Some criteria of quasiconformality and Landau type. theorems for harmonic mappings in the unit disk ( 単位円板上の調和写像の擬等角性の判定条件とランダウ型の定理 )

Transcription

1 Some criteria of quasiconformality and Landau type theorems for harmonic mappings in the unit disk ( 単位円板上の調和写像の擬等角性の判定条件とランダウ型の定理 ) Jian-Feng Zhu ( 朱剑峰 )

2 Some criteria of quasiconformality and Landau type theorems for harmonic mappings in the unit disk ( 単位円板上の調和写像の擬等角性の判定条件とランダウ型の定理 ) 平成 5 年度 Jian-Feng Zhu ( 朱剑峰 )

3 Contents 1 Introduction 1 On Heinz s inequality 1.1 An asymptotically sharp variant of Heinz s inequality Heinz s inequality for harmonic quasiconformal mappings in the small dilatation Estimate for the dilatations of harmonic quasiconformal selfmappings of the unit disk 8 4 Harmonic quasiconformal mappings between the unit disk and bounded convex domains Our boundary condition and the C 1,µ condition Examples Landau type theorems for bounded harmonic mappings 46 6 Two subclasses of harmonic mappings 5

4 Acknowledgments 61 References 6

5 1 Introduction A real-valued function u on an open set D C is harmonic if it is C on D and satisfies Laplace s equation: u = u xx + u yy =. A complex-valued function w = u + iv on D is said to be harmonic if both u and v are harmonic on D. Define complex derivatives of w(z) as follows: w z := 1 (w x iw y ) and w z := 1 (w x + iw y ), (1.1) where z = x + iy. A direct calculation shows that the Laplacian of w is w = 4w z z. Thus for a function w with continuous second partial derivatives, it is clear that w is harmonic if and only if w z is analytic. The derivatives of harmonic mappings w(z) in polar coordinates can be expressed as w θ := i (zw z zw z ) and w r := 1 r (zw z + zw z ), (1.) where z = re iθ D. It is easy to see that w θ and rw r are both harmonic mappings and it is known that rw r is the harmonic conjugate of w θ. Let U = {z C : z < 1} be the unit disk, and let w(z) be a harmonic mapping defined in U. According to [14] we know that w(z) has the representation w(z) = h(z) + g(z), where h(z) = a n z n and g(z) = b n z n n= 1

6 are analytic in U. For z U, let and Λ w (z) = λ w (z) = max w z(z) + e iα w z (z) = w z (z) + w z (z) α π min w z(z) + e iα w z (z) = w z (z) w z (z). α π According to Lewy s Theorem [33] we know that w(z) is locally univalent and sense-preserving in U if and only if its Jacobian satisfies J w (z) = w z (z) w z (z) > for every z U. Suppose that w(z) is a sense-preserving univalent harmonic mapping of U onto a domain Ω C. Then w(z) is a harmonic K-quasiconformal mapping if and only if w z (z) + w z (z) K(w) := sup z U w z (z) w z (z) K. Both harmonic mappings and quasiconformal mappings are natural generalizations of conformal mappings. Many mathematicians have studied some properties and extremal problems of harmonic quasiconformal mappings and obtained many interesting results(cf.[7]-[3], [39]-[43], [51]-[57]). In 195 E.Heinz discovered the following lemma. Heinz s Lemma ([19]). Let w(z) be a univalent harmonic mapping of the unit disk onto itself, normalized by w() =. Then w z () + w z () c (1.3) for some absolute constant c >.

7 The sharp value of c is 7 4π and it was finally verified by R.R.Hall [17] in 198. Under the additional assumption that w(z) is a K-quasiconformal mapping, in 5 D.Partyka and K.Sakan obtained an asymptotically sharp variant of Heinz s inequality as follows. Theorem ([41]). Let w(z) be a harmonic K-quasiconformal mapping of U onto itself satisfying w() =. Then the inequalities w x (z) + w y (z) 1 (1 + 1 K ) max{ 4 π, L K}, (1.4) w z (z) K + 1 K max{ π, L K}, hold for every z U, where L K := π 1 d ( Φ 1/K (s) ) s 1 s (1.5) is a strictly decreasing function of K. For L >, Φ L (s) is the Hersch- Pfluger distortion function defined by the equalities Φ L (s) := µ 1 (µ(s)/l), < s < 1 ; Φ L () :=, Φ L (1) := 1, where µ(s) stands for the module of Grötzsch s extremal domain U\[, s]. The above theorem can be seen as an asymptotically sharp variant of Heinz s inequality for harmonic quasiconformal mappings. However, it is meaningful only when we find out the range of K such that L K. In 7, π by improving the estimate of the function L K we obtained an asymptotically sharp estimate of (a variant of) Heinz s inequality which improved D.Partyka and K.Sakan s result. 3

8 Theorem 1 (Theorem.). Let L K be the function of K defined by (1.5). If 1 K 1.14, then L K π. If L K π, then K [1, log π]. Theorem (Theorem.3). Let w(z) be a harmonic K-quasiconformal self-mapping of U satisfying w() =. If 1 K 1.14, then w z (z) K + 1 (K 1)3 K 1 and w x (z) + w y (z) 1 ( ) K K (K 1) 3 K. Moreover, the lower bound 1 ( ) K K is asymptotically sharp (K 1) 3 K as K is close to 1. Subsequently, in 11 we proved the following theorem. Theorem 3 (Theorem.6). Let w(z) be a harmonic K-quasiconformal self-mapping of U satisfying w() =. Then the following inequality holds, where M K = 3 3 π 16 π w z () + w z () max{m K, 7 4π } { /K A /K K B K + 1 1/K 3 1+1/K 4 + /K A /K = π/4 cos(x) sin /K (x)dx and B K = π/3 π/4 cos(x) sink (x)dx. Moreover, lim K 1 + M K = M 1 = 1. }, Further we prove that M K is a continuous decreasing function of K and that if 1 K 1.3, then M K > 7 4π (see Theorem.7). 4

9 Let F be a sense-preserving homeomorphism of the unit circle T. Then w(z) = P [F ](z) = π P (r, θ t)f (e it ) dt, z = re iθ U, (1.6) is harmonic in U and has a continuous extension to U such that w(e it ) = F (e it ) on T, where P (r, t) = 1 π 1 r 1 r cos t + r is the Poisson kernel. In what follows we write F (t) instead of F (e it ) for the boundary function. Radó-Kneser-Choquet Theorem ([14]). Let Ω C be a bounded convex domain and let F be a homeomorphism of T onto the boundary of Ω. Then w = P [F ](z) = π is a univalent harmonic mapping of U onto Ω. P (r, θ t)f (t)dt (1.7) Assume that F is a sense-preserving homeomorphism of T onto itself. In, M.Pavlović proved the following theorem. Theorem ([44]). The mapping w = P [F ](z) is quasiconformal if and only if F is absolutely continuous and satisfies (1) ess inf F (θ) >, θ [,π] () ess sup F (θ) <, θ [,π] 1 (3) ess sup π θ [,π] π F (θ+t) F (θ t) tan(t/) dt + <. In section 3 we first obtain estimates for the Jacobian of w, then as an application we estimate the maximal dilatation of w and w z as follows. 5

10 Theorem 4 (Theorem 3.3). Assume that F is a sense-preserving absolutely continuous homeomorphic self-mapping of T. If (I). ess sup F (θ) = A < ; θ [,π] (II). ess inf F (θ) = B > ; θ [,π] π F (θ+t) F (θ t) (III). H[F 1 ] = ess sup π θ [,π] tan(t/) dx + = M <, then w(z) = P [F ](z) is a harmonic K-quasiconformal self-mapping of U and k 1 B, where k = K 1 (A+M) K+1. Theorem 5 (Theorem 3.4). Assume that w = P [F ](z) is a harmonic K-quasiconformal self-mapping of U satisfying w() =. Then for every z = re iθ U we have w z (z) π A 3 4(1 k ) and for a.e. θ [, π] 1 π F (θ + x) F (θ x) π dx tan(x/) k π A 3 1 k B, where A = K 3K 5(K 1 K )/, B = 5(1 K )/ (K +K 1) K and k = K 1 K+1. Assume that Ω C is a bounded convex domain, and that F is a sense-preserving homeomorphism of T onto the boundary of Ω. According to Radó-Kneser-Choquet theorem we know that the Poisson integral w = P [F ](z) of F is a sense-preserving univalent harmonic mapping of U onto Ω. It is interesting to study such a topic : under what conditions on F is w = P [F ](z) a harmonic quasiconformal mapping? A Jordan curve Γ C is said to has a C 1,µ parametrization for some 6

