorem 1. The above proof of Theorem 1 also yields the following. Theorem. Let f(z) be holomorphic in a hyperbolic domain Ω and convex in each Δ c (z), z Ω. Then the estimate ρωc (z) n 1 f (n) (z) λ Ω (z) f (z) n!n 1 (.) holds for n 8 and z Ω. If the equality holds in (.) at a point z Ω and for n with n 8, then the following items (I) and (II) hold. (I) There exist complex constants Q 0 and R such that Ω = { z C: Re(Qz + R) > 1/ } ; (.4) in particular, ρ Ωc (z) 1. (II) The function f(z)is of the form S(R + Qz) f(z)= + T, (.5) 1 + R + Qz where S 0 and T are complex constants. Conversely, suppose that f of (.5) is given in Ω of (.4). Then the equality holds in (.) at each point of the half line L given by (.) and for each n, whereas the inequality (.) is strict at each point of Ω \ L and for each n. Proof. With the same notation as above, we see that if f(z) is convex in each Δ c (z), z Ω, then the function F(w) defined in (.5) is a normalized convex function. It follows from (.8) and (.1) that ρc n 1 f (n) (0) n 1 n! B n,k n! = n! n 1, (.6) for n 8. The equality in this case gives (.1) with F(w)replaced by F(w)= w/(1 αw). Hence the same discussion yields Theorem. Note that, geometrically speaking, the extremal function in (1.6) is a conformal map of a half plane onto a plane slit in a half line, whereas the extremal function in (.) is a conformal map of a half plane onto a half plane (see [7,8]). The counterexample given by Kirwan and Schober [9] illustrates that (.11) is not true for n 10 and (.1) does not hold for certain n and k. The approach here cannot be used to deal with the case n 10 directly. On the other

8 588 J.-L. Li / J. Math. Anal. Appl. 9 (007) hand, Yamashita [16,17], Chua [5], Li, Srivastava and Zhang [10] as well as Li and Wei [11] only obtained partial results of Theorems 1 and. The equality condition in [5] is not complete enough as pointed out in [17]. The extension of Chua s result in [10,11] is in a simply connected convex domain and contains no discussion of the equality condition.. Applications to Avkhadiev Wirths conjecture The above results come back to the usual simply connected domains in the complex plane C will give partial solution to the conjecture of Avkhadiev and Wirths, which will be discussed below. Let Ω and Π be two simply connected domains in the complex plane C, which are not equal to the whole plane C, and let H(Ω,Π) denote the set of functions f : Ω Π holomorphic in Ω. Forn N, we consider the quantities C n (Ω, Π) defined by f (n) (z) λ Π (f (z)) C n (Ω, Π) := sup sup f H(Ω,Π) z Ω n!(λ Ω (z)) n, (.1) see Avkhadiev and Wirths [1 ]. Since λ D = (1 z ) 1 for z D, the classical Schwarz Pick lemma indicates that C 1 (D, D) = 1 and in turn C 1 (Ω, Π) = 1, for any pair (Ω, Π) of simply connected domains. For n, Ruscheweyh [14,15] and Yamashita [18] showed that n 1 C n (D, D) = n 1, C n (D, Λ) = n 1, and C n (Ξ, Λ) =, (.) n where Λ := {z C: Re(z) > 0} and Ξ := C \[1/4, ). Recently, Avkhadiev and Wirths [1] proved that (i) C n (D, Π) = n 1 for any convex domain Π and n ; (ii) C n (Ω, Π) 4 n 1 for all simply connected domains Ω and Π in C and n. The equality occurs in (ii) if and only if Ω = Π = Ξ. Here we do not pay attention to linear transformations, because the equation (.) C n (Ω, Π) = C n (aω + b,cπ + d) (.4) is valid with aω + b ={az + b: z Ω} and cπ + d ={cz + d: z Π} for some constants a,c C \{0}, b,d C. In [1], Avkhadiev and Wirths formulated the following two conjectures. Conjectrue 1. C n (Ω, Π) n 1 for all simply connected domains Ω and Π in C. Conjectrue. C n (Ω, Π) = n 1 if and only if Ω and Π are convex. Applying the results of Section, we shall generalize the above known results by giving the sharp upper bounds for the quantities C n (Ω, Π) in the case when n 8 and Ω is convex. The result may be summarized as the following.

