Harmonic quasiconformal mappings between $\mathscr{C}^1$ smooth Jordan domains

We prove the following result. If $f$ is a harmonic quasiconformal mapping between two Jordan domains $D$ and $\Omega$ having $\mathscr{C}^1$ boundaries, then the function $f$ is globally H\"older continuous for every $\alpha<1$ but it is not Lipschitz in general. This extends and improves a classical theorem of S. Warschawski for conformal mappings.


Introduction
Let U and V be two domains in the complex plane C. We say that a twice differentiable mapping f = u + iv : U → V is harmonic if ∆f := ∆u + i∆v = 0 in U .Any harmonic homeomorphism is by Lewy's theorem a diffeomorphism.If its Jacobian J f = |f z | 2 − |f z | 2 is positive, then it is a sense-preserving.
We say that a function u : D → R is ACL (absolutely continuous on lines) in the region D, if for every closed rectangle R ⊂ D with sides parallel to the x and y-axes, u is absolutely continuous on a.e.horizontal and a.e.vertical line in R. Such a function has partial derivatives u x , u y a.e. in D.
A sense-preserving homeomorphism w : D → Ω, where D and Ω are subdomains of the complex plane C, is said to be K-quasiconformal (K-q.c), with K 1, if w is ACL in D in the sense that its real and imaginary part are ACL in D, and (1.1) |Dw| Kl(Dw) a.e. on D, (cf.[1], pp.[23][24].Here A = D(w) is the formal differential matrix defined by The class of quasiconformal harmonic mappings has been firstly considered by O. Martio in [26].The class of q.c.harmonic mappings contains conformal mappings, and this is why the class has shown a large interest for experts in geometric function theory.
We should mention here the following result of Pavlović [31] which states that a harmonic quasiconformal mapping of the unit disk D onto itself is bi-Lipschitz continuous.In order to explain the importance of his result let us state the following two separate results.If we assume that the mapping f : D → D is merely quasiconformal, then it is only Hölder continuous with the Hölder coefficient α = 1−k 1+k .This is the celebrated Mori's theorem.On the other hand, if f : D → D is merely a harmonic diffeomorphism, then by a result of Hengartner and Schober it has a continuous extension up to the boundary (see [11,Theorem 4.3] or [8,Sec. 3.3]).However, in view of Radó-Kneser-Choquet theorem, this is the best regularity that such a mapping can have at the boundary.
We define the Poisson kernel by For a mapping f ∈ L 1 (T), where T is the unit circle, the Poisson integral is defined by The well-known Radó-Kneser-Choquet theorem states.If f is a homeomorphism of the unit circle onto a convex Jordan curve γ, then its Poisson integral is a harmonic diffeomorphism of the unit disk D onto the Jordan domain Ω bounded by γ.
A special case is when γ = T. E. Heinz has proved that, if f is a harmonic diffeomorphism of the unit disk onto itself, then the Hilbert-Schmidt norm of its derivative: where c > 0 depends only on f (0).It follows from (1.2), that the inverse of a quasiconformal harmonic mapping of the unit disk onto itself is Lipschitz continuous.So the main achievement of Pavlović in [31] (see also [33]), was to prove that a harmonic quasiconformal mapping of the unit disk onto itself is Lipschitz continuous on the closure of the domain.
In [21], the author proved that, every quasiconformal harmonic mapping between Jordan domains with C 1,α boundaries is Lipschitz continuous on the closure of domain.Later this result has been extended to Jordan domains with only Dini smooth boundaries [15].
A bi-Lipschitz property for harmonic quasiconformal mappings of the half-plane onto itself has been established by the author and Pavlović in [16].
Further it has been shown in [14] that a quasiconformal harmonic mappings between C 1,1 (not-necessarily convex) Jordan domains is bi-Lipschitz continuous.The same conclusion is obtained in [5] by Božin and Mateljević for the case of C 1,α Jordan domains.Further results in two dimensional case can be found in [19].Some results concerning the several-dimensional case can be found in [3], [20] and [28].For a different setting concerning the class of quasiconformal harmonic mappings we refer to the papers [7,25,30].For example the article [25] deals with the following problem of the class of quasiconformal harmonic mappings.
