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RADIAL LENGTH, RADIAL JOHN DISKS AND K-QUASICONFORMAL HARMONIC MAPPINGS 7 1 SHAOLIN CHEN AND SAMINATHAN PONNUSAMY 0 2 n Abstract. Inthisarticle,wecontinueourinvestigationsoftheboundarybehav- a ior of harmonic mappings. We first discuss the classical problem on the growth J of radial length and obtain a sharp growth theorem of the radial length of K- 6 quasiconformal harmonic mappings. Then we present an alternate characteriza- 2 tionofradialJohndisks. Inaddition,weinvestigatethelinearmeasuredistortion ] andthe Lipschitz continuity onK-quasiconformalharmonic mappings ofthe unit V diskontoaradialJohndisk. Finally,usingPommerenkeinteriordomains,wechar- C acterize certain differential properties of K-quasiconformalharmonic mappings . h t a m 1. Introduction and statement of main results [ 1 This paper continues the study of previous work of the authors [6] and is mainly v motivatedbythearticlesofBeardonandCarne[3],CarrollandTwomey[4], Chuaqui 7 et al. [10], Pommerenke [29], and the monograph of Pommerenke [30]. In order to 3 5 state our first result concerning the growth of the radial length of K-quasiconformal 7 harmonic mappings (see Theorem 1), we need to recall some basic definitions and 0 . some results which motivate the present work. 1 0 Let f be a complex-valued and continuously differentiable function defined in the 7 unit disk D = z : z < 1 and let ℓ (θ,r) be the length of the f-image (with f 1 { | | } counting multiplicity) of the radial line segment [0,z] from 0 to z = reiθ D, where : v ∈ θ [0,2π] is fixed and r [0,1). Then (cf. [5]) Xi ∈ ∈ r r r ℓ (θ,r) := ℓ f([0,z]) = df(ρeiθ) = f (ρeiθ)+e 2iθf (ρeiθ) dρ. a f z − z Z0 Z0 In [21], Keogh sh(cid:0)owed tha(cid:1)t if f is(cid:12)(cid:12)a bound(cid:12)(cid:12)ed, anal(cid:12)(cid:12)ytic and univalent func(cid:12)(cid:12)tion in D, then, for each θ [0,2π], ∈ 1/2 (1.1) ℓ (θ,r) = O log(1/(1 r)) as r 1 . f − − → (cid:16) (cid:17) Throughout the discussion, we l(cid:0)et (cid:1) 1/2 (1.2) ψ(r) = log(1/(1 r)) for 0 < r < 1. − (cid:0) (cid:1) File: Ch-P-Kqc-2016_final.tex, printed: 27-1-2017, 1.42 2010 Mathematics Subject Classification. Primary: 30C62, 30C75; Secondary: 30C20, 30C25, 30C45, 30F45, 30H10. Key words and phrases. K-quasiconformal harmonic mapping, radial John disk, radial length, Pommerenke interior domain. This second author is on leave from IIT Madras. 