AHLFORS DAVID REGULAR SETS AND BILIPSCHITZ MAPS

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1 Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 34, 2009, AHLFORS DAVID REGULAR SETS AND BILIPSCHITZ MAPS Pertti Mattila Pirjo Saaranen University of Helsinki, Department of Mathematics Statistics P.O. Box 68, FI University of Helsinki, Finl; Haaga-Helia University of Applied Sciences Hietakummuntie 1 A, FI Helsinki, Finl; [email protected] Abstract. Given two Ahlfors David regular sets in metric spaces, we study the question whether one of them has a subset bilipschitz equivalent with the other. 1. Introduction In this paper we shall study Ahlfors David regular subsets of metric spaces. Throughout (X, d) (Y, d) will be metric spaces. For E, F X x X we shall denote by d(e) the diameter of E, by d(e, F ) the distance between E F, by d(x, E) the distance from x to E. The closed ball with center x radius r is denoted by B(x, r) Definition. Let E X 0 < s <. We say that E is s-regular if it is closed if there exists a Borel (outer) measure µ on X a constant C E, 1 C E <, such that µ(x \ E) = 0 r s µ(b(x, r)) C E r s for all x E, 0 < r d(e), r <. Observe that this implies that the right h inequality holds for all x E, r > 0, µ(b(x, r)) 2 s C E r s for all x X, r > 0. We would get an equivalent definition (up to the value of C E ), if we would use the restriction of the s-dimensional Hausdorff measure on E, with r s on the left h side replaced by r s /C E. When we shall speak about a regular set E, µ will always st for a measure as above. We remark that closed bounded subsets of regular sets are compact, see Corollary 5.2 in [DS2]. Self similar subsets of R n satisfying the open set condition are stard examples of regular sets, see [H]. A map f : X Y is said to be bilipschitz if it is onto there is a positive number L, called a bilipschitz constant of f, such that d(x, y)/l d(f(x), f(y)) Ld(x, y) for all x, y X. The smallest such L is denoted by bilip(f). Evidently any bilipschitz image of an s-regular set is s-regular. But two regular sets of the same dimension s need not be bilipschitz equivalent. This is so even for very simple Cantor sets in R, see [FM], 2000 Mathematics Subject Classification: Primary 28A75. Key words: Regular set, bilipschitz mapping. Part of these results were included in the licentiate thesis of Pirjo Saaranen at the University of Jyväskylä in Both authors have been supported by the Academy of Finl.

2 488 Pertti Mattila Pirjo Saaranen [RRX] [RRY] for results on the bilipschitz equivalence of such Cantor sets, for [DS2] for extensive analysis of bilipschitz invariance properties of fractal type sets. The main content of this paper is devoted to the following question: suppose E is s-regular F is t-regular. If s < t, does F have a subset which is bilipschitz equivalent to E? In this generality the answer is obviously no due to topological reasons; E could be connected F totally disconnected. We shall prove in Theorem 3.1 that the answer is yes for any 0 < s < t if E is a stard s-dimensional Cantor set in some R n with s < n. We shall also prove in Theorem 3.3 that the answer is always yes if s < 1. In Section 4 we show that if E F as above are subsets of R n s is sufficiently small, then a bilipschitz map f with f(e) F can be defined in the whole of R n. We don t know if this holds always when 0 < s < 1. In the last section of the paper we shall discuss sub- supersets of regular sets. It follows from Theorem 3.1 that an s-regular set contains a t-regular subset for any 0 < t < s. In the other direction we shall show that if E X is s-regular X is u-regular, then for any s < t < u there is a t-regular set F such that E F X. On the other h there are rather nice sets which do not contain any regular subsets: we shall construct a compact subset of R with positive Lebesgue measure which does not contain any s-regular subset for any s > 0. Regular sets in connection of various topics of analysis are discussed for example in [DS1] [JW]. 2. Some lemmas on regular sets In this section we shall prove some simple lemmas on regular sets Lemma. Let 0 < s < let E X be s-regular. For every 0 < r < R d(e), R <, p E there exist disjoint closed balls B(x i, r), i = 1,..., m, such that x i E B(p, R), (5 s C E ) 1 (R/r) s m 2 s C E (R/r) s E B(p, R) m B(x i, 5r). Proof. By a stard covering theorem, see, e.g., Theorem 2.1 in [M], we can find disjoint balls B(x i, r), i = 1, 2,..., such that x i E B(p, R) the balls B(x i, 5r) cover E B(p, R). There are only finitely many, say m, of these balls, since the disjoint sets B(x i, r) have all µ measure at least r s, they are contained in B(p, 2R) which has measure at most C E (2R) s. More precisely, we have m mr s µ(b(x i, r)) µ(b(p, 2R)) C E (2R) s, whence m 2 s C E (R/r) s, mc E 5 s r s whence m (5 s C E ) 1 (R/r) s. m µ(b(x i, 5r)) µ(b(p, R)) R s, For less than one-dimensional sets we can get more information:

