ESI The Erwin Schrodinger International Boltzmanngasse 9 Institute for Mathematical Physics A-090 Wien, Austria Lipschitz Image of a Measure{null Set Can Have a Null Complement J. Lindenstrauss E. Matouskova D. Preiss Vienna, Preprint ESI 736 (999) August 5, 999 Supported by Federal Ministry of Science and Transport, Austria Available via anonymous ftp or gopher from FTP.ESI.AC.AT or via WWW, URL: http://www.esi.ac.at
LIPSCHITZ IMAGE OF A MEASURE-NULL SET CAN HAVE A NULL COMPLEMENT J. LINDENSTRAUSS, E. MATOUSKOV A, AND D. PREISS Abstract. We give two examples that in innite dimensional Banach spaces the measure-null sets are not preserved by Lipschitz homeomorphisms. There exists a closed set D `2 which contains a translate of any compact set in the unit ball of `2 and a Lipschitz isomorphism F of `2 onto `2 so that F (D) is contained in a hyperplane. Let X be a Banach space with an unconditional basis. There exists a Borel set A X and a Lipschitz isomorphism F of X onto itself so that the sets X n A and F (A) are both Haar null.. Introduction In innite dimensional Banach spaces there is no natural analogue of Lebesgue null sets. Various measure-theoretic notions of null sets were introduced there in order to generalize the theorem of Rademacher to innite dimensional Banach spaces: if f is a Lipschitz function on a separable Banach space then it is G^ateaux dierentiable \almost everywhere" (see [BL] for a survey of the topic). More generally, if f is a Lipschitz mapping from a separable Banach space to a Banach space with the Radon-Nikodym property (reexive Banach spaces are an example), then it is G^ateaux dierentiable \almost everywhere". Let X and Y be separable, and, say, reexive Banach spaces, let F : X! Y be a Lipschitz isomorphism; that is, F is surjective and both F and F? are Lipschitz. It is an open question if there is an x 2 X so that the G^ateaux derivative of F at x is surjective. If the answer was positive, Lipschitz isomorphic separable reexive Banach spaces would be linearly isomorphic. In Example 3. we show that even in `2 the set of points where the derivative of a Lipschitz isomorphism of `2 is not surjective can be rather large. In this case it contains a translate of any compact set in the unit ball of `2; in particular it is not Aronszajn, and not even Haar null. 99 Mathematics Subject Classication. Primary: 46B20, Secondary: 28C20. Key words and phrases. Lipschitz isomorphism, Haar null, Banach space. The authors were partially supported by the Erwin Schrodinger Institute of Vienna; the second author also by GAAV-A09705.
2 J. LINDENSTRAUSS, E. MATOU SKOV A, AND D. PREISS Lebesgue null sets in R n are preserved by Lipschitz isomorphisms. Even a weaker version of this in innite dimensional Banach spaces (image of a complement of a null set can not be null), would provide us with points where the derivative is surjective (see [BL] for the details). In Example 3.3 we show that in every separable Banach space with an unconditional basis there is a Borel set A X and a Lipschitz isomorphism F of X onto itself so that both the set X n A and F (A) are Haar null. Moreover, with A xed, we can take F to be Frechet dierentiable and arbitrarily close to identity in both the supremum and Lipschitz norm. It would be of interest to know if this can happen also for some more restrictive \G^ateaux dierentiability null sets", for example, the Aronszajn null sets. Finally we make a small observation of a dierent avor: if X is an innite dimensional normed linear space then all nets in X are Lipschitz equivalent. This is unlike the nite dimensional Banach spaces; already in R 2 there is a net which is not Lipschitz equivalent to Z 2 ([BuK], [Mc]). 2. Preliminaries A Borel subset A of a separable Banach space X is Haar null if there is a Borel probability measure on X so that (A + x) = 0 for all x 2 X. The set A is Aronszajn null if for every sequence (x i ) in i= X whose S closed linear span is X there exist Borel sets A i X such that A = A i= i and the intersection of A i with any line in the direction x i has a one-dimensional Lebesgue measure zero, for each i 2 N. In innite dimensional Banach spaces the family of Haar null sets contains all compact sets, and is therefore strictly larger than the family of Aronszajn null sets. Aronszajn null sets coincide by [C] with Gaussian null and cube null sets. For other properties of null sets see [BL]. For Example 3. we will need some results from [IP] (see also [BL]) which we list below. Lemma 2.. [IP] Let E be a Banach space and let fa k g k= be subsets of E such that A k A k?. Let ff k g be Lipschitz isomorphisms of E onto itself so that all F k 's and their inverses are L-Lipschitz for some constant L. Assume also that (i) The restriction of F k to A k is an isometry of A k onto itself. (ii) F k (x) = x for every x 2 E n A k?. (iii) There is a sequence k! 0 such that dist (x; E n A k ) k for every x 2 A k.