1 2 S.L. Chen and S. Ponnusamy Keogh also gave some examples to show that the exponent 1/2 in (1.1) can not be decreased. Jenkins improved on these examples in [16], and Kennedy [20] presented further examples by showing that ℓ (θ,r) = O(µ(r)ψ(r)) as r 1 f − → isfalseingeneralforeverypositivefunctionµin[0,1)satisfying µ(r) 0asr 1 . − → → In [4], Carroll and Twomey established certain refinements and extension of these results without the boundedness condition in the following form. Theorem A. Suppose that f(z) = a z + a z2 + is univalent in D. Then, for 1 2 ··· any fixed θ [0,2π], there is a constant C > 0 such that 1 ∈ (1.3) ℓ (θ,r) C max f(ρeiθ) ψ(r) for r (0.5,1). f 1 ≤ ρ [0,r]| | ∈ ∈ If, further, f(reiθ) = O(1) as r 1 , then (1.1) holds. − → Later, Beardon and Carne [3] gave a relatively simple argument to Theorem A in hyperbolic geometry and provided with further examples. It is worth pointing out here two results which strengthened (1.3) and was inspired by the work of Sheil- Small [33] and Hall [14]. If f is starlike, i.e. f(D) contains the line segment ∈ S [0,w] whenever it contains w, then (see [19]) ℓ (θ,r) f(reiθ) (1+r) < 2 f(reiθ) for r (0,1) f ≤ | | | | ∈ and the inequality of course is not sharp for all r, but the bound 2 sharp as the Koebe function k(z) = z/(1 z)2 shows and is attained when r approaches 1 (see − [14, 33]). Later in 1993, Balasubramanian et al. [1] showed that if f is convex, i.e. f(D) is a convex domain, then ∈ S ℓ (θ,r) f(reiθ) r 1arcsinr for r (0,1) f − ≤ | | ∈ and the inequality is sharp as the convex function f(z) = z/(1 z) shows. Note − that ϕ(r) = r−1arcsinr is increasing on (0,1) and ϕ(r) limr 1−ϕ(r) = π/2 and ≤ → thus, the conjecture of Hall [15] was settled (see also [2]). The first aims of this paper is to extend Theorem A for the case of harmonic quasiconformal mappings (see Theorem 1 below). We need some preparation to state this result. For a real 2 2 matrix A, we use the matrix norm A = sup Az : z = 1 and the matrix×function l(A) = inf Az : z = 1 . Forkz =k x+iy{| C,| th|e |forma}l {| | | | } ∈ derivative of the complex-valued function f = u+iv is given by the Jacobian matrix u u x y D = , f vx vy ! so that D = f + f and l(D ) = f f , f z z f z z k k | | | | | |−| | where fz = (1/2) fx ify and fz = (1/2) fx + if(cid:12)y . Let Ω(cid:12) be a domain in C, − (cid:12) (cid:12) with non-empty boundary. A sense-preserving homeomorphism f from a domain Ω (cid:0) (cid:1) (cid:0) (cid:1) Radial length, radial John disks and K-quasiconformal harmonic mappings 3 onto Ω, contained in the Sobolev class W1,2(Ω), is said to be a K-quasiconformal ′ loc mapping if, for z Ω, ∈ D (z) 2 K detD (z) , i.e., D (z) Kl D (z) , f f f f k k ≤ k k ≤ where K 1 and detD is t(cid:12)he determi(cid:12)nant of D (cf. [18, 2(cid:0)2, 35, 3(cid:1)6]). ≥ f (cid:12) (cid:12) f Let denote the family of sense-preserving planar harmonic univalent mappings H f = hS+g in D, with the normalization h(0) = g(0) = 0 and h(0) = 1. Recall that ′ f is sense-preserving if the Jacobian J of f given by f J := detD = f 2 f 2 = h 2 g 2 f f z z ′ ′ | | −| | | | −| | is positive. Thus, f is locally univalent and