A Quantitative Implicit Function Theorem for Lipschitz Functions

Transcription

1 A Quantitative Implicit Function Theorem for Lipschitz Functions Jonas Azzam, University of Washington-Seattle (MSRI) work with Raanan Schul September 11, 2011

2 A Dumb Example Motivation Fibers Implicit Function Theorem A Non-Example Let f = P 1 : R 2 R, the orthogonal projection into the first coordinate. The fibers are all perfectly aligned 1-dimensional subspaces. Our goal is to determine, quantitatively, how much this quality holds for general Lipschitz maps.

3 Rectifiablilty of fibers Motivation Fibers Implicit Function Theorem A Non-Example Theorem (Federer) Suppose f : R n+m R n is Lipschitz. Then f 1 (y) is m-rectifiable for almost every y R n. Theorem (Reichel, 09) Suppose M n is a σ-finite H n -rectifiable metric space and f : R n+m M n is Lipschitz. Then f 1 (x) is H m -rectifiable for H n -a.e. x M n.

4 Fibers Implicit Function Theorem A Non-Example A Classical Fiber-StraighteningTheorem Theorem (Implicit Function Theorem) Let f : R n R m R n be C 1 and suppose for some x 0 R n+m, rankd x0 f = n. Then there is g : R n+m R n+m so that f g 1 (x,y) = x for (x,y) in a neighborhood around x 0. R 2 f E g(e) f g 1 R Locally, the fibers of a smooth function can be straightened into m-planes, and f may be straightened into a projection map.

5 Fibers Implicit Function Theorem A Non-Example How well does this also hold for f : R n R m M n? f g 1 g E f f([0,1] 2 ) g(e) When does there exist E R n+m and g : R n+m R n+m L-bi-Lipschitz s.t., for (x,y) g(e), f g 1 behaves like a projection: 1 f g 1 (x,y) is constant in y 2 f g 1 (x,y) is bi-lipschitz in x. Can we estimate E and L?

6 Fibers Implicit Function Theorem A Non-Example Quantitative Inverse Function Theorems Theorem (Lip-bi-Lip Theorem (David 88, Jones 88, Schul 09)) Let h : [0,1] n M be 1-Lipschitz. For each ǫ > 0 there is M(ǫ,n) and a disjoint partition E 1,...,E M,G of [0,1] n s.t. 1 h Ei is 1 ǫ -bi-lipschitz, 2 H n (h(g)) < ǫ. Corollary If ǫ < 1 2 Hn (f([0,1]n )), there is C = C(ǫ) > 0 and E [0,1] n s.t. 1 E C 2 h E is 1 ǫ -bi-lipschitz. The condition H n (f([0,1]n ) > 0 substitutes Jf > 0.

7 Obstacle Motivation Fibers Implicit Function Theorem A Non-Example The following example cannot be equivalent to an orthogonal projection on a large subset: Theorem (Kaufman, 81) There is a C 1 function f : [0,1] 3 [0,1] 2 SURJECTIVE s.t. rankd x f 1 everywhere. We will briefly produce a simpler Kaufman-type example of an a.e. rank-one Lipschitz function f : [0,1] 4 [0,1] 2.

8 Kaufman-type example Fibers Implicit Function Theorem A Non-Example Let G R 2 denote the Garnett example, rotated and scaled so that P 1 (G) = [0,1]. G G is another Cantor set so that P 1,3 (G G) = [0,1] 2.

9 Fibers Implicit Function Theorem A Non-Example Let T R 4 be a quasiconvex tree whose leaves are G G. Let r : R 4 T be a Lipschitz retract. Then f = P 1,3 r : [0,1] 4 [0,1] 2 is s.t. rankd x f = 1 a.e.. Why is this an obstacle?

10 Moral Motivation Fibers Implicit Function Theorem A Non-Example If we want a function f δ(f) that guarantees f g 1 E = P for some E with E δ(f), we should have δ(kaufman) = 0.

11 A Definition Define the (n,m)-content of a function f : Q R n+m M H n,m (f,q) = inf j H (f(q n j )) l(q j ) n l(q j ) n+m where the infimum is over all collections of disjoint dyadic Q j s.t. Q\ Q j = 0. (Note the similarity to the coarea formula.)