LIPSCHITZ IMAGES 3 Then F n F converge pointwise to a Lipschitz isomorphism F of E onto itself. For 0 t let T t be the clockwise rotation of the plane R 2 by the angle (? t)=2 and let T t be the identity for t. Let (e k ) be the usual orthonormal basis of k= `2 and I be the identity mapping on `2. For each k = 2; 3; : : : let E k = span fe ; e k g, let P k be (for the sake of Example 3.) the orthogonal projection on E k, and let Q k = I? P k. Identifying E k with R 2 we dene linear operators T k;t on `2 by T k;t x = T t P k x + Q k x. Notice that if l =2 f; kg, then ht k;t x; e l i = hx; e l i. The following lemma appears in [IP] with more specic constants. Lemma 2.2. Suppose c; d > 0, k 2 N and A `2 satises (i) T k;t A = A for every t 0. (ii) dist (x; A) d implies kp k xk cd. Then the mapping f dened by f(x) = T k; d dist (x;a) (x) is the isometric rotation f(a ; a 2 ; : : : ) = (a k ; a 2 ; : : : ; a k? ;?a ; a k+ ; : : : ) on A, and it is identity when dist (x; A) d. Moreover, it is a Lipschitz isomorphism with the Lipschitz constants depending only on c. Lemma 2.3. [IP] If C is a convex set in a normed linear space X, Y is a metric space, f : C! Y is continuous and C can be covered by countably many sets on each of which the Lipschitz constant of f does not exceed L, then the Lipschitz constant of f is at most L. 3. Null sets with large Lipschitz images In [IP] it was shown that for any Hilbert cube Q in `2 (for example, for the set fx = (x n ) 2 `2 : jx n j 2?n g) there is a Lipschitz isomorphism of `2 which maps Q into a hyperplane. In [M] it is shown that in every separable Banach space X there is a closed set A which contains a translate of any compact in the unit ball of X and a Lipschitz isomorphism f of X so that every line in X intersects f(a) in a set of Lebesgue measure zero. The following example is a combination of the two above. It would be of interest to know if it can be sharpened: is it possible to map the complement of an Aronszajn null set into a countable union of hyperplanes by a Lipschitz isomorphism? Example 3.. There exists a closed set D `2 which contains a translate of any compact set in the unit ball of `2 and a Lipschitz isomorphism F of `2 onto `2 so that F (D) is contained in a hyperplane. Proof. For n 2 N 0 choose positive numbers n, n so that P i=n+ i < 2 n, n? 0 n > 0, n? c n for some constant c, and 4 n n?.
4 J. LINDENSTRAUSS, E. MATOU SKOV A, AND D. PREISS (For example, n = 2?5n and n = n would do.) For x 2 `2 denote x n = hx; e n i, for the reason of notation put k 0 = 0 and g 0 = I. Let k < < k n 2 N. Dene g k0 ;k ;:::;k n : `2! `2 by g k0 ;k ;:::;k n (x) = (T 2kn+;0 T 2k +;0)x Notice, that () = x + x 2kn+e? nx i=0 x 2ki +e 2ki +? nx i= x 2ki? +e 2ki +: if l 6= 2k i + ; i 2 f0; : : : ; ng then hg k0 ;k ;:::;k n (x); e l i = hx; e l i; and, similarly, (2) if l =2 f; 2k + g then hi? P 2k+ (x); e l i = hx; e l i: = B(0; 0 ) and dene inductively the following sets and a map- Set V k0 pings V k0 ;k ;:::;k n =(span fe ; : : : ; e kn g + nx i= i e 2ki + B(0; n )) (3) \ V k0 ;k ;:::;k n? A k0 ;k ;:::;k n =((I? P 2kn+) g k0 ;k ;:::;k n? )(V k0 ;k ;:::;k n ) + B E2kn + (0; 2 n?) + B(0; 3 n ) ~A k0 ;k ;:::;k n =A k0 ;k ;:::;k n + B(0; n ) f k0 ;k ;:::;k n (x) =T 2kn+; n dist (x;ak 0 ;k ;:::;kn ) (x); for x 2 `2: S We put D n = V k0 ;k ;:::;k n, where the union T is taken over all choices of 0 = k 0 < k < < k n, and D = D n. First we show that D contains a translate of any compact set in B(0; 0 =2). For a compact K B(0; 0 =2) choose k < k 2 < : : : so that Put z = P i= ie 2ki. Since k P K span fe ; : : : ; e kn g + B(0; n =2): i=n+ ie 2ki k < 2 n for n 2 N 0, we have K + z B(0; 0 ) = V k0 and then, inductively, K + z V k0 ;k ;:::;k n? \ (span fe ; : : : ; e kn g + nx i e 2ki + B(0; 2 n) + X i= i=n+ i e 2ki ) V k0 ;k ;:::;k n Hence K + z D.
