Sobolev homeomorphic extensions from two to three dimensions
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This is a self-archived version of an original article. This version may differ from the original in pagination and typographic details. Author(s): Title: Year: Version: Copyright: Rights: Rights url: Please cite the original version: CC BY 4.0 https://creativecommons.org/licenses/by/4.0/ Sobolev homeomorphic extensions from two to three dimensions © 2024 The Authors. Published by Elsevier Inc. Published version Hencl, Stanislav; Koski, Aleksis; Onninen, Jani Hencl, S., Koski, A., & Onninen, J. (2024). Sobolev homeomorphic extensions from two to three dimensions. Journal of Functional Analysis, 286(9), Article 110371. https://doi.org/10.1016/j.jfa.2024.110371 2024
Journal of Functional Analysis 286 (2024) 110371 Contents lists available at ScienceDirect Journal of Functional Analysis journal homepage: www.elsevier.com/locate/jfa Regular Article Sobolev homeomorphic extensions from two to three dimensions ✩ Stanislav Hencl a, Aleksis Koski b,∗, Jani Onninen c,d aCharles University, Department of Mathematical Analysis, Sokolovská 83, 186 00, Prague 8, Czech Republic bDepartment of Mathematics and Systems Analysis, P.O. Box 11100, FI-00076, Aalto University, Finland cDepartment of Mathematics, Syracuse University, Syracuse, NY 13244, USA dDepartment of Mathematics and Statistics, P.O. Box 35 (MaD), FI-40014, University of Jyväskylä, Finland a r t i c l e i n f o a b s t r a c t Article history: Received 1 August 2022 Accepted 28 January 2024 Available online 15 February 2024 Communicated by Guido De Philippis MSC: primary 46E35, 58E20 Keywords: Sobolev homeomorphisms Sobolev extensions L1-Beurling-Ahlfors extension We study the basic question of characterizing which boundary homeomorphisms of the unit sphere can be extended to a Sobolev homeomorphism of the interior in 3D space. While the planar variants of this problem are well-understood, completely new and direct ways of constructing an extension are required in 3D. We prove, among other things, that a Sobolev homeomorphism ϕ:R2onto −−→ R2in W1,p loc (R2, R2)for some p ∈[1, ∞)admits a homeomorphic extension h:R3onto −−→ R3in W1,q loc (R3, R3)for 1 ⩽q<3 2p. Such an extension result is nearly sharp, as the bound q=3 2pcannot be improved due to the Hölder embedding. The case q= 3 gains an additional ✩S. Hencl was supported by the grant GAČR P201/21-01976S A. Koski was supported by the Academy of Finland grant number 307023, the ERC Advanced Grant 834728, received financial support from the Spanish Ministry of Science and Innovation through the Severo Ochoa Programme for Centres of Excellence in R&D (CEX2019-000904-S and MTM2017-85934-C3-2-P2), and from the CAM through the line of excellence for University Teaching Staff between CM and UAM. J. Onninen was supported by the NSF grant DMS- 2154943. *Corresponding author. E-mail addresses: [email protected]ff.cuni.cz (S. Hencl), [email protected] (A. Koski), jk[email protected] (J. Onninen). https://doi.org/10.1016/j.jfa.2024.110371 0022-1236/© 2024 The Authors. Published by Elsevier Inc. This is an open access article under the CC BY license (http://creativecommons .org /licenses /by /4 .0/).
2S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 interest as it also provides an L1-variant of the celebrated Beurling-Ahlfors quasiconformal extension result. © 2024 The Authors. Published by Elsevier Inc. This is an open access article under the CC BY license (http:// creativecommons .org /licenses /by /4 .0/). 1. Introduction Throughout this paper Bdenotes the unit ball in R3and S=∂B. We study the following 3D–Sobolev homeomorphic extension problem. Problem 1. Suppose that a homeomorphism ϕ:Sonto −−→ Sadmits a continuous extension to Bin the Sobolev space W1,q(B, R3)for some q∈[1, ∞). Does the map ϕalso admit a homeomorphic extension to Bof class W1,q(B, R3)? Every boundary homeomorphism ϕ:Sonto −−→ Sextends as a homeomorphism to the ball B. On the other hand, according to a famous result of Gagliardo [13], for 1 <q<∞, the mapping ϕis the Sobolev trace of some (possibly non-homeomorphic) mapping in W1,q(B, R3)if and only if it belongs to the fractional Sobolev space W1−1 q,q(S, R3); that is, S S |ϕ(x)−ϕ(y)|q |x−y|q+1 dxdy<∞.(1.1) Note that the 2D result [31]that every boundary homeomorphism ϕ:∂Donto −−→ ∂Dextends as a W1,q-homeomorphism, q<2, to the unit disk D⊂R2has no counterpart in higher dimensions. Indeed, there are boundary homeomorphisms from Sonto itself that do not even admit a continuous Sobolev extension in W1,q(B, R3)for any q>1, see Example 3.1. First we give a discrete variant of (1.1); that is, we characterize the boundary homeomorphisms that admit a Sobolev extension in W1,q(B, R3)when q>2. Theorem 1.1. Let ϕ:Sonto −−→ Sbe a homeomorphism and q∈(2, ∞). Suppose that ˜ Dkis a dyadic decomposition of Sinto closed bi-Lipschitz squares of diameter c2−k. Then ϕ satisfies (1.1)if and only if ∞ k=1 2k(q−3) ˜ Qj∈˜ Dkdiam ϕ(˜ Qj)q<∞.(1.2) For the precise definition of ˜ Dkwe refer to Definition 2.1. The corresponding 2D–Sobolev homeomorphic extension problem [22]has an easy answer thanks to the available analytic methods of constructing 2D-Sobolev homeomorphisms. Indeed, let Dbe the unit disk in R2and q∈[1, ∞)then a boundary homeomorphism ϕ:∂Donto −−→ ∂Dadmits a homeomorphic extension to Din W1,q(D, R2)
S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 3 if and only if it admits a continuous extension to Din W1,q(D, R2). This follows from the Radó-Kneser-Choquet (RKC) theorem [11]for q⩽2. The RKC theorem asserts that a homeomorphic boundary value ϕ:∂Donto −−→ ∂Dadmits a homeomorphic harmonic extension of D. The harmonic extension belongs to W1,q(D, R2) for all q<2 and to W1,2(D, R2)exactly when is in the trace space of W1,2(D, R2). Similarly the q-harmonic variants of the RKC theorem [2]solve the 2D extension problem for q>2. An analogous approach fails in higher dimensions. Indeed, Laugesen [23] constructed a self-homeomorphism of the sphere Sin R3whose harmonic extension to the ball Bis not injective. Thus, the 3D extension problem requires new methods of constructing Sobolev homeomorphisms. Our main result tells us that the searched homeomorphic extension exists if the boundary homeomorphism satisfies a strengthened version of the condition (1.2). Theorem 1.2. Let q∈(1, ∞). Suppose that ˜ Dkis a dyadic decomposition of Sinto closed bi-Lipschitz squares of diameter c2−k. If a homeomorphism ϕ:Sonto −−→ Ssatisfies ∞ k=1 2k(q−3) ˜ Qj∈˜ DkH1ϕ(∂˜ Qj)q<∞,(1.3) then it admits a homeomorphic extension h:Bonto −−→ Bin W1,q(B, R3). Here H1stands for 1-dimensional Hausdorff measure and so H1ϕ(∂˜ Qj)measures the length of the curve ϕ(∂˜ Qj). For a Sobolev homeomorphism ϕ:Sonto −−→ Sthe trivial radial extension h(x) =|x|ϕ(x) produces a self homeomorphism of Bwhich has the same Sobolev regularity as the given boundary map ϕ. Clearly, such an extension is far from being optimal. Our next result, however, nearly characterizers the first order Sobolev spaces that admit a Sobolev homeomorphic extension to B. Theorem 1.3. Let ϕ:Sonto −−→ Sbe a homeomorphism in W1,p(S, R3)for some p ∈[1, ∞). Then ϕadmits a homeomorphic extension h:Bonto −−→ Bin W1,q(B, R3)for 1 ⩽q<3 2p. For the sharpness of this result we refer to the general embedding result by Sickel and Triebel [28, Theorem 3.2.1]. Namely for p ∈(1, ∞)we have W1,p(S, R3) ⊂W1−1 q,q(S, R3) if and only if q⩽3 2p. Even assuming that the mappings are homeomorphisms does not improve the inclusion at least when p ⩾2, see Example 3.2. We do not know if one can take q=3 2pin Theorem 1.3. Theorem 1.3 follows from Theorem 1.2. On the contrary there are self homeomorphisms of Swhich satisfy (1.3)and do not belong to any Sobolev class W1,p(S, R3), p ⩾1, see Example 3.3. In topology and analysis, a number of extension problems have been studied. A demand for Sobolev homeomorphic extension problems comes from the variational