3 Ahlfors David regular sets bilipschitz maps Lemma. Let 0 < s < 1, C 1, R > 0, let E X be closed bounded let µ be a Borel measure on X such that µ(x \ E) = 0 that µ(b(x, r)) Cr s for all x E, r > 0, µ(b(x, r)) r s for all x E, 0 < r < R. Let D = (3C2 s ) 1/(1 s) + 1. For every 0 < r < R/(2D) there exist disjoint closed balls B(x i, r), i = 1,..., m, positive numbers ρ i, r ρ i Dr, such that m Cd(E) s /r s, x i E, x j B(x i, ρ i ) for i < j, E m B(x i, ρ i ) E B(x i, ρ i + r) \ B(x i, ρ i ) =. Proof. Let x 1 E. Denote A 0 = B(x 1, r), A i = B(x 1, (i + 1)r) \ B(x 1, ir), i = 1, 2,.... If E A 1 =, denote ρ 1 = r. Otherwise, let l be the largest positive integer such that 2lr < R E A i for i = 1,..., l, say y i E A i. Then B(y i, r) A i 1 A i A i+1 B(x 1, 2lr). Therefore lr s l µ(b(y i, r)) l µ(a i 1 A i A i+1 ) 3µ(B(x 1, 2lr)) 3C2 s l s r s, whence l 1 s 3C2 s, since s < 1, l (3C2 s ) 1/(1 s) = D 1. As 2(l + 1) 2D < R/r, we conclude that E A l+1 = by the maximality of l. Let ρ 1 = (l + 1)r. Then r ρ 1 Dr E B(x 1, ρ 1 + r) \ B(x 1, ρ 1 ) =. Let x 2 E \ B(x 1, ρ 1 ) = E \ B(x 1, ρ 1 + r). Then the balls B(x 1, r) B(x 2, r) are disjoint. Repeating the same argument as above with x 1 replaced by x 2 we find ρ 2 such that r ρ 2 Dr E B(x 2, ρ 2 + r) \ B(x 2, ρ 2 ) =. After k 1 steps we choose x k E \ k 1 B(x i, ρ i ), if this set is non-empty. As in the proof of Lemma 2.1 this process ends after some m steps when E is covered by the balls B(x i, ρ i ), i = 1,..., m. Also, as before, m satisfies the required estimate m Cd(E) s /r s. The following lemma will be needed to get bilipschitz maps in the whole R n Lemma. Let C 1 λ 9. There are positive numbers s 0 = s 0 (C, λ), 0 < s 0 < 1, D = D(C, λ) > 1, depending only on C λ, with the following property. Let 0 < s < s 0, let E X be closed bounded, let R > 0 let µ be a Borel measure on X such that µ(x \ E) = 0 that µ(b(x, r)) Cr s for all x E, r > 0, µ(b(x, r)) r s for all x E, 0 < r < R.

4 490 Pertti Mattila Pirjo Saaranen For every 0 < r < R/D there exist disjoint closed balls B(x i, λρ i /3), i = 1,..., m, such that x i E, r ρ i Dr, m Cd(E) s /r s, m E B(x i, ρ i ) E B(x i, λρ i ) \ B(x i, ρ i ) =. Proof. The function s (1 3Cλ 2s (λ s 1)) 1/s is positive increasing in some interval (0, s 1 ), so it is bounded in some interval (0, s 0 ). We choose s 0 D so that λ(1 3Cλ 2s (λ s 1)) 1/s D for 0 < s < s 0. Set c = log D/ log λ. Let x E denote A 0 = B(x, r), A i = B(x, λ i r) \ B(x, λ i 1 r), i = 1, 2,.... If E A 1 =, denote r(x) = r. Otherwise, let l be the largest positive integer such that l c that E A i for 1 = 1,..., l. Then for i = 1,..., l there is y i E A i with B(y i, λ i 2 r) A i 1 A i A i+1. By the choice of c, λ l 2 r < Dr < R. Hence r s λ s λsl 1 λ s 1 = rs This gives whence l λ s(i 2) l µ(b(y i, λ i 2 r)) 3µ(E B(x, λ l+1 r)) 3Cλ s(l+1) r s. (1 3Cλ 2s (λ s 1))λ sl 1, l µ(a i 1 A i A i+1 ) λ l+1 λ(1 3Cλ 2s (λ s 1)) 1/s D. Thus l + 1 c we conclude that E A l+1 =. Let r(x) = λ l r. We have now shown that for any x E there is r(x), r r(x) Dr, such that E B(x, λr(x)) \ B(x, r(x)) =. Let M 1 = sup{r(x) : x E}. Choose x 1 E with r(x) > M 1 /2, then inductively j x j+1 E \ B(x i, r(x i )) with r(x j+1 ) > M 1 /2 as long as possible. Thus we get points x i E radii r(x i ), r r(x i ) Dr, for i = 1,..., k 1 such that r(x i )/2 r(x j ) 2r(x i ), x j B(x i, r(x i )) for i < j, k 1 {x E : r(x) > M 1 /2} B(x i, r(x i )). If for some l = 1, 2,... the points x 1,..., x kl have been selected there is some x E \ k l B(x i, r(x i )), let k l M l+1 = sup{r(x) : x E \ B(x i, r(x i ))},