LIPSCHITZ IMAGES 5 Notice that T 2kn+;t(A k0 ;k ;:::;k n ) = A k0 ;k ;:::;k n for each t 0 and that kp 2kn+(x)k 2 n? + 4 n (2c + 4) n for each x 2 Ak0 ~ ;k ;:::;k n. Therefore, by Lemma 2.2, f k0 ;k ;:::;k n is an isometric rotation of A k0 ;k ;:::;k n onto A k0 ;k ;:::;k n and the identity outside of Ak0 ~ ;k ;:::;k n. Moreover, all the mappings f ::: and their inverses are Lipschitz with some constant L independent of the indexes. For n 2 N we dene F n (x) = ( f k0 ;k ;:::;k n (x) if x 2 ~ Ak0 ;k ;:::;k n ; x if x 2 `2 n S ~ Ak0 ;k ;:::;k n : Each F n is well dened, and it is a Lipschitz isomorphism of `2; the same Lipschitz constant works for all n. This follows from Lemma 2.3 after we verify that for each n 2 N the sets A ~ k0 ;k ;:::;k n are pairwise disjoint. Let (k ; : : : ; k n ) 6= (k; 0 : : : ; kn); 0 let i be the smallest index for which k i 6= ki, 0 for example, k i > ki. 0 By (), (2), and (3) we have for any u 2 Ak0 ~ ;k ;:::;k n and v 2 Ak0 ~ ;k 0 that ;:::;k0 n ku? vk hu? v; e 2ki i i? 2 i? 8 n > 0: We dene the mapping F as the pointwise limit of F n F n? F. The mapping F is a Lipschitz isomorphism of `2 by Lemma 2.. Indeed, the condition (i) S of the lemma is satised since each F n is an isometry of the set A n = A k0 ;k ;:::;k n onto itself. The condition (iii) is satised since dist (x; `2 n A n ) (2c + 3) n for each x 2 A n. To verify (ii), we observe that F n = I outside of the set An ~ S = Ak0 ~ ;k ;:::;k n. We will prove that A ~ k0 ;k ;:::;k n A k0 ;k ;:::;k n?. This will imply in particular that A k A k?. Suppose z 2 A ~ k0 ;k ;:::;k n. Then z = (I? P 2kn+)(g k0 ;k ;:::;k n? (x)) + (x 0 e + x 0 2k n+ e 2k n+) + u for some x 2 V k0 ;k ;:::;k n, kx 0 e + x 0 2k n+ e 2k n+k 2 n? and kuk 3 n. Hence since z =(I? P 2kn? +)(g k0 ;k ;:::;k n?2 (x)) + (x 0 e? x 2kn?2 +e 2kn? +) + ((x 0 2k n+? x 2kn+)e 2kn+ + u) 2A k0 ;k ;:::;k n? ; kx 0 e? x 2kn?2 +e 2kn? +k 2 n? + n?2 2 n?2 k(x 0 2k n+? x 2k n+)e 2kn+ + uk 2 n? + n + 3 n 3 n? : and It remains to show that F (D) Ker e. Let x 2 D. Since for each n 2 N the sets V k0 ;k ;:::;k n are pairwise disjoint and V k0 V k0 ;k : : :, there are some k < k 2 < : : : so that x 2 T n2n V k 0 ;k ;:::;k n. We will
6 J. LINDENSTRAUSS, E. MATOU SKOV A, AND D. PREISS show that (4) F (x) = x? X i=0 It suces to show for all n that (5) and (6) x 2ki +e 2ki +? X i= x 2ki? +e 2ki +: (F n? F )(x) = g k0 ;k ;:::;k n? (x) 2 A k0 ;k ;:::;k n ; jhg k0 ;k ;:::;k n? (x); e ij n? : We will prove this by induction on n. If n =, then x = (I? P 2k +)(x) + x e + x 2k +e 2k + 2 A k0 ;k ; since kx e + x 2k +e 2k +k kxk 0 ; we have also jhx; e ij kxk 0. Now suppose that (5) and (6) hold, we will show that they also hold if n is replaced by n +. By (5) we have that (F n F n? F )(x) = (T 2kn+;0 g k0 ;k ;:::;k n? )(x) = g k0 ;k ;:::;k n (x): By () and (3) hence Finally, jhg k0 ;k ;:::;k n? (x); e 2kn+ij = jhx; e 2kn+ij n ; jhg k0 ;k ;:::;k n (x); e ij = jhg k0 ;k ;:::;k n? (x); e 2kn+ij n : g k0 ;k ;:::;k n (x) =(I? P 2kn+ +)(g k0 ;k ;:::;k n (x)) + hg k0 ;k ;:::;k n (x); e ie + x 2kn+ +e 2kn+ + 2 A k0 ;k ;:::;k n+ ; since khg k0 ;k ;:::;k n (x); e ie + x 2kn+ +e 2kn+ +k n + n+ 2 n. Remark. The set of points where the mapping F in