11 < µ 1, if there exists an injective continuous function G(e is ) of T onto Γ such that F (s) := G(e is ) is of class C 1, F (s) = and F (s) F (t) sup s,t [,π], s t s t µ <. (1.8) In 8 D.Kalaj proved the following theorem. Theorem ([9]). Let w = P [F ](z) be a harmonic mapping of U onto a bounded convex domain Ω C such that the boundary curve Γ = Ω has a C 1,µ parametrization for some < µ 1 and suppose that F is a sense-preserving homeomorphism of T onto Γ. Then w is a harmonic quasiconformal mapping if and only if F is absolutely continuous and satisfies (i) ess inf F (θ) >, θ [,π] (ii) ess sup F (θ) <, θ [,π] (iii) H[F 1 ] = ess sup π θ [,π] π + F (θ+t) F (θ t) tan(t/) dt <. We prove the following Theorem 6 in section 4, which gives a criterion of quasiconformality for w = P [F ](z) by means of some conditions on F. This result partially improves D.Kalaj s theorem. Theorem 6 (Theorem 4.1). Suppose that F is a sense-preserving homeomorphism of T onto the boundary of a bounded convex domain Ω C. If (1) ess inf θ [,π] F (θ) = q >, () there exist constants M > and α > 1 such that F (θ + x) + F (θ x) F (θ) Mx α, (1.9) holds for every θ [, π] and x >, then w = P [F ](z) is a harmonic 7

12 quasiconformal mapping of U onto Ω. We prove in section 4.1 that our boundary condition (1.9) is weaker than the condition that the boundary function is C 1,µ. Moreover, we give an example in section 4. to show that in our condition (1.9) α > 1 is indispensable. Let F be the family of analytic functions f defined in U satisfying f() = and f () = 1. The classical Landau theorem states that if f F and f(z) < M on U for some constant M 1, then f is univalent in the disk U r := { z < r } with r = 1 M+ M 1, and f(u r ) contains the disk U R with R = Mr. For bounded harmonic mappings f defined in U, under some suitable assumptions we can obtain similar results. Suppose that f = h + ḡ is a bounded harmonic mapping of U where h(z) = z + a n z n and g(z) = b n z n (1.1) n= n= are analytic in U and satisfy the coefficients condition a n + b n c, n =, 3, (1.11) for some positive constant c. Then we obtain Landau type theorems for such mappings f and L(f), where L(f)(z) := zf z (z) zf z (z) = if θ (z). Theorem 7 (Theorem 5.1). Suppose that f = h + ḡ is a bounded harmonic mapping of U which has the form (1.1) and satisfies the coefficients condition (1.11). Then 8

13 (1) f(z) is univalent in a disk U r and f(u r ) contains a disk U R, where r = 1 c 1+c and R = r cr 1 r ; () L(f)(z) = zf z (z) zf z (z) is univalent in a disk U ρ and L(f)(U ρ ) contains a disk U σ, where ( ρ = 1 3 c c + 1 ( 3 c c c c + 1 [ ] 1 and σ = ρ cρ (1 ρ 1. ) c c + 1 ) + 1 ( ) c 3 7 c + 1 ) + 1 ( ) c 3. 7 c + 1 Moreover, if f is a harmonic K-quasiconformal self-mapping of U and satisfies only (1.1) without (1.11), then we obtain a univalent radius of L(f) by means of K. Theorem 8 (Theorem 5.3). Suppose that f = h + ḡ is a harmonic K- quasiconformal self-mapping of U which has the form (1.1). Then L(f) Λ K 1 is univalent in a disk U ρk, where ρ K = 1 Λ K +Λ K 1 and L(f)(U ρ K ) ( ) contains a disk U σk, where σ K = ρ K Λ K 1 ρ K Λ K 1 ρ K and Λ K := Kπ K 9K/ 15(K 1/K)/4. Finally we introduce two subclasses of harmonic mappings and discuss quasiconformality of these two classes. This also gives some criteria of quasiconformality for harmonic mappings. 9

14 where Assume that f(z) = h(z) + g(z) is a harmonic mapping defined in U, h(z) = z + a n z n and g(z) = n= b n z n (1.1) are analytic in U with b 1 < 1. For µ < 1, denote by HS(µ) the class of such harmonic mappings f which satisfy the condition (n µ)( a n + b n ) (1 µ)(1 b 1 ). (1.13) n= In section 6 we prove the following theorem. Theorem 9 (Theorem 6.). Let < µ < 1. Then every f HS(µ) is a bi-lipschitz harmonic mapping. Furthermore, the following inequalities µ(1 b 1 ) µ hold for every z 1, z U. z 1 z f(z 1 ) f(z ) 4 + µ( b 1 3) z 1 z, µ Let Ũ = {z C : z > 1} be the exterior of U. Assume that f(z) = h(z) + g(z) is a harmonic mapping defined in Ũ where h(z) = αz + a n z n and g(z) = βz + b n z n (1.14) are analytic in Ũ with α > β. For µ < 1, denote by Σ H(µ) the class of such harmonic mappings f which satisfy the condition (n µ)( a n + b n ) (1 µ)( α β ). (1.15) For f Σ H (µ) we prove the following theorem. Theorem 1 (Theorem 6.3). Assume that f Σ H (µ). Then f is a 1

15 sense-preserving univalent harmonic mapping. Furthermore, if < µ < 1 and f(z) αz + a 1 z + β z + b 1 z, then f is a harmonic K-quasiconformal mapping, where K is a constant determined by µ. Furthermore we give some examples to show that µ > is indispensable in Theorem 9 and Theorem 1. 11

16 On Heinz s inequality Let w(z) be a harmonic K-quasiconformal self-mapping of U satisfying w() =. In 5 D.Partyka and K.Sakan proved the following theorems. Theorem A ([41]). Let w be a harmonic K-quasiconformal self-mapping of U satisfying w() =. If F is the boundary limiting valued function of w, then d F 1 K max{ π, L K}, (.1) where d F := ess inf x [,π] F (x). Moreover, the right-hand side in (.1) is a decreasing and continuous function of K 1 with values in (,1]. Here L K = π 1 d ( Φ 1/K (s) ) = 4 π Φ 1/K( 1 ) + π s 1 s (.) 1 1 s s (1 s ) 3/ Φ 1/K(s) ds, and for L >, Φ L (s) is the Hersch-Pfluger distortion function defined by the equalities Φ L (s) := µ 1 (µ(s)/l), < s < 1 ; Φ L () :=, Φ L (1) := 1, where µ(s) stands for the module of Grötzsch s extremal domain U\[, s]. By [41, (1.)] and [41, (1.38)] we know that x 1/K Φ K (x) 4 1 1/K x 1/K, x 1, K 1; (.3) 4 1 K x K Φ 1/K (x) x K, x 1, K 1. (.4) For the function L K, D.Partyka and K.Sakan proved the following theorem (cf.[41, Lemma1.4]). 1

17 Theorem B ([41]). L K is a strictly decreasing function of K such that lim L K = L 1 = 1, K 1 lim L K =, K as well as where L = 4 ( ln ). π L K L K1 L K K 1 K 1, K 1, Using the above Theorem A and Theorem B, they proved the following theorem. Theorem C ([41]). Let w(z) be a harmonic K-quasiconformal self-mapping of U satisfying w() =. Then the inequalities w x (z) + w y (z) 1 (1 + 1 K ) max{ 4 π, L K}, (.5) w z (z) K + 1 K max{ π, L K}, hold for every z U. 13

18 .1 An asymptotically sharp variant of Heinz s inequality The above Theorem C can be seen as an asymptotically sharp variant of Heinz s inequality for harmonic K-quasiconformal mappings. However, it is meaningful only when we find out the range of K such that L K π. To this purpose, in this section we are interested in finding a constant K > 1 such that L K π whenever K [1, K ]. First, we improve Theorem B as follows. Theorem.1([51, Theorem.1]). Let L K be the function of K defined by (.). Then for 1 K 1 K, L K satisfies the following bi-lipschitz condition: 64 ln π 5K K 1 K L K1 L K In particular, if 1 K 1 K, then 4(1 + ln ) K K 1. (.6) π ln 16π (K K 1 ) L K1 L K 4 π (1 + ln )(K K 1 ). (.7) Proof. It follows from (.) that L K1 L K = 4 π [Φ 1/K 1 ( 1 ) Φ 1/K ( 1 ) ] (.8) + π 1 1 s [Φ s (1 s ) 3 1/K1 (s) Φ 1/K (s) ]ds. (I).If K 1 K K 1, then we set R = K 1 K. Note that 16 1 R t R t is an 14