sense-preserving in D if and only if J (z) > 0 in D; or equivalently if h = 0 in D and the dilatation ω = g /h has the f ′ ′ ′ property that ω(z) < 1 in D (see [611, 12, 23]). The family together with a H | | S few other geometric subclasses, originally investigated in detail by [11, 34], became instrumental in the study of univalent harmonic mappings (see [12, 31]) and has attracted the attention of many function theorists. If the co-analytic part g is identically zero in the decomposition of f = h+g, then the class reduces to the H S icnlaDss.icIaflfa0m=ilySf =ofhal+lngormalize:dga(n0a)l=yti0cu,ntihveanletnhtefufanmctiiloyns0h(izs)b=otzh+nor∞nm=a2laannzdn SH { ∈ SH ′ } SH P compact. See [11] and also [8, 6, 12, 31]. Theorem 1. For K 1, let f be a K-quasiconformal harmonic mapping. H ≥ ∈ S Then, for any fixed θ [0,2π], there is a constant C > 0 such that 2 ∈ ℓ (θ,r) C max f(ρeiθ) ψ(r) for r (0.5,1). f 2 ≤ ρ [0,r]| | ∈ ∈ If, further, f(reiθ) = O(1) as r 1 , then − → ℓ (θ,r) = O(ψ(r)) as r 1 , f − → and the exponent1/2 in ψ(r) definedby (1.2) cannotbe replacedbya smallernumber. First we remark that if K = 1, then Theorem 1 coincides with Theorem A. Sec- ondly, the proof of Theorem 1 is substantially harder than the proof of Theorem A. This is because Beardon and Carne’s argument of Theorem A in [3] is not applicable in the proof of Theorem 1. We need further notation and terminology before stating our second result. Let d (z) be the Euclidean distance from z to the boundary ∂Ω of Ω. If Ω = D, then Ω we set d(z) := dD(z). Definition 1. A bounded simply connected plane domain G is called a c-John disk for c 1 with John center w G if for each w G there is a rectifiable arc γ, 0 1 ≥ ∈ ∈ called a John curve, in G with end points w and w such that 1 0 σ (w) cd (w) ℓ G ≤ for all w on γ, where γ[w ,w] is the subarc of γ between w and w, and σ (w) is the 1 1 ℓ Euclidean length of γ[w ,w] (see [6, 13, 17, 28, 30]). 1 4 S.L. Chen and S. Ponnusamy Remark 1. If f is a complex-valued and univalent mapping in D, G = f(D) and, for z D, γ = f([0,z]) in Definition 1, then we call c-John disk a radial c-John disk, ∈ where w = f(0) and w = f(z). In particular, if f is a conformal mapping, then we 0 call c-John disk a hyperbolic c-John disk. It is well known that any point w G 0 ∈ can be chosen as a John center by modifying the constant c if necessary. When we do not wish to emphasize the role of c, then we regard the c-John disk simply as a John disk in the natural way (cf. [6, 13, 17, 28]). Unless otherwise stated, throughout the discussion we consider the following terminology. Denote by (K) if f and is a K-quasiconformal harmonic mapping in D, where K 1. AFlso, we de∈noFte by (K,Ω) if f (K) and f maps D onto ≥ F ∈ F Ω. We prove several results mainly when equals one