5 Ahlfors David regular sets bilipschitz maps 491 choose x kl+1 E \ k l B(x i, r(x i )) with r(x kl+1 ) > M l+1 /2, so on. This process will end for some l = p. Thus we get points x 1,..., x m E, m = k p, such that, with ρ i = r(x i ), r ρ i Dr, for i < j, x j B(x i, ρ i ) r j 2ρ i, E m B(x i, ρ i ) E B(x i, λρ i ) \ B(x i, ρ i ) =. To show that the balls B(x i, λρ i /3) are disjoint, let i < j. Then ρ j 2ρ i x j E (R n \ B(x i, ρ i )) = E (R n \ B(x i, λρ i )). So d(x i, x j ) > λρ i (λ/3)(ρ i + ρ j ) λρ i < d(x i, x j ), which implies that B(x i, λρ i /3) B(x j, λρ j /3) =. The required estimate m Cd(E) s /r s follows as before. 3. Bilipschitz maps In this section we begin to prove the bilipschitz equivalences mentioned in the introduction. It is easy to get explicit bounds for the bilipschitz constants of the maps from the proofs. In Theorem 3.1 bilip(f) is bounded by a constant depending only on s, t, n C E. In Theorems , if C E, C F d(e)/d(f ) (interpreted as 0 if F is unbounded) are all C, then bilip(f) L where L depends only on s, t C, also on n in Theorem 4.2. If E F are bounded, this dependence on the diameters is seen by first observing that we may assume that d(f ) d(e); otherwise F can be replaced in the proofs by F B(p, d(e)/2) for any p F. Secondly, changing the metrics to d E (x, y) = d(x, y)/d(e) d F (x, y) = d(x, y)/d(f ), we have d(e) = d(f ) = 1, the regularity constants don t change a bilipschitz constant L changes to Ld(E)/d(F ). For any 0 < t < n we shall define some stard t-dimensional Cantor sets in R n. Define 0 < d < 1/2 by 2 n d t = 1. Let Q R n be a closed cube of side-length a. Let Q 1,..., Q 2 n Q be the closed cubes of side-length da in the corners of Q. Continue this process. Then C(t, a) is defined as C(t, a) = Q i1...i k, i 1...i k k=1 where i j = 1,... 2 n each Q i1...i k is a closed cube of sidelength d k a such that Q i1...i k i, i = 1,..., 2 n, are contained in the corners of Q i1...i k. It is well known easy to prove that C(t, a) is t-regular, it is also a particular case of a self similar set satisfying the open set condition as considered in [H] Theorem. Let E X be a bounded s-regular set 0 < t < s. Then there is a t-regular subset F of E a bilipschitz map f : F C(t, d(e)) where C(t, d(e)) is a Cantor subset of R n with t < n as above. Moreover, C F C where C depends only s, t, n C E. Proof. We may assume that d(e) = 1. Choose a sufficiently large integer N so that denoting d = 2 Nn/t, i.e., 2 Nn d t = 1, we have d < 1/3 d s t < (15 s C E ) 1 =: c. Then we can write C(t, 1) as C(t, 1) = Q i1...i k k=1 i 1...i k

6 492 Pertti Mattila Pirjo Saaranen where each Q i1...i k, 1 i j 2 Nn, is a closed cube of side-length d k such that Q i1...i k i k+1 Q i1...i k. By Lemma 2.1 we can find disjoint balls B(x i, 3d), x i E, i = 1,..., m, such that m cd s > d t = 2 Nn. Now we keep the first 2 Nn points x i forget about the others. Repeating this argument with E replaced by E B(x i, d) so on, we can choose points x i1...i k i k+1 E B(x i1...i k, d k ), 1 i j 2 Nn, such that the balls B(x i1...i k i, 3d k+1 ), i = 1,..., 2 Nn, are disjoint contained in the ball B(x i1...i k, 3d k ). Then for 1 l < k, k 1 k 1 (3.2) d(x i1...i l, x i1...i k ) d(x i1...i j, x i1...i j+1 ) d j < 2d l as d < 1/2. Denote F = B(x i1...i k, 3d k ). k=1 i 1...i k Then F E. Let y i1...i k be the center of Q i1...i k denote Define the maps j=l j=l F k = {x i1...i k : i j = 1,..., 2 Nn, j = 1,..., k} C k = {y i1...i k : i j = 1,..., 2 Nn, j = 1,..., k}. f k : F k C k by f(x i1...i k ) = y i1...i k. We check now that f k is bilipschitz with a constant depending only on s, t, n C E. Let x = x i1...i k, x = x j1...j k F k with x x. Let l 1 be such that i 1 = j 1,..., i l = j l i l+1 j l+1 ; if i 1 j 1 the argument is similar. Then by (3.2) x B(x i1...i l i l+1, 2d l+1 ) B(x i1...i l, 2d l ) x B(x j1...j l j l+1, 2d l+1 ) B(x i1...i l, 2d l ). Since the balls B(x i1...i l i l+1, 3d l+1 ) B(x j1...j l j l+1, 3d l+1 ) are disjoint, we get that d l+1 d(x, x ) 4d l. Letting y = y i1...i k y = y j1...j k we see from the construction of C(t, 1) that (1 2d)d l y y nd l. Hence d(f k (x), f k (x )) = y y ( n/d)d(x, x ) d(f k (x), f k (x )) = y y ((1 2d)/4)d(x, x ). Denote L = max{ n/d, 4/(1 2d)}. If x F there is a unique sequence (i 1, i 2,... ) such that x B(x i1...i k, 3d k ) for all k = 1, 2,.... Let y C(t, 1) be the point for which y Q i1...i k for all k = 1, 2,.... Then y = lim k y i1...i k = lim k f k (x i1...i k ). We define the map f : F C(t, 1) by setting f(x) = y. If also x = lim k x j1...j k y = lim k y j1...j k we have d(f(x), f(x )) = lim k d(f k (x i1...i k ), f k (x j1...j k )) lim k Ld(x i1...i k, x j1...j k ) = Ld(x, x ) similarly d(f(x), f(x )) d(x, x )/L. Obviously, f(f ) = C(t, 1). The last statement, C F C, of the theorem follows immediately from the fact that L depends only s, t, n C E. Next we turn to study less than one-dimensional sets.