19 increasing function of t [, 1]. Using (.3) and (.4) we have Φ 1/K1 (s) Φ 1/K (s) = Φ 1/R (Φ 1/K (s)) Φ 1/K (s) 16 1 R (Φ 1/K (s)) R Φ 1/K (s) 16 1 R s K 1 s K. (.9) Since 16 x 1 6x holds for every x [, 1 ], we see that 161 R 1 6(1 R). We conclude from (.9) and [41, (1.47)] that Φ 1/K1 ( 1 ) Φ 1/K ( 1 ) 16 1 R K 1 K = K 1 (16 1 R 1) + K 1 K K 1 6(K K 1 ) + ln (K K 1 ) (6 + ln )(K K 1 ). Since 1 s (1 s ) 3 1 holds for every s [, 1 ], applying (.9) we have = 1 K 1 1 s [Φ s (1 s ) 3 1/K1 (s) Φ 1/K (s) ]ds 1 s [Φ 1/K 1 (s) Φ 1/K (s) ]ds (16 1 R s K 1 s K )ds K 1 1 (161 R 1) + ( K1 K 1 1 K K 1 ) 1 K 1 K 1 1 6(K K 1 ) + ( + ln )(K K 1 ) (8 + ln )(K K 1 ). 15

20 We conclude from (.8) that L K1 L K 4 ln (6 + π )(K K 1 ) + π (8 + ln )(K K 1 ) = 4 π (1 + ln )(K K 1 ). (.1) (II). If K 1 < K, then m K K 1 for some natural number m. Let R := ( K K 1 ) 1/m. Then we see that R. Replacing K 1 by R n 1 K 1 and K by R n K 1 in (.1), we obtain L K1 L K = m (L Rn 1 K 1 L Rn K 1 ) m 4 π (1 + ln )( R n K 1 R n 1 K 1 ) = 4 π (1 + ln )( K 1 + R m K 1 ) = 4 π (1 + ln )(K K 1 ). This completes the proof of the right-hand inequality. For the left-hand inequality, let R := K 1 K. Then < R 1. Applying 16

21 (.3) and (.4) we obtain Φ 1/K1 (s) Φ 1/K (s) = Φ 1/R (Φ 1/K (s)) Φ 1/K (s) Φ 1/K (s) R Φ 1/K (s) = t R t (t := Φ 1/K (s) ) = t ξ ln t(r 1) (by Lagrange s mean-value theorem, R ξ 1) t ln t K 1 K K 16 1 K s K ln s K K 1 K K >. This shows that L K1 L K 4 π [Φ 1/K 1 ( 1 ) Φ 1/K ( 1 ) ] 4 π 161 K K ln (K K 1 ). In particular, if 1 K 1 K, then ln 16π (K K 1 ) L K1 L K 4 π (1 + ln )(K K 1 ). This completes the proof. Remark.1. We point out here that in our proof we use a sharp estimate 1 s (1 s ) 3/ 1 for s [, 1 ] and finally improve Theorem B. In fact, in [41] the authors used the following estimate 1 s (1 s ) 3/. 17

22 By using the Lagrange s mean-value theorem we prove that L K is co-lipschitz. Theorem.([51, Theorem.]). Let L K be the function of K defined by (.). If 1 K 1.14, then L K π. If L K π, then K [1, log π]. Proof. According to (.) and (.4) we have L K = 4 π Φ 1/K( 1 ) + π 4 π 161 K ( 1 )K + π 1/ 1/ = π 31 K + π 161 K 1/ = π 31 K + π 161 K (K = π 31 K + π 161 K (K 1 s s (1 s ) 3/ Φ 1/K(s) ds 1 s s (1 s ) 3/ 161 K s K ds s K 1 d( s 1 s ) 1/ π/4 s K 1 s ds 1 K ) Since sin x π x holds for every x [, π ], we see that 4 K π/4 (sin x) K dx 1 K ). (.11) π/4 (sin x) K dx K ( π x)k dx = 1 K π 4 K K 1. (.1) From (.11) and (.1) we obtain L K π 31 K + ( ) π π 161 K 1 K 4 K K 1 1 K = K 1 31 K. (.13) Calculating by Mathematica (see Figure 1) we see that if 1 K 1.14, 18

23 1..8 Π then L K.. Figure 1: A lower bound of L K K K 1 31 K. This completes the first part of Theorem π For the second part, by Theorem B we know that lim K 1 L K = L 1 = 1. We conclude from (.) that 1 1 s (1 s ) 3/ ds = π 1. Since sk ( 1 ) K = 1 K holds for every s 1, using (.) and (.4) we obtain L K 4 π ( 1 ) K + π 1/ π [1 K + 1 K ( π 1)] = 1 K. 1 s (1 s ) 3/ sk ds By the assumption that L K π, we see that π L K 1 K. This implies that K log π. This completes the proof. 19

24 Now we prove our main result as follows. Theorem.3([51, Theorem.3]). Let w(z) be a harmonic K-quasiconformal self-mapping of U satisfying w() =, and let F be the boundary limiting valued function of w. If 1 K 1.14, then and d F w z (z) 1 (K 1)3 K 1, K + 1 (K 1)3 K 1, w x (z) + w y (z) 1 (1 + 1 K ) K (K 1) 3 K. Moreover, the lower bound 1 (1 + 1 K ) K sharp as K is close to 1. (K 1) is asymptotically 3K Proof. According to (.13), if 1 K 1.14, then L K K K 1 31 K. Applying Theorem A and Theorem C we can obtain the results directly: π and d F 1 K max{l K, π } = L K K 1 (K 1)3 K 1, w z (z) K + 1 K max{l K, π } = K + 1 K L K w x (z) + w y (z) 1 (1 + 1 K ) max{l K, 4 π } K + 1 (K 1)3 K 1, = 1 (1 + 1 K ) L K 1 (1 + 1 K ) K (K 1) 3 K. This completes the proof.

25 . Heinz s inequality for harmonic quasiconformal mappings in the small dilatation In this part we improve Heinz s inequality as w z () + w z () max { M K, 7 4π }, where M K is a strictly decreasing function of K. Moreover, if K [1, 1.3], then the previous corresponding results in [41] are improved. The following two auxiliary results play a key role in the proofs of our main theorems. Lemma.4([53, Lemma.1]). Let w = P [F ](z) be a harmonic quasiconformal self-mapping of U with the boundary function F (e it ) = e iγ(t). For every z 1 = e i(s+t), z = e i(s t) T, let θ = γ(s + t) γ(s t). Then F (z 1 ) = e iθ F (z ) and the following inequalities hold for every s, t π. 1 1K sin K (t) sin θ 1 1/K sin /K (t) Proof. According to the quasi-invariance of harmonic measure (see [41, (1.9)]), we have Φ 1/K (cos t ) cos θ 4 Φ K(cos t ) (.14) hold for every s, t π and θ = γ(s + t) γ(s t). Since Φ K (x) + Φ 1/K ( 1 x ) = 1 holds for every x 1, this shows that Φ 1/K (sin t ) sin θ 4 Φ K(sin t ). (.15) Using Hübner inequalities (cf. (.3) and (.4)), we see that s 1/K Φ K (s) 4 1 1/K s 1/K and 4 1 K s K Φ 1/K (s) s K. Applying (.14), (.15) and the above two inequalities, we have 1 1K sin K (t) sin θ 1 1/K sin /K (t) 1

26 hold for every s, t π and θ = γ(s + t) γ(s t). This completes the proof. Lemma.5([53, Lemma.]). Assume that α > and set A α := B α := π/4 π/3 cos(x) sin α (x)dx, (.16) cos(x) sin α (x)dx. (.17) Then A α B α ( π/ α/ )α ( [ 1 ( ). )α απ ] 8(1 + α) ( )α, Proof. Applying integration by parts we see that A α = π/4 Hence A α = 1 1+α/ cos(x) sin α (x)dx = 1 π/4 ( )α α sin α x cos xdx = 1 ( )α α ( π/4 1 ( )α α sin α xdx α π/4 sin α xdx ) π/4 sin α x cos(x)dx.. Since π x sin x x