4S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 approach to Geometric Function Theory (GFT) [4,15,21,26]and mathematical models of Nonlinear Elasticity (NE) [3,6,9]. Both theories enquire into homeomorphisms h:X onto −−→ Yof smallest stored energy EX[h]= X E(x, h, Dh)dx, E:X×Y×Rn×n where the so-called stored energy function Echaracterizes the mechanical and elastic properties of the material occupying the domains. In a pure displacement setting, typically an orientation-preserving boundary homeomorphism ϕ:∂X onto −−→ ∂Yis given. The class of admissible deformations consists of Sobolev homeomorphisms or just Sobolev mappings h:Xonto −−→ Ywith non-negative Jacobian determinant Jh(x) =detDh(x) ⩾0 (an axiomatic assumption in NE) which coincides with ϕon the boundary and having a finite stored energy. In such variational problems, a first issue to address is the nonemptiness of the class of admissible deformations; that is, to solve the corresponding Sobolev homeomorphic extension problem. Note that an arbitrary orientation-preserving Sobolev homeomorphism hneed not be strictly orientation-preserving in the sense that Jh(x) =detDh(x) >0almost everywhere. For every q<3, there even exists a homeomorphism h:Bonto −−→ Bin W1,q(B, R3) with Jh(x) =0for almost every x ∈B, see [14]. However, the homeomorphic extensions h:Bonto −−→ Bconstructed in Theorem 1.3 and Theorem 1.2 are piecewise linear. Thus, they are strictly orientation-preserving provided that the given boundary homeomorphism itself preserves the orientation. In particular, these homeomorphisms have finite distortion. The theory of mappings of finite distortion arose out of a need to extend the ideas and applications of the classical theory of quasiconformal mappings to the degenerate elliptic setting [15,21]. We recall that a homeomorphism h:X onto −−→ Yof Sobolev class W1,1 loc (X, Rn) defined on a domain X ⊂Rnhas finite distortion if |Dh(x)|n⩽K(x)Jh(x) (1.4) for some measurable function 1 ⩽K(x) <∞. Here, |Dh(x)|is the operator norm of the weak differential Dh(x): X →Rnof hat a point x ∈X. We obtain quasiconformal mappings if K∈L∞(X). There are several other distortion functions of great interest in GFT. Each of them is designed to measure the deviation from conformality of a given mapping h:X →Rnin terms of the tangent linear map Dh(x): Rn→Rn. The most interesting, from the applied point of view, is the inner distortion function. In NE one is typically provided information not only on the differential matrix, but also on its (n −1) ×(n −1)–minors; that is, the cofactor matrix Dhcalled co-differential of h. Now, for a homeomorphism h ∈W1,1 loc (X, Rn)of finite distortion we introduce its inner distortion function, to be the smallest KI(x) =KI(x, f) ⩾1 satisfying |Df(x)|n=KI(x)·Jf(x)n−1
S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 5 The most pronounced extension result in GFT is the Beurling-Ahlfors quasiconformal extension theorem [7]. It states that a self-homeomorphism of the unit disk Dis quasiconformal if and only if the boundary correspondence homeomorphism ϕ:∂Donto −−→ ∂D is quasisymmetric. The Beurling-Ahlfors result has found a number of applications in Teichmüller theory, Kleinian groups, conformal welding and dynamics, see e.g. [4,19]. It has generalized to the n-dimensional quasiconformal maps as well, first for n =3 by Ahlfors [1]and then for n =4by Carleson [8]. A full n-dimensional version of the Beurling-Ahlfors extension is due to Tukia and Väisälä [30]. Their extension uses, among other things, Sullivan’s theory [29]of deformations of Lipschitz embeddings. Moreover, Astala, Iwaniec, Martin and Onninen [5], as a part of their studies of deformations with smallest mean distortion, characterizes self homeomorphisms of the unit circle that admit a homeomorphic extension to the unit disk Dwith integrable distortion. This L1–Beurling-Ahlfors extension theorem enjoys the following 3D-variant. Theorem 1.4. Let ψ:Sonto −−→ Sbe an orientation-preserving homeomorphism. Suppose that the inverse ψ−1=ϕsatisfies (1.3)with q=3. Then ψadmits a homeomorphic extension f:Bonto −−→ Bwith integrable inner distortion. Theorem 1.4 is actually a relatively straightforward consequence of Theorem 1.2, thanks to an important connection between the conformal energy of a homeomorphism and the inner distortion function of the inverse mapping. Indeed it is easy to see, at least formally, that the pullback of the 3-form KI(y, f) dy∈∧ 3Bby the inverse mapping f−1:Bonto −−→ Bis equal to |Df−1(x)|3dx ∈∧ 3B. This observation is the key to the identity, B |Dh(x)|3dx= B KI(y,f)dy, where h=f−1:Bonto −−→ B.(1.5) The optimal Sobolev regularity of deformations to guarantee the identity is wellunderstood today, [10,16,17,24]. In particular, if a homeomorphism h:Bonto −−→ Bof finite distortion belongs to the Sobolev class W1,3(B, R3), then the inverse f=h−1has integrable inner distortion. Thus, Theorem 1.4 simply follows from Theorem 1.2. It is worth noting that the borderline case in Theorem 1.3 (p = 3 and q = 2), if true, would have an interesting corollary. Namely, a homeomorphism ψ:R2onto −−→ R2of locally integrable distortion would then admit a homeomorphic extension f:R3onto −−→ R3with locally integrable inner distortion. Acknowledgements. We would like to thank the referee for their many insightful comments and suggestions which particularly helped in improving the presentation of the paper considerably.
6S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 2. A discrete characterization, proof of Theorem 1.1 Let I=[a, b]2be an initial square in R2. The standard dyadic decomposition of I consists of closed squares ˜ Q⊂Iwith sides parallel to the sides of Iand of side length l(˜ Q) =2 −k(b −a), k=1, 2, 3, ...; refers to the k-th generation in the construction. That is, the squares in the k-th generation have the form ˜ Qj=2 −k(I+vj)⊂I,for some vj∈R2. They cover Iand have side length 2−k(b −a). The collection of the k-th generation squares are denoted by ˜ Dk. There are 22ksquares in ˜ Dk. The interiors of the squares in the same generation ˜ Dkare pairwise disjoint. Let Q3=[0, 1]3be the unit cube in R3. We define the k-th generation dyadic decomposition of ∂Q3as follows: first we divide each of the six faces of ∂Qinto the k-th generation squares and then the k-th generation dyadic decomposition of ∂Q3simply consists of the union of these closed squares. Now, since Bis a bi-Lipschitz equivalent with Q3, defining a k-th generation dyadic decomposition of ∂B=Scan be easily induced from the above case. Definition 2.1. Let Φ: R3→R3be a bi-Lipschitz map which takes Q3onto B. Then the k-th generation dyadic decomposition of S, denoted by ˜ Dk, consists of Φ( ˜ Qj), where ˜ Qj is a k-th generation dyadic square of ∂Q3. Theorem 2.2. Let ϕ:R2→R2be a homeomorphism, IR=[−R, R]2⊂R2for R>0 and let N∈N. Denote the collection of k-th generation dyadic squares of INby ˜ DN k. Then, for 2 <q<∞we have IR IR |ϕ(x)−ϕ(y)|q |x−y|q+1 dxdy<∞for every R>0 (2.1) if and only if ∞ k=1 2k(q−3) ˜ Qj∈˜ DN kdiam ϕ(˜ Qj)q<∞for every N∈N.(2.2) Proof. First we assume the condition (2.1)with R=2 12. Now, the mapping ϕ:R2→R2 admits an extension f:R3→R3in W1,p(IR×[−R, R], R3)which is continuous and agrees with ϕon R2×{0}(see (1.1)and the paragraph before). It suffices to prove (2.2) with N=1. Fix ˜ Qk,j ∈˜ D1 kfor some k∈Nand j∈{1, ..., 22k}. We denote the centre of ˜ Qk,j ⊂R2 by x◦. Let B3 Rbe the 3-dimensional ball in R3centred at x◦with radius R>0and B2 R=B3 R∩(R2×{0}).(2.3)
S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 7 Choose η∈(2, q). According to the Sobolev imbedding theorem on spheres [15, Lemma 2.19] there is a constant C>0such that for a.e. s ∈(0, R)we have diam f(∂B3 s)⩽Cs 1−2 η⎛ ⎜ ⎝ ∂B3 s |Df|η⎞ ⎟ ⎠ 1 η . This is the moment where we used the assumption q>2. By (2.3)we always have diam f(∂B2 s)⩽diam f(∂B3 s). Since ϕ:R2onto −−→ R2is a homeomorphism we get diam ϕ(B2 s)=diamϕ(∂B2 s). For fixed r∈(0, R/2), the above estimates give diam ϕ(B2 r)⩽Cs 1−2 η⎛ ⎜ ⎝ ∂B3 s |Df|η⎞ ⎟ ⎠ 1 η for a.e. s∈(r, R) and diam ϕ(B2 r)η 2r r ds sη−2⩽C B3 2r\B3 r |Df|η.(2.4) Thus diam ϕ(B2 r)⩽Cr1−3 η⎛ ⎜ ⎝ B3 2r |Df|η⎞ ⎟ ⎠ 1 η and diam ϕ(˜ Qk,j)⩽C2−k(1−3/η)⎛ ⎜ ⎝ B3 23−k |Df|η⎞ ⎟ ⎠ 1 η .(2.5) The k-th dyadic decomposition ˜ Dk={˜ Qk,j :k∈N, j=1, ..., 22k}of I1⊂R2defines a corresponding Whitney decomposition of I1×[0, 2] ⊂R3, Wk={˜ Q3 k,j :k∈N,j=1,...,22k}
8S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 where ˜ Q3 k,j =˜ Qk,j ×[2−k+1,2−k+2]. Let x ∈˜ Q3 k,j and c =2 11. Then B3 c2−k(x) =B3(x, c2−k) ⊃B3 23−kand so diam ϕ(˜ Qk,j)⩽C2−k(1−3/η)⎛ ⎜ ⎝ B3 c2−k(x) |Df|η⎞ ⎟ ⎠ 1 η by (2.5). In particular, we have diam ϕ(˜ Qk,j)⩽C2−kMc|Df|η(x)1 ηfor all x∈˜ Q3 k,j .(2.6) Here Mcdenotes the Hardy-Littlewood maximal operator, Mc|Df|η(x)=sup r<c 1 |B3 r(x)| B3 r(x) |Df|η. Raising the estimate (2.6)to the power qand then integrating it over the cube ˜ Q3 k,j we have 2−3kdiam ϕ(˜ Qk,j)q⩽C2−qk ˜ Q3 k,j Mc|Df|η(x)q η. Thus, ∞ k=1 22k j=1 2k(q−3)diam ϕ(˜ Qk,j)q⩽C ∞ k=1 22+2k j=1 Q3 k,j Mc|Df|η(x)q η =C I1×[0,2] Mc|Df|η(x)q η. Since q/η > 1we can use the boundedness of the Hardy-Littlewood maximal function in Lq ηfor the function |Df|ηto obtain ∞ k=1 22k j=1 2k(q−3)diam ϕ(˜ Qk,j)q⩽C Ic×[−2c,2c] |Df|q as claimed.