7 Ahlfors David regular sets bilipschitz maps Theorem. Let E X be s-regular F Y t-regular with 0 < s < 1 s < t. Suppose that either E is bounded or both E F are unbounded. Then there is a bilipschitz map f : E f(e) F. Proof. We shall first consider the case where both E F are bounded. By the remarks in the beginning of this section, we then may assume that d(e) = d(f ) = 1. Let D = (3C E 2 s ) 1/(1 s) + 1. Choose d so small that 0 < d t s < (2 s 15 t D s C E C F ) 1 2Dd < 1. We shall show that there exist s-regular sets E i1...i k, points x i1...i k E, y i1...i k F radii ρ i1...i k where 1 i j m i0...i j 1, j = 1,..., k, with m i0...i j 1 C E 2 s D s /d s, i 0 = 0, such that for all k = 1, 2,..., E = i 1...i k E i1...i k, E i1...i k i k+1 E i1...i k, d k ρ i1...i k Dd k, x i1...i k E i1...i k B(x i1...i k, ρ i1...i k ), E B(x i1...i k, ρ i1...i k + d k ) \ B(x i1...i k, ρ i1...i k ) =, d(e i1...i k, E j1...j k ) d k if i k j k, y i1...i k i k+1 F B(y i1...i k, d k ), B(y i1...i k i k+1, 2d k+1 ) B(y i1...i k, 2d k ), B(y i1...i k, 3d k ) B(y j1...j k, 3d k ) = if i k j k. By Lemma 2.2 we find x i E ρ i, d ρ i Dd, with i = 1,..., m 0, m 0 C E /d s, such that the balls B(x i, d) are disjoint, x j B(x i, ρ i ) for i < j, m 0 E B(x i, ρ i ) E B(x i, ρ i + d) \ B(x i, ρ i ) =. By Lemma 2.1 we find y i F with i = 1,..., n 0, n 0 (15 t C F d t ) 1 C E /d s m 0 such that the balls B(y i, 3d) are disjoint. We define E 1 = E B(x 1, ρ 1 ) E i = E B(x i, ρ i ) \ i 1 j=1 E j for i 2. Then the required properties for k = 1 are readily checked. Suppose then that for some k 1, E i1...i k, x i1...i k E, y i1...i k F ρ i1...i k have been found with the asserted properties. Fix i 1... i k. We shall apply Lemma 2.2 with E = E i1...i k, R = d k, r = d k+1 C = C E, recall that 2Dd < 1. Since d(e i1...i k, E \ E i1...i k ) d k, we have E B(x, r) = E i1...i k B(x, r) for x E i1...i k 0 < r < d k, so this is possible. Thus we obtain x i1...i k i E i1...i k ρ i1...i k i, i = 1,..., m i0...i k,

8 494 Pertti Mattila Pirjo Saaranen such that m i0...i k C E d(e i1...i k ) s /d (k+1)s C E 2 s D s /d s, the balls B(x i1...i k i, d k+1 ) are disjoint, x i1...i k j B(x i1...i k i, ρ i1 ki) for i < j, Define d k+1 ρ i1...i k i Dd k+1, E i1...i k m i0...i k B(x i1...i k i, ρ i1...i k i) E B(x i1...i k i, ρ i1...i k i + d k+1 ) \ B(x i1...i k i, ρ i1...i k i) =. E i1...i k 1 = E i1...i k B(x i1...i k 1, ρ i1...i k 1) E i1...i k i = E i1...i k B(x i1...i k i, ρ i1...i k i) \ i 1 j=1 E i1...i k j for i 2. Applying Lemma 2.1 we find points y i1...i k i F B(y i1...i k, d k ), i = 1,..., n i0...i k, with n i0...i k (15 t C F d t ) 1 C E 2 s D s /d s m i0...i k such that the balls B(y i1...i k i, 3d k+1 ), i = 1,..., n i0...i k, are disjoint. Then the required properties are easily checked. Set A k = {x i1...i k : i j = 1,..., m i0...i j 1, j = 1,..., k} Define the maps B k = {y i1...i k : i j = 1,..., m i0...i j 1, j = 1,..., k}. f k : A k B k by f(x i1...i k ) = y i1...i k. We check now that f k is bilipschitz with a constant depending only on s, t, C E C F. Let x = x i1...i k, x = x j1...j k A k with x x. Let l 1 be such that i 1 = j 1,..., i l = j l i l+1 j l+1 ; if i 1 j 1 the argument is similar. Then, as in (3.2) in the proof of Theorem 3.1, x E i1...i l i l+1 B(x i1...i l, 2Dd l ) x E j1...j l j l+1 B(x i1...i l, 2Dd l ). Since d(e i1...i l i l+1, E j1...j l j l+1 ) d l+1, we get that d l+1 d(x, x ) 4Dd l. Letting y = y i1...i k y = y j1...j k, we have y B(y i1...i l i l+1, 2d l+1 ) B(y i1...i l, 2d l ) y B(y j1...j l j l+1, 2d l+1 ) B(y i1...i l, 2d l ). Hence, as B(y i1...i l i l+1, 3d l+1 ) B(y j1...j l j l+1, 3d l+1 ) =, d l+1 d(y, y ) 4d l, d(f k (x), f k (x )) = d(y, y ) (4/d)d(x, x ) d(f k (x), f k (x )) = d(y, y ) (d/(4d))d(x, x ). Denote L = 4D/d > 4/d. As in the proof of Theorem 3.1 we define the map f : E f(e) F by f(x) = lim k f k (x i1...i k ) when x = lim k x i1...i k. Then bilip(f) L. If E is bounded F unbounded, the same proof works with F replaced by F B(p, 1) for any p F. Suppose E F are unbounded, let p E. Using the proof of Lemma 2.2 we find R k, (2D) k R k D(2D) k, k = 1, 2,..., such that E B(p, R k + (2D) k ) \ B(p, R k ) =.

