Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 30, 2005, 385–392

LINEAR DILATATION AND ABSOLUTE CONTINUITY

P. Koskela and S. Rogovin

University of Jyväskylä, Department of Mathematics and Statistics, P.O. Box 35 FI-40014 University of Jyväskylä, Finland; pkoskela@maths.jyu.fi, sakallun@maths.jyu.fi

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}, n ≥ 2, and f: Ω → Ω′ ⊂ R^{n} a homeomorphism. For x ∈ Ω and 0 < r < dist(x, ∂Ω) we set

\begin{aligned}L_{f}(x, r) & = max\{|f(x) − f(y)| : |x − y| ≤ r\}, \\ l_{f}(x, r) & = min\{|f(x) − f(y)| : |x − y| ≥ r\}. \end{aligned}

The linear lim sup-dilatation of f at x is defined as

H_{f}(x) = lim sup_{r→0} H_{f}(x, r)

where H_{f}(x, r) = L_{f}(x, r)/l_{f}(x, r). Similarly we can define the linear lim inf - dilatation h_{f}(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 H_{f}(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 h_{f}(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 m_{2}({H_{f}(x) > K}) < CK^{−α}, for some α > 3 and K ≥ K_{0} , 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 H_{f} ∈ L_{loc}^{α}(Ω), α > n/(n−1), guarantees the ACL-property. In the above results, it was also assumed that H_{f} be finite outside a set of σ -finite (n − 1)-measure, which seems crucial, see Remark 1.2(b).

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.

In this paper we will show that already H_{f} ∈ L_{loc}^{1}(Ω) is sufficient for the ACLproperty. Before stating our results, let us introduce the following dilatations:

K_{f}(x) = lim sup_{r→0}(\frac{diam(f(B(x, r)))^{n}}{|fB(x, r)|} )^{1/(n−1)},

and

k_{f}(x) = lim inf_{r→0}(\frac{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 H_{f} and h_{f}, respectively. It is also easy to see that K_{f} and k_{f} are Borel functions.

Our main theorem is the following result.

Theorem 1.1. Let f: Ω → Ω′, where Ω, Ω′ ⊂ R^{n} are domains, be a homeomorphism for which k_{f}(x) < ∞ outside a set S of σ -finite (n − 1)-measure, and suppose that k_{f} ∈ L_{loc}^{1}(Ω). Then f belongs to W_{loc}^{1,1}(Ω, R^{n}).

The above theorem gives us the ACL-property, since a continuous W_{loc}^{1,1} - 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^{n−1} → f(]0, (1/e)[×R^{n−1}),

f(x) = (1/ log(1/x_{1}), x_{1} sin(1/x_{1}) + x_{2}, x_{3}, …, x_{n}).

This mapping is a non-ACL homeomorphism of R^{n} which satisfies k_{f}(x) ∈ L_{loc}^{s} 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 k_{f}(z) = 1 almost everywhere, but f is not ACL.

The following corollary summarizes the conclusions obtained from Theorem 1.1 for various distortion functions.

Corollary 1.3. Let f: Ω → Ω′, where Ω, Ω′ ⊂ R^{n} 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_{loc}^{1,1}(Ω, R^{n}).

(1) K_{f}(x) < ∞ outside S and K_{f} ∈ L_{loc}^{1}(Ω),

(2) k_{f}(x) < ∞ outside S and k_{f} ∈ L_{loc}^{1}(Ω),

(3) H_{f}(x) < ∞ outside S and H_{f} ∈ L_{loc}^{1}(Ω),

(4) h_{f}(x) < ∞ outside S and h_{f} ∈ L_{loc}^{n/(n−1)}(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 h_{f} ∈ L_{loc}^{1} could be sufficient. This would follow if it were true that the requirement h_{f}(x) < ∞ outside S and h_{f} ∈ L_{loc}^{1}(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_{loc}^{1,1}(Ω, R^{n}) 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 ⋐ Ω whose sides are parallel to the coordinate axes. Assume that Q = Q_{0} × J_{0} , where Q_{0} is (n − 1)-interval in R^{n−1} , and J_{0} = [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 x_{n} -axis.

Next, for a Borel set A ⊂ Q_{0} , write

Φ(A) := |f(A × [a − d, b + d])| ≤ |f(Q + d)| < ∞,

where d = \frac{1}{10} dist(Q, ∂Ω) and Q + d = \{x ∈ Ω : dist(x, Q) ≤ d\}. Then Φ is a finite Borel measure on Q_{0} , and hence it has a finite derivative Φ′(y) for almost all y ∈ Q_{0} . Denote by E_{0} the set where Φ′ does not exists or is not finite.

Next we consider the set 𝒜 = {I ⊂ J_{0} : 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 ∈ Q_{0} , we know by the Fubini theorem that

∫_{{y}×[a−d,b+d]} k_{f}(z) dz_{n} < ∞.