S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 15 Fig. 1. The cube Uk,j and its image set Vk,j defined as a region spanned by the curve Γk,j ×{2−(k−1)}and its corresponding curve ˆ Γk,j ×{2−k}on the next level. We aim to define the extension hso that it maps each horizontal section of Uk,j to the horizontal section of Vk,j of the same height. The horizontal sections of Vk,j will still need to be defined, however, and to do this we will need to construct an appropriate homotopy between the curve Γk,j to the curve ˆ Γk,j which we define as the outer boundary of 4 m=1 ˆ Γ(m) k,j , i.e. the curve corresponding to Γk,j on the next dyadic level. In terms of estimating the Sobolev norm of h, our main goal is to show the following. Goal: The map h :Uk,j →Vk,j will be a Lipschitz mapping. The Lipschitz constant of the map should be estimated from above by a uniform constant times the quantity (H1(Γk,j) +4 m=1 H1(ˆ Γ(m) k,j ))2k, or possibly this quantity added together with the same quantity over all of the neighbours of Γk,j. After Sections 5to 7we will have defined the monotone extension hon each dyadic cube Uk,j so that the goal estimate above holds, and this extension is further modified into an injective extension hin Section 8with the same estimates still holding. The W1,q-norm of hcan then be estimated by estimating the differential |Dh|above by the Lipschitz constant of h. Combined with the goal estimate this gives Uk,j |Dh(z)|qdz ⩽2k(q−3) H1(Γk,j)+ 4 m=1 H1ˆ Γ(m) k,j q . Combined with (4.1)this will yield that hbelongs to the Sobolev space W1,q([0, 1]3)as desired. The proof of Theorem 1.2 is then finished in Section 9where we explain the slight changes in the arguments needed for the spherical case. 5. Decomposition of the domain and target side In this section we start with the standard dyadic decomposition ˜ Dkof the boundary and define a modification of it in order to control the lengths of the image curves of the image grid under the given boundary map ϕ. Furthermore, we will define piecewise
16 S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 linear replacements of these image curves. These divisions on the domain and target side will be used in later sections to assist in defining the extension map we use to prove our main result, Theorem 1.2. We also show in this section that Theorem 1.3 follows from Theorem 1.2. Lemma 5.1. Let ˜ Dk={˜ Qk,j :k∈N, j=1 ...22k}be the dyadic decomposition of the unit square Q0=[0, 1]2into closed squares of side length 2−kfor each fixed k. Let p >1 and ϕ :Q0→Q0be a homeomorphism in the space ϕ ∈W1,p(3Q0, R2). Then there exists a set of closed quadrilaterals Dk={Qk,j :k∈N, j=1 ...22k}such that (1) For each point ˜v∈Q0which is a vertex of a dyadic square of side length 2−k in ˜ Dk, there exists exactly one corresponding point v∈Q0which is a vertex of a quadrilateral from Dk. The vertices vof a quadrilateral Qk,j in Dkare exactly the points which correspond to the vertices ˜vof the dyadic square ˜ Qk,j. Moreover, for the coordinates of these points v=[v1, v2]and ˜v=[˜v1, ˜v2]we have (see Fig. 2) v1−˜v1∈2−k 10 −2−k 40 ,2−k 10 and v2−˜v2∈2−k 10 −2−k 40 ,2−k 10 (5.1) for all pairs of corresponding vertices. (2) The quadrilaterals Qk,j for each fixed level kare thus mutually disjoint apart from their boundaries. (3) If we inherit the parent-child relation between dyadic squares from ˜ Dto D, then the following holds. The children Q1, ...Q 4∈D k+1 of a given square Q ∈D k(i.e. Q =Q1∪Q2∪Q3∪Q4) need not be contained in Qnor does their union need to cover Q. However, for ˆ Q=∪4 i=1Qithe boundaries ∂Q and ∂ˆ Qalways intersect exactly at two points. (4) For each k, jwe have the inequality 2−k ∂Qk,j |Dϕ(t)|pdt ⩽C 2Qk,j |Dϕ(z)|pdz. (5.2) Proof. (1) and (4): Let us first explain that it is possible to choose the grid so that (1) is satisfied and we have the key inequality (5.2). This follows essentially from [18, Section 4.2] and therefore we only explain how to apply this approach here: All of our cubes in the r=2 −kgrid are of type A since we can freely move points outside of Q0. We would like to apply analogy of [18, Lemma 4.9] for M=0and ε =1 10 . The only difference is that in [18, Lemma 4.9] they choose [v1,v 2]∈Iε=[˜v1+t, ˜v2+t]:|t|⩽ε2−k
S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 17 Fig. 2. Given a dyadic cube ˜ Qk,j with vertices ˜v1, ˜v2, ˜v3, ˜v4we construct a quadrilateral Qk,j with vertices v1, v2, v3, v4. Each viis close to ˜vi, it is slightly shifted to the top and to the right from ˜vi. Fig. 3. Boundaries of Qand ˆ Q=∪4 i=1Qiintersect at two points Sand T. Note here that Q1, ..., Q4refer to quadrilaterals which form the set ˆ Qwhich is the (almost square) octagon in the middle. but we would like to make this choice in the subset of Iε(of length 1/8times the original length) [v1,v 2]∈I=[˜v1+t, ˜v2+t]:t∈[1 10 2−k−1 40 2−k,1 10 2−k]. This does not change anything substantial in the proof there, it only affects some multiplicative constants -use 8225 εr instead of 25 εr in the definition of Γ(A, B, M)and then the proof carries through with obvious minor modifications. Then we can finish this step by applying analogy of [18, Lemma 4.13 and Lemma 4.16] (again with slightly increased multiplicative constant) to get our (5.2). (2): This is easy to see from the definition of vertices of Qk,j in step (1) (see Fig. 2). (3): Let Qand ˆ Q=4 i=1 Qibe as in the statement part (3) (see Fig. 3). Let us define notation for certain vertices here, consult Fig. 3for specific positions. Here v˜ Qis a vertex of ˜ Q, v1 Qand v2 Qare vertices of Qand v1 ˆ Q, v2 ˆ Q, v3 ˆ Qare vertices of ˆ Q (in fact the corresponding part of ˆ Qis given by two segments v1 ˆ Qv2 ˆ Qand v2 ˆ Qv3 ˆ Q). From (5.1)we obtain for the x-coordinates of these points that (v1 Q)1−(v˜ Q)1,(v2 Q)1−(v˜ Q)1∈2−k 10 −2−k 40 ,2−k 10
18 S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 and similarly from (5.1)for the choice of Dk+1 (v1 ˆ Q)1−(v˜ Q)1,(v2 ˆ Q)1−(v˜ Q)1,(v3 ˆ Q)1−(v˜ Q)1∈2−(k+1) 10 −2−(k+1) 40 ,2−(k+1) 10 . It follows that the distance of this side of Q(=segment v1 Qv2 Q) and this side of ˆ Q(=union of segments v1 ˆ Qv2 ˆ Qand v2 ˆ Qv3 ˆ Q) is at least 2−k 10 −2−k 40 −2−(k+1) 10 =2−k 40 and thus these two sides do not intersect. By a similar reasoning on other sides we obtain that ∂Q and ∂ˆ Q intersect at exactly two points Sand Tas in Fig. 3. Let us also note that the distance of Sand v1 Q(and similarly distance of Sand vˆ Q1) is at least 2−k 40 and thus these intersection points are not too close to the vertices of ∂Q and ∂ˆ Q. Definition 5.2. Note that conditions (1)-(3) above do not involve the boundary map ϕ. Hence we may define that any set Dkof quadrilaterals Qk,j satisfying the conditions (1)-(3) is called a good modification of the standard dyadic decomposition of Q0. Proof of Theorem 1.3.Note that the statement is obvious if p ⩾qas we can use the trivial radial extension. In the following we thus assume that p <q. Given a homeomorphism ϕ ∈W1,p loc (R2, R2)we were able to find in Lemma 5.1 a good modification Dkof the dyadic grid so that (5.2)holds. We could start with a homeomorphism ϕ ∈W1,p(S, S)and some analogy of dyadic grid on S. Analogously to the proof of Lemma 5.1 we can find a good modification Dkof this grid on Sso that an analogy of (5.2)holds for ϕ. In fact the whole statement can be also obtained locally using a bilipschitz change of variables. Given k, our dyadic grid Dkcontains bi-Lipschitz squares of diameter ≈2−kand of perimeter H1(∂Qk,j) ≈2−k. Moreover, there are approximately 22ksuch squares, let us denote by nkhere the total amount of bi-Lipschitz squares in Dk. In view of Theorem 1.2 it is now enough to show finiteness of (1.3). Using Hölder’s inequality, (5.2), q/p ⩾1and p >2 3qwe obtain ∞ k=1 nk j=1 2−(3−q)kH1(ϕ(∂Qk,j))q⩽ ∞ k=1 nk j=1 2−(3−q)k ∂Qk,j |Dϕ|q ⩽ ∞ k=1 nk j=1 2−(3−q)k ∂Qk,j |Dϕ|p1 p(2−k)1−1 pq ⩽C ∞ k=1 2−(3−q)k2−k(q−q p) nk j=12k 2Qk,j |Dϕ|p1 pq ⩽C ∞ k=1 2−k(3−q p)2kq p nk j=1 2Qk,j |Dϕ|p
S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 19 ⩽C ∞ k=1 2−k(3−2q p)<∞. The aim of the next lemma is to consider the modified dyadic grid given by Lemma 5.1. For each level k, we then look at the image of the grid of level kunder ϕ(specifically the set ϕ(∪j∂Qk,j)). The aim is to modify this “image grid” so that instead of general Jordan curves it consists of curves which are piecewise linear. It is necessary to preserve both the topology of the image grid and the lengths of the image curves. This piecewise linear approximation will simplify future computations. Lemma 5.3. Let p ⩾1and ϕ :Q0→Q0be a homeomorphism in the space ϕ ∈ W1,p(Q0, R2). Let Dkbe the set of modified dyadic quadrilaterals given by Lemma 5.1. In particular, the Jordan curves ϕ(∂Qk,j)for each Qk,j ∈D keach have finite length. Then for each quadrilateral Qk,j there exists a corresponding closed Jordan curve Γk,j ⊂Q0 on the image side such that. (1) Each of the curves Γk,j is piecewise linear. (2) Each point on the curve Γk,j is of distance at most 2−kfrom the set ϕ(∂Qk,j). (3) The inequality H1(Γk,j) ⩽H1(ϕ(∂Qk,j)) holds. (4) Γk,j passes through the four points ϕ(v), where vranges over the four vertices of the quadrilateral Qk,j. These four points are called the vertices of Γk,j. (5) If two quadrilaterals Qk,j, Qk,j∈D kshare a common side with endpoints v1, v2, then the subarcs of their corresponding image curves Γk,j, Γk,jwith endpoints at the common vertices ϕ(v1)and ϕ(v2)are the same. (6) Apart from the cases where two curves Γk,j, Γk,jat the same level kshare either a single vertex or a single subarc between two vertices as before, these Jordan curves are mutually disjoint (for each fixed level k). (7) For every Qk,j ∈D kand Qk+1,j∈D k(see Fig. 3) we know that Γk,j ∩Γk+1,j=ϕ(∂Qk,j)∩ϕ(∂Qk+1,j). That is each Γk,j passes not only through its vertices but also through its intersection with grids of step k+1and k−1, i.e. images of boundaries of Dk+1 and Dk−1. Proof. In this proof we use ideas of [12]and [18](see also [20]and [25]) where a similar piecewise linear approximation of curves was used. The idea is to do this approximation in three steps: First we linearize around vertices of the image grid, secondly linearize between intersection points of levels kand k+ 1 (to ensure that (7) is satisfied), and lastly to linearize the remaining non-intersecting curves. Step 1. Linearization near vertices: Fix kfor a moment, and denote by Vk=ϕ(v): vis a vertex of some Qk,j
