Motivation
Test for Nontriviality
Main Theorem
Tools
A Quantitative Implicit Function Theorem for
Lipschitz Functions
Jonas Azzam, University of Washington-Seattle (MSRI)
work with Raanan Schul
September 11, 2011
Jonas Azzam, University of Washington-Seattle (MSRI) work with
A Raanan
Quantitative
Schul Implicit Function Theorem for Lipschitz Functio
Motivation
Test for Nontriviality
Main Theorem
Tools
Fibers
Implicit Function Theorem
A Non-Example
A Dumb Example
Let f = P1 : R2 → 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.
Jonas Azzam, University of Washington-Seattle (MSRI) work with
A Raanan
Quantitative
Schul Implicit Function Theorem for Lipschitz Functio
Motivation
Test for Nontriviality
Main Theorem
Tools
Fibers
Implicit Function Theorem
A Non-Example
Rectifiablilty of fibers
Theorem (Federer)
Suppose f : Rn+m → Rn is Lipschitz. Then f −1 (y ) is m-rectifiable
for almost every y ∈ Rn .
Theorem (Reichel, ’09)
Suppose Mn is a σ-finite Hn -rectifiable metric space and
f : Rn+m → Mn is Lipschitz. Then f −1 (x) is Hm -rectifiable for
Hn -a.e. x ∈ Mn .
Jonas Azzam, University of Washington-Seattle (MSRI) work with
A Raanan
Quantitative
Schul Implicit Function Theorem for Lipschitz Functio
Motivation
Test for Nontriviality
Main Theorem
Tools
Fibers
Implicit Function Theorem
A Non-Example
A Classical Fiber-StraighteningTheorem
Theorem (Implicit Function Theorem)
Let f : Rn × Rm → Rn be C 1 and suppose for some x0 ∈ Rn+m ,
rankDx0 f = n. Then there is g : Rn+m → Rn+m so that
f ◦ g −1 (x, y ) = x for (x, y ) in a neighborhood around x0 .
R2
f
g (E )
f ◦ g −1
E
R
Locally, the fibers of a smooth function can be straightened into
m-planes, and f may be straightened into a projection map.
Jonas Azzam, University of Washington-Seattle (MSRI) work with
A Raanan
Quantitative
Schul Implicit Function Theorem for Lipschitz Functio
Motivation
Test for Nontriviality
Main Theorem
Tools
Fibers
Implicit Function Theorem
A Non-Example
How well does this also hold for f : Rn × Rm → Mn ?
f ◦ g −1
f ([0, 1]2 )
g
E
f
g (E )
When does there exist E ⊆ Rn+m and g : Rn+m → Rn+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?
Jonas Azzam, University of Washington-Seattle (MSRI) work with
A Raanan
Quantitative
Schul Implicit Function Theorem for Lipschitz Functio
Motivation
Test for Nontriviality
Main Theorem
Tools
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 E1 , ..., EM , G of [0, 1]n s.t.
1
h|Ei is 1ǫ -bi-Lipschitz,
2
n (h(G )) < ǫ.
H∞
Corollary
n (f ([0, 1]n )), there is C = C (ǫ) > 0 and E ⊆ [0, 1]n s.t.
If ǫ < 12 H∞
1
|E | ≥ C
2
h|E is 1ǫ -bi-Lipschitz.
n (f ([0, 1]n ) > 0 substitutes Jf > 0.
The condition H∞
Jonas Azzam, University of Washington-Seattle (MSRI) work with
A Raanan
Quantitative
Schul Implicit Function Theorem for Lipschitz Functio
Motivation
Test for Nontriviality
Main Theorem
Tools
Fibers
Implicit Function Theorem
A Non-Example
Obstacle
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.
rankDx 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 .
Jonas Azzam, University of Washington-Seattle (MSRI) work with
A Raanan
Quantitative
Schul Implicit Function Theorem for Lipschitz Functio
Motivation
Test for Nontriviality
Main Theorem
Tools
Fibers
Implicit Function Theorem
A Non-Example
Kaufman-type example
Let G ⊆ R2 denote the Garnett example, rotated and scaled
so that P1 (G ) = [0, 1].
G × G is another Cantor set so that P1,3 (G × G ) = [0, 1]2 .