Example 3. is G^ateaux dierentiable but the derivative is not onto contains a translate of any compact set from a xed ball. To see this we will follow the notation of the proof of Example 3.. Let Q = 0 fx 2 4 `2 : jx n j 2?n g. If K B(0; 0 =4) is compact, then K + Q B(0; 0 =2) is also compact T and there is some z 2 `2 and k < k 2 < : : : so that K + Q + z n2n V k 0 ;k ;:::;k n for some k < k 2 < : : :. Then by (4) the mapping F on K + Q + z is a restriction of a non surjective linear mapping. Therefore it is G^ateaux dierentiable on K + z with the operator dened by (4) as the derivative (see e.g. Section 7.4 of [BL]). We pass now to an example of a dierent kind. Let X be a Banach space with an unconditional normalized basis (e n ). Recall, that if x = P x n e n 2 X, then kjxkj = supfk P a n e n k : ja n j jx n jg is an
LIPSCHITZ IMAGES 7 equivalent norm on X; the basis (e n ) is again normalized. For the purpose of the following example we will always assume X to be equipped with this norm. We will need the following simple observation. Lemma 3.2. Let X be a Banach space with an unconditional basis (e n ) renormed as above. Let " > 0 and f n : R! R be surjective functions so that (i) f n (0) = 0 and ( + ")? js? tj jf n (s)? f n (t)j ( + ")js? tj for all n 2 N and s; t 2 R. Then the mapping F : X! X dened by F ( P x n e n ) = P f n (x n )e n is a Lipschitz isomorphism; the Lipschitz constants of both F and F? are at most + ". Suppose, moreover, that P (ii) for some a n 0 with N a n " we have jf n (t)? tj a n for all t 2 R and n 2 N, and (iii) the functions f n are uniformly dierentiable, that is, for every > 0 there exists > 0 so that if s 2 R, 0 < jtj <, and n 2 N, then j (f t n(s + t)? f n (s))? fn(s)j 0 <. Then kf (x)? xk " for all x 2 X and both F and F? are continuously Frechet dierentiable. Proof. Since jf n (t)j (+")jtj, we have k P f n (x n )e n k (+")k P x n e n k < for any x = (x n ) 2 X, and F is well dened. Similarly, ~x = (f?(x n n)) 2 X, and F (~x) = x, hence F is surjective. The mapping F is bi-lipschitz, since kx?yk +" =( + ")? k X (x n? y n )e n k k X (f n (x n )? f n (y n ))e n k Further, ( + ")k X (x n? y n )e n k = ( + ")kx? yk: kf (x)? xk = k X (f n (x n )? x n )e n k X a n ": To see that F is Frechet dierentiable at a given z = (z n ) 2 X, dene a linear mapping D z : X! X by D z (v) = P f 0 n (z n)v n e n. Since jf 0 n(z n )j ( + "), D z is continuous. For a given > 0 choose > 0 as in (iii). Then for any v 2 X with kvk we have kvk kf (z + v)? F (z)? D z(v)k = kvk kp (f n (z n + v n )? f n (z n )? f 0 n (z n)v n )e n k = kvk kp fn: v n6=0g ( fn(zn+vn)?fn(zn) v n? f 0 n(z n ))v n e n k kvk kp v n e n k = :
8 J. LINDENSTRAUSS, E. MATOU SKOV A, AND D. PREISS If s; 2 R and js? j < then (iii) implies that jf 0 n(s)? f 0 n()j < 2 for each n 2 N. If y; z 2 X and ky? zk <, then kd z (v)? D y (v)k = k X (f 0 n(z n )? f 0 n(y n ))v n e n k 2k X v n e n k 2; for any v 2 X with kvk. Hence F is continuously Frechet differentiable and by the inverse mapping theorem the same is true for F?. In R n consider the family of axis-parallel unit cubes centered at points with integer coordinates. Remove from the center of each cube of the family an