20 S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 Fig. 4. We replace original curve near vertices (see dotted curves) by segments near vertices. the set of images of vertices of Dk. Let us also define the image grids G0=∅and Gk= j ϕ(∂Qk,j). Let Wk=Gk∩Gk+1 denote the set of intersection points between image grids of successive levels. Analogously to the reasoning in the proof of Lemma 5.1 (3), we see that both Vk and Wkare finite. We now choose a collection of small balls Bkwith centres at each point in Vk. More precisely, for each vertex vof some Qk,j we choose r>0small enough so that the balls B(ϕ(v), 2r)are pairwise disjoint and that these balls do not contain any of the points in Wkor Vk+1. Due to the latter property we may also assume that the balls in Bkand the balls in Bk+1 do not intersect either, as for each kwe may first choose the balls in Bkand then later choose the balls in Bk+1 small enough to not intersect the previous set of balls. Furthermore, we may use the uniform continuity of ϕ−1and ϕto assume that |ϕ(x)−ϕ(v)|<2−k,∀x∈Bv,diam(ϕ−1(B(ϕ(v),r)).(5.3) For each vertex vof the grid Dkwe have four sides S1, S2, S3and S4of some Qk,j that have vas their endpoint (see Fig. 4). On each of these sides we choose points si∈Siso that pi=ϕ(si) ∈∂B(ϕ(v), r)and so that siis furthest away from vwith this property (e.g. on S3in Fig. 4we have three points whose image intersects ∂B(ϕ(v), r)). Now we replace ϕon each segment [si, v]by a segment [pi, ϕ(v)] and we leave ϕthe same outside of these four segments (see Fig. 4). In this way we replace ϕ(∂Qk,j)by a curve Γ(∗) k,j which is piecewise linear close to the vertices. It is easy to see that this new curve Γ(∗) k,j satisfies an analogy of (2) by (5.3)and it is not difficult to see that these new curves are one-to-one (see Fig. 4), i.e. they intersect only at original vertices v. These new curves have also length shorter or equal to the original H1(ϕ(∂Qk,j)). We proceed to do the linearization process of this step on each level k=1, 2, 3, ..., replacing the collection of all curves ϕ(∂Qk,j)by a new set of curves Γ(∗) k,j. To reiterate,
S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 21 Fig. 5. We replace curves γmon the sides (see dotted curves) by piecewise linear curves. We may need to choose a one-to-one shortening of these replacements, i.e. we ignore some dashed part of the replacement of γ3. on each level kthese curves are now linear around the points Vk, but were unchanged near the set of intersection points Wk. The properties (2) −(7) are preserved in this process and we may continue the linearization to achieve (1) later. Step 2. Linearization at the intersection points Wk:In the previous step, we avoided making any changes near the set of intersection points Wkbetween curves of level kand k+1. In this step we will, for each level k, linearize the curves Γ(∗) k,j around the points Wk. This process can be done quite analogously to Step 1. We choose a new set of balls B kwhich are centred around points in Wk, and may again assume that the balls within each collection and between each successive collection (B kand B k+1) are disjoint. We then apply the same linearization process of Step 1 in each of these balls, linearizing each of the four parts (two from level kand two from k+1) which meet at the centre of each ball. This replaces the curves Γ(∗) k,j by another set of curves Γ(∗∗) k,j which are now also piecewise linear near the points in Wk. In this modification the properties (2) −(7) are again preserved for the whole collection of curves. Step 3. Linearization of sides: Now we need to linearize the curves Γ(∗∗) k,j in the remaining parts which consist of simple Jordan curves between the balls in Bkand B k. We call γk,m the parts of Γ(∗∗) k,j where our curve is not piecewise linear yet, these correspond to image by ϕof segments of Qk,j (minus some small segments near vertices of Dkand intersection points of Dkand Dk+1). These γk,m are pairwise disjoint and we can choose 0 <δ<2−kso that γk,m + B(0, 2δ)are pairwise disjoint. Furthermore, we may choose δsmall enough so that the sets γk,m +B(0, 2δ)do not contain points from the curves γk+1,mby the fact that the sets of curves γk,m and γk+1,mare mutually disjoint. We choose enough division points in γk,m and we connect them by segments (see Fig. 5) so that the union of these segments approximates the original curve. We definitely include two endpoints aγk,m and bγk,m in these division points and we assume that we have so many division points so that the union of these segments lies inside γk,m +B(0, δ). It follows that these segments for different γk,m do not intersect. However, it may happen that they intersect (see γ3in Fig. 5) for a given γk,m. In this case we simply choose a shortest path in the union of these segments between the
22 S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 endpoints aγk,m and bγk,m and we replace the union of these segment by this shortest path (see the right side of Fig. 5). It is not difficult to see that by this replacement we get a one-to-one piecewise linear curve that replaces γk,m. Now we call Γk,j the corresponding piecewise linear approximation of Γ(∗∗) k,j . It is now easy to see that we have (1), (2) (using δ<2−k), (3), (4), (5) and (6) for our Γk,j. Property (7) comes from our treatment of intersection points in Step 2, and the fact that in this step we chose δsmall enough to not intersect the curves γk+1,m. Parametrization of Γk,j:We have constructed a piecewise linear curve Γk,j that approximated ϕ(∂Qk,j)and passes through the same image vertices Vkand intersection points Wk=Gk∩Gk+1. We know that there are four y∈V ksuch that y=ϕ(v)for some vertex of Qk,j. Further, there are at most 8points in Gk+1 ∩ϕ(∂Qk,j)=Gk+1 ∩Γk,j as on the image of each side of Qk,j there are at most two (see Fig. 3and the proof of Lemma 5.1 (3)). Furthermore, we have at most two points in Gk−1∩ϕ(∂Qk,j), see Lemma 5.1 (3). Note also that analogously to the proof of Lemma 5.1 (3), there is C>0 with such that |ϕ−1(y)−ϕ−1(z)|≥C2−k,(5.4) for any two distinct points y, z∈V k∪W k∪W k−1. Thus the distance between the preimages of these points is comparable to the sidelength of Qk,j, i.e. 2−k. Now we divide Γk,j into at most 4 +8 +2 = 14 pieces Piby these points in Vk∪ Wk∪W k−1. For points x ∈ϕ−1(Vk∪W k∪W k−1)we define p(x) =ϕ(x)so that our parametrization phas the same value as original mapping ϕon these “vertices” and intersection points. We parametrize the pieces Piby a constant speed parametrization pthere, i.e. on each of those pieces it has constant speed which might be different for each piece. Since the length of these pieces is bounded by H1(ϕ(Qk,j)), we obtain using (5.4)that |Dp|⩽CH1(ϕ(Qk,j)) 2−kon the whole Qk,j. 6. The 2D extension Let Sbe the square with vertices at {(1, 0), (0, 1), (−1, 0), (0, −1)}and Ybe a Jordan domain with piecewise linear boundary. Suppose that a boundary homeomorphism ϕ : ∂S →∂Yis given. We now describe a way to extend ϕas a homeomorphism of Sto Y with Lipschitz-continuity controlled by the boundary map. First, we describe an extension Hϕof ϕwhich is a monotone map from Sto Y, meaning it is continuous and the preimage of every point is connected. The final homeomorphic extension will be obtained via an arbitrarily small modification of Hϕas we are able to
S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 23 describe the points where it fails to be injective and fix them accordingly. However, this modification will be done only later in Section 8. The extension Hϕwill also be called the shortest curve extension of ϕ. To define Hϕ, we let lsdenote the horizontal line segment which is obtained as the intersection between the line {(x, y) :y=s}and S. This segment lshas two endpoints asand bs(from left to right) on ∂S. We let As=ϕ(as), Bs=ϕ(bs), and define Lsas the shortest curve in ϕ(S)which connects Asto Bs. The map Hϕis now given by defining it to map each horizontal segment lsto the corresponding shortest curve Lsvia constant speed parametrization. It is simple to verify that this mapping is continuous. Lemma 6.1. If ϕ :∂S →∂Yis Lipschitz with constant L, then the shortest curve extension Hϕis also Lipschitz with constant at most CL for a uniform constant C. Proof. Case 1. Lipschitz continuity in the horizontal direction. We show that Hϕsatisfies the required Lipschitz-continuity on each of the horizontal segments ls. For this, note that the constant speed parametrization on each of these segments implies that we only need to show that |Ls| ⩽2L|ls|, where | ·|denotes length. The endpoints of lsseparate ∂S into two connected components, the shorter of which we may call γs. Since Ltis the shortest curve from Asto Bs, we find that |ϕ(γs)| ⩾|Ls|. However, due to the Lipschitz-continuity of ϕwe must have that |ϕ(γs)| ⩽L|γs|. Thus |Ls|⩽|ϕ(γs)|⩽L|γs|⩽2L|ls|, where the last inequality is due to the fact that lsis the hypotenuse of a right-angled triangle with sides given by γs. Case 2. Lipschitz continuity in the vertical direction. Let us fix s ∈(−1, 1) and pick a point z∈ls. For small δwe let zδ=z+(0, δ)and our aim is to show that |Hϕ(zδ) −Hϕ(z)| ⩽CLδ. As Lipschitz-continuity is a local property, we may assume that δis arbitrarily small. In fact, to simplify calculations we assume that δis very small compared to |ls|, which lets us assume that the trapezium bounded by the segments lsand ls+δis actually a rectangle with longer sides of length |ls|due to the fact that these two shapes are bilipschitz-equivalent with a uniform constant (say 2) for small enough δ. Consider the curves Lsand Ls+δ. By choosing δsmall enough, we may assume that the endpoints Asand As+δlie on the same line segment of the piecewise linear boundary ∂Y. The same may be assumed for Bsand Bs+δ. Now basic geometry dictates that the curves Lsand Ls+δmust each consist of three parts as follows (for a detailed argument, see [18]). See also Fig. 6.