27 holds for every x [, π/4], we have ( 1 A α = α/ α/ ( [ 1 ( )α α ( π )α απ 8(1 + α) ( (π/4)1+α )α 1 + α )α ]. ) Applying integration by substitution we have B α = π/3 cos(x) sin α (x)dx π/4 = This completes the proof. π/1 sin(x) sin α (x + π/4)dx π/1 ( )α sin(x)dx 3 = ( )α ( 4 1 ). Using the above Lemmas, we can prove the following main results. Theorem.6([53, Theorem 3.1]). Let w(z) be a harmonic K-quasiconformal self-mapping of U satisfying w() =. Then the following inequality holds, where M K = 3 3 π 16 π w z () + w z () max{m K, 7 4π } (.18) { /K A /K K B K + ( } 3 1 1/K 4 )1+1/K + /K 3

28 satisfies lim K 1 M K = M 1 = 1. Proof. Assume that F (t) = e iγ(t) is the boundary function of w. Here we again write F (t) instead of F (e it ) for the boundary function. Let w(z) = h(z) + g(z), where h(z) = a n z n and g(z) = b n z n are analytic in U. An application of Parseval s relation (cf. [1]) leads to the expression 1 π π e i[γ(s+t) γ(s t)] ds = a n e int + b n e int for arbitrary t R. Taking real parts, we arrive at the formula where have Let 1 J(t) = J(t) = 1 π π ( a n + b n ) cos(nt), (.19) sin ( γ(s + t) γ(s t) ) ds. Since w(z) is a harmonic quasiconformal mapping, by Lemma.4 we Applying (.19), we obtain 8 π π/ 1 1K sin K (t) J(t) 1 1/K sin /K (t). cos t cos ( π 3 + t), t π 6 M(t) = cos t cos ( π 3 t), π 6 t π 3 π, 3 t π. M(t)(1 J(t))dt = a 1 + b π 1 ( a3n 9n + b 3n ) 1 a 1 + b 1. (.) 4

29 It is easy to see that 8 π π/ In view of the definition of M(t), we have π/ = M(t)J(t)dt = π/4 π/3 cos(t)j(t)dt (1 1/K) π/4 3 (1 1/K) M(t)dt = 3 3 π. (.1) [ ] cos(t) + sin t cos t J(t)dt π/3 π/4 3 cos(t)j(t)dt + π/3 cos(t) sin /K tdt (1 1K) π/3 cos t sin (/K+1) tdt = /K A /K K B K + sin t cos tj(t)dt π/3 π/4 cos(t) sin K tdt 3 1 1/K (3/4)1+1/K + /K. Applying the above inequality, (.) and (.1), we have w z () + w z () 8 π/ M(t)dt 16 π/ π π [ 3 3 π 16 3 π 4 1 1/K A /K K B K + M(t)J(t)dt 3 ] (3/4)1+1/K 1 1/K + /K = M K. (.) According to [17] and the above inequality, we have w z () + w z () max{m K, 7 4π }. Particularly if K 1, then A = 1 4 π 16 and B =

30 π We see that M K M 1 = 3 3 π 16 π ( π 16 ) = 1 as K 1. Here the lower bound 1 is the best possible. Theorem.7([53, Theorem 3.]). Let M K be the function of K defined in the Theorem.6. Then M K is a continuous decreasing function of K. Moreover if 1 K 1.3, then M K > 7 4π. Proof. Using B K <, we can easily obtain that 1 1/K A /K, 1 1K B K and 1 1/K (3/4) 1+1/K +/K are all strictly increasing functions of K. Hence M K is a strictly decreasing function of K. According to Lemma.5, we have [ M K = 3 3 π 16 3 π 4 1 1/K A /K K B K + + /K 3 3 π 16 π { /K K (1/) K ( := N K. ( 1 + 1/K /) + (1/) 1+1/K π(1/)1/k 4(K + ) 1 1/K (3/4)1+1/K + /K } 1 1/K (3/4)1+1/K ) ] Calculating by Mathematica we obtain that if 1 K 1.3, then M K N K > 7 4π. This completes the proof. Finally, set P K = 1 { K } max { 4 π, L K}, we compare our result (.18) with [41] by the following table. K P K M K

31 From the above table we can see that M K is closer to 1 than P K as K is close to 1. This shows that our result is better than (.5) at z =. 7

32 3 Estimate for the dilatations of harmonic quasiconformal self-mappings of the unit disk Assume that F (e ix ) = e iγ(x) is a sense-preserving homeomorphism of T onto itself, where γ(x) is an increasing homeomorphism of R satisfying γ() < π and γ(x + π) = π + γ(x). Then w = P [F ](z) is a sensepreserving univalent harmonic mapping of U onto itself. In this section, we again write F (x) instead of F (e ix ) for the boundary function. In M.Pavlović proved the following theorem. Theorem D ([44]). The mapping w = P [F ](z) is quasiconformal if and only if F is absolutely continuous and satisfies (1) ess inf F (θ) >, θ [,π] () ess sup F (θ) <, θ [,π] 1 (3) ess sup π θ [,π] π F (θ+t) F (θ t) tan(t/) dt + <. In this part we find the relationship between K and the conditions (1), () and (3). Furthermore, we obtain some estimates of w z (z) and J w. The following two Lemmas play a key role in proving our main results. Lemma 3.1([5, Lemma 1]). Suppose that F is a sense-preserving absolutely continuous homeomorphic self-mapping of T satisfying F (x) B for a.e. x [, π] where B is a positive constant. If w = P [F ](z) satisfies w() =, then J w (z) B for every z U. 8

33 Proof. Let w(z) = P [F ](z) = u + iv for z = re iθ U. Then and w r (re iθ ) = u(re iθ ) = v(re iθ ) = π π w θ (re iθ ) = 1 r P (r, x θ) cos γ(x) dx, P (r, x θ) sin γ(x) dx, π π P (r, x θ)f (x) dx, (3.1) P (r, x θ)f (x) sin(x θ) dx. (3.) According to [7, Theorem.8] we know that the radial limits of w θ (re iθ ) and w r (re iθ ) exist almost everywhere, and lim w θ(re iθ ) = F (θ) = iγ (θ)e iγ(θ) r 1 a.e. θ [, π]. Hence lim u θ(z) = γ (θ) sin γ(θ), r 1 lim v θ(z) = γ (θ) cos γ(θ). r 1 Using u(e iθ ) = cos γ(θ) and v(e iθ ) = sin γ(θ), we obtain that for a.e. e iθ T, lim J w(re iθ u r v θ u θ v r ) = lim r 1 r 1 r = lim r 1 + lim r 1 = lim r 1 { u(e iθ ) u(re iθ ) 1 r { v(e iθ ) v(re iθ ) π 1 r K(x, θ) 9 P (r, x θ) 1 r } (γ (θ) cos γ(θ)) } (γ (θ) sin γ(θ)) dx, (3.3)

34 where K(x, θ) = γ (θ) γ (θ) cos(γ(θ) γ(x)). Since γ is an increasing homeomorphism of R, this implies that K(x, θ). Applying the following inequality P (r, x θ) 1 r = 1 + r π(1 + r r cos(θ x)) 1 π(1 + r) 1 4π and w() = we obtain that for a.e. e iθ T, lim r 1 π K(x, θ) P (r, x θ) 1 r dx lim r 1 = π π γ (θ) [1 cos(γ(θ) γ(x))] γ (θ) 1 4π dx B. 1 4π dx Thus lim J w(re iθ ) B r 1. Applying [7, Corollary.9] we have J w(z) = w z (z) w z (z) B, for every z U. This completes the proof. Lemma 3.([5, Lemma ]). Suppose that F is a sense-preserving absolutely continuous homeomorphic self-mapping of T satisfying F (x) A for a.e. x [, π] where A is a positive constant. If w = P [F ](z) satisfies w() =, then lim r 1 J w(re iθ ) π A 3 4 for a.e. θ [, π]. 3