of , 0 , and , and Ω F SH SH SH2 equals either radial John disk or Pommerenke interior domain. Further, for z D, we define ∈ (1.4) B(z) := ζ : z ζ < 1, argz argζ π(1 z ) . { | | ≤ | | | − | ≤ −| | } In the following, we continue our previous work of [6], and give an another charac- terization of the radial John disk. Theorem 2. Let f 0 (K). Then the following are equivalent: ∈ SH (i) Ω = f(D) is a radial John disk. (ii) There is an x (0,1) such that ∈ (1 ρ2) D (ρζ) x+r sup sup − k f k < 1 for ρ = . (1 r2) D (rζ) 1+xr ζ=1r (0,1) f | | ∈ − k k f(z) f(w) (iii) sup | − | < . z D, w B(z) (1 z 2) Df(z) ∞ ∈ ∈ −| | k k Next, we establish the linear measure distortion on K-quasiconformal harmonic mappings of D onto a radial John disk. Theorem 3. Let f = h+g 0(K,Ω), where Ω is a radial John disk. Then, for a ,a D with B(a ) B(a∈),SthHere is a positive constant C such that 1 2 1 2 3 ∈ ⊂ diamf(B(a )) ℓ(B(a ) ∂D) α 1 1 C ∩ . diamf(B(a )) ≤ 3 ℓ(B(a ) ∂D) 2 (cid:18) 2 ∩ (cid:19) where α = supf H |h′′2(0)| and B(z) is defined by (1.4). ∈S knWowenre(smeear[k6,t9h,a1t22,≤34]α).=Wseupdfis∈cSuHss|h′t′2(h0e)|L<ip∞sch,ibtuztcothnetinshuaitrypovnalKue-qoufaαsiicsosntfiollrmuna-l harmonic mappings of D onto a radial John disk, which is as follows. Theorem 4. Let f = h+g 0(K,Ω), where Ω is a radial John disk. Then, for z D with z 1 and ζ ,ζ∈ SHB(z), there are constants δ (0,1) and C > 0 ∈ | | ≥ 2 1 2 ∈ 1 ∈ 4 such that ζ ζ δ1 1 2 f(ζ ) f(ζ ) C d (f(z)) | − | . 1 2 4 Ω | − | ≤ 1 z (cid:18) −| | (cid:19) Radial length, radial John disks and K-quasiconformal harmonic mappings 5 Let f (K,G), where G is domain. For 0 < r < 1, let D = z : z < r and H r ∂D deno∈teSthe boundary of D . Now, for w ,w f (∂D ), let γ{ be|t|he sm}aller r r 1 2 r r subarc of f (∂D ) between w and w , and let ∈ r 1 2 d (w ,w ) = infdiamΓ, Gr 1 2 Γ where Γ runs through all arcs from w to w that lie in G = f(D ) except for their 1 2 r r endpoints. If ℓ γ [w ,w ] r 1 2 (1.5) sup sup < , d (w ,w ) ∞ 0<r<1(w1,w2∈γr (cid:0)Gr 1 2(cid:1)) then we call G a Pommerenke interior domain (cf. [6, 29]). In particular, if G is bounded, then we call G as a bounded Pommerenke interior domain. Given a sense-preserving harmonic mapping f = h+g in D, fix ζ D andperform ∈ a disk automorphism (also called Koebe transform F of f) to obtain f z+ζ f(ζ) 1+ζz − (1.6) F(z) = =: H(z)+G(z). h(cid:16)(ζ)(1(cid:17) ζ 2) ′ −| | A calculation gives, H (0) 1 h (ζ) ′′ = (1 ζ 2) ′′ 2ζ . 