24 S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 Fig. 6. The shortest curves Lsand Ls+δ, split into three parts. (1) αsand αs+δ: Curves which start from Asand As+δand do not intersect except at their common other endpoint. In fact, if δis assumed small enough these curves may be assumed to be line segments. (2) A common part of lsand Ls, which is a piecewise linear curve we denote by γ. (3) βsand βs+δ: Analogously to the first part, these can be assumed to be line segments from Bsand Bs+δrespectively which meet at a common point (the other endpoint of γ). We may assume that Hϕ(z) lies on either αsor γas the case where it lies on βsis handled by symmetry. Let Ddenote the line segment between Asand As+δ. Then since ϕis L-Lipschitz-continuous on ∂S, we find that |D| ⩽Lδ. By the triangle inequality we obtain that ||αs| −|αs+δ|| ⩽Lδ and using the same argument for the β-curves gives ||Ls| −|Ls+δ|| ⩽2Lδ. Let also ddenote the distance between zand as, which is also the distance from zδto as+δ. Suppose first that Hϕ(z) lies on γ. The length of the part of Lsbetween Asand Hϕ(z) may now be calculated in two ways. The constant speed parametrization tells us that it is equal to |Ls|d/|ls|. On the other hand, it is also equal to |αs| +|γ|, where γdenotes the part of γbetween αsand Hϕ(z). Thus |αs|+|γ|=|Ls|d |ls|. If Γ denotes the part of Ls+δbetween Hϕ(z)and Hϕ(zδ), then we may calculate the length of the part of Ls+δbetween as+δand Hϕ(zδ)in two ways similarly as above to obtain that |αs+δ|+|γ|±|Γ|=|Ls+δ|d |ls|. The ±in this equation is there to account for the two cases on which side of Ls+δthe point Hϕ(zδ) lies in comparison to Hϕ(z). In either case, we find by combining the above two equalities that
S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 31 Pto create the Jordan curve that bounds ˆ Y. Now the situation is dealt with the same arguments as the previous case. Lemma 6.5. Suppose that ϕ0, ϕ1:∂S →R2are two piecewise linear embeddings of the square ∂S into R2. Let Y0and Y1be the Jordan domains bounded by the respective image curves ϕ0(∂S)and ϕ1(∂S). Suppose that ϕ0(z) =ϕ1(z)for all z∈∂S−and both maps have constant speed on S+. Suppose that the curves ϕ0(S+)and ϕ1(S+)do not intersect except for their endpoints. Suppose also that both embeddings ϕ0and ϕ1 are Lipschitz-continuous with constant L. Then there exists a homotopy ϕt, t ∈[0, 1] of piecewise linear curves which is simple, CL-Lipschitz in (z, t), and ϕtlies within the region bounded by ϕ0and ϕ1. Proof. Let γ0=ϕ0(S+)and γ1=ϕ1(S+). We first describe a homotopy γtbetween these two curves, which will then be used to construct ϕtby setting ϕt(S+) =γtand fixing a parametrization. On S−we naturally set ϕt≡ϕ0. The curve γtis defined as follows. Let the mutual endpoints of γ0and γ1be Aand Band the domain between these curves be denoted by ˆ Y. Let γ1/2be the shortest path from Ato Bwithin the closure of ˆ Y. We now need to only describe how to deform γ0 to γ1/2as the case from γ1/2to γ1will be handled in the same way. For t ∈[0, 1/2], note that 2tvaries from 0to 1. We choose γtas follows. First, travel along γ0starting from Auntil we have travelled a curve of length 2t|γ0|. We have arrived at a point of γ0which we shall call Pt. For the remainder of the parametrization, we take the shortest curve from Ptto Bwithin the closure of ˆ Y. This defines γtup to parametrization, and the exact parametrization of γtwill be defined now. We divide the time interval [0, 1/2] into intervals [tn, tn+1)so that for all t ∈[tn, tn+1) the curve ϕtis obtained from ϕtnvia simple modification, at least as long as we now guarantee that the parametrization aligns with the requirements in Definition 6.3. For a fixed parameter t, the curves γtand γ0agree on the initial part of γ0of length 2t|γ0|. For those s ∈S+for which ϕ0(s)is on this initial part, we also set ϕt(s) =ϕ0(s). Let s2t∈S+be defined so that Pt=ϕt(s2t), and recall that the curves γtnand γtfor t ∈[tn, tn+1)only differ by moving Ptnto Pt. Let t>tbe so that P=ϕtn(s2t)is the next vertex after Ptnon this curve, so that Pis also the next vertex after Ptfor ϕt. Now if the angle ∠PtPtnPis concave (above π) towards the interior, then the curves γtnand γtare the same and we may also set the parametrizations ϕtnand ϕtto be exactly the same. In the case where the angle is convex (less than π), we set ϕt(s) =ϕtn(s)for all s ⩾2t. It remains to define ϕton (2t, 2t) assuming by induction that ϕtnis given. Let the union of the segments PtPtnand PtnPbe U1and let U2denote the segment PtP. We choose a constant speed map Ψt:U1→U2, and this constant is smaller than one because U2is shorter than U1. Then we define ϕt(s) =Ψ t(ϕtn(s)) for s ∈(2t, 2t). This shows that the Lipschitz constant of ϕtin sdecreases as tincreases. It remains to obtain estimates in t. It is enough to show that |ϕt(s) −ϕtn(s)| ⩽CL|t −tn|for
32 S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 Fig. 10. Moving the point Pto Qthrough a point P∗via two simple modifications. s ∈(2t, 2t). For this, through some simple geometry we see that the distance between the points ϕt(s)and ϕtn(s), which lie on the sides of the triangle ΔPtPtnP, can be estimated from above by the length of the side PtPtn. But |Pt−Ptn| =|ϕtn(2t) −ϕtn(2tn)| ⩽ 2L|t −tn|, which finishes the proof. Definition 6.6. A homotopy ϕt:∂S →R2, t ∈[0, 1] of piecewise linear Jordan curves is called a 2-simple homotopy if for all t1and t2>t 1sufficiently close to t1, the curve ϕt2 may be obtained from ϕt1via two successive simple modifications on the same vertex P. The difference between one and two simple modifications is that in a simple modification the point Pis only moving along the ray −−→ P1P, while after two simple modifications the point Pmay technically move to any other in the plane. In our case, some further restrictions will apply as we must also maintain injectivity during this process. Lemma 6.7. If ϕt:S→R2, t ∈[0, 1] is a 2-simple homotopy of piecewise linear Jordan curves and Lipschitz-continuous in (z, t)with constant L, then the shortest curve extensions Hϕtare also Lipschitz-continuous in (z, t)with constant CL for a uniform constant C. Proof. Fix t1and let t2>t 1be close to t1. Then Definition 6.6 implies that there is a simple modification which turns ϕt1into another curve ϕ∗and another simple modification which turns ϕ∗into ϕt2. It is enough to show that we may choose ϕ∗so that the estimate |ϕt2(s) −ϕ∗(s)| ⩽CL|t1−t2|is satisfied, as then the two simple modifications ϕt1→ ϕ∗and ϕ∗→ ϕt2can be seen to be CL-Lipschitz-continuous in (z, t)and we may finish by applying the proof of Lemma 6.4 to obtain the desired result. Let Pbe the vertex on the curve ϕt1being moved to the vertex Qon ϕt2, and let P1 and P2be their shared neighbouring vertices. Let us pick t2close enough to t1so that Pand Qare on the same side of the segment P1P2, eliminating Case 3 in Fig. 10. We may assume that the ray −−→ P1Qintersects the segment P2Pat a point P∗(otherwise we consider the intersection of −−→ P2Qand P1P, or switch the roles of Pand Q). Now due to the assumption that the homotopy ϕtis Lipschitz continuous in twith constant L, we have that dist(Q, P1P∪PP2) ⩽L|t1−t2|(at least for t2close enough to t1so that there is no interference from the rest of the curve). Due to some elementary geometry