The quasi-hyperbolic metric d h in a domain D of complex plane is defined as follows.For each z 1 , z 2 ∈ D, where the infimum is taken over all rectifiable arcs γ joining x 1 and x 2 in D. V. Manojlović in [25] proved the following theorem: if f : D → D ′ is a quasiconformal and harmonic mapping, then it is bi-Lipschitz with respect to quasihyperbolic metrics on D and D ′ .
In order to formulate the main theorem of this paper let us define the chord-arc curves.A rectifiable Jordan curve γ is a B−chord-arc curve if L γ (z 1 , z 2 ) B|z 1 − z 2 | for all z 1 , z 2 ∈ γ, where L γ (z 1 , z 2 ) denotes the length of the shortest arc of γ joining z 1 and z 2 .Here B 1.
Theorem 1.1.Let D and Ω be Jordan domains having C 1 boundaries and assume that a ∈ D and b ∈ Ω. Assume that ω D (ω Ω ) is the modulus of continuity of the derivative of arc-length parametrisation of ∂D (∂Ω).Assume further that ∂D and ∂Ω satisfy B−arc-chord condition for some B 1. Then for every α ∈ (0, 1) and k ∈ [0, 1), there is a constant Moreover for every p > 0, there is a constant B p , that depends on the same parameters as M α so that In other words g ′ , h ′ belong to the Bergman space A p for every p > 0.Here λ is the Legesgue's measure in the plane.
Remark 1.2.In Theorem 1.1 we consider the mappings between Jordan domains.The same conclusion can be made for multiply-connected domain bounded by finite number of C 1 Jordan curves.We also expect that a similar conclusion can be made for non-bounded domains, but we did not pursue this question seriously.
1.1.The organization of the paper.We continue this section with some immediate corollaries of the main result.We prove that a K−quasiconformal mapping between C 1 domains is β−Hölder continuous for every β < 1/K.In particular we prove that a conformal mapping is β−Hölder continuous for every β < 1.In the second section we prove a variation of the main result which will be needed to prove to prove Theorem 1.1 in the full generality.The proof of Theorem 1.1 is presented in the last section.The proof depends on a two-side connection between the α−Hölder constant and the so-called α−Bloch type norm of the holomorphic function defined on the unit disk expressed in Lemma 1.4.By using this connection, and by a subtle application of C 1 smoothness of the boundary curve of the image domain, we first find an a priori estimate of the α−Hölder constant of a harmonic quasiconformal mapping of the unit disk onto a C 1 Jordan domain having C 1 extension up to the boundary.Then we use an approximation argument to get an estimate of α−Hölder constant for a harmonic q.c.mapping which has not necessary smooth extension up to the boundary.To deal with the mappings whose domain is not the unit disk is a simple matter having proved the results from the second section.
Corollary 1.3.[24] If f is a univalent conformal mapping between two Jordan domains D and Ω with C 1 boundaries, then f is α Hölder continuous for every 0 < α < 1.Moreover, if ∂D and ∂Ω satisfy B−arc-chord condition for some B 1, then for every α ∈ (0, 1) and every a ∈ D and b = f (a) ∈ Ω, for every z, w ∈ D. Then for every β < 1/K, there is a constant In connection to Theorem 1.4, we want to mention that some more general results are known under some more general conditions on the domains but they do not cover this result.For example O. Martio and R. Näkki in [27] showed that if f induces a boundary mapping which belongs to Lip α (∂D), then f is in Lip β (D), where β = min{α, 1/K}; the exponent β is sharp.We also want to refer to the papers [22] and [29] which also consider the global Hölder continuity of quasiconformal mappings.Concerning the integrability of the derivative of a quasiconfromal mapping and its connection to the global Hölder continuity we refer to the paper by Astala and Koskela [2].
Remark 1.5.Similar result can be shown for multiply connected domains in the complex plane having a C 1 boundary.If f a conformal mapping of the unit disk onto a Jordan domain with merely C 1 boundary, then f is not necessarily Lipschitz continuous.See an example given by Lesley and Warschawski in [24] as well as the example given in the Pommerenke book [34], which is a conformal diffeomorphism of the unit disk onto a Jordan domain with merely C 1 boundary.Then |f ′ 0 (z)| is not bounded and thus f 0 is not Lipschitz continuous.The content of Corollary 1.3 is not new (see for example [23]).See also Warschawski [37,Corollary,p. 255] for a related result.We should also mention the paper by Brennan, [6] where the famous Brannen conjecture comes from.Theorem 3 of that paper contains a short proof of special case of (1.4) for Ω = D and f being conformal.