35 Proof. According to (3.3) we have for a.e. e iθ T, lim J w(re iθ u r v θ u θ v r ) = lim r 1 r 1 r = lim r 1 = lim r 1 + lim r 1 π π π K(x, θ) P (r, x θ) 1 r dx ( ) γ(θ) γ(θ x) P (r, x) γ (θ) sin 1 r dx ( ) γ(θ) γ(θ + x) P (r, x) γ (θ) sin 1 r dx. Since the inequality sin( x ) x π holds for any x π, we see that for a.e. e ix T, P (r, x) 1 r = 1 + r π[(1 r) + 4r sin (x/)] 1 + r π π 4rx. According to the assumption that F is absolutely continuous and F (x) A for a.e. x [, π], we see that γ(x) γ(y) A x y (3.4) holds for any x, y [, π]. Using (3.4) and the inequality sin x x, we have for a.e. e iθ T π lim J w(re iθ A x 1 + r π ) 4A lim r 1 r 1 4 π 4rx dx = π A 3 4. This completes the proof. In [44], M.Pavlović proved that if w = P [F ](z) is a harmonic K-quasiconformal self-mapping of U, then w z (z) is bounded. The following theorems show 31

36 the relationship between K, w z and the conditions (1), () and (3) in Theorem D. Theorem 3.3([5, Theorem 1]). Assume that F is a sense-preserving absolutely continuous homeomorphic self-mapping of T. If (I). ess sup F (θ) = A < ; θ [,π] (II). ess inf F (θ) = B > ; θ [,π] π F (θ+t) F (θ t) (III). H[F 1 ] = ess sup π θ [,π] tan(t/) dx + = M <, then w(z) = P [F ](z) is a harmonic K-quasiconformal self-mapping of U and k 1 B, where k = K 1 (A+M) K+1. Proof. According to [7, Theorem.8] we know that the radial limits of w θ (re iθ ) and w r (re iθ ) exist at e iθ for a.e. θ [, π]. Applying (3.1) and (3.) we obtain lim w θ(re iθ ) = F (θ), lim w r(re iθ ) = H[F ](e iθ ) a.e. θ [, π], r 1 r 1 and w z (z) = e iθ ( w r (z) i w ) θ(z). r Since F is a function of bounded variation, using [1, Theorem 3.5] we know that the radial limit of w z (z) exists almost everywhere and lim w z(re iθ ) 1 r 1 lim r 1 ( w r(re iθ ) + w θ(re iθ ) r ) A + M ( a.e. θ [, π]). Applying [1, Theorem.1] we have w z (z) A + M for every z U. 3

37 Since ess inf θ [,π] F (θ) = B >, according to Lemma 3.1 we have J w (z) = w z (z) w z (z) B. Then w z (z) w z (z) 1 B w z (z) 1 B (A + M). We see that w(z) is a harmonic K-quasiconformal mapping with K 1 K + 1 B 1 (A + M). This completes the proof. Theorem 3.4([5, Theorem ]). Assume that w = P [F ](z) is a harmonic K-quasiconformal self-mapping of U satisfying w() =. Then for every z = re iθ U we have w z (z) π A 3 4(1 k ) and for a.e. θ [, π] 1 π F (θ + x) F (θ x) π dx tan(x/) k π A 3 1 k B, where A = K 3K 5(K 1 K )/, B = 5(1 K )/ (K +K 1) K and k = K 1 K+1. Proof. Since w = P [F ](z) is a harmonic K-quasiconformal mapping, we have w z (z) w z (z) k < 1. Then J w (z) = w z (z) w z (z) w z (z) (1 k ). According to [4] we know that B = 5(1 K )/ (K + K 1) K F (x) K 3K 5(K 1 K )/ = A. 33

38 Using (3.1) and Lemma 3. we obtain that for a.e. e iθ T, lim w θ(re iθ ) = r 1 F (θ) and lim J w(re iθ ) π A 3 r 1 4. This implies that for a.e. e iθ T lim w z(re iθ ) J w lim r 1 r 1 1 k π A 3 4(1 k ). Using [1, Theorem.1 and Theorem 3.5] again we have w z (z) π A 3 4(1 k ) for every z U. On the other hand, since w z (z) + w z (z) = 1 have w z (z) + w z (z) w z (z) (1 + k ) 1+k 1 k π A 3 4. Thus for a.e. θ [, π] lim r 1 w r(re iθ ) = 1 π π + F (θ + x) F (θ x) tan(x/) ( w r (z) + w θ(z) r ), we dx = lim r 1 ( w z(re iθ ) + w z (re iθ ) ) F (θ) 1 + k π A 3 1 k B. This completes the proof. Remark 3.1. If w is a harmonic quasiconformal mapping of U, then w z (z) is not necessarily a bounded analytic function in general. For example, let w(z) be a conformal mapping of U onto a rectangle. Then w z (z) is no longer bounded. 34

39 4 Harmonic quasiconformal mappings between the unit disk and bounded convex domains Suppose that F (e ix ) = ρ(x)e iγ(x) is a sense-preserving homeomorphism of T onto the boundary Γ of a bounded convex domain in C, where γ(x) is an increasing homeomorphism of R and ρ(x) is a non-negative continuous function of R such that γ() < π, γ(x + π) = γ(x) + π and ρ(x + π) = ρ(x). In this section we again write F (x) instead of F (e ix ) for the boundary function. Definition 4.1. A Jordan curve Γ C is said to has a C 1,µ parametrization for some < µ 1, if there exists an injective continuous function G(e is ) of T onto Γ such that F (s) := G(e is ) is of class C 1, F (s) = and F (s) F (t) sup s,t [,π], s t s t µ <. (4.1) In 8 D.Kalaj proved the following theorem. Theorem E ([9]). Let w = P [F ](z) be a harmonic mapping of U onto a bounded convex domain Ω C such that the boundary curve Γ = Ω has a C 1,µ parametrization for some < µ 1 and suppose that F is a sense-preserving homeomorphism of T onto Γ. Then w is a harmonic quasiconformal mapping if and only if F is absolutely continuous and satisfies (i) ess inf F (θ) >, θ [,π] (ii) ess sup F (θ) <, θ [,π] (iii) H[F 1 ] = ess sup π θ [,π] π + F (θ+t) F (θ t) tan(t/) dt <. 35

40 In this part we give a criterion of quasiconformality for w = P [F ](z) by means of some conditions on F. This result partially improves D.Kalaj s theorem. Theorem 4.1([55, Theorem 1]). Suppose that F is a sense-preserving homeomorphism of T onto the boundary of a bounded convex domain Ω C. If (1) ess inf θ [,π] F (θ) = q >, () there exist constants M > and α > 1 such that F (θ + x) + F (θ x) F (θ) Mx α, (4.) holds for every θ [, π] and x >, then w = P [F ](z) is a harmonic quasiconformal mapping of U onto Ω. Proof. Without loss of generality, we may assume that w() =. Since F is a sense-preserving homeomorphism, it follows from Radó-Kneser-Choquet theorem [14] that w(z) = P [F ](z) is a univalent sense-preserving harmonic mapping of U onto Ω. For each z = re iθ U, we have w(z) = P [F ](z) = h(z)+g(z) = 1 π and w z (z) = 1 π e ix π (e ix z) F (x)dx = e iθ π π π e ix F (x) e ix z dx+ 1 π e ix π e iθ (e ix F (θ + x)dx + r) π π zf (x) e ix z dx, (4.3) e ix (e ix F (θ x)dx. r) 36

41 Since π ie ix (e ix r) dx =, this shows that π e ix π (e ix r) dx + e ix (e ix dx =. (4.4) r) In view of this observation, the last expression for w z (z) may be rewritten as w z (z) = e iθ π + e iθ π = e iθ π e iθ π π π π π e ix (e ix [F (θ + x) F (θ)] dx r) e ix (e ix [F (θ x) F (θ)] dx r) e ix (e ix [F (θ + x) + F (θ x) F (θ)] dx r) [ e ix (e ix r) e ix ] (e ix r) [F (θ x) F (θ)] dx. According to [1, Theorem. and Theorem 3.5], we see that w z (z) H p for all < p < 1 and its radial limit exists for a.e. e iθ T. Thus for a.e. e iθ T π lim w z(re iθ ) = e iθ e ix r 1 π (e ix [F (θ + x) + F (θ x) F (θ)] dx. 1) Now allowing r 1 in the modulus of the right-hand side, we conclude 37

