Diffeomorphic approximation of Planar Sobolev Homeomorphisms in rearrangement invariant spaces

Let $\Omega\subseteq\mathcal{R}^2$ be a domain, let $X$ be a rearrangement invariant space and let $f\in W^{1}X(\Omega,\mathcal{R}^2)$ be a homeomorphism between $\Omega$ and $f(\Omega)$. Then there exists a sequence of diffeomorphisms $f_k$ converging to $f$ in the space $W^{1}X(\Omega,\mathcal{R}^2)$.


Introduction and main results
Recently, motivated by applications in non-linear elasticity and in geometric function theory, a great deal has been devoted in understanding the question of approximating homeomorphisms f : Ω ⊂ R n → f (Ω) ⊂ R n with either diffeomorphisms or piece-wise -affine homeomorphisms. This problem is not trivial because the usual approximation techniques like mollification or Lipschitz extension using maximal operator destroy, in general, the injectivity.
In variational models of nonlinear elastic deformations of solid flexible bodies we search for minimisers of energy functionals (often) of the form where W : R n×n → R is a stored-energy functional satisfying We require that our model respects the law of non-interpenetration of matter and, assuming that the body does not fracture or break, it is therefore natural to look for a minimiser among homeomorphisms. Therefore we minimise the functional over Sobolev homeomorphisms satisfying given boundary values. Intuition gives the impression that the minimising deformation should be in essence a diffeomorphism (say up to a null set). A naive perception is that a Sobolev homeomorphism is essentially a diffeomorphism. In fact the question of the regularity of minimisers and the question of the behaviour of Sobolev homeomorphisms are somewhat inter-related. A key step to proving the regularity of minimisers (see [4,5]) is to show that any Sobolev homeomorphism can be approximated arbitrarily well by diffeomorphisms. This is the so-called Ball-Evan's approximation question and is currently a topic of much interest. The initial breakthrough in the planar case were the papers [15] and [16], planar case f ∈ W 1,p , p > 1. This was followed by [14], planar homeomorphisms in W 1,1 . The latter techniques have further been developed in [19] 2000 Mathematics Subject Classification. 46E35. The first and fourth authors were supported by the grant GACR 20-19018Y. L.G. and R.S. are members of Gruppo Nazionale per l'Analisi Matematica, la Probabilità e le loro Applicazioni (GNAMPA) of INdAM. The research of R.S. has been funded by PRIN Project 2017JFFHSH..
Naturally, given that one can approximate homeomorphisms by diffeomorphisms in the Orlicz-Sobolev sense (see [8]), the question of approximation in other classes of function spaces such as Lorentz Sobolev spaces, or Grand Sobolev spaces arises. These classes are an important tool in studying the regularity of solutions of certain PDEs and variational problems and usually provide sharper results for existence and regularity a of solution. To provide results for all these important classes at once we will study the question of approximation in general Banach function space. The result from [8] gives us a strong indication that a similar result should hold under a more general context. On the other hand in general r.i. spaces one lacks the explicit norm expression utilised for that result and there are several obstacles that must be overcome.
Let us just recall that by a r.i. space we mean a Banach function space on Ω endowed with a norm · X(Ω) such that u X(Ω) = v X(Ω) whenever u * = v * where u * and v * denote the decreasing rearrangements of the functions u, v. The Sobolev space over X is defined as For the full definition see Definition 2.1.
In order to introduce our main result we include the definition of the Lebesgue point property. This property is stronger than the absolute continuity (see (2.6)) of the norm and has been thoroughly characterised in [9]. We refer to the points x for which (1.2) hold as Lebesgue points in X.

Theorem 1.2.
Let Ω ⊆ R 2 be a domain. Let X(Ω) be a rearrangement-invariant Banach function space satisfying the Lebesgue point property (see Definition 1.1). Let f ∈ W 1 X(Ω, R 2 ) be a homeomorphism. For arbitrary ε > 0 there exists a diffeomorphismf such that Df − Df X(Ω) < ε and f −f L ∞ (Ω) < ε Remark 1.3. Not only can we approximate a Sobolov homeomorphism by diffeomorphisms but by locally finite piece-wise affine homeomorphisms in r.i. spaces satisfying the Lebesgue point property. Moreover, if ∂Ω is a polygon on which f is piece-wise linear then we can approximate f by finitely piece-wise affine homeomorphisms.
It is not hard to observe that the absolute continuity of the norm of X is a necessary condition for diffeomorphic approximation of homeomorphisms in W 1 X and by [9] it is also necessary for the Lebesgue point property. Our technique relies heavily on estimates derived directly from the Lebesgue point property. The authors hypothesise that the Lebesgue point property is in fact necessary for the approximation of f ∈ W 1 X by smooth functions (independent of injectivity).