2 2 −| | h(ζ) − (cid:26) ′ (cid:27) Now, we consider the class of all harmonic mappings f = h+g satisfying SH2 ∈ SH h (z) (1.7) sup (1 z 2) ′′ 2z < 4. z D(cid:12) −| | h′(z) − (cid:12) ∈ (cid:12) (cid:12) Thisinequality obviouslyholds(cid:12)ifh andhisnot(cid:12)theKoebefunctionz/(1 eiθz)2, θ R. Note that for the Koeb(cid:12)e fun∈ctiSon the suprem(cid:12)um turns out to be 4. O−ur next ∈ two results are extension of [29, Theorem 3]. Theorem 5. Let f (K,G), where G is a bounded Pommerenke interior domain. If there are po∈sitSivHe2 constants δ (0,1) and C such that, for each ζ ∂D 2 5 ∈ ∈ and for 0 ρ ρ < 1, 1 2 ≤ ≤ 1 ρ2 δ2−1 (1.8) D (ρ ζ) C − D (ρ ζ) , f 2 5 f 1 k k ≤ 1 ρ k k (cid:18) − 1(cid:19) then 1 D (ξ) 1 ζ 2 f sup k k −| | dξ < . ζ∈D 2π Z∂D kDf(ζ)k|ξ −ζ|2| | ∞ We remark that if K = 1, then Theorem 5 coincides with [29, Theorem 3]. By using similar reasoning as in the proof of Theorem 5, one can easily get the following result which replaces the assumption f by a more general condition ∈ SH2 f and thus, we omit its proof. H ∈ S 6 S.L. Chen and S. Ponnusamy Theorem 6. Let f (K,G), where G is a bounded Pommerenke interior do- H ∈ S main. If there are constants C > 0, C > 0, δ > 0 and δ (0,1) such that, for 6 7 3 4 each ζ ∂D and for 0 ρ ρ < 1, ∈ 1 2 ∈ ≤ ≤ 1 ρ1 δ3−1 1 ρ2 δ4−1 C − D (ρ ζ) D (ρ ζ) C − D (ρ ζ) , 6 f 1 f 2 7 f 1 1 ρ k k ≤ k k ≤ 1 ρ k k (cid:18) − 2(cid:19) (cid:18) − 1(cid:19) then 1 D (ξ) 1 ζ 2 f sup k k −| | dξ < . ζ∈D 2π Z∂D kDf(ζ)k|ξ −ζ|2| | ∞ Also, the following result easily follows from Theorem 5 and [6, Theorem 1]. Corollary 1. For K 1, let f 0 be a K-quasiconformal harmonic mapping from D onto a≥bounded Po∈mmSHer2e∩nkSeHinterior domain G. If G is a radial John disk, then 1 D (ξ) 1 ζ 2 f sup k k −| | dξ < . ζ∈D 2π Z∂D kDf(ζ)k|ξ −ζ|2| | ∞ The proofs of Theorems 1-5 will be presented in Section 2. 2. The proofs of the main results Let λD stand for the hyperbolic distance (or Poincaré distance) on the unit disk D. We have dz z z λD(z1,z2) = inγfZγ 1−| |z||2 = tanh−1(cid:12)11−−z1z22(cid:12), where the infimum is taken over all smooth curves γ (cid:12)(cid:12)in D conn(cid:12)(cid:12)ecting z1 D and z D (cf. [30]). In [34], Sheil-Small proved that if f =(cid:12) h+g (cid:12) , then ∈ 2 H ∈ ∈ S (1 z )α 1 (1+ z )α 1 − − (2.1) −| | h(z) | | ′ (1+ z )α+1 ≤ | | ≤ (1 z )α+1 | | −| | and h (0) ′′ α := sup | | < . 2 ∞ f H ∈S Unless otherwise stated, the number α will be used throughout the discussion and is indeed called the order of the linear invariant family (see [34]). H S Lemma 1. Suppose that f (K). Then, for z ,z D, H 0 1 ∈ S ∈ 1 f(z ) f(z ) K 1 e 2αλD(z1,z0) | 1 − 0 | e2αλD(z1,z0) 1 . − α(1+K) − ≤ (1 z 2) f (z ) ≤ α(1+K) − 0 z 0 −| | | | (cid:2) (cid:3) (cid:2) (cid:3) In particular, 1 K (2.2) 1 e 2αλD(z,0) f(z) e2αλD(z,0) 1 , z D. − α(1+K) − ≤ | | ≤ α(1+K) − ∈ (cid:2) (cid:3) (cid:2) (cid:3) Radial length, radial John disks and K-quasiconformal harmonic mappings 7 Proof. By assumption f = h+ g is a K-quasiconformal harmonic mapping, H where h and g are analytic in D. T∈huSs, by (2.1), we have 2K 2K (1+ z )α 1 − (2.3) D (z) h(z) | | f ′ k k ≤ K +1| | ≤ K +1(1 z )α+1 −| | and