42 that lim z(re iθ ) r 1 1 π 1 π 4 sin F (θ + x) + F (θ x) F (θ) dx (x/) 1 8π π Mx α (x/π) dx = Mπα 8(α 1) ( for a.e. eiθ T ). Applying [1, Theorem.1] we have w z (z) Mπα 8(α 1) ( for every z U ). (4.5) Since q = ess inf θ [,π] F (θ) >, using [7, Lemma.9] we have J w (z) = w z (z) w z (z) qr Γ, (4.6) where Γ = Ω, r Γ = dist(γ, ) >. Now using (4.5) and (4.6) we obtain w z (z) w z (z) 1 qr Γ w z (z) 1 3qr Γ(α 1) M π α < 1. This shows that w(z) is a harmonic quasiconformal mapping of the unit disk U onto Ω. The proof is completed. Remark 4.1. In the proof of Theorem 4.1, we can see that w z H is a bounded analytic function. We point out here that under the assumption of (4.), H[F ] is also bounded. In fact, for z = re iθ U, according to 38

43 (4.3) we see that w r (z) = 1 π = 1 π + 1 π π π π (1 + r ) cos x 4r (1 r cos x + r F (θ + x)dx ) (1 + r ) cos x 4r (1 r cos x + r F (θ + x)dx ) (1 + r ) cos x 4r (1 r cos x + r F (θ x)dx. ) Applying (4.4) we have Hence, w r (z) = 1 π π Since and π (1 + r ) cos x 4r (1 r cos x + r dx =. ) (1 + r ) cos x 4r (1 r cos x + r [F (θ + x) + F (θ x) F (θ)] dx. ) w z (z) = e iφ w θ (z) = π ( w r (z) i w ) θ(z) H, r P (r, θ x)f (x) dx, (4.7) using [1, Theorem 3.5 and Theorem.] we know that the radial limit of w r (z) and w θ (z) exist for a.e. e iθ T. Then lim w θ(re iθ ) = F (θ) and r 1 lim w r(re iθ ) = H[F ](e iθ ) for a.e. e iθ T. Thus for a.e. e iθ T r 1 lim w r(re iθ ) = 1 π r 1 π F (θ + x) + F (θ x) F (θ) sin dx (4.8) (x/) 39

44 and lim r(re iθ ) r 1 1 π 4π Mπ α 4(α 1). F (θ + x) + F (θ x) F (θ) (x/π) dx This shows that H[F ] Mπα 4(α 1). Corollary 4.([55, Corollary]). Let w(z) = P [F ](z) be a harmonic mapping of U onto a bounded convex domain Ω C satisfying w() = such that the boundary function F is a sense-preserving homeomorphism of T onto the boundary of Ω. If there exist constants M > and α > 1 such that F (θ + t) + F (θ t) F (θ) Mt α holds for every θ [, π] and t >, then w(z) is a harmonic quasiconformal mapping if and only if ess inf F (θ) = q >. θ [,π] Proof. Proof of only if part: Let w = P [F ](z) = h(z)+g(z) be a harmonic K-quasiconformal mapping of U onto Ω. According to [9, Proposition 3.3], we have w z (z) every z U ρ Ω (1+k), where ρ Ω = dist(, Ω) and k = K 1 K+1. Hence for w z (z) w z (z) (1 k) w z (z) 1 k (1 + k) ρ Ω. (4.9) Using (4.7) we have lim w θ(re iθ ) = F (θ) for a.e. e iθ T. On the r 1 [ ] other hand, since w θ (z) = izw z (z) + izw z (z), it follows that F (θ) = [ ] lim izw z (z) + izw z (z). Using (4.9) we obtain that F (θ) ρ Ω r 1 K >. The proof of if part is shown in the Theorem 4.1. This completes the proof. 4

45 4.1 Our boundary condition and the C 1,µ condition First, we point out that our boundary condition (4.) is weaker than the condition that the boundary function is C 1,µ. In fact, we can prove the following proposition. Proposition 4.3. If F gives a C 1,µ parametrization of Γ = Ω, then F satisfies the condition (4.). Proof. Since F is of C 1,µ ( < µ 1), according to (4.1) we have N := F (s) F (t) sup s,t [,π], s t s t µ <. This shows that F (s) F (t) N s t µ where s, t [, π] and s t. For any θ [, π] and x > 1 F (θ + tx) dt = 1 [F (θ + x) F (θ)], x Hence, 1 F (θ tx) dt = 1 [F (θ x) F (θ)]. x 1 F (θ + x) + F (θ x) F (θ) = x [F (θ + tx) F (θ tx)] dt, (4.1) 41

46 and F (θ + x) + F (θ x) F (θ) = x x x [F (θ + tx) F (θ tx)] dt F (θ + tx) F (θ tx) dt N tx µ dt = µ N 1 + µ x1+µ := Mx α for α = 1 + µ. This shows that F satisfies condition (4.) and the proof is completed. Zygmund class of functions. A π-periodic function F on R is said to belong to the Zygmund class if there exists a positive constant M such that the inequality F (x + h) + F (x h) F (x) Mh, (4.11) holds for all x R and h >. The class was introduced by A.Zygmund. We see that our condition (4.) is stronger than (4.11) when < h < 1. However, α > 1 in our condition (4.) is indispensable. In next section we will give an example to show this. 4

47 4. Examples The following Example 4.1 shows that our boundary condition (4.) is easy to be judged. Example 4.1([55, Example 1]). Assume that a > b >, consider the boundary function F (x) = a cos x+ib sin x = a + b eix + a b e ix. Then we have F (x) = a sin x + b cos x = (a b ) sin x + b b >. For x > and θ [, π], there exist θ < ξ 1, ξ < θ + x and θ x < ξ 3, ξ 4 < θ such that F (θ + x) F (θ) = ax sin ξ 1 + ibx cos ξ, F (θ x) F (θ) = ax sin ξ 3 ibx cos ξ 4. Hence there exist ζ between ξ 3 and ξ 1, η between ξ and ξ 4 such that F (θ + x) + F (θ x) F (θ) = ax(ξ 3 ξ 1 ) cos ς ibx(ξ ξ 4 ) sin η, and thus F (θ + x) + F (θ x) F (θ) ax ξ 3 ξ 1 + bx ξ ξ 4 (a + b)x. Applying Theorem 4.1 we see that w(z) = P [F ](z) = π p(r, x θ)f (x)dx, z = re iθ U, is a harmonic quasiconformal mapping of U onto the ellipse : x a + y 1. In fact, b w(z) = π p(r, x θ) Thus k = sup w z (z) w z (z) = a b a + b < 1. z U [ a + b eix + a b ] e ix dx = a + b z + a b z. 43

48 Figure : Example 4. (figure of the function φ) The following Example 4. (cf. [39]) shows that in the condition (4.) α > 1 is indispensable. Example 4.. Let φ(x) be the continuous increasing piecewise linear function satisfying φ(x + π) = φ(x) + π and 1 + ( ) π x π x < φ(t) = 1 + ( 1 1 ) π x x π. Then φ (x) exists for a.e. x R. Now put F (x) := e iφ(x). Then F exists for a.e. x R, and F (x) = φ (x) π. Thus for any x R and t > F (x + t) F (x) = e iφ(x+t) e iφ(x) cos φ(x + t) cos φ(x) + sin φ(x + t) sin φ(x) φ (η 1 ) t + φ (η ) t (x < η 1, η < x + t) ( ) t. π 44

49 This shows that F (x + t) + F (x t) F (x) 4 ( ) π t and F satisfies the condition (4.) for α = 1. However, according to [39, Example 4.3] we know that H[F ] = and w = P [F ](z) is not a harmonic quasiconformal mapping. Therefore in Theorem 4.1, α > 1 is indispensable. 45

50 5 Landau type theorems for bounded harmonic mappings Let M be a positive constant. Denote by H M (U) the class of all harmonic mappings f(z) = h(z) + g(z) defined in U satisfying f(z) M, where h(z) = a n z n and g(z) = b n z n are analytic in U. In [7, Theorem 1.1] Chen et al. showed that if f H M (U), then for each n 1 a n + b n 4M π (5.1) holds. The estimate (5.1) is sharp for each n 1. Suppose that f = h + ḡ is a bounded harmonic mapping of U where h(z) = z + a n z n and g(z) = b n z n (5.) n= n= are analytic in U and satisfy the coefficients condition a n + b n c, n =, 3, (5.3) for some positive constant c. Then we obtain Landau type theorems for such mappings f and L(f), where L(f)(z) := zf z (z) zf z (z) = if θ (z). Moreover if f is a harmonic K-quasiconformal self-mapping of U and satisfies only (5.) without (5.3), then we obtain a univalent radius of L(f) by means of K. Theorem 5.1([57, Theorem 1]). Suppose that f(z) is a bounded harmonic mapping defined in U satisfying f() = f z () = f z () 1 = and the 46