Jonas Azzam, University of Washington-Seattle (MSRI) work with
A Raanan
Quantitative
Schul Implicit Function Theorem for Lipschitz Functio
Motivation
Test for Nontriviality
Main Theorem
Tools
Fibers
Implicit Function Theorem
A Non-Example
Let T ⊆ R4 be a quasiconvex tree whose leaves are G × G .
Let r : R4 → T be a Lipschitz retract.
Then f = P1,3 ◦ r : [0, 1]4 → [0, 1]2 is s.t. rankDx f = 1 a.e..
Why is this an obstacle?
Jonas Azzam, University of Washington-Seattle (MSRI) work with
A Raanan
Quantitative
Schul Implicit Function Theorem for Lipschitz Functio
Motivation
Test for Nontriviality
Main Theorem
Tools
Fibers
Implicit Function Theorem
A Non-Example
Moral
If we want a function f 7→ δ(f ) that guarantees f ◦ g −1 |E = P for
some E with |E | ≥ δ(f ), we should have δ(Kaufman) = 0.
Jonas Azzam, University of Washington-Seattle (MSRI) work with
A Raanan
Quantitative
Schul Implicit Function Theorem for Lipschitz Functio
Motivation
Test for Nontriviality
Main Theorem
Tools
Definition
A Test for Nontriviality
Define the (n,m)-content of a function f : Q ⊆ Rn+m → M
n,m
H∞
(f , Q) = inf
X Hn (f (Qj ))
∞
ℓ(Qj )n+m
ℓ(Qj )n
j
where the infimum is over all collections of disjoint dyadic Qj s.t.
[ Q\ Qj = 0.
(Note the similarity to the coarea formula.)
Jonas Azzam, University of Washington-Seattle (MSRI) work with
A Raanan
Quantitative
Schul Implicit Function Theorem for Lipschitz Functio
Motivation
Test for Nontriviality
Main Theorem
Tools
Moral
Suppose there is a set E and g : Rn+m → Rn+m L-bi-Lipschitz s.t.
f ◦ g −1 |E is a projection map.
2,2
Kauffman Example: H∞
(f , [0, 1]n+m ) = 0
n,m
Orthogonal projection: H∞
(f , [0, 1]n+m ) = 1
n,m
ǫ-scaling: H∞
(f , [0, 1]n+m ) = ǫn .
The content will determine lower and upper bounds for the size of
E and L respectively.
Jonas Azzam, University of Washington-Seattle (MSRI) work with
A Raanan
Quantitative
Schul Implicit Function Theorem for Lipschitz Functio
Motivation
Test for Nontriviality
Main Theorem
Tools
Theorem (Simplified version)
n,m
Let f : Rn × Rm → Rn be 1-Lipschitz, δ = H∞
(f , [0, 1]n+m ) > 0.
Then ∃ g : Rn+m → Rn+m and E ⊆ [0, 1]n+m s.t.
(a) Hn+m (E ) ≥ η(n, m, δ) > 0, and
(b) g is CLip -bi-Lipschitz, CLip = CLip (n, m, δ) > 1,
(c) f ◦ g −1 |g (E ) = PRn .
R2
f
g (E )
f ◦ g −1
E
R
Jonas Azzam, University of Washington-Seattle (MSRI) work with
A Raanan
Quantitative
Schul Implicit Function Theorem for Lipschitz Functio
Motivation
Test for Nontriviality
Main Theorem
Tools
Theorem (Quantitative Implicit Function Theorem, (A., Schul))
Let f : Rn × Rm → M be 1-Lipschitz, 0 < Hn (M) ≤ 1, and
n,m
δ = H∞
(f , [0, 1]n+m ) .
Then ∃ g : Rn+m → Rn+m and E ⊆ [0, 1]n+m s.t. if F = f ◦ g −1 :
(i) Hn+m (E ) ≥ η(n, m, δ) > 0
(ii) g is CLip -bi-Lipschitz, CLip = CLip (n, m, δ) > 1,
(iii) for x ∈ Rn , F (x, ·) is constant on ({x} × Rm ) ∩ g (E ),
(iv) for y ∈ Rm , F (·, y ) is CLip -bi-Lipschitz on (Rn × {y }) ∩ g (E ).