9 Ahlfors David regular sets bilipschitz maps 495 Let E k = E B(p, R k ). We check that E k is s-regular with C Ek (2D) s C E. To see this, let x E k 0 < r d(e k ) (2D) k+1. If r (2D) k, then E k B(x, r) = E B(x, r), so µ(e k B(x, r)) r s. If r > (2D) k, we have µ(e k B(x, r)) (2D) ks (2D) s r s. These facts imply that C Ek (2D) s C E. Since the sets E k are bounded we can find bilipschitz maps f k : E k f(e k ) F with bilip(f k ) L where L depends only on s, t, C E C F. Using Arzela Ascoli theorem we can extract a subsequence (f ki ) such that the sequence (f ki ) ki k converges on E k for every k = 1, 2,.... Then f = lim i f ki : E f(e) F is bilipschitz with bilip(f) L. 4. Mappings in R n In this section we prove for small dimensional sets in R n that we can find bilipschitz mappings of the whole R n. The following lemma may be well known, but we have not found a suitable reference in literature Lemma. Let 0 < δ < c(n), where c(n) < 1/2 is a positive constant depending only on n determined later. Let p, q R n R > 0. For i = 1,..., m let δr r i R/3 x i B(p, R) y i B(q, R) with B(x i, 3r i ) B(x j, 3r j ) = B(y i, 3r i ) B(y j, 3r j ) = for i j. Then there is a bilipschitz map f : R n R n such that f(x) = x p+q for x R n \B(p, 2R) f(x) = x x i +y i for x B(x i, r i ). Moreover, bilip(f) L where L depends only on n δ. Proof. We may assume that p = q = 0 R = 1. Let ε = δ 2n+3. It is enough to construct a bilipschitz map f : R n R n with bilip(f) L, L depending only on n δ, such that f(x) = x for x > 3 n f(x) = x x i +y i for x B(x i, ε). To see this, consider bilipschitz maps g, h: R n R n with bilipschitz constants depending only on n δ such that g(x) = (ε/r i )(x x i ) + x i for x B(x i, r i ), g(x) = x for x B(0, 3/2)\ m B(x i, 2r i ), h(y) = (ε/r i )(y y i )+y i for y B(y i, r i ), h(y) = y for y B(0, 3/2) \ m B(y i, 2r i ), g(x) = h(x) for x > 2 g(b(0, 2)) = h(b(0, 2)) = B(0, 3 n). Then h 1 f g has the required properties. For the rest of the proof we assume that n 2, for n = 1 a much simpler argument works. Denote Q = [ 2, 2] n 1. Let a, b B(0, 1) R n 1. For v B(a, ε) denote by v the single point in Q {t(v a) + a : t 1}. Let g(a, b): Q Q be the bilipschitz map such that g(a, b)(x) = x a + b for x B(a, ε) for v B(a, ε) g(a, b) maps the line segment [v, v ], affinely onto the line segment [v a + b, v ]. Then g(a, b)(x) = x for x Q g(a, a) is the identity map. Moreover, g(a, b) has a bilipschitz constant depending only on n. Now we show that there exists a unit vector θ S n 1 such that θ (x i x j ) > 5ε θ (y i y j ) > 5ε for i j. To see this, let σ denote the surface measure on S n 1. We have by some simple geometry (or one can consult [M], Lemma 3.11) σ({θ S n 1 : θ (x i x j ) 5ε}) C 1 (n) x i x j 1 ε C 1 (n)δ 2n+2, similarly for y i, y j. There are less than C 2 (n)δ 2n pairs (x i, x j ) (y i, y j ), whence σ({θ S n 1 : θ (x i x j ) 5ε or θ (y i y j ) 5ε for some i j}) < δ,

10 496 Pertti Mattila Pirjo Saaranen if C 1 (n)c 2 (n)δ < 1, which we have taking c(n) (C 1 (n)c 2 (n)) 1 in the statement of the theorem. Taking also c(n) σ(s n 1 ) our θ exists. We may assume that θ = (0,..., 0, 1). Let t i u i, i = 1,..., m, be the n th coordinates of x i y i, respectively, let t 0 = u 0 = 2, t m+1 = u m+1 = 2. We may assume that t i < t i+1 u i < u i+1 for i = 0,..., m. Then t i t j > 5ε u i u j > 5ε for i j, i, j = 0,..., m + 1. For x = (x 1,..., x n ) R n, let x = (x 1,..., x n 1 ). Let Q 0 = [ 2, 2] n for i = 1,..., m, R i = {x Q 0 : x n t i ε}, S i = {y Q 0 : y n u i ε}. We shall define f in Q 0 with the help of the maps g(a, b) in such a way that it maps R i onto S i translating B(x i, ε) onto B(y i, ε). Between R i R i+1 f is defined by simple homotopies changing f R i to f R i+1, similarly in Q 0 below R 1 above R m. Finally f can be extended from Q 0 to all of R n rather trivially. We do this now more precisely. Let x Q 0 1 i m + 1. We set f(x) = (g( x i, ỹ i )( x), x n t i + u i ) if x n t i ε i m, f(x) = (g((2ε x n t i )/ε) x i, (2ε x n t i )/ε)ỹ i )( x), x n + u i t i ) if ε x n t i 2ε i m, f(x) = ( x, xn t i 1 2ε t i t i 1 4ε (u i 2ε) + t i 2ε x n t i t i 1 4ε (u i 1 + 2ε)) if t i 1 + 2ε x n t i 2ε, f(x) = x if 2 x n 2 + 2ε or 2 2ε x n 2. Then f : Q 0 Q 0 is bilipschitz with a constant depending only on n δ, f(x) = x x i + y i for x B(x i, ε), f(x) = x for x Q 0 with x n = 2 or x n = 2, at the other parts of the boundary of Q 0 f is of the form f(x) = ( x, φ(x n )) where φ: [ 2, 2] [ 2, 2] is strictly increasing piecewise affine. It is an easy matter to extend f to a bilipschitz mapping of R n with a bilipschitz constant depending only on n δ with f(x) = x for x R n \ B(0, 3 n). For example, setting x = max{ x 1,..., x n 1 }, we can take f(x) = ( x, (3 x )φ(x n ) + ( x 2)x n ) when 2 x 3 x n 2, f(x) = x when x > 3 or x n > Theorem. Let C 1 let s 0 = s 0 (C, 18), 0 < s 0 < 1/6, be the constant of Lemma 2.3. Let 0 < s < s 0 s < t < n, let E R n be s-regular F R n t-regular with C E, C F C. Suppose that either E is bounded or both E F are unbounded. Then there is a bilipschitz map f : R n R n such that f(e) F. Proof. We assume that E F are bounded. The remaining case can be dealt with as at the end of the proof of Theorem 3.3. We can then assume that E, F B(0, 1) with d(e) = d(f ) = 1/2. Let c(n) D = D(C, 18) be as in Lemma 2.3, choose d such that d < c(n), 12Dd < 1 0 < d t s < (2 s 60 t C E C F D t ) 1.