Denote the set where the above fails by E_{1} . Let us define for every I ⊂ 𝒜 a function g_{I}: Q_{0} → R,

g_{I}(y) = ∫_{{y}×I} k_{f}(z) dz_{n}.

By the Fubini theorem, g_{I} ∈ L^{1}(Q_{0}), and thus for almost every y ∈ Q_{0}

lim_{r→0} ∫−_{B^{n−1}(y,r)}g_{I}(z) dz = g_{I}(y).

Denote by E_{I} the set where this is not true. Now E = E_{0} ∪ E_{1} ∪ (⋃_{I∈𝒜} E_{I}) has measure zero, because it is a countable union of sets of measure zero.

Fix y ∈ Q_{0} \E. We will prove that f is absolutely continuous on the segment {y} × J_{0} which will prove the claim.

Let {I_{j}}_{j=1}^{l}, I_{j} = [a_{j}, b_{j}], be a union of closed intervals on J_{0} , whose interiors are mutually disjoint, and whose endpoints are rational numbers. Since f is continuous, for every j = 1, …, l there is δ_{j} such that

|f(y, a_{j}) − f(x)| < \frac{|f(y, a_{j}) − f(y, b_{j})|}{4} when |(y, a_{j}) − x| < δ_{j}

and

|f(y, b_{j}) − f(x)| < \frac{|f(y, a_{j}) − f(y, b_{j})|}{4} when |(y, b_{j}) − x| < δ_{j}.

Denote δ = min_{j}{δ_{j}}.

Let 0 < r < δ and let ε > 0. For each k = 0, 1, 2, …, write

A_{k} = {x ∈ B^{n−1}(y, r) × ⋃_{j} I_{j} : 2^{k} ≤ k_{f}(x) < 2^{k+1}}.

Then A_{k} is a Borel set,

B^{n−1}(y, r) × ⋃_{j} I_{j} \ S = ⋃_{k} A_{k}

and for every k there exists open U_{k} such that A_{k} ⊂ U_{k} and

|U_{k}| ≤ |A_{k}| + \frac{ε}{2^{2k}} .

Fix k. Now for every x ∈ A_{k} there is r_{x} > 0 such that

(i) 0 < r_{x} < \frac{1}{10} min\{r, d, |a_{j} − b_{j}|\},

(ii) diam(fB_{x})^{n/(n−1)} < 2^{k+1}|fB_{x}|^{1/(n−1)} , and (iii) B_{x} ⊂ U_{k}.

Here B_{x} = B(x, r_{x}).

By the Besicovitch covering theorem we find balls \overline{B}_{1}, \overline{B}_{2}, … from balls \overline{B}(x, r_{x}) so that

B^{n−1}(y, r) × ⋃_{j} I_{j} ⊂ ⋃_{j} \overline{B}_{j} ⊂ B^{n−1}(y, 2r) × [a − d, b + d]

and ∑_{j} χ_{\overline{B}_{j}}(x) ≤ C(n) for every x ∈ R^{n}.

Let us define

ϱ(x) = (∑_{j=1}^{l} |f(y, a_{j}) − f(y, b_{j})|)^{−1} ∑_{i} \frac{diam(fB_{i})}{diam(B_{i})}χ_{2B_{i}}(x).

The function ϱ is a Borel function, because it is a countable sum of (simple) Borel functions. In the following we denote G = ∑_{j=1}^{l} |f(y, a_{j}) − f(y, b_{j})|.

We wish to estimate the volume integral of ϱ. First of all

∫_{B^{n−1}(y,r)×∪_{j}I_{j}} ϱ(x) dx ≥ G^{−1} ∫_{B^{n−1}(y,r)} ∫_{∪_{j}I_{j}} ∑_{B_{i}∩({z}×∪_{j}I_{j})≠∅}

× \frac{diam(fB_{i})}{diam(B_{i})}χ_{2B_{i}}(z, x_{n}) dx_{n} dz.

Notice that

∫_{∪_{j}I_{j}} χ_{2B_{i}}(z, x_{n}) dx_{n} ≥ \frac{diam(B_{i})}{2}

for the balls B_{i} for which B_{i} ∩ ({z} × ⋃_{j} I_{j}) ≠ ∅. Moreover, for almost every z ∈ B^{n−1}(y, r), the sets fB_{i} cover the f({z} × ⋃_{j} I_{j}) up to a countable set, because S has σ -finite (n − 1)-measure (see Theorem 30.16 in [10]). Thus, since r < δ, we have that

(1)  ∑_{B_{i}∩(\{z\}×∪_{j}I_{j})≠∅} diam(fB_{i}) ≥ \frac{G}{4}

for almost every z ∈ B^{n−1}(y, r).

So

∫_{B^{n−1}(y,r)×∪_{j}I_{j}} ϱ(x) dx ≥ C(n)r^{n−1}.