thus, for z D, we obtain ∈ f(z) D (ζ) dζ f | | ≤ k k| | Z[0,z] α 2K z (1+ρ)α 1 K 1+ z | | − (2.4) dρ = | | 1 . ≤ K +1 (1 ρ)α+1 α(K +1) 1 z − Z0 − " −| |! # On the other hand, let Γ be the preimage under f of the radial segment from 0 to f(z). Again, because 2 2 (1 z )α 1 − l(D (z)) h(z) −| | , f ≥ K +1| ′ | ≥ K +1(1+ z )α+1 | | it follows that α 1 1 z (2.5) f(z) l(D (ζ)) dζ 1 −| | . f | | ≥ | | ≥ α(K +1) − 1+ z ZΓ " | |! # Let z = z1 z0 so that z = z+z0 , where z ,z D. Then, by assumption, 1 −z0z1 1 1+z0z 0 1 ∈ − f(z) f(z ) 0 F(z) = − (1 z 2)h(z ) ∈ SH 0 ′ 0 −| | and is a K-quasiconformal harmonic mapping, i.e. F (K). Applying (2.4) and H ∈ S (2.5) to F gives us the desired result if we take into account of the fact that z z e2λD(z1,z2) 1 1 2 1 −z z = e2λD(z1,z2) −+1 = tanhλD(z1,z2). (cid:12) − 1 2(cid:12) (cid:12) (cid:12) (cid:3) The proof of the lemma is complete. (cid:12) (cid:12) (cid:12) (cid:12) Lemma 2. Assume that f (K). Then H ∈ S C f(z) D (z) z 8| | for z D, f k k| | ≤ 1 z ∈ −| | where z (1+ z )α 1 − (2.6) C = 2αKsup | | | | K. 8 z∈D (cid:26)[(1+|z|)α −(1−|z|)α](cid:27) ≥ Proof. Suppose that f = h+g (K), where h and g are analytic in D. Next, for H fixed ζ D, consider the Koeb∈eStransform F of f given by (1.6). By assumption, ∈ F and is also a K-quasiconformal harmonic mapping. By letting z = ζ in H ∈ S − (1.6) and applying (2.2) to F, we obtain (since f(0) = 0) f(ζ) 1 [(1+ ζ )α (1 ζ )α] F( ζ) = | | | | − −| | | − | (1 ζ 2) h(ζ) ≥ α(1+K) (1+ ζ )α ′ −| | | | | | 8 S.L. Chen and S. Ponnusamy which gives h(ζ) (1+ ζ )α 1 α ′ − | | | | . (1+K) f(ζ) ≤ [(1+ ζ )α (1 ζ )α] · 1 ζ | | | | − −| | −| | Since this follows for each ζ D, by the first inequality in (2.3), we easily have ∈ D (z) z 2K h(z) z C f ′ 8 k k| | | || | , f(z) ≤ K +1 f(z) ≤ 1 z | | | | −| | where C is given by (2.6). (cid:3) 8 Lemma 3. Let f (K) and, for any fixed θ [0,2π], set H ∈ S ∈ m (r,θ) = max f(ρeiθ) , f ρ [0,r]| | ∈ where r [0,1). Then, for 0 < ρ r < 1 and 0 ρ r, there is a constant 0 ∈ ≤ ≤ ≤ C > 0 which depends only on ρ such that 9 0 f(ρeiθ) (2.7) | | C m (r,θ), 9 f ρ ≤ where ρ is a constant. 0 Proof. Without loss of generality, we assume that θ = 0. Clearly, (2.2) yields that f(ρ) K e2αλD(ρ,0) 1 2K lim | | lim − = , ρ 0+ ρ ≤ α(1+K) ρ 0+ ρ 1+K → → which implies that f(ρ)/ρ is bounded in [0,ρ ], where ρ is a constant such that 0 0 0 < ρ r < 1. Hence there is a constant C > 0 such that 0 10 ≤ f(ρ) (2.8) C m (r,0) for ρ [0,ρ ], 10 f 0 ρ ≤ ∈ where r [ρ ,1). For r [0,1), let 0 ∈ ∈ K T(r) = 1 e 2αλD(r,0) . − α(1+K) − (cid:2) (cid:3) Then T is increasing in [0,1), which, together with (2.2), yields that f(ρ) f(ρ) 1 (2.9) 0 < T(ρ ) T(ρ) f(ρ) m (r,0) for ρ [ρ ,r], 0 f 0 ≤ ≤ ≤ ρ ≤ ρ ≤ ρ ∈ 0 0 where r [ρ ,1). Therefore, (2.7) follows from (2.8) and (2.9). (cid:3) 0 ∈ Lemma 