Linear Dilatation and Absolute Continuity

Abstract

We show that already the local integrability of the linear dilatation of a homeomorphism guarantees that the homeomorphism is absolutely continuous on almost all lines parallel to the coordinate axes, under the assumption that the linear dilatation be finite outside a set of σ -finite (n − 1) -measure. 1. Background and statement of results Let Ω be a domain in R , n ≥ 2, and f : Ω → Ω ⊂ R a homeomorphism. For x ∈ Ω and 0 < r < dist(x, ∂Ω) we set Lf (x, r) = max{|f(x)− f(y)| : |x − y| ≤ r}, lf (x, r) = min{|f(x)− f(y)| : |x − y| ≥ r}. The linear lim sup-dilatation of f at x is defined as Hf (x) = lim sup r→0 Hf (x, r) where Hf (x, r) = Lf (x, r)/lf(x, r). Similarly we can define the linear lim inf dilatation hf (x) by replacing lim sup with lim inf . A well-known result of Gehring [2], [3] says that if a homeomorphism f has its linear lim sup-dilatation Hf (x) uniformly bounded a.e. in Ω\E , where E has σ finite (n− 1)-measure, then f is a quasiconformal mapping. Similarly it has been showed in [6] that if the linear lim inf -dilatation hf (x) is uniformly bounded a.e. in Ω \E , then f is still quasiconformal. For earlier results, see [4], [5]. In particular, in both cases f is ACL in Ω, which means that f is absolutely continuous on almost every line segment parallel to the coordinate axes in Ω. In [9], Tukia conjectured that the condition m2 ( {Hf (x) > K} ) < CK , for some α > 3 2000 Mathematics Subject Classification: Primary 30C62, 30C65. The second author was supported by the Academy of Finland. Part of the research was done while the second author was visiting at the University of Bern. She would like to thank the mathematics department for its hospitality. 386 P. Koskela and S. Rogovin and K ≥ K0 , is sufficient for the ACL-property of a plane homeomorphism f . This was proven in [1] together with a space analogue. Furthermore, it was later showed in [7] that Hf ∈ L α loc(Ω), α > n/(n−1), guarantees the ACL-property. In the above results, it was also assumed that Hf be finite outside a set of σ -finite (n − 1)-measure, which seems crucial, see Remark 1.2(b). In this paper we will show that already Hf ∈ L 1 loc(Ω) is sufficient for the ACLproperty. Before stating our results, let us introduce the following dilatations: Kf (x) = lim sup r→0 ( diam ( f ( B(x, r) ))n |fB(x, r)| )1/(n−1) , and kf (x) = lim inf r→0 ( diam ( f ( B(x, r) ))n |fB(x, r)| )1/(n−1) , where |A| denotes the Lebesgue measure of a set A . We noticed during our studies that these are more natural (and so more useful) for proving absolute continuity. At the points of differentiability with l ( Df(x) ) = min{|Df(x)e| : |e| = 1} > 0 these dilatations are comparable to Hf and hf , respectively. It is also easy to see that Kf and kf are Borel functions. Our main theorem is the following result. Theorem 1.1. Let f : Ω → Ω , where Ω, Ω ⊂ R are domains, be a homeomorphism for which kf (x) < ∞ outside a set S of σ -finite (n − 1) -measure, and suppose that kf ∈ L 1 loc(Ω) . Then f belongs to W 1,1 loc (Ω,R ) . The above theorem gives us the ACL-property, since a continuous W 1,1 loc mapping is ACL (see Proposition I.1.2 in [8]). Remarks 1.2. (a) To see that Theorem 1.1 is sharp, consider the mapping f : ]0, (1/e)[×R → f(]0, (1/e)[×R), f(x) = ( 1/ log(1/x1), x1 sin(1/x1) + x2, x3, . . . , xn ) . This mapping is a non-ACL homeomorphism of R which satisfies kf (x) ∈ L s loc for any s < 1. (b) The condition of σ -finiteness of S is crucial. For example, if g: [0, 1] → [0, 1] is the Cantor staircase function, then f : ]0, 1[×]0, 1[→]0, 2[×]0, 1[ defined by f(x, y) = ( g(x) + x, y ) is a homeomorphism with kf (z) = 1 almost everywhere, but f is not ACL. The following corollary summarizes the conclusions