9 J.-L. Li / J. Math. Anal. Appl. 9 (007) Theorem. For any convex domains Ω and Π in C, the assertion C n (Ω, Π) n 1 (.5) is valid for n 8. Theorem 4. For any convex domain Ω and any simply connected domain Π in C, the assertion C n (Ω, Π) (n + 1) n (.6) is valid for n 8. Proof of Theorems and 4. For the convex domain Ω and the simply connected domain Π, let ξ = f(z) H(Ω,Π), z 0 Ω and ξ 0 = f(z 0 ) Π. Denote by φ Ω (respectively φ Π ) the conformal map of D onto Ω (respectively Π) with φ Ω (0) = z 0 (respectively φ Π (0) = ξ 0 ) and φ Ω (0) = 1/λ Ω(z 0 )>0 (respectively φ Π (0) = 1/λ Π(ξ 0 )>0). Then the function z = Φ Ω (w) := φ Ω(w) φ Ω (0) φ Ω (0), w D, (.7) is a normalized convex function, and the corresponding function ξ = Φ Π (w) := φ Π (w) φ Π (0) φ Π (0), w D, is a normalized univalent function. Let ( Φ 1 Ω (z)) k = B n,k z n, k N, (.8) n=k as in (.6). Then B k,k = 1 (k N) and ( φ 1 Ω (z)) k B n,k = (φ Ω (z z 0) n, k N. (.9) (0))n n=k Consider the function g H(D,Π) defined by We have g(w) := f ( φ Ω (w) ) = f(z) f(z 0 ) = a k w k, w D. (.10) k=0 a k ( φ 1 Ω (z)) k, z Ω, or from (.9) f (n) (z 0 ) (z z 0 ) n B n,k = a k n! (φ (z z 0) n n=1 n=k Ω (0))n { } B n,k = a k (φ Ω (z z 0 ) n, (.11) (0))n n=1

10 590 J.-L. Li / J. Math. Anal. Appl. 9 (007) which yields f (n) (z 0 ) n! 1 = (φ Ω (0))n a k B n,k. (.1) We shall estimate a k. First note that g(d) Π. The function g(w) g(0) φ Π (0) = λ Π (ξ 0 )a k w k defined in accordance with (.10) is subordinate to the univalent function Φ Π (w) [8, Theorem.1]. Since de Branges celebrated proof of the Bieberbach conjecture implies the Rogosinski conjecture (see [6], [7, p. 196]), this leads to the inequalities λ Π (ξ 0 ) a k k, k N. (.1) Furthermore, if Π is convex, then Φ Π (w) is a convex univalent function. The subordination principle and [7, Theorem 6.4, p. 195] imply λ Π (ξ 0 ) a k 1, k N. (.14) It follows from (.1), (.14) and (.1) that for n 8, f (n) (z 0 ) n! ( λ Ω (z 0 ) ) n (λ Ω(z 0 )) n λ Π (f (z 0 )) n a k B n,k n 1 = (λ Ω(z 0 )) n λ Π (f (z 0 )) n 1, (.15) which yields (.5) for n 8. It follows from (.1), (.1) and (.1) that for n 8, f (n) (z 0 ) n! ( λ Ω (z 0 ) ) n (λ Ω(z 0 )) n λ Π (f (z 0 )) n a k B n,k n 1 k = (λ Ω(z 0 )) n λ Π (f (z 0 )) (n + 1)n, (.16) which yields (.6) for n 8. Acknowledgments I am indebted to the referees for their valuable suggestions.

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