1.1. A brief description of the proof of Theorem 1.2. In this subsection we outline the basic plan of our proof of Theorem 1.2. As suggested above the general concept is similar to that in [14] and [8]. Assume that we have a homeomorphic and locally-finite piece-wise affine approximation of f . We can then approximate these homeomorphisms by diffeomorphisms using [18]. The diffeomorphisms from this result coincide with the original piece-wise affine homeomorphisms up to a tiny set. Further they have the same Lipschitz constant as the approximated map up to a bounded multiplicative constant. The combination of the above two facts with the absolute continuity of the norm of X means that the diffeomorphisms given by [18] also converge to the piece-wise affine homeomorphisms in W 1 X(Ω). The entire argument is in Lemma 2.7 and thanks to this, the question reduces to approximating by piece-wise affine homeomorphisms.
In Lemma 2.8 we separate Ω into open disjoint sub-sets Ω i which are further split into squares of a given size so small that the following holds • all the squares have the same size and fill all of Ω i except a set so small that the norm of the restriction of f to this set is bounded by 2 −i ε, • thanks to the Lebesgue point property for Df , f is very close to an affine function except for some squares whose union has measure so small that the restriction of f to this set is bounded by 2 −i ε. In general it is necessary to know that the behaviour of f is reasonable on the boundary of the squares. This is not automatically true but by slightly moving the boundaries of the squares it is true, which is achieved in Lemma 2.9.
On the squares where f is very close to a nice affine map (Jacobian not too small, derivative not too big) the map which is affine on each of the pair of triangles of the square approximates f well. If the Jacobian is zero we use the result Theorem 2.6 to approximate. Now either the map is close to a constant on the given square or is not close to any linear map. In either case we use Theorem 2.4 to define our piece-wise affine approximation. In the last two cases the smallness either of Df or of the set is enough to make sure that the error is small.
Finally we have Lemma 2.11 to fill the small space around the boundary of Ω i . Because the size of the set is so small we get that the norm of the map and the approximation is less than 2 −i ε.

Preliminaries
In this section we shortly list the basic notation that will be used throughout the paper. The set Q(c, r) = {(x, y) ∈ R 2 : |x − c 1 | ≤ r, |y − c 2 | ≤ r} will denote the square centred at c with side length 2r. Similarly, B(c, r) is the ball centred at c with radius r. For the ease of notation, for t > 0 we will denote tQ(c, r) = Q(c, tr), and tB(c, r) = B(c, tr).
Sometimes we will work on 1-dimensional objects in R 2 , which can be parametrised by a Lipschitz curve ϕ : [0, 1] → R 2 , for example segments, and various polygons. We may assume that our ϕ is one-to-one and |ϕ | is constant almost everywhere. For almost all t ∈ (0, 1) there exists a vector ϕ (t) |ϕ (t)| which we call the tangential vector at the point ϕ(t) and denote this vector as τ = τ (ϕ(t)). If a mapping f is defined on ϕ([0, 1]) and f • ϕ is absolutely continuous, then we call tangential derivative along the curve ϕ.
We take advantage of standard denotation of average integrals using the symbol − . Since we integrate with respect to different measures, we emphasise the fact that we divide the integral by the measure of the set we integrated over, where we measure the set with the same measure used in the integral.

Rearrangement invariant function spaces.
Here we collect all the background material that will be used in the paper.
Let (E, µ) be a measure space of finite measure. We set From now on we shall identify functions (P4) Let G ⊂ E be a set of a finite measure. Then (P5) Let G ⊂ E be a set of a finite measure. Then there exists a constant C G depending only on the choice of the set G for which (P6) f X(E) = g X(E) whenever f * = g * (rearrangement invariance) If a functional · X enjoys the properties (P1)-(P5) we call it a Banach function norm. If it also enjoys (P6) we call it a rearrangement invariant Banach function norm. Let · X be (rearrangement-invariant) Banach function norm then we call set endowed with the norm · X(E) a (rearrangement-invariant) Banach function space.
Following the properties (P2),(P4), (P5) one can observe that if |E| < ∞ for arbitrary Banach function space, the following holds true where → stands for a continuous embedding. Given a r.i. Banach function space X(E) and 0 ≤ s ≤ µ(E) one may define a fundamental function of X(E) by where G ⊂ E is an arbitrary subset of E of measure s. The properties of r.i. norms guarantee that the fundamental function is well defined. This function is non-decreasing on [0, µ(E)) concave and ϕ X(E) (0) = 0.
Given a Banach function space X(E) define an associated Banach function space X (E) as a subspace of measurable functions endowed by associated norm given by Note, that the associated space of a Banach function space is also a Banach function space and the following Hölder inequality holds Let us remind the reader that for arbitrary rearrangement-invariant Banach function space we have that The proof of this classical result called Hardy-Littlewood-Polya inequality may be found for instance in [7,Theorem 4.6]. Given a Banach function norm · X(E) and a normed linear space Y we shall define Let f : R n → R k be a measurable function we define the maximal operator of such a function by Let us also recall that if E ⊂ R n and µ is the Lebesgue n-dimensional measure and f : R n → R k , we have by the Riesz-Herz equivalence that with constant independent of t and f . For proof of this result see [7,Theorem 3.8].