obtained from Theorem 1.1 for various distortion functions. Linear dilatation and absolute continuity 387 Corollary 1.3. Let f : Ω → Ω , where Ω, Ω ⊂ R are domains, be a homeomorphism and suppose that S has σ -finite (n− 1) -measure. Now each of the conditions below guarantees that f ∈ W 1,1 loc (Ω,R ) . (1) Kf (x) < ∞ outside S and Kf ∈ L 1 loc(Ω) , (2) kf (x) < ∞ outside S and kf ∈ L 1 loc(Ω) , (3) Hf (x) < ∞ outside S and Hf ∈ L 1 loc(Ω) , (4) hf (x) < ∞ outside S and hf ∈ L n/(n−1) loc (Ω) . Remarks 1.4. The mapping discussed in the first part of Remarks 1.2 shows the sharpness of the integrability assumptions in (1), (2), (3). Regarding (4), we do not know if already hf ∈ L 1 loc could be sufficient. This would follow if it were true that the requirement hf (x) < ∞ outside S and hf ∈ L 1 loc(Ω) would guarantee a.e. differentiability. We do not know if this could be the case. In any case, (4) already substantially improves on the known results from [6]. 2. Proofs Proof of Theorem 1.1. To prove that f ∈ W 1,1 loc (Ω,R ) we first show that f is ACL. After that we show the local integrability of the partial derivatives, whose existence is guaranteed by the ACL-property. Pick a closed cube Q b Ω whose sides are parallel to the coordinate axes. Assume that Q = Q0 × J0 , where Q0 is (n − 1)-interval in R n−1 , and J0 = [a, b] ⊂ R . In order to show that f is ACL it suffices to show that f is absolutely continuous on almost every line segment in Q , parallel to the coordinate axes, and by symmetry it is sufficient to consider segments parallel to the xn -axis. Next, for a Borel set A ⊂ Q0 , write Φ(A) := |f(A× [a − d, b + d])| ≤ |f(Q + d)| < ∞, where d = 1 10 dist(Q, ∂Ω) and Q + d = {x ∈ Ω : dist(x, Q) ≤ d} . Then Φ is a finite Borel measure on Q0 , and hence it has a finite derivative Φ (y) for almost all y ∈ Q0 . Denote by E0 the set where Φ ′ does not exists or is not finite. Next we consider the set A = {I ⊂ J0 : I is a finite union of closed intervals, whose interiors are mutually disjoint and whose end points are rational} . This set is countable: (i) If we take all the intervals whose end points are rational, there is just a countable number of intervals. (ii) If we take all the pairs of intervals, whose endpoints are rational, there is just a countable number of pairs. (iii) If we take all triples of the same type we again have a countable numbers of triples, etc. Thus A is a countable union of countable sets and so countable. Now, for almost every y ∈ Q0 , we know by the Fubini theorem that ∫ {y}×[a−d,b+d] kf (z) dzn < ∞. 388 P. Koskela and S. Rogovin Denote the set where the above fails by E1 . Let us define for every I ⊂ A a function gI : Q0 → R , gI(y) = ∫ {y}×I kf (z) dzn. By the Fubini theorem, gI ∈ L (Q0), and thus for almost every y ∈ Q0 lim r→0 ∫ − Bn−1(y,r) gI(z) dz = gI(y). Denote by EI the set where this is not true. Now E = E0 ∪E1 ∪ (⋃ I∈A EI ) has measure zero, because it is a countable union of sets of measure zero. Fix y ∈ Q0 \E . We will prove that f is absolutely continuous on the segment {y} × J0 which will prove the claim. Let {Ij} l j=1 , Ij = [aj , bj] , be a union of closed intervals on J0 , whose interiors are mutually disjoint, and whose endpoints are rational numbers. Since f is continuous, for every j = 1, . . . , l there is δj such that |f(y, aj) − f(x)| < |f(y, aj) − f(y, bj)| 4 when |(y, aj) − x| < δj and |f(y, bj) − f(x)| < |f(y, aj) − f(y, bj)| 4 when |(y, bj) − x| < δj . Denote δ = minj{δj} . Let 0 < r < δ and let ε > 0. For each k = 0, 1, 2, . . ., write Ak = { x ∈ B(y, r)× ⋃ j Ij : 2 k ≤ kf (x) < 2 k+1 } . Then Ak is a Borel set, B(y, r)× ⋃

Cite this paper

@inproceedings{KoskelaLinearDA, title={Linear Dilatation and Absolute Continuity}, author={Pekka Koskela and Sari Rogovin} }