S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 33 the distance |QP∗|from Qto P∗must be comparable to the distance from Qto the segments P1Pand PP2, giving that |QP∗| ⩽CL|t1−t2|. Now let us compose ϕt2with a piecewise linear map which is otherwise the identity but sends the segments P1Qand QP2to P1P∗and P∗P2respectively. This is a simple modification of ϕt2which we call ϕ∗. Each point on the curve ϕt2is moved at most a distance of |QP ∗|, which gives the desired estimate |ϕt2(s) −ϕ∗(s)| ⩽|QP∗| ⩽CL|t1−t2|. Moreover, it is clear that ϕt1is a simple modification of ϕ∗as P∗lies on P2P. Thus the proof is complete. 7. The 3D extension We now proceed to the construction of the extension hinto the upper half space, continuing the proof of Theorem 1.2 along the lines described at the start of Section 4. The main goal here is to define hprecisely on each Uk,j. Recall the definition of the sets Uk,j, ˜ Qk,j and curves Γk,j from Section 4. Step 1. We define hon the sides of the top and bottom faces of Uk,j. We wish to map the top sides ∂˜ Qk,j ×{2−(k−1)}to the Jordan curve Γk,j and the bottom sides ∂˜ Qk,j ×{2−k}to ˆ Γk,j. Note that here and what follows we abuse ∂to mean the 1D boundary of these sets rather than taking the topological boundary of the sets in 3D space. Step 2. We define hon the top and bottom faces of Uk,j. To simplify notation, we set Ut=˜ Qk,j ×{t}. Furthermore, let top := 2−(k−1) and bot := 2−kso that Utop is the top face and Ubot is the bottom one. Similarly we set ϕt=h|∂Utand ht=h|Ut, although only ϕtop and ϕbot have been defined so far. On Utop, we simply define htop as the shortest curve extension of ϕtop. Note that this choice also forces us to define hbot on Ubot in a specific way to avoid discontinuity. Indeed, the bottom side Ubot is in fact the union of four top sides of dyadic cubes of the form Uk+1,jon the next level. Thus on Ubot the map hbot is defined separately in each of the four squares as the shortest curve extension of the corresponding boundary values. Step 3. Let mid := 2−k+2−k−1be the middle point of [2−k, 2−(k−1)]so that Umid is the middle level of the cube Uk,j. On the sides of Umid and for every parameter t ∈[bot, mid], we define ϕtequal to ϕbot. On Umid we define hmid as the shortest curve extension of ϕmid. Hence for t ∈[bot, mid], the mapping hthas the same boundary values on each level Utbut is a different map on the faces Umid and Ubot. We return to this part in a later step and describe how to define htfor t ∈(bot, mid)to give the correct isotopy between the maps hmid and hbot. Step 4. For t ∈[mid, top], we will define htas the shortest curve extension of ϕt. However, we have not yet defined ϕtfor these parameters. Note that the image of ϕtop is Γk,j and the image of ϕmid is ˆ Γk,j. Thus we must define a homotopy ϕtbetween these two curves which is what we will do now. The left part of Fig. 11 depicts the curves Γk,j and ˆ Γk,j. Since the curves Γk,j (respectively ˆ Γk,j) form a grid topologically equivalent with a dyadical grid, we may abuse terminology here and talk about vertices and edges of Γk,j when considered as a topo-
34 S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 Fig. 11. On the left, the curve Γk,j and its corresponding curve ˆ Γk,j on the next level. On the right, Γk,j has been modified to ˜ Γk,j . (For interpretation of the colours in the figure(s), the reader is referred to the web version of this article.) Fig. 12. The plus-shaped region whose boundary consists of two crosses and curves from the points mito ˆmi. logical square. As in the figure, let us label the vertices of these curves by vjand ˆvj, j=1, 2, 3, 4in corresponding order. We pick one pair of such vertices, say v1and ˆv1. The vertex v1is the intersection point of two edges of Γk,j as well as two other edges in the same grid, for a total of four. We let the midpoint of the edges meeting at v1be mj, j=1, 2, 3, 4, see Fig. 12. We similarly define four points ˆmjas the midpoints of the edges in the grid formed by the curves ˜ Γk,j which meet at ˆv1, numbered correspondingly to the points mj. We now connect each of the points mjwith ˆmjthrough a piecewise linear curve gjwhich does not intersect either of the grids and has length comparable to the infimal length of such curves. Our aim now is to deform the cross formed by the curves with endpoints at m1, ..., m4 and intersecting at v1, to a cross with the same endpoints but middle point at ˆv1instead.
S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 35 Naturally we wish to introduce no new intersection points during this homotopy and keep the deformation within the plus-shaped region pictured in Fig. 12. At each point in time the cross we are considering meets four different dyadic regions in the image side, and we wish to create this deformation between crosses in a way where we can apply Lemma 6.7 for each of these four regions to obtain the required interior Lipschitz-estimates. Thus it is necessary to form the homotopy in a way that with respect to each four regions the part of the border that is deforming behaves as a 2-simple homotopy (see Definition 6.6). A fixed number of reparametrizations of curves is also needed in the arguments used here, but we recall that Lemma 6.2 allows us to do so while still maintaining the required interior estimates. We first connect the points m1and ˆv1with a piecewise linear Jordan curve α1which does not intersect any of the other considered curves and has distance comparable to the sum of the length of the curve g1from m1to ˆm1and the curve from ˆm1to ˆv1which is part of ˆ Γk,jfor some j. This can be done for example by choosing a curve sufficiently close to those two curves but not intersecting them or itself. Similarly, we define a curve α2from m2to ˆv1, see again Fig. 12. Let ψ0be the union of the curves from m1to v1and from v1to m2, parametrized on [0, 1]. Similarly, let ψ1be the union of α1and α2. We may assume that ψ0(1/2) =v1 and ψ1(1/2) =ˆv1. Using the method of Lemma 6.5 we connect ψ0to ψ1via a homotopy ψt. We define a curve from v1to ˆv1by Ψ(t) =ψt(1/2). This homotopy from ψ0to ψ1gives one part of the sought homotopy between the two crosses. Let β1denote the curve from m3to v1and β2the curve from m4to v1. We denote by ψ∗ 0the union of β1and β2, parametrized again on [0, 1] with ψ∗ 0(1/2) =v1. We wish to construct another simple homotopy ψ∗ twith ψ∗ t(0) =m3, ψ∗ t(1/2) =ψt(1/2) =Ψ(t), ψ∗ t(1) =m4, and so that the curve ψ∗ thas no additional intersection points with ψt. At each time twe must find curves from m3and m4to Ψ(t). In order to do this we first describe the properties of the curve Ψ(t), as this curve may not be injective. Following the construction done in Lemma 6.5, the domain bounded by the two curves ψtand ψ1is decreasing as a function of t. Thus it is not possible for the curve Ψ(t)to form a proper loop to intersect itself, but a priori it can be constant on some interval and it can also travel backwards along itself. For the moment, let us describe the construction of ψ∗ twhile assuming that Ψ(t)does not intersect itself or ψ∗ 0. The idea of the construction of the homotopy ψ∗ tis to add to the initial curve ψ∗ 0a part which follows close to the curve Ψto a certain point and then returns back along another path close to Ψ. At t =1we will travel the full length of the curve Ψto the point ˆv1and back. Let us suppose that the homotopy ψ∗ thas been defined up to a point tnwhere Pn:= Ψ(tn)is a vertex on the piecewise linear curve given by Ψ. Let Pn+1 be the next vertex after Pnon Ψ, and let P1 n−1and P2 n−1denote the two neighbouring vertices of Pnon the curve ψ∗ tn. The aim now is to “open up” a part of the segment PnPn+1 into two segments P1 nQtand P2 nQt, but some care must be made to not cause intersections, see the rightmost part of Fig. 13 to illustrate this process.