axis-parallel cube of the volume =n 2 ; what remains in R n we will call A n. The slightly smaller set D n we dene similarly: we remove from the center of each unit cube of the family an axis-parallel cube of volume =n. The set A n can be mapped onto the set D n in a bi-lipschitz manner simply by blowing up the cubes of volume =n 2 a little bit (while xing the boundaries of the unit cubes). The sets in the following example will be dened as products of the sets n A n, respectively n D n, for some suitable scalars n > 0. The Lipschitz isomorphism F of the Banach space X will be dened on blocks of coordinates; there it will correspond to blowing up the smaller axis-parallel cubes to the bigger ones. Recall, that we always consider a Banach space with an unconditional basis (e n ) to be equipped with the norm satisfying kxk = supfk P a n e n k : ja n j jx n jg. For example, `2 with the usual norm is such a space. Example 3.3. Let X be a Banach space with a normalized unconditional basis and a norm as above. There exists a Borel set A X and a Lipschitz isomorphism F of X onto itself so that both the set X n A and F (A) are Haar null. Moreover, there exists a xed Borel set A X and a xed Borel set D X so that both X n A and D are Haar null and for each " > 0 we can choose F with F (A) = D so that kf (x)? xk ", ( + ")? kx? yk kf (x)? F (y)k ( + ")kx? yk for each x; y 2 X and both F and F? are continuously Frechet dierentiable. Proof. Choose some n & so that P =n n converges. (If we just wish that F is a Lipschitz homeomorphism, it is enough to put n = 2 for all n 2 N.) Let r n = ( n ) n, and A n =R n n (Z n + [? 2 rn n ; 2 rn n ] n ); D n =R n n (Z n + [? 2 r n; 2 r n] n ):
We write X as LIPSCHITZ IMAGES 9 X = R R 2 R 3 : : : ; here we consider R as span fe g, R 2 as span fe 2 ; e 3 g, : : :. By P n we denote the projection on R P n, that is on the n-th block of the basis. Choose some n > 0 with n n < and dene A n =fx 2 X : P n (x) 2 n A n g; A = [ k= ( \ A n ); D n =fx 2 X : P n (x) 2 n D n g; D = [ k= ( \ D n ): Later on we will dene also a Lipschitz isomorphism F of X so that F (A) = D. To show that the set D is Haar null, we will observe that all the sets T D n, k 2 N are Haar null. Let n be the Lebesgue measure on R n restricted to n [0; ] n and normalized. The periodicity of the sets A n and D n implies that (7) n ( n A n + x) =? n n and n ( n D n + x) =? n for any x 2 R P n. Let = 2 3 : : :. Since n n <, is a probability measure on X. For any z 2 X and n 2 N we have that D n + z = fx 2 X : P n (x) 2 P n (z) + n D n g. Hence for any z 2 X and k 2 N we get by (7) that \ Y Y (z + D n ) = n (P n (z) + n D n ) = (? n ) = 0: Now we show that the set X n A is Haar null. If z 2 X is arbitrary, then A n + z = fx 2 X : P n (x) 2 P n (z) + n A n g and by (7) (z + A) sup k2n (z + = sup k2n Y \ A n ) = sup k2n (? n n ) = : Y n (P n (z) + n A n ) To dene the mapping F we rst dene ' n : R! R which will stretch the coordinates in R n so that the cubes (k ; : : : ; k n )+[? 2 rn n ; 2 rn n ] n will be mapped on the cubes (k ; : : : ; k n ) + [? r 2 n; r 2 n] n, where k ; : : : ; k n 2