Auxiliary results
The starting point of this section is the theorem of Warschawski for conformal mappings which states the following.Assume that f is a conformal mapping of the unit disk onto a Jordan domain Ω with a C 1 boundary γ.Assume that g is the arc-length parametrisation of γ, and assume that ω = ω g ′ is modulus of continuity of g ′ .Assume also that γ satisfies B−chordarc condition for some constant B > 1.Then for every p ∈ R, there is a constant A p , depending only on Ω, ω, B, p and f (0) so that (2.1) We first give an extension of (1.4), and prove a variation of the main result needed in the sequel.
Theorem 2.1.If f = g + h is a Kq.c.harmonic mapping of the unit disk D onto a domain Ω with C 1 boundary, so that h has holomorphic extension beyond the boundary of the unit disk, then g ′ , 1/g ′ ∈ H p (D) for every p > 0. Moreover where F p is a constant that depends on the same parameters as E p in (2.1) as well as on k.

Now recall the Morrey inequality.
Proposition 2.2 (Morrey's inequality).Assume that 2 < p ∞ and assume that U is a bounded domain in R 2 with C 1 boundary.Then there exists a constant C depending only on p and U so that for every u ∈ C 1 (U ) ∩ L p (U ), where Here W 1,p (U ) is the Sobolev space.
Corollary 2.3.Under the conditions of the previous theorem, for every α < 1, f and f −1 are α−Hölder continuous.The result is optimal since, f is not necessarily Lipschitz in general.
Remark 2.4.If h ≡ 0, then Theorem 2.1 reduces to the classical result of Warschawski ([36]), see also a similar result by Smirnov [35] and Goluzin [13,Theorem 7,p. 415].We include the proof of Theorem 2.1 for the completeness of the argument.
Proof of corollary 2.3.Let α < 1 and prove that f is α−Hölder continuous.We have As h is smooth in D, it follows that g is α−Hölder continuous in T. By using the well-known Hardy-Littlewood theorem [13, Theorem 4, p.413], we get that g is α−Hölder continuous on D. Thus f is α−Hölder continuous on D.
To prove that f −1 is α−Hölder continuous, observe that for w = f (z), Thus Here λ is the Lebesgue measure in the plane.Therefore by using the isoperimetric inequality for holomorphic functions we get From (2.3) we infer that u = f −1 is α−Hölder continuous and the corollary is proved.
Proof of Theorem 2.1.We use the following proposition Proposition 2.5.[17] If f (z) = P[f * ](z) is a quasiconformal harmonic mapping of the unit disk onto a Jordan domain bounded by a curve γ, then the function is a well defined and smooth in D * := D\{0} and has a continuous extension to T if and only if γ ∈ C 1 .Furthermore, there holds where β(ϕ) is the tangent angle of γ at f * (e iϕ ).
By the assumption we have that h(z) = ∞ j=0 b j z j for |z| < ρ, where ρ is a certain constant bigger than 1.
Therefore, the mapping is well defined holomorphic function in the domain D 1 = {z : |z| > 1/ρ}.Since Γ = ∂Ω is rectifiable, for z = re it , we have that (see e.g.[18,31]).Therefore, by having in mind the quasiconformality, we get that g ′ , h ′ ∈ H 1 (D).In particular, there exist non-tangential limits of those functions almost everywhere on T. We recall that h 1 (D) and H 1 (D) are the Hardy classes of harmonic and holomorphic functions, respectively, defined in the unit disk D. Let Then, for almost every t ∈ [−π, π], we have Then there is a set of points 0 < ϕ 1 < ϕ 2 < ϕ 3 < ϕ 4 < 2π so that (2.4) lim r→1 H(re iϕ j ) = H(e iϕ j ), exist for every j = 1, 2, 3, 4. Let 1 < R < ρ and let R, 1)} and let w = Φ j (z) be a conformal mapping of the unit disk onto the region S j so that Define the holomorphic mapping K j (z) = H(Φ j (z)), z ∈ D, j = 1, 2. In view of (2.4), we have that H is bounded on the boundary arcs I j = [1/R, 1]e iϕ j , j = 1, 4 of S. Also it is clear that it is bounded in the inner arc.Therefore K j is a non-vanishing bounded analytic function defined in the unit disk.Let L j (z) = log K j (z).Then for j = 1, 2 v j (z) = ℑL j (z) = arg(K j (z)), is a bounded harmonic function, so that lim r→1 v j (re it ) = v j (e it ) is a continuous function on the unit circle.
To show that v is a bounded well-defined function, observe that and so First of all for |z| close to 1, the function where k is the constant of quasiconformality.
On the other hand, in view of Proposition 2.5, i(g ′ − zh ′ /z) = f t (e it )/z has a continuous argument in the punctured disk 0 < |z| 1.Since ℜ(1 − zh ′ /(zg ′ )) > 0, we obtain that arg(g ′ ) is well-defined and bounded function close to the boundary of the unit disk.
We can also choose R close enough to 1 so that the variation of the argument: (2.6) ∆ T argK j (e it ) 1 + ∆ T argH j (e is ).