36 S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 Fig. 13. Opening up the curve Ψ to create a homotopy of Jordan curves. More precisely, let us suppose that the angle ∠P1 n−1PnPn+1 (interpreted as the smaller angle of the two choices) is smaller or equal than ∠P2 n−1PnPn+1 (again, the smaller choice). We pick another point P1 non the segment PnP1 n−1which may be chosen arbitrarily close to Pn. We may let P2 n:= Pnin this case, if the size of the two angles ∠P1 n−1PnPn+1 and ∠P2 n−1PnPn+1 is reversed then so is the role of P1 nand P2 n. For a point tn+1 >t nto be chosen later, we will now define ψ∗ tfor t ∈(tn, tn+1]. For t ∈(tn, tn+1]let Xtdenote a point parametrized linearly on PnPn+1 so that Xtn=Pn and Xtn+1 =Pn+1. For each t ∈(tn, tn+1]we now define ψ∗ tby mapping the preimage of the segment P1 nPnto P1 nXtand the preimage of P2 nPnto P2 nXt. This simply corresponds to moving the point Pnalong the segment PnPn+1 to the point Xtwhile keeping the parametrization consistent, see Fig. 13. By choosing P1 nclose enough to Pnwe can guarantee that no new intersection points are created during this process (since by assumption Ψdoes not intersect itself), and that the added length is comparable to the length of Ψ. Let us elaborate a bit further on the parametrization of the curves ψ∗ tused here. We pick one constant speed parametrization Θfrom I:= [1/4, 3/4] to the final curve between P1 0and P2 0defined by the process above. This final curve travels arbitrarily close to Ψ all the way up to ˆv1and then back along a similar curve to P2 0. Let us first reparametrize the initial curve ψ∗ 0in order to guarantee that a small part is not mapped to Θin the end. We choose ψ∗ 0to map the intervals [1/4, 1/2] and [1/2, 3/4] to the two segments P1 0P0and P0P2 0, keeping the relation ψ∗ 0(1/2) =P0=v1. The exact parametrization can be inherited backwards from the final parametrization Θ, so that the preimage of the segments P1 nXtand XtP2 nunder each curve ψ∗ tfor t ∈[tn, tn+1)is the same set as the preimage of the part of Θ between P1 nto P2 n. As the latter image curve is longer we may guarantee that the Lipschitz-constant of ψ∗ ton [1/4, 3/4] is controlled by the length of Θ. The parametrization in the time variable tcan also be chosen based on Θ. In fact, as long as we pick the time intervals [tn, tn+1)to have comparable length to the total length of the preimage of the segments P1 n−1P1 nand P2 n−1P2 nunder Θ, the Lipschitz constant in the time direction will be bounded from above by a constant times the length of Θ. Thus the boundary curves ψ∗ thave the correct Lipschitz bounds, and we turn our attention to interior estimates. Note that there are four different regions meeting at the
S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 37 cross with centre Ψ(t). Let us denote the region which only meets ψtby V1 t, the region which only meets ψ∗ tby V4 t, and let V2 tand V3 tbe the two regions which meet one half of both of these curves. We let Ui tdenote the corresponding dyadic squares on the domain side (which, if interpreted as planar sets, are the same set for each t), whose boundaries are all identified with Sfor the sake of constructing the shortest curve extension to Vi t. In each of the sets Vi t, one part of the boundary is fixed while the deformation of the other part is dictated by the homotopies ψtand ψ∗ t. Whichever domain Vi tis chosen, locally in tthe deformation only consists of moving around the single vertex Ψ(t). Hence as long as the preimage (in Ui t) of the part being deformed corresponds to being either contained completely in S+or completely in S−, this homotopy induces a homotopy on ∂V 1 twhich is at worst a 2-simple homotopy (see Definition 6.6). The preimage being contained entirely in S+or S−happens exactly when the preimage of v1happens to be identified with the vertices (0, ±1) on S, while the opposite is true when Ψ(t)is identified with (±1, 0). If the homotopy of ∂V i tis indeed 2-simple, then Lemma 6.7 implies that the shortest curve extension satisfies the required interior Lipschitz bounds. We need hence address the case where Ψ(t)is identified with (±1, 0). Note that in the definition of the shortest curve extension which is now applied inside the diamond shaped domain Ui t, there is an implicit choice of horizontal/vertical direction based on which two opposing vertices we pick as the top and bottom vertices. If we choose the direction where the horizontal lines point towards the preimage of Ψ(t), then the condition of the deformation being contained inside S+or S−in Definition 6.3 is satisfied. Naturally we cannot a priori choose the orientation to always satisfy this condition as exactly two of the vertices of Ui tsatisfy this condition and two do not, and eventually we will need to repeat this argument with respect to crosses with centres at each of the four vertices of Ui t. We take care of this issue with the following trick. Let ρdenote a bilipschitz map from the square domain bounded by Sto the unit disk, and let νt(z) =eiπt/2zdenote a rotation map on the unit disk. Let ˜ H0:Ui t→Vi tdenote the shortest curve extension of a boundary map ˜ϕ0:∂Ui t→∂V i t. We then define a new map ˜ Hton Ui tby making a change of variables on the domain side in Ui t(identified with S) via the map ρ−1◦νt◦ρ, and instead of extending ˜ϕfrom ∂Ui twe extend the map ˜ϕt:= ˜ϕ0◦ρ−1◦ν−t◦ρvia shortest curve extension. Thus ˜ Htand ˜ H0have the same boundary values but differ in the interior. In essence, ˜ Htcorresponds to “rotating” the horizontal lines in Ui tby an angle πt/2and constructing the shortest curve extension based on these new curves. But we only need to know that for t =1the map ˜ H1corresponds to constructing the shortest curve extension with the horizontal lines in Ui treplaced by vertical lines, which can be done by choosing the bilipschitz map ρaccordingly. The homotopy ˜ Htcan be seen to be Lipschitz continuous in (z, t)with constant CL, where Lis the Lipschitz constant of ˜ϕ0. This follows from the Lipschitz continuity of ρ, νtand their inverses, and an application of Lemma 6.2 since ˜ϕtsatisfies the correct bounds in (z, t). The homotopy ˜ Htcan be used to temporarily change the direction of horizontal lines in Ui tto suit our purposes,
38 S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 showing that we may reduce to the previous case where the homotopy on the boundary is 2-simple. Let us now address the fact that in general the curve Ψmay intersect itself. Perhaps the easiest way to deal with this is to make a slight modification on the construction of Lemma 6.5, as the homotopy of curves γtparametrized on [0, 1] constructed in that lemma defines Ψby the relation γt(1/2) =Ψ(t). We will now make a slight perturbation of the curves γtto make them mutually non-intersecting, which will guarantee that Ψ(t) becomes injective. Note that if two of the curves γtdo intersect, they in particular intersect at a vertex Pof ∂ˆ Y, where ˆ Ydenotes the Jordan domain bounded by the curves γ0and γ1. At any such vertex Pwe attach to it a small segment PVPfacing the interior of ˆ Yand bisecting the angle of ∂ˆ Yat P. Now for each such segment we consider all the curves γtwhich pass through PVPand let the intersection point of γtwith this segment be Pt. Thus for those parameters tthe map t →Ptdefines either an increasing or decreasing parametrization of PVP, which is not strictly monotone as some interval of parameters is sent to the point P. However, we may make an arbitrarily small modification to this parametrization to make it strictly monotone, replacing each point Ptwith another point P∗ ton PVP. This gives us a way to replace each of the piecewise linear curves γtby another curve γ∗ twhich, for each segment PVPthat intersects γt, passes through the point P∗ tinstead of Pt. As this modification may be done in an arbitrarily small way we may assume that the Lipschitz estimates we obtained before for ϕtand for Hϕtalso hold after the modification up to a multiplicative constant arbitrarily close to 1. Thus although the new homotopy induced by the curves γ∗ tis not necessarily simple, it gives the desired Lipschitz-estimates inside and all of the curves γ∗ tare mutually nonintersecting. For further details also see Section 8where a similar construction is explained in more depth. This concludes the construction of the homotopy of the two crosses with centres v1 and ˆv1. After doing this process for every vertex vjand every curve Γk,j on level k, we have replaced the curve Γk,j with another curve ˜ Γk,j with the same vertices as ˆ Γk,j but not intersecting it, see Fig. 11. The homotopy between ˜ Γk,j and ˆ Γk,j is now easy to construct. Between each pair of neighbouring vertices, say ˆv1and ˆv2, we deform the part of ˜ Γk,j into ˆ Γk,j via the method explained in Lemma 6.5. After deforming each four parts in succession we have deformed ˜ Γk,j into ˆ Γk,j. Still in the situation of Fig. 11, we provide a few more details regarding parametrization and estimates happening here. We may divide the interval [mid, top]into two halves, on one of which we deform Γk,j into ˜ Γk,j and on the other ˜ Γk,j into ˆ Γk,j. To offer more details on what happens in the first half, we divide the first half further into four intervals so that on each we move one of the vertices vjto the corresponding point ˆvj, j=1, 2, 3, 4. In the first half, the length of the relevant curves is always controlled from above by |Γk,j| +|ˆ Γk,j|, plus the same quantity over the neighbours of Γk,j. As the initial curves are parametrized with constant speed we know by Lemma 6.4 that the Lipschitz-constant of
S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 39 the shortest curve extension hin the (z, t)-variables is thus controlled by 2k(|Γk,j| +|ˆ Γk,j|) added with this quantity over the neighbours. In the second half, each part of ˜ Γk,j having two of the ˆvias endpoints is deformed to the part of ˆ Γj,k with the same endpoints. Here we are again using Lemma 6.4 and therefore the Lipschitz-constant is estimated from above by 2k(|Γk,j| +|ˆ Γk,j|). Step 5. For t ∈[bot, mid], the situation is as follows. The maps hmid and hbot have already been defined. We interpret these maps as planar maps, identifying the horizontal sections Utof the cube Uk,j on the domain side with the same square domain which we call U. Both maps hmid and hbot are hence interpreted to be defined on Uand as they have the same boundary map ϕmid =ϕbot, we may interpret them to map Uinto the same target domain Vbounded by the piecewise linear Jordan curve ϕmid(∂U). The difference between these two maps is that hmid is defined by the shortest curve extension of ϕmid and hbot is defined as the shortest curve extension of its boundary values in each of the four child squares of U. Let us denote by Cthe cross formed by the two segments between opposing midpoints of the sides of U. Hence the way hmid maps Cis determined by the shortest curve extension and we denote the image cross by Tmid =hmid(C). The way hbot maps C is predetermined by the piecewise linear approximations of the original boundary map defined in Section 5. We denote Tbot =hbot(C). A key point to note is the following. Let Udenote one of the four children of U. Then we claim that hmid restricted to Uis actually the shortest curve extension of its boundary value on ∂U. Let denote one of the horizontal line segments inside U (the meaning of ‘horizontal’ here is as it was used in the definition of the shortest curve extension), with aand bbeing its endpoints. Then is part of a horizontal segment of Uand is mapped to a curve under hmid which is the shortest such curve between its endpoints. This must mean also that the curve is the shortest curve from hmid(a)to hmid(b)inside U. Moreover, since hmid maps each horizontal segment in Uto its target curve with constant speed, hmid must also have constant speed on . This cements the fact that hmid on Uis the shortest curve extension of its boundary values. However, the above argument has the following minor defect. In Section 6, the shortest curve extension was defined for a boundary map from a square to a piecewise linear Jordan domain. But the map hmid might not map the two line segments making up C to true Jordan curves as the shortest curve extension may fail to be injective and thus the image cross Tmid may touch the boundary in V. Nevertheless, these curves are still piecewise linear and are given by a uniform limit of Jordan curves. There is no issue defining the notion of shortest curves and shortest curve extensions to areas bounded by such degenerate Jordan curves as well, and the estimates we have established before in results such as Lemma 6.2 and Lemma 6.4 extend naturally to this setting as well. This can be seen by verifying that the proofs go through in the degenerate case as well. From now the strategy to define a homotopy htfor t ∈[bot, mid]is as follows. For each such t, the map hton ∂Uwill have the same boundary values ϕmid. Moreover, we