Let X(E) be a Banach function space. We say that X(E) has locally absolutely continuous norm if for any finite measure set M ⊂ E and any function f ∈ X(E) one has Note that if µ(E) < ∞ this property implies the ε − δ-continuity of the norm. This means that for any f ∈ X and ε > 0 one can find δ > 0 such that (2.6) Let Ω ⊂ R n be an open set and let X(E) be a Banach function space. We define the Sobolev space over X(E) by where we use the standard operator norm to determine the size of |Df |. Now follows a prepatory lemma. Then for every M > 0 andε > 0 there exists aδ > 0 such that for all u ∈ X(G) with u L ∞ (G) ≤ M and u L 1 (G) <δL n (G) one has u X(G) <εL n (G).
Proof. For arbitrary D > 0 we have where C L ∞ →X stands for the optimal constant of imbedding of L ∞ (G) into X(G). The last inequality follows from First choose D such that (2) <ε L n (G) 2 . Then chooseδ such that (1) <ε L n (G) 2 . Remark 2.3. If X has the Lebesgue point property, we have by [9, Proposition 3.1], that the norm is locally absolutely continuous. Therefore one has lim t→0+ ϕ X (t) = 0, and Lemma 2.2 is valid for such a space.
2.2. The reformulation of known extension results. In this section our aim is to prove the following extension theorem, which will allow us to construct homeomorphisms from boundary values and gives us a useful control on their modulars. Theorem 2.4. There exists a C > 0 such that for any r > 0 and any finitely piecewise linear and one-to-one function ϕ : ∂Q(0, r) → R 2 we can find a finitely piece-wise affine homeomorphism h :

7)
and where the L ∞ space on the left is with respect to the two dimensional Lebesgue measure L 2 and the L ∞ space on the right is with respect to the one dimensional Hausdorff measure H 1 .
Proof. The construction is precisely that of Hencl Thus the construction given in [14] is precisely the one we need.
The following Corollary is an immediate result of the previous theorem. It is the fact that we get an L ∞ bound from the L 1 space that our approach works in the relatively general setting.
Corollary 2.5. There exists a constant C > 0 such that for every r > 0 and ϕ : ∂Q(0, r) → R 2 finitely piece-wise linear and one-to-one function with |D τ ϕ| constant on each side of Q = Q(0, r) ⊂ R 2 , there exists a piece-wise affine homeomorphism h : Q → R 2 such that Proof. If |D τ ϕ| is constant on sides, then D τ ϕ L ∞ (∂Qr) ≤ 4 r ∂Qr |D τ ϕ| dH 1 . Further we will reformulate [8, Theorem 3.7] to fit in with our current setting better. The question is how to approximate a map which is close to a degenerate linear map Φ (up to a roation in the pre-image we may assume that Φ = Theorem 2.6. Let d, δ > 0, let r 0 ∈ (0, 1) and let Q = Q(0, r 0 ). Then for every ϕ : ∂Q → R 2 finitely piece-wise linear and one-to-one mapping with and D τ ϕ ∞,∂Q ≤ d + 2δ, where τ is the unit tangential vector to ∂Q, there exists a finitely piece-wise affine homeomorphism g : Q → R 2 and a set W ⊂ Q such that Proof. Calling A i the vertices of Q and A the centre of Q, we define Then it follows that |Q \ W (δ)| ≤ Cδr 2 0 . In steps 2 and 3 of [8, Theorem 3.7] a mapping g was defined on W so that the estimates (given in step 4 of that theorem) hold (see (3.19) and (3.23) from [8]). Let us observe that the construction of g from [8, Theorem 3.7] gives a map that is Cd Lipschitz continuous on Q \ W . In step 5 the annulus Q \ W is separated into small square-like areas. On each of these squares Theorem 2.4 is applied. The Lipschitz bound of the extension g, of the boundary values is controlled by the Lipschitz constant of the boundary values themselves. We control the Lipschitz constant of the boundary values as follows. On those sides, which are subsets of ∂Q, we have g = ϕ and by assumption this is bounded by d + 2δ. On sides contained in ∂W we bound Dg by the size of Dg on W , which is less than 2d. On sides connecting ∂Q and ∂W we have a fixed K independent of d such that |Dg| < K by (3.29) of [8]. Therefore we have |Dg| < C(d + 1) on Q \ W by Theorem 2.4.
Letf ∈ W 1 X(Ω, R 2 ) be a countably piece-wise affine homeomorphism. Then for every ε > 0 there exists a diffeomorphismf such that Moreover, iff is continuous up to the boundary of Ω, thenf can be chosen to be continuous up to the boundary of Ω andf =f on ∂Ω.
Proof. We denote byf the piece-wise homeomorphism and byf a diffeomorphism constructed as in [18]. The claim that Follows immediately from Theorem A there. We only need to show that the arguments in [18] extend to handling the norm · X(Ω) instead of · L p (Ω) . We refer to the end of the proof of Theorem A, on pg 1422 of [18]. Let {T i } i∈I be the triangulation of Ω. For each T i , by construction we have that there exists a constant M i such that We can assume that

2.3.
Preliminary results on grids and approximations on grids. Let us remind the reader that by dyadic squares we mean the family D = k∈Z D k , where For a domain Ω we say that an open set G separates components of ∂Ω if any continuous curve connecting different components of ∂Ω has to intersect G. We say that the domain Ω ⊂ R 2 is finitely connected if R 2 \Ω has finite number of components.