4. For r (0,1), let Ω be the Stolz-type domain consisting of the interior r of the convex hull ∈of the point r and the disk D . Then, for z = ρeiη Ω D , r/4 r r/4 ∈ \ 4π 4π η (r ρ) < (1 ρ). | | ≤ r√15 − r√15 − Radial length, radial John disks and K-quasiconformal harmonic mappings 9 C E = ρ eiη 1 z = ρeiη η O A D B Figure 1. Stolz-type domain Proof. Assume without loss of generality that η 0. Let A, D, and E represent the points r, r/4, and ρ eiη (see Figure 1), respecti≥vely. As ∠OCA = π/2, it is clear 1 that sin∠COA = √15, cos∠COA = 1, sin∠COE = ρ21 − 1r62 and cos∠COE = r . 4 4 q ρ 4ρ 1 1 Then, because (sinη)/η 2/π for η < π/2, it follows that for η 0 | | ≥ | | ≥ 2η sinη = sin(∠COA ∠COE) π ≤ − = sin∠COAcos∠COE cos∠COAsin∠COE − √15r2 16ρ2 r2 = − 1 − 16ρ p1 r2 ρ2 = − 1 ρ √15r + 16ρ2 r2 1 1 − Note that r < ρ < ρ < r and(cid:16), becausep (cid:17) 4 1 r2 ρ2 4 − 1 < r2 ρ2 < 8(r ρ ) < 8(r ρ), ρ r − 1 − 1 − 1 (cid:0) (cid:1) the last above relation clearly implies that 2η 8(r ρ) < − π r√15 which gives the desired conclusion. Observe that 8 (r ρ) is less than 6/√15 from r√15 − which we also deduce that η < 3π/√15. (cid:3) | | 10 S.L. Chen and S. Ponnusamy Proof of Theorem 1. Assume without loss of generality that θ = 0. For r (0,1), ∈ we use Ω to denote the Stolz-type domain, where Ω is same as in Lemma 4. Let r r z = ρeiη Ω D . Then, by Lemma 4, there is a constant C > 0 which depends r r/4 11 ∈ \ only on r such that (2.10) η < C (1 ρ). 11 | | − Suppose that f = h+g (K). By calculations, we get H ∈ S f(ρeiη) f(ρ) η ρeith(ρeit) ρe itg (ρeit) (2.11) log log = i ′ − − ′ 1 dt. ρeiη − ρ f(ρeit) − Z0 ! Taking real part of (2.11) on both sides, and then using (2.10), (2.11) and Lemma 2, we see that there is a constant C such that 12 f(ρeiη) f(ρ) η ρ D (ρeit) f log | | log | | k k dt ρ − ρ ≤ f(ρeit) Z0 | | η dt C C C , 12 11 12 ≤ 1 ρ ≤ Z0 − which gives that (2.12) f(z) = f(ρeiη) eC11C12 f(ρ) . | | | | ≤ | | By (2.2), weseethat(2.12)alsoholdsforz D . Then, by(2.12), thereisconstant r/4 C such that f(Ω ) is contained in D ∈ , which yields that 13 r C13mf(r,0) (2.13) J (ζ)dA(ζ) C2 m2(r,0), f ≤ 13 f ZΩr where ζ = x+iy, dA = dxdy/π and m (r,θ) is defined as in Lemma 3. f By [24, Theorem 2], there is a constant C such that 14 1 (2.14) (1 ρ) H (ρ) 2dρ C H (z) 2dA(z), ′ 14 ′ − | | ≤ | | Z0 ZΩ1 where H(z) is analytic in D. For r (0,1), let H(z) = h(rz) for z D. Then, by ∈ ∈ (2.14), we obtain r (r ρ) h(ρ) 2dρ C h(z) 2dA(z), ′ 15 ′ − | | ≤ | | Z0 ZΩr which implies that r (r ρ) D (ρ) 2dρ C K J (z)dA(z), f 15 f − k k ≤ Z0 ZΩr (2.15) C2 C Km2(r,0) (by (2.13)), ≤ 13 15 f where C > 0 is a constant. 15 By Lemmas 2 and 3, for r (1/2,1) and ρ [0,r], there is a constant C such 16 ∈ ∈ that r r m2(r,0) r (2.16) D (ρ) 2dρ C f dρ = C m2(r,0) . k f k ≤ 16 (1 ρ)2 16 f 1 r Z0 Z0 − −