40 S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 will define a homotopy of crosses Ttbetween the two crosses Tmid and Tbot. Once such a homotopy has been defined and parametrized as a map Φt:C→Tt, for each child U of Uwe define hton Uas the shortest curve extension of its boundary values on ∂U. Thus htwill be equal to ϕmid on ∂Uand to Φton C. To construct the homotopy between the two crosses, we would like to apply the same argument from Step 4 which was used to create a homotopy between the crosses depicted in Fig. 12. However, in the argument from Step 4 it was essential that the two crosses only had two intersection points (on the curves between v1, m1and v1, m2). In our case, the crosses Tmid and Tbot may have arbitrarily many intersection points. To address this issue, we define another cross Tfix which satisfies this property respective to both the crosses Tmid and Tbot, and then simply deform first Tmid to Tfix and then to Tbot. Due to Lemma 6.2, the exact nature of the parametrization Φtdoes not play a role here and we may assume for example that on each of the four arms of Cthe parametrization always has constant speed. Before defining Tfix, we make a small modification to Tmid in order to replace it with a cross Tmid∗which does not intersect the boundary except at the four endpoints. Since the cross Tmid consists of piecewise linear curves, this modification can be done by moving each of its vertices that touch the boundary (except for the four endpoints) by an arbitrarily small amount towards the interior of Vso that the resulting cross does not intersect itself nor ∂V. This modification provides a homotopy from Tmid to Tmid∗ which we may, for example, dedicate the first quarter of the interval [bot, mid] towards in t. The fact that this modification to the cross may be done in an arbitrarily small way guarantees that the Lipschitz estimates (in t) both on Cand for the shortest curve extensions to the four regions of Vcan be controlled by above with a constant of our choice. It now remains to define Tfix. Since neither of the crosses Tmid∗and Tbot touch the boundary ∂Vexcept at their common four endpoints, we may choose Tfix for example as follows. We pick a point Pin Vclose enough to an image point of a corner of Uunder ϕmid so that Pbelongs to hmid∗(U) ∩hbot(U)for one of the children Uof U. Then we connect Pto the four endpoints of Tmid∗via piecewise linear curves to form the cross Tfix. These curves, if chosen to run sufficiently close along the boundary ∂V, may be assumed to satisfy the necessary properties of not intersecting themselves or each other. Moreover, they can be chosen so that two of them intersect Tmid∗and Tbot exactly once and two of them do not intersect these crosses (apart from the endpoints). See Fig. 14. This means that the crosses Tfix and Tmid∗are in the same configuration as the crosses in Step 4, and the same goes for Tfix and Tbot. Hence we may repeat the argument to find a homotopy between these crosses, and extend the boundary values defined by this via the shortest curve extension to the whole of U. For each t, we lift the copy of Uand the map htto the appropriate horizontal section at height tin Uk,j and Vk,j in order to fully define our extension there. We have thus defined the extension has a monotone map on each set Uk,j to the image set Vk,j. We now return to our original goal of controlling the Lipschitz constant of hin
S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 47 Fig. 16. The construction of refined dyadic quadrilaterals on the side and top faces. we rescale the parametrization on the interval (tk+1, tk]on the domain and target side so that if Mdenotes the midpoint of this interval, we scale (t∗ k, tk]to (M, tk]and (tk+1, t∗ k] to (tk+1, M]. The length of the interval (M, tk]is hence comparable to 2−k, which means that the Lipschitz constant of the map for parameters t ∈(M, tk]on Uk,j is controlled by 2k|ˆ Γk,j|as we have wanted. This finishes the construction and the proof. 9. Extending a boundary map of the sphere In this section we describe how to modify the local extension method constructed in Sections 5to 8to obtain a proof of Theorem 1.2. We go through the arguments in order and explain the changes needed in each part. Proof of Theorem 1.2.First we must define a dyadic decomposition of the unit sphere. For this purpose we embed the boundary of the unit cube smoothly onto the sphere and inherit the dyadic decomposition from each face of the unit cube. Thus the dyadic decomposition of the sphere splits into six dyadic decompositions of squares, which correspond to the six faces of the unit cube, and we may label the respective sets on the sphere as four side faces and one top and bottom face. The key difference in the spherical case lies in Lemma 5.1, where the dyadic decomposition is refined on each level. The main issue is that in Lemma 5.1 the vertices of the refined quadrilaterals Qk,j were positioned in the same direction (to the right and up) with respect to the original dyadic squares, whereas no such uniform direction can be chosen on the sphere. Instead we do as follows. For each dyadic quadrilateral Qk,j belonging to one of the side faces, we apply the same arguments as in Section 5and choose the vertices of its four children in the direction of east and north on the sphere, see Fig. 16. Thus on the side faces the construction can proceed as usual. We turn our attention to the top face. Let us fix a dyadic level kand suppose that the choice of quadrilaterals Qk,j has been made. Let us denote by {vm}the collection of points that are either vertices of the quadrilaterals Qk,j, midpoints of their sides, or intersections of two segments between opposing midpoints. The points {ˆvm}will denote
48 S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 Fig. 17. Picking vertices ˆvmin three stages. vertices of the quadrilaterals Qk+1,j which we must now choose. Let us define two sets of vertices Oand ˆ Oby saying that vm∈Oif the vertex vmis on the outer boundary of the union of all Qk,j on the top face, and likewise ˆvm∈ˆ Oif ˆvmis on the outer boundary of the union of all Qk+1,j on the top face. The choices made on the side faces already fix the points ˆvm∈ˆ Oand imply that on each dyadic level k, vertices ˆvm∈ˆ Oare closer to the north pole than the vertices vm∈O, and thus belong inside the union of all Qk,j, see Fig. 16. On the bottom face this relation is reversed, but these two cases are analogous enough that we only need to describe the construction on the top face and the other case is done with similar arguments. For vertices vm/∈O, we must pick one of four possible directions in which to choose ˆvmin, corresponding to the four dyadic quadrilaterals meeting at vm. Supposing that k⩾2, we pick the vertices as follows. For each vertex vmfor which vmhas a neighbour vm∈O, we choose ˆvmto lie inside the same quadrilateral as ˆvm, see Fig. 17. There are four vertices near the corners where the choice of vmis not unique and thus we have two quadrilaterals to choose from: one in the corner and one adjacent to it. In this case we pick ˆvmin the quadrilateral adjacent to the corner. For vertices vmnot having neighbours in O, we can pick the direction in which to choose ˆvmarbitrarily. As we have now chosen the grids on the domain side, we proceed as usual to define curves Γk,j on the image side as piecewise linear approximations of the image curves of ∂Qk,j under ϕ. Topological information can be preserved here since ϕis a homeomorphism, which means that we can assume that the image grid formed by the Γk,j is topologically equivalent to the domain grid. Hence on each dyadic level kthe grid formed by the Γk,j and the grid on the next level formed by the children ˆ Γk,j can be assumed to have topologically the same intersection points as the respective grids on the domain side. Due to the appearance of some additional intersection points compared to the arguments in Section 7, we must explain how the homotopy between Γk,j and ˆ Γk,j is defined in our case. Denote by Vmand ˆ Vmthe vertices on the image side corresponding to vm and ˆvm, and abuse notation to define Vm∈Oif vm∈O. First we note that due to the
S. Hencl et al. / Journal of Functional Analysis 286 (2024) 110371 49 Fig. 18. Deforming the two crosses. choice of the vertices ˆvmbefore, if ˆvm∈ˆ Othen at these points we are in the topologically correct situation to apply the homotopy construction from Section 7. As in the argument presented there, we may deform the cross with centre ˆ Vminto a cross with centre Vm and having the same endpoints, see Stage 2 in Fig. 18. At the four vertices in the corners of the top face there is a special situation where only three curves meet at vmand ˆvminstead of four, so technically we can not apply the previous homotopy argument between crosses here. But the “cross” consisting of three curves is only easier to deform than one with four. For example, one can add an auxiliary curve to both configurations, use the previous argument for four curves, and then forget about the auxiliary curves altogether. Thus we may apply an initial homotopy at the points ˆ Vm∈ˆ Oand the side faces to replace the grid formed by the curves ˆ Γk,j with another grid ˆ Gwhose outer boundary curves and points align with the respective Γk,j and Vm. See Stages 2 and 3 in Fig. 19. We must then describe how to deform the parts of the two grids left over inside the top face to each other despite the existence of some extra intersection points. In order to do this we simply define an auxiliary grid with vertices at points we denote by Wmas follows. The points Wmwill be chosen in the same direction with respect to both points Vm/∈Oand ˆ Vm/∈ˆ O. Precisely we mean that if the grid ˆ Gis identified with a square grid of dimensions 2k×2k, then each point Wmlies in the square to, say, the lower right of its respective point ˆ Vm∈G. We may make this choice so that Wmalso lies to the lower right with respect to Vmin the original grid Gconsisting of the curves Γk,j. The points Wmcan then be connected by piecewise linear Jordan curves with lengths comparable to the total length of the respective curves Γk,j and ˆ Γk,j. This may be justified for example by travelling sufficiently close to either of the given grids Gand ˆ G. These curves form an auxiliary grid ˜ Gcontaining the points Wm, and we can moreover pick this grid so that each of the curves in ˜ Gbetween neighbouring points Wmonly intersects both grids Gand ˆ Gat most once.
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