11 Ahlfors David regular sets bilipschitz maps 497 By Lemma 2.3 we find x i E ρ i, d ρ i Dd, with i = 1,..., m 0, m 0 C E /d s, such that the balls B(x i, 6ρ i ) are disjoint, m 0 E B(x i, ρ i ) E B(x i, 18ρ i ) \ B(x i, ρ i ) =. By Lemma 2.1 we find y i F with i = 1,..., n 0, n 0 (5 t C F ) 1 (1/(12Dd)) t C E /d s m 0 such that the balls B(y i, 6Dd), j = 1,..., n 0, are disjoint. Next applying Lemma 2.3 with E replaced by E B(x i, ρ i ), R = d, r = d 2 C = C E, we find for every i = 1,..., m 0, x ij E B(x i, ρ i ) ρ ij, d 2 ρ ij Dd 2, with j = 1,..., m i, m i C E d(e B(x i, ρ i )) s /d 2s C E 2 s D s /d s, such that the balls B(x ij, 6ρ ij ) are disjoint, E B(x i, ρ i ) m i j=1 B(x ij, ρ ij ) E B(x ij, 18ρ ij ) \ B(x ij, ρ ij ) =, by Lemma 2.1 we find y ij F B(y i, d), j = 1,..., n i, n i (5 t C F ) 1 (d/(6dd 2 )) t 2 s C E /d s m i such that the balls B(y ij, 6Dd 2 ) are disjoint. Continuing this we find for all k = 1, 2,..., x i1...i k, ρ i1...i k y i1...i k such that for all i j = m i0...i j 1, j = 1,..., k, k = 1, 2,..., with i 0 = 0, E i 1...i k B(x i1...i k, ρ i1,...,i k ), B(x i1...i k, 6ρ i1...i k ) B(x j1...j k, 6ρ i1...i k ) = if i k j k, x i1...i k i k+1 E B(x i1...i k, ρ i1...i k ), d k ρ i1...i k Dd k, B(x i1...i k i k+1, 4ρ i1...i k i k+1 ) B(x i1...i k, 2ρ i1...i k ), E B(x i1...i k, 18ρ i1...i k ) \ B(x i1...i k, ρ i1...i k ) =, y i1...i k i k+1 F B(y i1...i k, d k ), B(y i1...i k, 6Dd k ) B(y j1...j k, 6Dd k ) = if i k j k. Using Lemma 4.1 we find a bilipschitz map f 1 : R n R n such that f 1 (x) = x for x > 2 f 1 (x) = x x i + y i for x B(x i, 2ρ i ), bilip(f) L where L depends only on s, t, n C. Let B k = i 1...i k B(x i1...i k, 2ρ i1,...,i k ). Then B k+1 B k for all k E = k=1 B k. We use Lemma 4.1 to define inductively f k : R n R n such that f k+1 (x) = f k (x) for x R n \ Bk o, where Bo k is the interior of B k, f k+1 B(x i1...i k, 2ρ i1,...,i k ) is L-bilipschitz f k+1 (x) = x x i1,...,i k+1 + y i1,...,i k+1 for x B(x i1...i k+1, 2ρ i1,...,i k+1 ). We check now by induction that (4.3) x y /L f k (x) f k (y) L x y for all x, y R n.