Let Ω ⊂ R 2 be a finitely connected bounded domain and let f ∈ W 1,1 loc (Ω, R 2 ) be a homeomorphism. Then there exists a strictly increasing sequence of sets Ω k Ω k+1 Ω such that Ω = k Ω k , also Ω k separates components of ∂Ω and ∂Ω k is piece-wise linear and parallel to coordinate axes and ∂Ω k has the same number of components as ∂Ω. Further, for every k ∈ N and ε k , δ k > 0 and any set A k with iii) for each 1 ≤ j ≤ K k the square Q k j shares at least two of its sides with other consists of segments and there is a one-to one correspondence between the endpoints of these segments and vertices of ∂( The following construction depends on a parameter M ∈ N, which we fix in the first step of the construction. For every l ∈ N we define W −l as the set of squares Q ∈ D −l such that 16Q ⊂ Ω. As l tends to infinity these sets fill Ω and because Ω is finitely connected there exists an M such that the set separates components of ∂Ω (which means that any continuous curve connecting different components of ∂Ω has to intersectΩ 1 ). The setΩ 1 is not necessarily connected. Assume that there are N 1 components, we choose a point y 1 n in each of the components. Each pair y 1 1 and y 1 n , n = 2, . . . , N 1 are path-wise connected in Ω. Call γ 1 n a path connecting y 1 n with y 1 1 inside Ω. Now set Define the set Note that the set  Now let us suppose that we have defined l k−1 and Ω k−1 . Set Call N k the number of components ofΩ k and let γ k n , n = 2, 3, . . . , N k be the paths in Ω connecting each component ofΩ k to one of the given components. We find an l k > l k−1 such that we can cover each γ k n with dyadic squares from W −M −l k . We put for the squares connecting the components along the paths) and for squares that were surrounded by C k in order to define Thus we define Ω k , k ∈ N inductively. This decomposition satisfies our requirements for Ω k . Now let us choose k ∈ N and construct the collection of cubes {Q k i } from the claim. Either Ω is simply connected and Ω k has one component, or Ω is multiply connected and Ω 1 has one component and for all k ≥ 2 the number of components of Ω k \ Ω k−1 equals the number of components of ∂Ω. In the latter case it suffices to consider each component separately and so with respect to this fact we may assume that Ω k is connected.
We have k fixed; for each m ∈ N we call where m ≥ M + l k + 8. Firstly notice that because the map f is uniformly continuous on Ω k and so for ε k there exists an m such that diam f (Q(x, 2 −m )) < ε k for all m ≥ m . By this condition we get point vii). Calling we get the existence of an m such that for all m ≥ m . This will be crucial for getting ii). Our next step will be to shift the squares to guarantee vi) holds, then we exclude the squares too close to ∂(Ω k \ Ω k−1 ) to give iv) and v). The other properties will follow quickly.
Almost every a ∈ Ω k \ Ω k−1 is a point of differentiability of f (for example see [13, Lemma A.28]) and a Lebesgue point of the derivative of f in X. (it follows from (2.1) that it is also a Lebesgue point of Df in classical sense). We have for almost every a ∈ Ω k \ Ω k−1 (because Ω k Ω we may assume that Q(a, 2 1−m ) ⊂ Ω for all m large enough). Therefore there exists an m and a set E k such that for all and then and therefore we can find a v ∈ Q(0, 2 −m k −1 ) such that be the set of indices of the bad squares. Now, note that since (2.13) holds, we have card(B k ) ≤ δ k 2 2m k −2 and thus In summary we have the following, each square Q(a i , 2 −m k ) either has i ∈ B k and then calculated above is L 2 ( i∈B k Q k i ) < δ k , otherwise i / ∈ B k and in this case a i / ∈ A k and I m k (a i ) < ε k 4 . From I m k (a i ) < ε k 4 we easily get the inequalities in vi).

It holds that
If necessary we change the indexing of the squares to get the set {Q k i ; 1 ≤ i ≤Õ k } and none of the squares Q k i intersects ∂(Ω k \ Ω k−1 ) for 1 ≤ i ≤Õ k . Simultaneously, (because diam ∞ Q k i = 2 1−m k ) for arbitrary x in the outer squares we have that We exclude the first layer of outer squares. Now for the elements of the outer squares of the remaining system We repeat this excluding process two more times and we get the final set There is only a corner of ∂ K k i Q k i near a corner of ∂(Ω k \ Ω k−1 ) because m k ≥ M + l k + 8 (which is iv)). Simultaneously, because the corners of ∂ K k i Q k i have distance greater than 32 · 2 1−m k , any single square of {Q k i } contains at most one corner of ∂ K k i Q k i , therefore no square has more than two external sides and so shares at least two sides with other squares of {Q k i }, which is iii). It is obvious that i) holds; the union of the squares in U k has no holes, the shifted squares are the same and no holes inside can be formed by removing squares from the edge. the set K k i=1 Q k i consists of all of the squares. The last point to check is ii). By the choice of m k ≥ m we can estimate by (2.10) and so we get ii).