Thus
The constant G p depends on the same parameters as the constant E p from (2.1) together with the constant of quasiconformality k, and this follows from the fact that Ψ(0) = a j,0 , (2.5), (2.9), (2.6) and a Cauchy type inequality for H(z) in the annulus 1/R < |z| < 1, where 1/R = (1/ρ + 1)/2.Here ρ is a given constant bigger than 1 as in the begging of the proof. Since The constant L p depends on the same parameters as E p from (2.1) and the quasiconformal constant k.
Thus H ∈ H p (D), and so f t ∈ h p (D).Since f is quasi-conformal, it follows that g ′ ∈ H p .Lemma 2.6.Let α ∈ (0, 1).Then there is a positive constant C(α) > 1 satisfying the following property.If f is a holomorphic function defined in the unit disk with continuous extension up to the boundary and if Remark 2.7.We want to mention that a result similar to Lemma 2.6 is probably valid for the more general classes of mappings such as, real harmonic functions, or quasiconformal harmonic mappings, but we do not need such results (see e.g.[32]).
Observe that δ, and so A depends on K, γ, α and modulus of continuity of f at the boundary, but not on a specific point z ∈ D.

Let us remove the assumptions f is
√ α− Hölder continuous and use Approximation argument.If p ∈ ∂Ω = γ and γ ∈ C 1 , then, after possible rotation and translation of Ω (similarly as in (3.1)), which preserves the harmonicity and the quasiconformal constant of the corresponding mapping, we can assume that p = 0 and the unit normal vector is N p = (1, 0).So we can find a sub-arc of γ containing p at its interior which is the graphic of a function defined as follows We also can assume that η > 0 is a positive constant that depends only on γ but not on the specific point p.Then we have φ ′ (0) = 0. Let Ω p ⊂ Ω be a Jordan domain bounded by a C 1 Jordan curve Γ p consisted of γ p (η/2) and an interior part, which we denote by χ p (η), which is subset of Ω and assume that a p ∈ Ω p be a fixed point.Then for small enough σ = σ(γ) > 0, the domain Ω Since T is compact, there is a finite family of Jordan domains Ω p j , j = 1, . . ., n so that T D) and Φ p j ,κ is α 1/2 −Hölder continuous because of Corollary 2.3.Further, in view of the first case, there is a constant A p j (see Lemma 2.6 and (3.9)) which depends only on Ω p j and α so that |f • Φ p j ,κ (e it ) − f • Φ p j ,κ (e is )| A p j |e is − e it | α 2/3 .Note that A p j also depends on the modulus of continuity of f • Φ p j ,κ where κ ∈ [0, σ], but this family is uniformly continuous, and we can choose modulus of continuity that does not depend on κ, so A p j will not depend on κ either.Namely the K−quasiconformal mappings G κ := κ + f • Φ p j ,κ , κ ∈ [0, σ], map the unit disk onto Ω p j ∈ C 1 and satisfy the condition G κ (0) = a p j .By letting κ → 0 we get |f • Φ p j ,0 (e it ) − f • Φ p j ,0 (e is )| A p j |e is − e it | α 2/3 .Therefore, by having in mind the fact that Φ −1 p j ,0 is α 1/3 −Hölder continuous on T j (in view of Corollary 2.3), we conclude that f is α−Hölder continuous in T ′ j ⊂ T j , where T ′ j is a little bit smaller arc, but so that T ⊂ n j=1 T ′ j .Thus, f is α−Hölder continuous in T. By the standard argument we now obtain that f is α−Hölder continuous in D, concluding the case a).
Notice that α > 0 is an arbitrary number smaller than 1, so f is also α 1/2 −Hölder continuous.
Hence, if we want to get more explicit estimate of A, then we repeat one more time the procedure proceed in the previous subsection, but with where 1/q + 1/q ′ = 1, and q = p + 1.

2000
Mathematics Subject Classification.Primary 30C62; Secondary 30C20, 31A20 .Key words and phrases.Harmonic mappings, quasiconformal mappings, smooth domains.where | • | is the Euclidean norm.Notice that the condition (1.1) can be written as |w z | k|w z | a.e. on D where k =

Proof of Corollary 1 . 3 .
Let a be a univalent conformal mapping of the unit disk D onto D and b be a univalent conformal mapping of the unit disk onto Ω.Then in view of Theorem 1.1, b and a −1 are √ α−Hölder continuous.Then f = b • a −1 , is α−Hölder continuous.Now we prove the following theorem which deals with Hölder continuity of quasiconformal mappings between smooth domains.Theorem 1.4.Assume that D and Ω are two Jordan domains with C 1 boundaries and assume that a ∈ D and b ∈ Ω. Assume further that ∂D and ∂Ω satisfy B−arc-chord condition for some B 1. Let K 1.

3 . 1 )
Proof of main result (Theorem 1.We divide the proof into two cases.a) D is the unit disk D, b) D is a general Jordan domain with a C 1 boundary.a) Since γ ∈ C 1 , γ has the following property.For every point p ∈ γ there are complex numbers |a| = 1 and b so that the parametrisation of the curve (3.1)