12 498 Pertti Mattila Pirjo Saaranen For k = 1 this was already stated. Suppose this is true for k 1 for some k 2 let x, y R n. If x, y R n \ B o k, (4.3) follows from the definition of f k the induction hypothesis. If x, y B(x i1...i k, 2ρ i1,...,i k ) for some i 1... i k, then (4.3) follows from the fact that f k is a translation in B(x i1...i k, 2ρ i1,...,i k ). Finally, let x B(x i1...i k, 2ρ i1,...,i k ) y R n \ B(x i1...i k, 2ρ i1,...,i k ). Let z B(x i1...i k, 2ρ i1,...,i k ) be the point on the line segment with end points x y. Then, by the two previous cases, f k (x) f k (y) f k (x) f k (z) + f k (z) f k (y) L x z + L z y = L x z. This proves the right h inequality of (4.3). A similar argument for f 1 k with the balls B(y i1...i k, 2ρ i1,...,i k ) gives the left h inequality. We have left to show that the limit lim k f k (x) = f(x) exists for all x R n. Then also f satisfies (4.3) f(e) F. First, if x R n \ E, then x R n \ B l for some l, so f k (x) = f l (x) for k l. If x E, there are i 1, i 2,..., such that x B(x i,...ik, 2ρ i1...i k ) for all k. Then f k (x) B(y i1...i k, 2Dd k ) lim k f k (x) = y = f(x) where y = lim k y i1...i k. 5. Sub- supersets In this section we shall consider the question whether a given regular set contains regular subsets of smaller dimension whether it is contained in higher dimensional regular sets Theorem. Let E X be s-regular 0 < t < s. For every x E 0 < r < d(e), E B(x, r) contains a t-regular subset F such that C F C d(f ) cr where C c are positive constants depending only on s, t C E. This can be proven with the same method as Theorem 3.1. In fact, that method gives that E B(x, r) has a t-regular subset which is bilipschitz equivalent with C(t, r) with a bilipschitz constant depending only on s, t C E. Observe that the regularity of E implies that d(e B(x, r)) C 1/s E r Theorem. Let 0 < s < t < u. Suppose that E X is s-regular that X is u-regular. Then there is a t-regular set F with E F X. Moreover, C F C where C depends only on s, t, C E C X. Proof. We shall only consider the case where X E are bounded. A slight modification of the proof works if X or both X E are unbounded. Recalling the remarks at the beginning of Section 3, we may assume that d(e) = 1. Let 0 < d < 1/30 be such that d u s < 4 s 30 u C 1 E C 1 X. By Lemma 2.1 there are for every k = 1, 2,..., disjoint balls B(x k,i, 6d k ), i = 1,..., m k, such that x k,i E the balls B(x k,i, 30d k ) cover E. Further, there are disjoint balls B(x k,i, 6d k ), i = m k +1,..., n k, such that x k,i X \ m k B(x k,i, 30d k ) the balls B(x k,i, 30d k ), i = 1,..., n k, cover X. Fix k i, 1 i m k. Denote J = {j {1,..., n k } : B(x k+1,j, d k+1 ) B(x k,i, 3d k )}, J = {j {1,..., n k } : B(x k+1,j, 30d k+1 ) B(x k,i, d k ) }, I = {j J : E B(x k+1,j, 6d k+1 ) },

13 Ahlfors David regular sets bilipschitz maps 499 let n, n m be the number of indices in J, J I, respectively. Then, as d < 2/31, J J so n n. Since B(x k,i, d k ) j J B(x k+1,j, 30d k+1 ), we have, comparing measures as in the proof of Lemma 2.1, that n n 30 u C 1 X d u. If j I there is z j E B(x k+1,j, 6d k+1 ) then, as also j J d < 1/7, B(z j, d k+1 ) B(x k+1,j, 7d k+1 ) B(x k,i, 4d k ). Then the balls B(z j, d k+1 ), j I, are disjoint md (k+1)s j I µ(b(z j, d k+1 )) µ(b(x k,i, 4d k )) 4 s C E d ks, whence m 4 s C E d s. Combining these inequalities recalling the choice of d, we find that m 4 s C E d s < 30 u C 1 X d u n. Thus we can choose some j J \ I. Let y k,i = x k+1,j B k,i = B(y k,i, d k+1 ). Denote also 2B k,i = B(y k,i, 2d k+1 ). Then for a fixed k the balls 2B k,i, i = 1,..., m k, are disjoint. If x 2B k,i, then, as j I, d(x, E) 4d k+1. On the other h, as j J, d(x, E) d(x, x k,i ) d(x, y k,i ) + d(y k,i, x k,i ) 2d k+1 + 3d k < d k 1. It follows that the balls 2B k,i 2B l,j with k l 2 are always disjoint. Hence any point of X can belong to at most two balls 2B k,i, i = 1,..., m k, k = 1, 2,.... By Theorem 5.1 we can choose for every k, i, 1 i m k, t-regular sets F k,i B k,i such that C Fk,i C d(f k,i ) cd k with C c depending only on t, u C X. Let ν k,i be the Borel measure related to F k,i as in Definition 1.1. We define m k F = E ν = k=1 F k,i m k ν k,i. k=1 Then F is a closed bounded subset of X containing E. We check now that F is t-regular. Let x F 0 < r d(f ). It is enough to verify the required inequalities for r < d, so we assume this. Let l be the positive integer for which d l+1 r < d l. Denote We have K = {(k, i) : i = 1,..., m k, k < l B k,i B(x, r) } L = {(k, i) : i = 1,..., m k, k l B k,i B(x, r) }. ν(b(x, r)) (k,i) K ν k,i (B k,i B(x, r)) + (k,i) L ν k,i (B k,i B(x, r)). If (k, i) K, then r < d k+1 B(x, r) 2B k,i. Since this can happen for at most two balls 2B k,i, K can contain at most two elements the first sum above is bounded by 2 t+1 Cr t. To estimate the second sum, let p k be the number of indices in I k = {i : (k, i) L}. Let (k, i) L. Then B k,i B(x, r), so d(x k,i, x)