The next lemma is needed to describe how to slightly move the vertices of a grid to obtain better boundary values. This result is proved in [8], but we give the proof for the reader's convenience.
and such that f is absolutely continuous on each side of ∂Q k i and for all i it holds that − (2.14) Further for those i / ∈ B k it holds that In both cases above, C > 0 is an absolute constant, a i is the center of Q k i and ε k > 0 was chosen in Lemma 2.8.
Proof. For each vertex V = (v 1 , v 2 ) of the grid, we let S V be the segment For each set of two neighbouring vertices V 1 and V 2 of the grid (i.e., V 1 and V 2 are endpoints of the same side of a square Q of the grid), the following estimates hold (2.17) Above, for X i ∈ S V i , i = 1, 2, [X 1 , X 2 ] denotes the segment whose endpoints are X 1 and X 2 . The integrals in left-hand side are meaningful, as a consequence of the wellknown property of the Sobolev mapping f of being a.c. on almost all lines. Moreover, inequalities (2.16) and (2.17) hold because H 1 (S V 1 ) ≈ r k and co(S V 1 ∪ S V 2 ) ⊂ 2Q. Furthermore, Q is any square of the grid such that V 1 , V 2 ∈ ∂Q.
For simplicity, we only estimate (2.14). It will be clear that we shall be able to guarantee also (2.15) for the 'good' quadrilaterals.
Fix λ > 4. By Chebyshev's inequality, from (2.16) we deduce that there exist a subset S(V 1 , V 2 ) of S V 1 and a subset S(V 2 , V 1 ) of S V 2 such that and for every X 1 ∈ S(V 1 , V 2 ) and X 2 ∈ S(V 2 , V 1 ) the following estimate holds Now, let V be a vertex of the grid. There are (at most) four vertices V 1 , V 2 , V 3 , V 4 of the grid which are neighbouring vertices of V and therefore, by the above construction we find four subsets S(V, V i ), i = 1, 2, 3, 4 of S V such that has positive H 1 -measure, since λ > 4 and thus it is not empty. We replace V with ã V ∈ S. To conclude, each square Q = co {V 1 , V 2 , V 3 , V 4 } of the grid will be replaced by the quadrilateral Q = co ¶Ṽ Lemma 2.10. Let Ω ⊂ R 2 be a finitely connected bounded domain and let f ∈ W 1 X(Ω, R 2 ) be a homeomorphism. Let k ∈ N and ε k , δ k > 0 be given numbers.
Let Ω k be the set chosen in Lemma 2.8 containing squares of type Q k i = Q(c i , 2 −m k ) determined by the given numbers ε k , δ k . Let Q k i be the quadrilaterals in Ω k \ Ω k−1 derived from Q k i by Lemma 2.9. Then there exists a piece-wise linear injective function h defined on Γ k = i ∂Q k i such that; i) it holds that f − h L ∞ (Γ k ) < 4ε k , ii) at each vertex x of each Q k i it holds that h(x) = f (x), iii) for any S a side of a Q k i and for H 1 a.e. x ∈ S we have Proof. The first step is to define a piece-wise linear approximation of f , (call itf n ) on each side S of every Q k i . We call x 1 , . . . , x n n evenly spaced points allong S with x 1 and x n being the two endpoints of S. Then we putf n (x j ) = f (x j ) andf n is linear on each segment between x j and x j+1 . Of coursef n converges uniformly to f on S and for n large enough is injective. The argument is rather simple and a detailed version can be found in [14], [8] or an alternative approach can be found in [12]. We define h so that h(S) =f (S) for each S, but h parametrizes its image from S at constant speed.
Point i) holds because of Lemma 2.8 point vii) and because the oscilation of h on any ∂Q k i is bounded by the oscillation of f on the ∂Q k i which is bounded by the oscillation of f on the 2Q k i . Point ii) is obvious. Point iii) holds because the piece-wise linear curve is not longer than the original curve i.e. When we denote L(t) as the segment L ⊂ ∂Q k i described above that contains the point t ∈ ∂Q k i we can easily calculate Now for any i / ∈ B k we can see by (2.15) of Lemma 2.9 that the above can further be estimated by ε k , which is exactly point iv).
We now prove point v). At any point x where J f (x) > 0 it holds that J f (x) = πλ 1 λ 2 , where λ 1 = max{|Df (x)v|; |v| = 1} and λ 2 = min{|Df (x)v|; |v| = 1}. Given that λ 1 < 1 √ ε k and J f (x) > 4π √ ε k it must follow that λ 2 > 4ε k . The distance therefore between the image of any two vertices of Q k i in the linear map Df (a i ) is at least 3·2 −m k ε k . On the other hand the error of f from the affine map f (a i )+Df (a i )(x−a i ) is less than 2 · 2 −m k ε k . Therefore any 2-piece-wise affine map coinciding with f on the vertices of Q k i must have positive orientation and be injective on Q k i . Lemma 2.11. Let Ω ⊂ R 2 be a finitely connected bounded domain and let f ∈ W 1 X(Ω, R 2 ) be a homeomorphism. Let k ∈ N 0 and ε k , δ k > 0. Let Ω k ⊂ Ω k+1 be a pair of sets chosen in Lemma 2.8 corresponding to ε k and δ k . Let Q k i be the quadrilaterals from Lemma 2.9 in Ω k and Q k+1 i be the quadrilaterals from Lemma 2.9 in Ω k+1 . Call S k the space between Q k i and Q k+1 i , i.e. the union of all components of ] which intersects ∂Ω k . Finally let h be the piece-wise linear function detemined by Lemma 2.10. Then there exists a finite collection of quadrilaterals and a finitely piece-wise affine homeomorphism g defined on S k such that g = h on ∂S k and (2.18) Proof. We start by assuming that Ω is simply connected and Ω k+1 \ Ω k has exactly one component. If this were not so, then we could deal with each component in the same way.