51 coefficients condition (5.3). Then (1) f(z) is univalent in a disk U r and f(u r ) contains a disk U R, where r = 1 c 1+c and R = r cr 1 r ; () L(f)(z) = zf z (z) zf z (z) is univalent in a disk U ρ and L(f)(U ρ ) contains a disk U σ, where ρ is given by (5.5) and σ = ρ cρ (1 ρ [ ] 1 1. ) Proof. (1) For any z 1 = r 1 e iθ 1, z = r e iθ we have f(z 1 ) f(z ) = (z 1 z ) + z 1 z z 1 z ( ( + n= c U, let r = max{r 1, r }. Then a n (z n 1 z n ) + n= + n= + n= n( a n + b n )r n 1 ) nr n 1 ) [ ( )] 1 = z 1 z 1 c (1 r) 1. b n (z n 1 zn ) ( ) 1 Let 1 c (1 r 1 ) univalent in U r. =. Then r = 1 c c+1. This shows that f(z) is For z U r, we have f(z) = + z + ( an z n + b n z n) z r c n= + n= + n= ( a n + b n ) z n r n = r cr 1 r = R. 47

52 Since c 1+c < c c+1, this shows that R >. We see that f(u r ) contains the disk U R. obtain () For any z 1 = r 1 e iθ 1, z = r e iθ U, let ρ = max{r 1, r }. Then we ( ) ( L(f)(z 1 ) L(f)(z ) = z 1 h (z 1 ) z 1 g (z 1 ) z h (z ) z g (z )) + + = (z 1 z ) + na n (z1 n z n ) nb n (z1 n zn ) n= n= ( ) + z 1 z 1 ( a n + b n )n ρ n 1 z 1 z [ 1 c n= + n= n ρ n 1 ] ( ( )) 1 + ρ = z 1 z 1 c (1 ρ) 3 1. as t 3 + ( ) Let 1 c 1+ρ (1 ρ 1 =. By setting t ) 3 = 1 ρ, we can simplify it c c+1 t c c+1 =. Solving this equation, then ( t = 3 c c c + 1 c + 1 ( + 3 c c c c + 1 ) + 1 ( ) c 3 7 c + 1 ) + 1 ( ) c 3 (5.4) 7 c + 1 and ( ρ = 1 3 c c + 1 ( 3 c c c c + 1 c c + 1 ) + 1 ( ) c 3 7 c + 1 ) + 1 ( ) c 3. (5.5) 7 c

53 This shows that L(f) is univalent in the disk U ρ. For any z U ρ, we obtain + L(f)(z) = z + ( nan z n nb n z n) z ρ c n= + n( a n + b n ) z n n= + nρ n n= ( ) 1 = ρ cρ (1 ρ ) 1 = σ. Let φ(t) := t 3 + c c c+1t Since φ( c c+1 ) <, this shows that t > see that f(u ρ ) contains the disk U σ. The proof is completed. c+1. Then φ(t) is an increasing function and φ(t ) =. c c+1 and therefore σ >. We Applying (5.1) and Theorem 5.1 we have Corollary 5.([57, Corollary 1]). Assume that f H M (U) satisfies f z () 1 = f z () =. Then (1) f is univalent in a disk U r, where r = 1 () L(f) is univalent in a disk D ρ, where and α = ρ = 1 3 α 4M 4M+π. α α3 3 α + 4M 4M+π. α α3 Theorem 5.3([57, Theorem ]). Suppose that f = h+ḡ is a harmonic K- quasiconformal self-mapping of U, satisfying f() = f z () 1 = f z () =. 49

54 Then L(f) is univalent in a disk U ρk, where ρ K = 1 L(f)(U ρk ) contains a disk U σk, where σ K = ρ K Λ K 1 Λ K is shown by (5.6). Λ K 1 Λ K +Λ K 1 ( ) and ρ K 1 ρ K and Λ K Proof. Since f is a harmonic K-quasiconformal self-mapping of U, according to Theorem 3.4 we know that h (z) π A 3 K+1, where k = 4(1 k ) K 1 and A = K 3K 5(K 1/K)/. Thus Λ f = h + g (1 + k) h KπA 3 := Λ K. (5.6) According to [34] we see that a n + b n Λ K 1 nλ K. For any z 1 = r 1 e iθ 1, z = r e iθ U, let ρ = max{r 1, r }. Following the proof of Theorem 5.1 we have L(f)(z 1 ) L(f)(z ) z 1 z ( 1 + n= ( a n + b n )n ρ n 1 ) [ ] z 1 z 1 Λ K 1 + nρ n 1 Λ K n= [ = z 1 z 1 Λ K 1 ( )] 1 Λ K (1 ρ) 1. Solving the equation 1 Λ K 1 Λ K 1 Λ K +Λ K 1. Λ K [ ] 1 (1 ρ K 1 ) =, we obtain ρ K = 1 5

55 For z U ρk, we have L(f)(z) z + n= ρ K Λ K 1 Λ K = ρ K Λ K 1 Λ K n( a n + b n ) z n + ρ n K n= ( ρ ) K = σ K. 1 ρ K Since Λ K 1 Λ K +Λ K 1 < Λ K 1 L(f)(U ρk ) contains the disk U σk. This completes the proof. Λ K +Λ K 1, this implies that σ K >. We see that 51

56 6 Two subclasses of harmonic mappings In this section we consider two subclasses of harmonic mappings defined in U and the exterior of U and discuss quasiconformality for these two classes. This also gives some criteria of quasiconformality for harmonic mappings. where Assume that f(z) = h(z) + g(z) is a harmonic mapping defined in U, h(z) = z + a n z n and g(z) = n= b n z n, (6.1) are analytic in U and b 1 < 1. For µ < 1, denote by HS(µ) the class of such harmonic mappings f which satisfy the condition (n µ)( a n + b n ) (1 µ)(1 b 1 ). (6.) n= In [37] M.Öztürk and S.Yalcin proved the following theorem. Theorem F ([37]). If f HS(µ), then f is a sense-preserving univalent harmonic mapping of U and satisfies the following inequalities f(z) z (1 + b 1 ) + (1 µ )(1 b 1 ) z and f(z) (1 b 1 ) ] [ z (1 µ ) z. Equalities are attained by the functions for properly chosen real θ. f θ (z) = z + b 1 e iθ z + 1 b 1 (1 µ ) z In this section for < µ < 1 we prove that f is a bi-lipschitz harmonic mapping. In particular, f is a quasiconformal mapping. 5

57 Theorem 6.1([54, Theorem 1]). Let < µ < 1. Then every f HS(µ) is a harmonic K-quasiconformal mapping, where K 1 is a constant determined by µ. Proof. According to Theorem F we know that f(z) is sense-preserving univalent in U. Applying (6.) we have ( a n + b n ) = 1 n= (n µ)( a n + b n ) + µ n= (1 µ)(1 b 1 ) + µ n= ( a n + b n ). n= + µ n ( a n + b n ) µ Hence ( a n + b n ) (1 µ)(1 b 1 ). (6.3) µ n= Using (6.), (6.3) and n( a n + b n ) = (n µ)( a n + b n ) + µ ( a n + b n ), n= n= n= we see that g (z) h (z) = = nb n z n 1 b 1 + n b n n= 1 + na n z n 1 1 n a n n= n= b 1 + n b n n= 1 (1 µ)(1 b 1 ) µ(1 µ)(1 b 1 ) µ + b 1 + b 1 + n b n n= n= n b n + (1 b 1 ) µ µ. n= n b n 53