12 Moral Motivation Suppose there is a set E and g : R n+m R n+m L-bi-Lipschitz s.t. f g 1 E is a projection map. Kauffman Example: H 2,2 (f,[0,1] n+m ) = 0 Orthogonal projection: H n,m (f,[0,1] n+m ) = 1 ǫ-scaling: H n,m (f,[0,1] n+m ) = ǫ n. The content will determine lower and upper bounds for the size of E and L respectively.

13 Theorem (Simplified version) Let f : R n R m R n be 1-Lipschitz, δ = H n,m (f,[0,1] n+m ) > 0. Then g : R n+m R n+m and E [0,1] n+m s.t. (a) H n+m (E) η(n,m,δ) > 0, and (b) g is C Lip -bi-lipschitz, C Lip = C Lip (n,m,δ) > 1, (c) f g 1 g(e) = P R n. R 2 f E g(e) f g 1 R

14 Theorem (Quantitative Implicit Function Theorem, (A., Schul)) Let f : R n R m M be 1-Lipschitz, 0 < H n (M) 1, and δ = H n,m (f,[0,1]n+m ). Then g : R n+m R n+m and E [0,1] n+m s.t. if F = f g 1 : (i) H n+m (E) η(n,m,δ) > 0 (ii) g is C Lip -bi-lipschitz, C Lip = C Lip (n,m,δ) > 1, (iii) for x R n, F(x, ) is constant on ({x} R m ) g(e), (iv) for y R m, F(,y) is C Lip -bi-lipschitz on (R n {y}) g(e). F g E f f([0,1] 2 ) g(e)

15 bi-lipschitz Extensions Carleson Estimates Typical bi-lipschitz extension results involve some sort of compromise: Increasing dimension Theorem (David and Semmes) If f : E R n is bi-lipschitz, E R d compact, then f may be extended to a bi-lipschitz injection f : R n R max{n,2d+1}. Theorem (MacManus, 95) f : E R n R n bi-lipschitz may be extended to a bi-lipschitz homeomorphism of R 2n onto itself.

16 bi-lipschitz Extensions Carleson Estimates Typical bi-lipschitz extension results involve some sort of compromise: Small bi-lipschitz constant Tukia and Väisälä ( 84, 86): Which sets E R D satisfy the following: There is L 0 = L 0 (E,D) > 1 s.t. if f : E R D is L-bi-Lipschitz, L [1,L 0 ), then f has a bi-lipschitz extension f : R D R D. (e.g. E = R n, S n, n < D) Instead of increasing dimension, or imposing constraints on our bi-lipschitz constant, we jettison a controlled amount of measure to permit a bi-lipschitz extension.

17 bi-lipschitz Extensions Carleson Estimates Theorem (Bi-Lipschitz Extension on Large Pieces (A., Schul)) Let D n and f : [0,1] n R D be 1-Lipschitz. For any ǫ satisfying 0 < ǫ < 1 2 Hn (f[0,1])n ), there is E [0,1] n and F : R n R D s.t. F is L-bi-Lipschitz L D 1 ǫ, if D = n, then F : R n R n is a homeomorphism. E η(n,d,ǫ) > 0 F E = f E.

18 bi-lipschitz Extensions Carleson Estimates In the case f : R n+m M n R D, we need a theorem that says there is a large cube where f is close to being affine. Theorem (Dorronsoro, 85) For F : R n R D Lipschitz, and Q R n, let F(x) A(x) α(q) = inf{sup : A : R n D is affine}. x Q diamq Then α(q)>ǫ,q R Q ǫ,n R. We need to develop an analogue for f : R n M n.

19 bi-lipschitz Extensions Carleson Estimates Theorem (Kirchheim 94) If f : R n M is Lipschitz, then for a.e. z R n, u z := lim r 0 f(z +ru) f(z) r defines a seminorm, called the metric differential, and f(x) f(y) = x y z +o( x z + y z ).

20 bi-lipschitz Extensions Carleson Estimates Theorem (A., Schul) For α > 0, F : R n+m M n, G = {Q : s.t. F(x) F(x Q)) x x Q < αl(q)}. Then Q ([0,1] n )\G Q α,n 1.