Step 1. Go from squares of size 2 −m k to squares of size 2 −m k+1 . We have the quadrilaterals Q k i from Lemma 2.9 which were constructed from the squares Q k i ∈ D v −m k from Lemma 2.8. Now we name the neighbours of Q k i . We index the finite set of neighbouring squares That is to say the squares Q k i , K k + 1 ≤ i ≤K 1 k are those shifted dyadic squares that are not contained in the set K k i=1 Q k i but share at least one common vertex with them. By v) of Lemma 2.8 and by diam ∞ Q(x, 2 −m k ) = 2 1−m k we have that for all x ∈ ∂Ω k . Similarly call the neighbours of the previously added squares for all x ∈ ∂Ω k . We divide each squareQ k , for each z a corner of Q(0, 1). We number the squares we get by dividingQ k i , for all x ∈ ∂Ω k . We now repeat this last operation. We call the neighbours of the previously added squares We divide the squares −m k −2 , for each z a corner of Q(0, 1) and call these squares As before we have that After m k+1 − m k + 1 steps we start adding squares of type Q(c i , 2 −m k+1 ). We call those to ∂Ω k is between 2 and 4 times 2 −m k+1 . By adding shifted dyadic squares of type Q ∈ D v k+1 −m k+1 to the grid in Ω k+1 \ Ω k we can guarantee that the distance from ∂ i Q k+1 i to ∂Ω k is also between 2 and 4 times 2 −m k+1 . Thus the distance between ∂ This means that we add two layers of squares of side length 2 1−m k −j before we start adding squares of side length 2 1−m k −(j+1) . Therefore 2Q k i intersects at most its neighbours and its neighbours' neighbours but no other squares further away. Figure 3. Filling in the left over space (in the picture shaded) between two neighbouring grids with rectangles uniformly piece-wise affine bi-Lipschitz equivalent with Q(0, 2 −m k+1 ). The line vertices X i are partnered with the line vertices Y i . The vertex A is a reflex vertex and C is an acute vertex. We connect A and A to the near vertex of ∂Ω k with a segment. The vertex C is connected with the next avaialble neighbour of A (in this case Y −1 ). The vertex Y 0 is connected with ∂Ω k .
Therefore there exists some finite number M − 1 of squares which 2Q k i intersects. Therefore for almost every x ∈ S k .
Step 2. Fill the rest of S k with quadrilaterals. What is left over is a 'tube' around ∂Ω k approximately 2 −m k+1 wide, which can be divided into quadrilaterals all uniformly bi-Lipschitz equivalent with a square of side length 2 1−m k+1 . Although this fact is obvious we describe one way how to do this in detail here.
We call those vertices X of squares Q k i such that X ∈ ∂ Ñ k i=1 Q k i outer vertices. Similarly we also call those vertices X of squares Q k+1 i such that X ∈ ∂ • of a small circular arc centered at the vertex lies inside Figure 3. In the other case we call a vertex of ∂ • of a small circular arc centered at the vertex lies inside Figure 3. We describe an outer vertex as a line vertex if it is not a corner of ∂ Figure 3.
Firstly we deal with the area around corners of ∂Ω k . We add no new segments going from a reflex corner A. We add a segment from the vertices A , A neighbouring A to the nearby corner of ∂Ω k , this creates a quadrilateral (see Figure 3). For an acute corner C with adjacent sides S 1 and S 2 (whose corresponding opposite sides areS 1 andS 2 ) we create new quadrilaterals by adding a segment from C to both of the second neighbours of A.
Now choose any side of ∂Ω k and consider the two corresponding sides S,S, with S a side of ∂ . Both S andS have endpoints very close to the same corners of ∂Ω k and so are very close to the same length and therefore the difference between the number of line vertices on S andS varies by at most 4. Thus it is possible to pair the line vertices of S and the line vertices ofS so that at most two line vertices of S (resp.S) close to a given corner of S, (resp. S) do not have a partner inS, (resp S) moreover the segment between a pair of partnered vertexes is nearly purpendicular to the corresponding segment of ∂Ω k . Let X 1 , X 2 be a pair of line vertices on S with partners Y 1 and Y 2 , their partners inS. Each of the segments X 1 Y 1 and X 2 Y 2 intersect ∂Ω k exactly once, call the points Z 1 and Z 2 respectively. Any of the quadrilaterals X 1 Z 1 Z 2 X 2 and Y 1 Z 1 Z 2 Y 2 formed by any two pairs of neighbouring line vertexes are uniformly bi-Lipschitz equivalent with Q(0, 2 −m k+1 ) by a 2-piece-wise affine map. Permitting a small bastardisation of the notation we call these quadrilaterals Q k i forÑ k + 1 ≤ i ≤Ñ k . If we have a left over line vertex X with no partner we create a new quadrilateral with a vertex at X by adding a segment from X to the point on ∂Ω k half way between the neighbouring vertices on ∂Ω k (see Y 0 in Figure 3). We call the entire collection of these quadrilaterals Q k i , 1 ≤ i ≤ N k .