58 Since n µ 1 µ n, for n =, 3,, we have Thus n( a n + b n ) n= n= g (z) h (z) (1 b 1 ) µ n µ 1 µ ( a n + b n ) 1 b 1. µ = k < 1. This shows that f is a harmonic K-quasiconformal mapping and K = 1+k 1 k is a constant determined by µ. The proof is completed. Theorem 6.([54, Theorem ]). Let < µ < 1. Then every f HS(µ) is a bi-lipschitz harmonic mapping. Furthermore, the following inequalities µ(1 b 1 ) µ hold for every z 1, z U. z 1 z f(z 1 ) f(z ) 4 + µ( b 1 3) z 1 z, µ Proof. For any z 1, z U, we have f(z 1 ) f(z ) = (z 1 z ) + a n (z1 n z n ) + b n (z1 n zn ) n= ( ) z 1 z 1 b 1 n( a n + b n ) n= ( z 1 z 1 b 1 (1 µ)(1 b 1 ) µ (1 µ)(1 b ) 1 ) µ = µ(1 b 1 ) z 1 z. µ 54

59 On the other hand, f(z 1 ) f(z ) = (z 1 z ) + a n (z1 n z n ) + b n (z1 n zn ) n= ( ) z 1 z 1 + b 1 + n( a n + b n ) n= ( z 1 z 1 + b 1 + (1 µ)(1 b 1 ) + µ (1 µ)(1 b ) 1 ) µ = 4 + µ( b 1 3) z 1 z. µ This completes the proof. Remark 6.1. Since n µ 1 µ is an increasing function of µ, we see that n µ 1 1 µ 1 n µ 1 µ holds for µ 1 µ < 1. This shows that HS(µ ) HS(µ 1 ) for µ 1 µ < 1. For µ =, f(z) may not be a quasiconformal mapping. Take f(z) = z + z 4 + z 4 for example. Since ( n µ 1 1 µ ( a n + b n ) = ) 1 b 1 = 1, 4 n= we see that f(z) HS(). However, its complex dilatation a(z) = f z(z) f z(z) = z +z and sup a(z) = 1. This shows that f(z) is not a quasiconformal mapping. Thus µ > is indispensable in Theorem 6.1 and Theorem z U 6.. Let Ũ = {z C : z > 1} be the exterior of U and f is a harmonic mapping defined in Ũ that maps to in the sense that lim f(z) =. z It is known that such a mapping can be represented by f(z) = A log z + h(z) + g(z) 55

60 where h(z) = z + a n z n and g(z) = b n z n n= n= and A C is a constant. Under the additional assumption that f is a sense-preserving univalent harmonic mapping, W.Hengartner and G.Schober proved the following theorem. Theorem G ([1]). Let f be a sense-preserving univalent harmonic mapping of Ũ satisfying f( ) =. Then f has the representation f(z) = A log z + αz + β z + a k z k + b k z k, (6.4) k= k=1 where β α and A C. In addition, a = f z /f z is analytic and satisfies a(z) < 1. On the other hand, if f is a harmonic mapping of Ũ which has the form (6.4), then is f univalent? In [1] the authors did not mention it. Denote by Σ H the class of all harmonic mappings f(z) = h(z) + g(z) of Ũ where h(z) = αz + a n z n, g(z) = βz + b n z n (6.5) are analytic in Ũ with α > β. JM.Jahangiri and H.Silverman proved in [5, Theorem ] that if f Σ H satisfies n( a n + b n ) α β, then f is a sense-preserving univalent starlike harmonic mapping of Ũ. For µ < 1, denote by Σ H (µ) the class of harmonic mappings f Σ H which satisfy the following condition (n µ)( a n + b n ) (1 µ)( α β ). (6.6) 56

61 In the following theorem, we give another criterion of quasiconformality for Σ H (µ). Theorem 6.3([54, Theorem 3]). Assume that f Σ H (µ). Then f is a sense-preserving univalent harmonic mapping. Furthermore, if < µ < 1 and f(z) αz + a 1 z +β z + b 1 z, then f is a K-quasiconformal mapping, where K is a constant determined by µ. Proof. Take any z 1, z Ũ, where z 1 z. According to (6.5) we have ( z n f(z 1 ) f(z ) = α(z 1 z )+β(z 1 z )+ a z1 n n z1 nzn Since z 1 > 1, z > 1 and n µ 1 µ f(z 1 ) f(z ) z 1 z n, we have ( ( z 1 z α β 1 z 1 z α β = z 1 z ( α β ) ) + b n ( z n z n 1 z n 1 zn ) n( a n + b n ) ) α β z 1 z ( 1 1 ) z 1 z >. ). This shows that f is univalent in Ũ. On the other hand, since ( h (z) g (z) α β ) n( a n + b n ) z n+1 ( ) α β 1 z n( a n + b n ) ( ( α β ) 1 1 ) z > for z Ũ, this implies that f is a sense-preserving harmonic mapping. 57

62 and According to (6.6) we know that (n µ)( a n + b n ) (1 µ)( α β ), (n µ)( a n + b n ) (1 µ)( α β a 1 b 1 ). n= Using +µ n µ 1 (for n ), we have ( a n + b n ) = 1 n= (n µ)( a n + b n ) + µ n= + µ n ( a n + b n ) µ ( a n + b n ). n= 1 µ ( α β a 1 b 1 ) + µ n= Thus ( a n + b n ) 1 µ µ ( α β a 1 b 1 ). (6.7) n= According to (6.6) and (6.7) we have n a n (1 µ)( α β ) + µ ( a n + b n ) n b n + = (1 µ)( α β ) n b n + µ( a 1 + b 1 ) µ(1 µ) µ ( α β a 1 b 1 ) (1 µ) µ ( α β ) + µ µ ( a 1 + b 1 ) n b n. 58

63 Applying n( a n + b n ) g (z) h (z) = β + n b n α n a n α + β + n µ 1 µ ( a n + b n ) α β, we have β + n b n n b n (1 µ) µ µ ( α β ) µ ( a 1 + b 1 ) β + n b n n b n + µ µ ( α β a 1 b 1 ) α α + µ µ ( α β a 1 b 1 ) = k. Since f(z) αz + a 1 z + β z + b 1 z, according to (6.5) and (6.6) we know that α β a 1 b 1 >. This shows that k < 1 and f is a K-quasiconformal mapping, where K = 1+k 1 k The proof is completed. is a constant determined by µ. Example 6.1. Let f(z) = z + 1 4z z 4 z + 1. Then 8 z n( a n + b n ) = a 1 + b 1 + ( a + b ) = 1 α β = 1. This implies that f Σ H (). However, since f z (z) = z f z (z) = 1 + 1, we have 4z 4z 3 f z (z) sup f z (z) = sup 1 + z 4z 3 z 1 = 1. z U z U + 1 4z 3 and This shows that f(z) is not a quasiconformal mapping. Thus in our Theorem 6.3, µ > is indispensable. 59

64 Example 6.. Let f(z) = z + 1 z + 1 z. Then n µ 1 µ ( a n + b n ) = a 1 + b 1 = 1 α β = 1. This implies that f Σ H (µ) for any µ < 1. However, since f z (z) = 1 z and f z (z) = 1 + 1, we have z f z (z) sup f z (z) = sup 1 z 1 = 1. z U z U This shows that f is not a quasiconformal mapping. Thus in our Theorem 6.3, the assumption that f(z) αz + a 1 z + β z + b 1 z is necessary. 6

65 Acknowledgments I would like to express my sincere gratitude to my supervisor Professor K.Sakan for his constant encouragement and advices. I am very grateful to Department of Mathematics, Graduate School of Science, Osaka City University for the admission of one year s stay of me. I would like to thank Professors X.Z.Huang, S.Ponnusamy and D.Kalaj for their helpful advices. June, 13 Jian-Feng Zhu 61

66 References [1] P. Ahuja: Planar harmonic univalent and related mappings. J. Inequal. Pure Appl. Math. 6(5), [] G.D. Anderson, M.K. Vamanamurthy and M. Vuorinen: Distortion function for plane quasiconformal mappings. Israel J.Math. 6(1988), [3] A.L. Beueling and V. Ahlfors: The boundary correspondence under quasiconformal mapping. Acta Math. 96(1956), [4] M. Bonk: On Bloch s constant. Proc. Amer. Math. Soc. 1(4) (199), [5] H. H. Chen, P. M. Gauthier and W. Hengartner: Bloch constants for planar harmonic mappings. Proc. Amer. Math. Soc. 18 (), [6] SH. Chen, S. Ponnusamy and X. Wang: Landau s Theorem and marden constant for harmonic v-bloch mappings. Bull. Aust. Math. Soc. 84 (11), [7] SH. Chen, S. Ponnusamy and X. Wang: Coefficient estimates and Landau-Bloch s constant for plannar harmonic mappings. Bull. Malays. Math. Sci. Soc. 34(11),