Next we establish an upper bound for this integral. Using the monotone convergence theorem, we obtain the estimate

∫_{B^{n−1}(y,r)×∪_{j}I_{j}} ϱ(x) dx ≤ C(n)G^{−1} ∑_{i} diam(fB_{i})|B_{i}|^{1−(1/n)}

≤ C(n)G^{−1}(∑_{i} |fB_{i}|)^{1/n}

× (∑_{i}(\frac{diam(fB_{i})^{n}}{|fB_{i}|} )^{1/(n−1)}|B_{i}|)^{(n−1)/n}.

The last inequality is obtained by Hölder’s inequality.

For the first term we have that

∑_{j} |fB_{j}| ≤ C(n)|f(B^{n−1}(y, 2r) × [a − d, b + d])| = C(n)Φ(B^{n−1}(y, 2r))

because the overlapping of the balls was bounded. The approximation of the second term is a little bit trickier. First notice that |B_{i}| = (|B_{i} ∩ A_{k}| + |B_{i} \ A_{k}|). Thus

\begin{aligned}∑_{i}(\frac{diam(fB_{i})^{n}}{|fB_{i}|} )^{1/(n−1)}|B_{i}| & ≤ 2 ∑_{k} ∑_{x_{i}∈A_{k}} 2^{k}|B_{i} ∩ A_{k}| + 2 ∑_{k} ∑_{x_{i}∈A_{k}} 2^{k}|B_{i} \ A_{k}| \\ & ≤ ∫_{B^{n−1}(y,r)×∪_{j}I_{j}} k_{f}(x) dx + ε. \end{aligned}

Here x_{i} is the center of the ball B_{i}.

Combining the lower bound and the upper bound and remembering that ε was arbitrary, we obtain the inequality

\begin{aligned}∑_{j=1}^{l} |f(y, a_{j}) − f(y, b_{j})| & ≤ C(n)(\frac{Φ(B^{n−1}(y, 2r))}{(2r)^{n−1}})^{1/n} \\ & × (∫−_{B^{n−1}(y,r)} ∫_{∪_{j}I_{j}} k_{f}(z, x_{n}) dx_{n} dz)^{(n−1)/n}. \end{aligned}

Letting finally r tend to zero, we arrive at

∑_{j=1}^{l} |f(y, a_{j}) − f(y, b_{j})| ≤ C(n)(Φ′(y))^{1/n}(∫_{{y}×∪_{j}I_{j}} k_{f}(x) dx_{n})^{(n−1)/n}.

This estimate holds for rational a_{i}, b_{i}. By continuity, it then holds for all a_{i}, b_{i} ∈ R. Thus f is absolutely continuous on {y} × J_{0} .

To see that partial derivatives, which now exists a.e., are locally integrable, notice that for a.e. x we have

|∂_{i}f_{j}(x)| ≤ lim inf_{r→0} \frac{diam fB(x, r)}{r},

when i, j ∈ {1, …, n}. Since

\frac{diam fB(x, r)}{r} ≤ (\frac{|fB(x, r)|}{r^{n}})^{1/n} \frac{diam fB(x, r)}{|fB(x, r)|^{1/n}},

we conclude that for a.e. x we have

|∂_{i}f_{j}(x)| ≤ μ_{f}(x)^{1/n}k_{f}(x)^{(n−1)/n}, i, j ∈ {1, …, n},

where μ_{f} is the measure derivative of the measure m(A) := |f(A)|. Since the measure derivative and k_{f} are locally integrable, it follows by Hölder’s inequality that the partial derivatives are locally integrable.

Remark 2.1. The assumption that S have σ -finite (n − 1)-measure was only used to guarantee that (1) holds. In fact, this estimate follows even if we only know that the one-dimensional measure of f(L ∩ S) is zero for (almost every) line L parallel to the coordinate axes. Thus, instead of an exceptional set S of σ -finite (n − 1)-measure, we could allow for any exceptional set S with this property.

Proof of Corollary 1.3. First, Part (2) is the previous theorem.

Next, Part (1) is a trivial corollary of (2).

For (3), we first notice that always, for every x, we have

K_{f}(x) ≤ C(n)H_{f}(x)^{n/(n−1)},

and thus K_{f} is finite outside a set of σ -finite (n − 1)-measure. Secondly, since H_{f} is finite almost everywhere, our mapping is differentiable almost everywhere; for this see for example [7]. Now, in the set where f is differentiable and

min_{|e|=1} |Df(x)e| > 0,

it is easy to see that K_{f}(x) ≤ C(n)H_{f}(x). Set

S̃ = S ∪ {x ∈ Ω : f differentiable at x with |Df(x)| = 0}.

Then the one-dimensional measure of f(S̃∩ L) is zero for each line L. The claim follows from Theorem 1.1 and Remark 2.1.

Part (4) follows from Theorem 1.1 and the fact that we always have k_{f} ≤ C(n)h_{f}^{n/(n−1)} .

392 P. Koskela and S. Rogovin

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Received 23 August 2004