Step 3. Move the corners.
It is possible by moving the corners of the quadrilaterals of Q k i K k ≤ i ≤ N k to get quadrilaterals Q k i which satisfy the estimate (2.14). The idea is exactly that of the proof of Lemma 2.9, the only difference here is that neighbouring squares may not have exactly the same side length, but the ratio is bounded by 2. This case is dealt with in detail in [8, Theorem 4.1, step 2] and the interested reader can check the details there. We move corners by at most 2 −2−m k −j , for Q k i in the j-th generation of added squares.
Step 4. Define g on the grid and extend. We define g on ∂Q k i for K k + 1 ≤ i ≤ N k exactly the same way as we defined h in Lemma 2.10 for 1 ≤ i ≤ K k + 1. We get a piece-wise linear and injective function g on N k i=K k +1 ∂Q k i . Because each each Q k i is uniformly bi-Lipschitz equivalent with a square by (uniformly bounded number of pieces) piece-wise affine maps we can apply Corollary 2.5 and get a finitely piece-wise affine homeomorphism g on S k , extending the original mapping and satisfying Further by applying the (2.14) type estimate we got from step 3 we get Since we move corners in step 3 by at most 2 −2−m k −j , for squares in the j-th generation, it holds for K k + 1 ≤ i ≤ N k that Q k i ⊂ Q(c, 5 4 2 −m k −j ) and therefore, for each y ∈ Ω and 0 < r < 2 −2−m k −j such that x ∈ Q(y, r) it holds that Q(y, r) ⊂ 2Q k i . Therefore by (2.20), for almost every x ∈ Q k i and for every 0 < r < 2 −2−m k −j we have Q(y, r) ⊂ 2Q k i and by − Q(y,r) Further for every Q(y, r) x such that r > 2 −2−m k −j we have and using (3.6) we get Therefore − Q(y,r) By taking supremum over all cubes on the left-hand side we obtain that By the Riesz-Herz equivalence (for details see (2.5) we have Now by the Hardy-Littlewood-Polya principle (see (2.4)) we obtain that 3. Proof of Theorem 1.2 Step 1. A grid of 'squares'. Let ε * > 0 be any fixed positive number, we want to find a piece-wise affine homeomorphic approximationf such that The first stage of this is to separate Ω into nested sets Ω k . This separation is the subject of Lemma 2.8. We want to apply Lemma 2.8 and get a grid of squares Q k i but in order to do this we need to choose ε k , δ k > 0 and a set A k . Using the 'ε − δ' continuity of the norm of X (recall that all spaces supporting the Lebesgue Point Property have this property) we choose δ k be a number so small that We find a number T k ≥ 1 such that Then we call A k the union of these sets, i.e. and L 2 (A k ) < δ k 32 . Now we apply Lemma 2.2 with G = Ω K , M = (2 + C (2.7) )T k , ε k = 2 −k ε * L 2 (Ω k ) . This gives us a numberδ k . We require where ϕ X (·) is the fundamental function of X. In each Ω k \ Ω k−1 , Lemma 2.8 gives a grid of squares Q k i which cover most (up to a set of measure δ k ) of Ω k . For the rest of the proof we fix k ∈ N 0 and work on Ω k . Finally at the end of the proof we sum over k.
We apply Lemma 2.9 to slightly alter the squares Q k i and call the resulting quadrilaterals Q k i . We deal with the rest of the set (S k : By Lemma 2.8 point vi) we get that if a i / ∈ B k then also a i / ∈ A k . Then for all i / ∈ B k (Df (a i ) = 0 or T −1 k < |Df (a i )| < T k ) and (J f (a i ) = 0 or J f (a i ) > 4πT −1 k ).
On the other hand by Lemma 2.8 point vi) because Q k i are pair-wise essentially disjoint. On all the quadrilaterals Q k i we have the estimates (2.14) and if i / ∈ B k also the estimate (2.15). We make the following categorisation of the quadrilaterals Q k i . We put Then every quadrilateral Q k i , 1 ≤ i ≤ K k belongs exactly to one of G k , N k , Z k or B k . As calculated in (3.4) (3.5) Finally we cover the rest of Ω k \ K k i=1 Q k i =: S k using Lemma 2.11. By the choice of the grid (i.e. all the squares in the grid in Ω k have the same size) then each square in Ω k has precisely 8 neighbours (where a neighbour is a square that shares at least one side or a vertex). The exception is when the square is at the edge of our grid in Ω k \ G k because these squares have neighbours of half their side length they can have as many as 12 neighbours. Because, for any x ∈ (Q k i ) • and any for almost every x ∈ i Q k i .
Step 2. Defining a piece-wise linear map on each ∂Q.