14 500 Pertti Mattila Pirjo Saaranen d(x k,i, y k,i ) + d(y k,i, x) 3d k + 2d k+1 + r < 5d l, which gives B(x k,i, d k ) B(x, 6d l ). Consequently, p k d ks i I k µ(b(x k,i, d k )) µ(b(x, 6d l )) C E 12 s d ls, so p k 12 s C E d (l k)s. Hence ν k,i (B k,i B(x, r)) 12 s C E d (l k)s C4 t d (k+1)t 12 s 4 t C E Cd ls (k,i) L k=l k=l d (t s)k = 12 s 4 t C E Cd lt 1 1 d (t s) 12s 4 t C E Cd t 1 1 d (t s) rt. This proves the upper regularity of ν. To prove the opposite inequality, suppose first that x E. Let k be the positive integer for which 33d k r < 33d k 1. Then for some i, 1 i m k, x B(x k,i, 30d k ). Since B k,i B(x k,i, 3d k ) we have that B k,i B(x, 33d k ) B(x, r). Thus ν(b(x, r)) ν k,i (B k,i ) d(f k,i ) t c t d kt c t d t 33 t r t. Suppose finally that x F k,i for some k i. If r 9d k, then d(f k,i ) cd k (c/9)r, whence ν(b(x, r)) ν(b(x, (c/9)r) (c/9) t r t. If r > 9d k, then d(x, x k,i ) 3d k < r/3, so B(x k,i, r/2) B(x, r). Since x k,i E, the required inequality follows from the case x E. In the next example note that lim r 0 L 1 (F B(x, r))/(2r) = 1 for L 1 almost all x F by the Lebesgue density theorem. However, F has no subset E with L 1 (E) > 0 for which L 1 (F B(x, r))/(2r) would be bounded below with a positive number uniformly for small r > Example. There exists a compact set F R with Lebesgue measure L 1 (F ) > 0 such that it contains no non-empty s-regular subset for any s > 0. Proof. Let a < b, 0 < λ < 1/2 0 < t < 1. We shall construct a family I ([a, b], λ, t) of closed disjoint subintervals of [a, b]. We do this for [0, 1] then define I ([a, b], λ, t) = {f(i) : I I ([0, 1], λ, t)} where f(x) = (b a)x + a. Let I 1,1 = [(1 λ)/2, (1 + λ)/2]. Then [0, 1] \ I 1,1 consists of two intervals J 1,1 J 1,2 of length (1 λ)/2. We select closed intervals I 2,1 I 2,2 of length λ(1 λ)/2 in the middle of them (that is, the center of I 2,i is the center of J 1,i ). Continuing this we get intervals I k,i, i = 1,..., 2 k 1, J k,i, i = 1,..., 2 k, such that d(i k,i ) = 2 1 k λ(1 λ) k 1 d(j k,i ) = 2 k (1 λ) k. Moreover, each I k,i is the mid-interval of some J k 1,j J k 1,j \ I k,i consists of two intervals J k,j1 J k,j2. Then l 2 k 1 d(i k,i ) = k=1 We choose l such that l λ(1 λ) k 1 = 1 (1 λ) l 1 as l. k=1 l 2 k 1 d(i k,i ) > t k=1

15 Ahlfors David regular sets bilipschitz maps 501 denote I ([0, 1], λ, t) = {I k,i : i = 1,..., 2 k 1, k = 1,..., l}. Then for any compact interval I R, d(j) > td(i). J I (I,λ,t) Let 0 < λ k < 1/2, 0 < t k < 1, k = 1, 2,..., such that lim k λ k = 0 t = k=1 t k > 0. Define I 1 = I ([0, 1], λ 1, t 1 ), inductively for m = 1, 2,..., The compact set F is now defined as For every m = 1, 2,... we have I m+1 = {J : J I (I, λ m+1, t m+1 ), I I m }. F = I. m=1 I I m I I m d(i) > t 1 t m > t, whence L 1 (F ) t. Suppose that s > 0 that E is an s-regular subset of F. Choose m so large that λ m < C s E /4. Let x E. Then x I for some I I m. Suppose that I would be one of the shortest intervals in the family I m. Then by our construction there is an interval J such that I is in the middle of J, I E = J E d(i) = λ m d(j). As B(x, d(j)/4) J we have by the regularity of E, 4 s d(j) s µ(b(x, d(j)/4)) = µ(b(x, d(i)) C E d(i) s = C E (λ m d(j)) s. Thus λ m C 1/s E /4. This contradicts with the choice of m. So E contains no points in the shortest intervals of I m. But then we can repeat the same argument with the second shortest intervals of I m concluding that neither can they contain any points of E. Continuing this we see that E =. [DS1] [DS2] [FM] References David, G., S. Semmes: Analysis of on uniformly rectifiable sets. - Surveys Monographs 38, Amer. Math. Soc., David, G., S. Semmes: Fractured fractals broken dreams. - Oxford University Press, Falconer, K. J., D. T. Marsh: On the Lipschitz equivalence of Cantor sets. - Mathematika 39, 1992, [F] Federer, H.: Geometric measure theory. - Springer-Verlag, [H] Hutchinson, J. E.: Fractals self similarity. - Indiana Univ. Math. J. 30, 1981, [JW] [M] Jonsson, A., H. Wallin: Function spaces on subsets of R n. - Harwood Academic Publishers, Mattila, P.: Geometry of sets measures in Euclidean spaces. - Cambridge University Press, 1995.

16 502 Pertti Mattila Pirjo Saaranen [RRX] Rao, H., H.-J. Ruan, L.-F. Xi: Lipschitz equivalence self-similar sets. - C. R. Math. Acad. Sci. Paris 342:3, 2006, [RRY] Rao, H., H.-J. Ruan, Y.-M. Yang: Gap sequence, Lipschitz equivalence box dimension of fractals. - Nonlinearity 21, 2008, Received 5 September 2008