F
f ([0, 1]2 )
g
E
f
g (E )
Jonas Azzam, University of Washington-Seattle (MSRI) work with
A Raanan
Quantitative
Schul Implicit Function Theorem for Lipschitz Functio
Motivation
Test for Nontriviality
Main Theorem
Tools
bi-Lipschitz Extensions
Carleson Estimates
Typical bi-Lipschitz extension results involve some sort of
compromise: Increasing dimension
Theorem (David and Semmes)
If f : E → Rn is bi-Lipschitz, E ⊆ Rd compact, then f may be
extended to a bi-Lipschitz injection f : Rn → Rmax{n,2d+1} .
Theorem (MacManus, ’95)
f : E ⊆ Rn → Rn bi-Lipschitz may be extended to a bi-Lipschitz
homeomorphism of R2n onto itself.
Jonas Azzam, University of Washington-Seattle (MSRI) work with
A Raanan
Quantitative
Schul Implicit Function Theorem for Lipschitz Functio
Motivation
Test for Nontriviality
Main Theorem
Tools
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 ⊆ RD satisfy the
following:
There is L0 = L0 (E , D) > 1 s.t. if f : E → RD is
L-bi-Lipschitz, L ∈ [1, L0 ), then f has a bi-Lipschitz extension
f : RD → RD .
(e.g. E = Rn , Sn , 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.
Jonas Azzam, University of Washington-Seattle (MSRI) work with
A Raanan
Quantitative
Schul Implicit Function Theorem for Lipschitz Functio
Motivation
Test for Nontriviality
Main Theorem
Tools
bi-Lipschitz Extensions
Carleson Estimates
Theorem (Bi-Lipschitz Extension on Large Pieces (A., Schul))
Let D ≥ n and f : [0, 1]n → RD be 1-Lipschitz. For any ǫ satisfying
n (f [0, 1])n ), there is E ⊆ [0, 1]n and F : Rn → RD s.t.
0 < ǫ < 12 H∞
F is L-bi-Lipschitz L .D 1ǫ ,
if D = n, then F : Rn → Rn is a homeomorphism.
|E | ≥ η(n, D, ǫ) > 0
F |E = f |E .
Jonas Azzam, University of Washington-Seattle (MSRI) work with
A Raanan
Quantitative
Schul Implicit Function Theorem for Lipschitz Functio
Motivation
Test for Nontriviality
Main Theorem
Tools
bi-Lipschitz Extensions
Carleson Estimates
In the case f : Rn+m → Mn ⊆ RD , we need a theorem that says
there is a large cube where f is close to being affine.
Theorem (Dorronsoro, ’85)
For F : Rn → RD Lipschitz, and Q ⊆ Rn , let
α(Q) = inf{sup
x∈Q
|F (x) − A(x)|
:
diamQ
A : Rn → D is affine}.
Then
X
|Q| .ǫ,n |R|.
α(Q)>ǫ,Q⊆R
We need to develop an analogue for f : Rn → Mn .
Jonas Azzam, University of Washington-Seattle (MSRI) work with
A Raanan
Quantitative
Schul Implicit Function Theorem for Lipschitz Functio
Motivation
Test for Nontriviality
Main Theorem
Tools
bi-Lipschitz Extensions
Carleson Estimates
Theorem (Kirchheim ’94)
If f : Rn → M is Lipschitz, then for a.e. z ∈ Rn ,
||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|).
Jonas Azzam, University of Washington-Seattle (MSRI) work with
A Raanan
Quantitative
Schul Implicit Function Theorem for Lipschitz Functio
Motivation
Test for Nontriviality
Main Theorem
Tools
bi-Lipschitz Extensions
Carleson Estimates
Theorem (A., Schul)
For α > 0, F : Rn+m → Mn ,
G = {Q : ∃|| · || s.t. |F (x) − F (xQ ))| − ||x − xQ || < αℓ(Q)}.
Then
X
|Q| .α,n 1.
Q∈∆([0,1]n )\G
Jonas Azzam, University of Washington-Seattle (MSRI) work with
A Raanan
Quantitative
Schul Implicit Function Theorem for Lipschitz Functio