Departments of Mathematics
Montana State University
Fall 2015
Prof. Kevin Wildrick
An introduction to non-smooth analysis and geometry
Lecture 7: The local Lipschitz constant and Stepanov’s Theorem
1. The Local Lipschitz constant
Rademacher’s theorem is clearly non-optimal in the sense that the Lipschitz condition is
global, while differentiation is a local issue. It is indeed possible to improve Rademacher’s
theorem to a more appropriate local statement.
The key quantity is a local version of the Lipschitz condition.
Definition 1.1. Let f : X → Y be a mapping between metric spaces, and let x ∈ X. We
define the (upper) local Lipschitz constant at x by
supy∈B(x,r) dY (f (x), f (y))
.
r
r→0
Similarly, define the lower local Lipschitz constant at x by
supy∈B(x,r) dY (f (x), f (y))
.
lip f (x) := lim inf
r→0
r
Theorem 1.2 (Stepanov’s Theorem). Let f : Rn → R be a mapping. Then f is differentiable at almost every point of the set
Lip f (x) := lim sup
Sf := {x ∈ Rn : Lip f (x)} < ∞.
2. Extension of Lipschitz functions
In order to prove Stepanov’s Theorem, we will first need another pillar of analysis: the
Lipschitz extension theorem. The proof of this result is easy, but there is a large body of
much more difficult results that vastly generalize it.
Theorem 2.1 (McShane-Whitney Lipschitz Extension). Let A be a non-empty subset of
an arbitrary metric space, and let f : A → R be an L-Lipschitz function. Then there is
an L-Lipschitz function F : X → R such that F |A = f .
Proof. Consider, for each a ∈ A, the globally defined function fa : X → R given by
fa (x) = f (a) + Ld(x, a).
Then, by the triangle inequality,
|fa (x) − fa (y)| = L|d(x, a) − d(y, a)| ≤ Ld(x, y),
and so fa is again L-Lipschitz. We will show that
F (x) := inf fa (x)
a∈A
is also L-Lipschitz, and restricts to f on A .
First note that if a ∈ A, then
F (a) ≤ fa (a) = f (a) < ∞,
0
and moreover, if a ∈ A is an arbitrary point of A, then
fa0 (a) = f (a0 ) + Ld(a, a0 ) ≥ f (a)
by the Lipschitz condition. This shows that F |A = f.
1
Now, for x, y ∈ X and a ∈ A, we see that
fa (y) ≥ fa (x) − Ld(x, y) ≥ F (x) − Ld(x, y).
Hence
F (y) ≥ F (x) − Ld(x, y),
which implies (using the finiteness of F )
F (x) − F (y) ≤ Ld(x, y).
Exchanging the roles of x and y yields the desired result.
3. Density points and derivatives
Lemma 3.1. Let (X, d, µ) be a doubling metric measure space, let A ⊆ X be a measurable
set, and let x0 be a density point of A. Then, for each > 0, there exists a radius r > 0
such that for each x ∈ B(x0 , r), there is a point y ∈ A with d(x, y) ≤ d(x, x0 ).
Proof. If not, then there is > 0 and a sequence of points {xj }j∈N Rn tending to x0 such
that
A ∩ B(xj , d(xj , x0 )) = ∅.
This implies that
µ(A ∩ B(x0 , (1 + )d(xj , x0 ))
µ(B(xj , d(xj , x0 ))
≤1−
µ(B(x0 , (1 + )d(xj , x0 ))
µ(B(xj , (1 + )d(xj , x0 ))
s
1
≤1−
<1
C 1+
where C and s are the constants associated to the homogeneity condition for µ (See
Lecture 3 Proposition 2.3). Letting xj tend to x0 now violates the density condition. 4. The proof of Stepanov’s Theorem
Proof of Stepanov’s Theorem. For j ∈ N, consider the sets
1
|f (x) − f (y)|
n
Cj := x ∈ R : |x − y| <
=⇒
≤j .
j
|x − y|
We first claim that
[
Cj .
Sf =
j∈N
If x ∈
/
S
j∈N Cj , then there is a sequence of points {yj }j∈N such that |x − yj | < 1/j and
|f (x) − f (yj )|
> j.
|x − yj |
Let rj = 2|x − yj |, then
Lip f (x) ≥ lim
supy∈B(x,rj ) |f (x) − f (y)|
j→∞
rj
j
,
j→∞ 2
≥ lim
showing that x ∈
/ Sf . On the other hand, if x ∈
/ Sf , then we may find a sequence of scales
rj & 0 and a sequence of points yj ∈ B(x, rj ) such that
|f (x) − f (yj )|
= ∞.
j→∞
rj
lim
By passing to a subsequence, we may assume that for each j ∈ N, it holds that rj < 1/j
and
|f (x) − f (yj )|
≥ j.
rj
Now, |x − yj | ≤ rj implies that
|f (x) − f (yj )|
≥j
|x − yj |
as well. This shows that x ∈
/ Cj for each j ∈ N.
Although this isn’t necessary for the proof, note that each Cj is closed. If {xk }k∈N ∈ Cj
tend to a point x, and |x − y| < 1/j, then |xk − y| < 1/j as well for sufficiently large k,
and hence
|f (x) − f (y)| ≤ lim |f (x) − f (xk )| + |f (xk ) − f (y)| ≤ lim j|x − xk | + j|xk − y| = j|x − y|.
k→∞
j→∞
This implies that x ∈ Cj .
This also gives us a hint as to how to proceed; if x and x0 are elements of Cj with the
property that |x − x0 | < 1/j, then by definition
|f (x) − f (x0 )| ≤ j|x − x0 |.
S
Hence, if we write Cj = k∈N Cj,k where each Cj,k has diameter at most 1/j, then f |Cj,k
is j-Lipschitz. To find such sets, one can extract a countable subcover of the cover
{B(x, 1/2j)}x∈Cj and intersect with Cj . This construction also implies that each Cj,k is
closed.
Now, using the McShane-Whitney extension theorem, we may extend each f |Cj,k to a
globally defined j-Lipschitz function Fj,k , which, by Rademacher’s theorem, is differentiable almost everywhere, and in particular, at almost every point of Cj,k . It remains to
show that f is also differentiable at almost all of the points of Cj,k ; we will show that f
is differentiable at each density pont of Cj,k .
To this end, let x0 be a density point of C := Cj,k at which F := Fj,k is differentiable,
and let > 0. By Lemma 3.1, there is a radius r > 0 such that if x ∈ B(x0 , r), then there
is a point y ∈ C with |x − y| < |x − x0 |. For r < 1/j, the definition of C implies that
|f (x) − f (y)| ≤ j|x − y| ≤ j|x − x0 |.
Hence,
|f (x) − f (x0 ) − ∇F (x0 )(x − x0 )|
|F (y) − F (x0 ) − ∇F (x0 )(y − x0 )|
≤ j+
(1 + j)
|x − x0 |
|y − x0 |
+ |∇F (x0 )|j.
If x → x0 , then y → x0 as well. Hence, the differentiability of F at x0 implies that
|f (x) − f (x0 ) − ∇F (x0 )(x − x0 )|
lim
≤ j + |∇F (x0 )|j.
x→x0
|x − x0 |
Letting tend to 0 shows that f is differentiable at x0 with ∇f (x0 ) = ∇F (x0 ).