We define an injective piece-wise linear functionf using Lemma 2.10 for each Q k i , 1 ≤ i ≤ K k we putf (x) = f (x) at each x vertex of Q. Especially we note thatf is linear on each side of each Q k i ∈ G by point v).
Step 3. Definingf , the piece-wise affine approximation of f . In each case Q ∈ G k , Q ∈ N k , Q ∈ Z k or Q ∈ B k we define an injective piece-wise affine extension off from ∂Q. The quadrilateral Q is the union of 2 triangles and (as was proved in Lemma 2.10 point v)) for Q ∈ G k we definef as linear on each side of each of these triangles and this definition is injective. In that casef extends as an affine map onto each triangle. If Q ∈ B k or if Q ∈ Z k then we apply a 2-piece-wise affine 2-bi-Lipschitz mapping Ψ, which maps Q onto Q(0, 2 −m k ) and there we use the extension Theorem 2.4, which gives us g. We definef = g • Ψ on Q. On Q ∈ N k we definef using Theorem 2.6. Notice that in each case we have definedf (see the use of Lemma 2.10 in step 3) on each side of ∂Q so that |D τf | ≤ C|Df (a i )| is constant on each side of ∂Q. (3.7) The last definition that needs to be made in Ω k is the application of Lemma 2.11 to getf on S k = Ω k \ K k i=1 Q k i .
It suffices to combine the estimates from Lemma 2.8 point vii) and point i) from Lemma 2.10 to get that f − f L ∞ ( i Q k i ) < 5ε k . The 'squares' Q k i , i = K k + 1, . . . , N k are even smaller than the squares Q k i for i = 1 . . . K k and so we may assume that the oscillation there has the same bound.
Step 5. Estimating the distance of Df from Df in X. In the following we refer to the 'centre' of the quadrilateral Q as a Q . That is if Q = Q k i then a Q = a k i . We estimate (3.8) By (3.1), (3.5) and Lemma 2.
The sum of the terms is immediately estimated by Lemma 2.8 iv), the finite overlap property (3.6) and (3.3) as follows (3.11) Because every square in S k has side length at most 2 1−m k it holds that N k i=K k +1 2Q k i ⊂ S k + Q(0, 2 −m k ).
Therefore the tube around ∂Ω k which has double the measure of S k is a superset of ∂Ω k + Q(0, 12 · 2 −m k ). Therefore the inclusion Therefore definingf = g on S k by Lemma 2.11 we get, using (2.18), (3.1) and the previous estimate, the following Having estimated the 'f '-terms, we proceed with the 'f '-terms. First we deal with ( Q∈B k χ Q + χ S k )Df X(Ω) . Notice that Q k i ⊂ Q(c, 5 4 2 −m k ) and therefore, for each y ∈ Ω and 0 < r < 2 −2−m k such that x ∈ Q(y, r) it holds that Q(y, r) ⊂ 2Q k i . Therefore, for almost every x ∈ Q k i and for every 0 < r < 2 −2−m k we have Q(y, r) ⊂  We estimate the term Q∈G k χ Q (Df − Df (a Q )) X(Ω) as follows. For each Q ∈ G k we have from Lemma 2.8 point iv) the estimate andf (x) = f (x) at each x, vertex of Q. Therefore on both of the triangles of Q we have |Df − Df (a Q )| < 4ε k . Therefore on all Q ∈ G k we have Df − Df (a Q ) L ∞ (Q) < 4ε k . This means, by (3.3), we can estimate < 4ε k ϕ(L 2 (Ω k )) < 2 −k ε * . (3.14) We estimate the term Q∈Z k χ Q Df X(Ω) as follows. For each Q ∈ Z k we have from Colorrary 2.5, (2.15) that Df L ∞ (Q) < C (2.7) − ∂Q |D τf |dH 1 < C (2.7) C (2.14) ε k − 2Q |Df |dL 2 < C (2.7) C (2.14) ε 2 k < ε k .
(3.15) Now let us estimate the Q∈N k χ Q (Df −Df (a Q )) X(Ω) term. We use Theorem 2.6 to definef on each Q ∈ N k . The extension on each Q has two parts, the W Q part and the Q \ W Q part. Now let us estimate (1 + C (2.7) )L 2 (Q)ε k + ε k L 2 (Q) · 4T k ã ≤ (1 + C (2.7) + 4T k )L 2 (Ω k )ε k <δ k L 2 (Ω k ) by (3.3). Of course however we have that for all Q ∈ N that |Df (a Q )| < T k and by (2.9) that Df L ∞ (Q) ≤ (1 + C (2.7) )T k . Therefore Summing this over k we get that From step 4 and the fact that ε k < 2 −k ε * (see (3.3)) we know that f −f L ∞ (Ω) < ε * .
Step 6. Smoothing piece-wise affine maps. From the above we may assume that we have af satisfying Df − Df X(Ω) < ε 2 and f −f L ∞ (Ω) < ε 2 . Now it suffices to apply Lemma 2.7 to prove Theorem 1.2.
Step 7. Finite-triangulation. The finite-triangulation part of Theorem 1.2 follows immediately from the calculations above and [8,Section 4.2].