0 J. LINDENSTRAUSS, E. MATOU SKOV A, AND D. PREISS Z. Namely, let ' n be piecewise linear and ' n (k + 2 rn n ) =k + r 2 n (8) ' n (k? 2 rn n ) =k? r 2 n; for k 2 Z. Then ' n (0) =0 (9) ' n (k + t) =k + ' n (t) for k 2 Z; t 2 [? ; ]: 2 2 There are some constants " n & 0 such that for any s; t 2 R (0) j' n (t)? tj < " n ( + " n )? js? tj < j' n (s)? ' n (t)j < ( + " n )js? tj: The latter holds, since (k + 2 r n)? (k? 2 r n) (k + 2 rn n )? (k? 2 rn n ) = r?n (k +? 2 r n)? (k + 2 r n) (k +? 2 rn n )? (k + 2 rn n n! (n! ) ) =? r n? r n n = e ln n n? n n e n ln n n?! (n! ) n n ln n ln n for any k 2 Z. To obtain a smooth mapping F we replace the piecewise linear functions ' n by a new family of increasing functions (which we again denote by (' n )) so that (8), (9), and (0) are still satised, and, moreover, ' n 2 C. Put n (t) = n ' n ( t n ), for t 2 R. Then n (0) = 0, j n (t)?tj " n n and (+" n )? js?tj j n (s)? n (t)j (+" n )js?tj for any s; t 2 R. Hence j n(s) 0? j " n for s 2 R. We will verify that the family ( n ) is uniformly dierentiable as required in Lemma 3.2, (iii). Let > 0 be given. Notice that each 0 n is continuous and periodic, hence uniformly continuous. By the mean value theorem, if n 2 N, s 2 R and t 6= 0 then j ( t n(s + t)? n (s))? 0 (s)j = j 0 (s + )? 0 n n n (s)j =: E for some suitable with jj jtj. If " n < =2 then jej <. For the nitely many n for which " n =2 we choose > 0 small enough so that jej < follows for jtj < from the uniform continuity of each 0. n If we dene a mapping F n : R n! R n by F n (x ; : : : ; x n ) = ( n (x ); : : : ; n (x n )); then F n ( n A n ) = n D n. Let " > 0 be given. Choose k 2 N so that " k < " and P n n" n < ". We dene F : X! X by F (x ; x 2 ; : : : ) = (P (x); : : : ; P k? (x); F k (P k (x)); F k+ (P k+ (x)); : : : );
LIPSCHITZ IMAGES that is, F leaves the coordinates in the rst (k?) blocs alone and acts as F n on the n-th block with n k. Then F (A n ) = A n if n < k and F (A n ) = D n if n k. Consequently, F (A) = D. The other properties of F follow from Lemma 3.2. Remark. Notice that neither the set X n A, nor the set F (A) in the previous lemma is Aronszajn null. This follows immediately after observing that each of them contains an ane, spanning copy of a Hilbert cube. 4. Nets An (a; b)-net in a metric space M is a subset N such that dist (x; y) a for every x 6= y in N and dist (x; N ) < b for every x 2 M. When a = and b we say that N is a -net. Two nets N and N 2 in a Banach space X are Lipschitz equivalent if there is T : N! N 2 bijective so that both T and T? are Lipschitz. For each n 2 there is a net in R n which is not Lipschitz equivalent to Z n ([BuK], [Mc]). Here we observe that this is unlike the situation in innite dimensional spaces: if X is an innite dimensional normed linear space then all nets in X are Lipschitz equivalent. We need only the following simple observation. Lemma 4.. Let M be a metric space and let N, N 2 be nets in M. Suppose c > 0 and T : N! N 2 is a bijection such that dist (x; T (x)) c for every x 2 N. Then both T and T? are Lipschitz. Proof. Denote by d the metric on M. Let N i be an (a i ; b i )-net. Due to the symmetry of the assumptions it is enough to show that T? is Lipschitz. Let x 6= y 2 N. We distinguish two cases. When d(x; y) 4c, then d(t (x); T (y)) a 2 = a 2 d(x; y) a 2 d(x; y): d(x;y) 4c When d(x; y) 4c, then d(t (x); T (y)) d(x; y)? d(t (x); x)? d(y; T (y)) d(x; y)? d(x; y) = d(x; y): 2 2 The Lipschitz constants of T and T? are bounded by maxf2; 4c a ; 4c a 2 g. Recall, that if N is a net in a normed linear space S then card N is equal to the density character dens X of X. Indeed, N is a dense n subset of X of the same cardinality as N. If N is a -net, say, then any dense subset of X has to intersect each of the pairwise disjoint balls B(x; ), x 2 N. 4
2 J. LINDENSTRAUSS, E. MATOU SKOV A, AND D. PREISS If N is a net in an innite dimensional normed linear space, then there exists r > 0 so that card (B(x; r) \ N ) = dens X for any x 2 X. Suppose for a contradiction that for each n 2 N there is x n 2 X and r n > n so that dens X > card (B(x n ; r n ) \ N ) and the latter set is not nite S (this we can assume, since X is innite dimensional). If D = (N? x n n), then D B(0; ) and card D < dens X which is a contradiction. Proposition 4.2. Let X be an innite dimensional normed linear space. If N and N 2 are nets in X then they are Lipschitz equivalent. Proof. To apply the lemma, we partition N [N 2 into pairwise disjoint sets A, 0 < dens X so that diam A 2c for some c > 0 and card (A \ N ) = card (A \ N 2 ) for each n 2 N. We can assume that both N and N 2 are -nets for some >. Choose a > 0 so that card (B(x; a)\n ) = dens X = card (B(x; a)\n 2 ) for any x 2 X. Put N = 3aN, S and c = 3a. We index N = fz : 0 < dens Xg and put D = X n z2n B(z; a). We dene inductively A 0 = (B(z 0 ; a) [ (B(z 0 ; c) \ D)) \ (N [ N 2 ) A = (B(z ; a) [ ((B(z ; c) n [ < A ) \ D) \ (N [ N 2 ): The sets A are pairwise disjoint and cover N [ N 2. Moreover, (N [ N 2 ) \ A (N [ N 2 ) \ B(z ; a), and A B(z ; c), so each A has the required properties. For each let T be some bijection of A \ N onto A \ N 2. Then T : N! N 2 dened by T (x) = T (x) for x 2 A is a bijection, and kt (x)? xk 2c for each x 2 N. Hence T is bi-lipschitz by Lemma 4.. Acknowledgment. We would like to thank the functional analysis group at the J. Kepler University in Linz for enabling us to attend the research semester at the Erwin Schrodinger Institute of Vienna in 999 where this paper was written. References [BL] [BuK] [C] [IP] Y. Benyamini, J. Lindenstrauss, Geometric non-linear functional analysis, Colloquium publication No. 48, Amer. Math. Soc., 999. D. Burago, B. Kleiner, Separated nets in Euclidean space and Jacobians of bilipschitz maps, Geom. Funct. Anal. 8, No. 2, 998, 273-282. M. Csornyei, Aronszajn null and Gaussian null sets coincide, to appear in Israel J. Math. D.J. Ives, D. Preiss, Not too well dierentiable Lipschitz isomorphism, to appear in Israel J. Math.
LIPSCHITZ IMAGES 3 [M] [Mc] E. Matouskova, Lipschitz images of Haar null sets, to appear in Bull. London Math. Soc. C.T. McMullen, Lipschitz maps and nets in Euclidean space, Geom. Funct. Anal. 8, No. 2, 998, 304-34. Department of Mathematics, The Hebrew University, Givat Ram, Jerusalem 9904, Israel Mathematical Institute, Czech Academy of Sciences, Zitna 25, CZ- 567 Prague, Czech Republic Department of Mathematics, University College London, Gower street, London WCE 6BT, Great Britain