Departments of Mathematics
Montana State University
Fall 2015
Prof. Kevin Wildrick
An introduction to non-smooth analysis and geometry
Lecture 8: An introduction to Cheeger Differentiability
1. An interpretation of differentiability
Let f : Rn → R be a function, and let z0 ∈ Rn . We say that f is differentiable at z0 if
there is a linear function Lz0 : Rn → R such that
lim
z→z0
|f (z) − f (z0 ) − Lz0 (z − z0 )|
= 0,
||z − z0 ||Rn
which we write in shorthand as
f (z) − f (z0 ) = Lz0 (z − z0 ) + o(||z − z0 ||) as z → z0 ,
and say “f is differentiable at z0 if there is a linear function approximating it to first order
at z0 .”
Let {e1 , . . . , en } be the standard basis of Rn . Consider the coordinate functions xj : Rn →
R defined by
xj (w) := hw, ej i
so that
w=
n
X
xj (w)ej .
j=1
Clearly, these functions, and linear combinations thereof, are linear function.
Using these functions, for any linear mapping L we may “express it in the standard
basis” as:
n
X
L(w) =
L(ej )xj (w).
j=1
Note that each L(ej ) is just a real number. Hence, we have characterized linear mappings
as linear combinations of the coordinate functions.
In this way, we can say that f is differentiable at z0 if there is a linear combination of
the coordinate functions approximating it to first order at z0 .
Moreover, we note that the coordinate functions are 1-Lipschitz. This leads us to
the first big idea: we can in fact characterize differentiability in Rn almost everywhere,
without using any thing more than the metric on Rn , by replacing the concept of coordinate
functions with the concept of Lipschitz functions!
Consider the following:
Definition 1.1 (Toy definition). Fix Lipschitz functions φ1 , . . . , φn : Rn → R. We say
that a function f : Rn → R is Cheeger differentiable at z0 ∈ Rn with respect to (φ1 , . . . , φn )
if there are constants c1 , . . . , cn ∈ R such that
!
n
X
f (z) − f (z0 ) =
cj (φj (z) − φj (z0 )) + o(||z − z0 ||) as z → z0 .
j=1
1
Definition 1.2 (Toy definition, metric version). Let (X, d, µ) be a metric measure space,
and let φ1 , . . . , φn : X → R be Lipschitz functions. We say that a function f : X → R
is Cheeger differentiable at z0 ∈ X with respect to (φ1 , . . . , φn ) if there are constants
c1 , . . . , cn ∈ R such that
!
n
X
f (z) − f (z0 ) =
cj (φj (z) − φj (z0 )) + o(d(z, z0 )) as z → z0 .
j=1
We have just shown above that in Euclidean space and for the choice φj = xj , j =
1, . . . , n, Cheeger differentiability and classical differentiability coincide.
Now, let φ1 , . . . , φn be arbitrary Lipschitz functions, and suppose that f : Rn → R
is Cheeger differentiable with respect to these functions at almost every z0 ∈ Rn . By
Rademacher’s theorem, all the functions c1 φ1 , . . . , cn φn are differentiable at z0 for almost
every z0 ∈ Rn , and hence, for almost every z0 ∈ Rn .
!
!
n
n
X
X
cj (φj (z) − φj (z0 )) =
cj ∇φj (z0 )(z − z0 ) + o(||z − z0 ||) as z → z0 ,
j=1
j=1
Hence we also have
f (z) − f (z0 ) =
n
X
!
cj ∇φj (z0 )(z − z0 )
+ o(||z − z0 ||) as z → z0 ,
j=1
Note that
w 7→
n
X
!
cj ∇φj (z0 )(w)
j=1
is a linear function on Rn , and so we see that f is classically differentiable almost everywhere as well.
Now, if we choose the functions φ1 , . . . , φn poorly, it may not be the case that a clasically
differentiable function is Cheeger differentiable. For example, suppose that φ1 = x1 and
φj = 0 for j = 2, . . . , n. Then the function f (x) = hx, e2 i is not Cheeger differentiable
with respect to these functions.
This leads us to the conclude that in order for Cheeger differentiability to be a reasonable concept, the choice of Lipschitz “coordinates” has to reflect the structure of the
space they are defined on.
This leads us to a deep question:
Question 1.3. Let (X, d, µ) be a metric measure space. Are there Lipschitz functions
φ1 , . . . , φn : X → R so that every Lipschitz function f : X → R is Cheeger differentiable
almost everywhere with respect to (φ1 , . . . , φn )?
2. Measurable Differentiable Structures
A metric measure space might be look rather different in different parts of the space.
For example, we might take two spaces and glue them together along a nice subset. Our
intuition says that apart from the set where we glued, the “almost everywhere differentiability” of Lipschitz functions on such a space should not be ruined under this gluing.
However, the structure of the two spaces might be very different. We are led to the
following definition.
Definition 2.1 (Cheeger). A measurable differentiable structure on a metric measure
space (X, d, µ) is a countable collection {(Xα , φα )} of pairs, called coordinate charts,
where
• : each Xα is a measurable subset of (X, d, µ) with µ(Xα ) > 0, and
!
[
µ X\ Xα ) = 0.
α
• for each α, there is a number N (α) ∈ N, called the dimension of the coordinate
chart, so that φα : X → RN (α) is a Lipschitz function; in other words, each φα is
N (α)
a tuple (φ1α , . . . , φα ) of Lipschitz functions from X to R,
• for each α, the coordinate function φα spans the differentials on Xα , i.e., for every
Lipschitz function f : X → R, there is a measurable function df α : Xα → RN (α)
so that for µ-almost every z0 ∈ Xα ,
(2.1)
f (z) − f (z0 ) = df α (z0 ) · (φα (z) − φα (z0 )) + o(d(z, z0 )) as z → z0 ,
• if g : Xα → RN (α) is another measurable function satisfying (2.1) for µ-almost
every z0 ∈ Xα , then g = df α , µ-almost everywhere,
• the dimensions N (α) are bounded above independently of α.
The dimension of the measurable differentiable structure is defined to be the quantity
max N (α).
α
The main goal of this course will be to find a class of metric measure spaces so that
(1) each space in the class posses a measurable differentiable structure,
(2) the class contains many classical smooth spaces,
(3) the class is stable under appropriate limits,
We can begin by giving some examples of spaces that posses a measurable differentiable
structure.
First, and of course most importantly, is Euclidean space Rn . Here we may choose
(Rn , id : Rn → Rn ) as the single coordinate chart. Then the condition that id span the
differentials is exactly the assertion of Rademacher’s theorem.
Next, it’s a simple definition chase to show that if (M, dg , µ) is a Riemannian manifold
equipped with its volume measure, then any covering collection of smooth charts also
forms a measurable differentiable structure. Again we will need to apply Rademacher’s
theorem, in the charts.
Now, suppose that E ⊆ Rn is any subset of positive Ln -measure. Then (E, || − ||Rn , Ln )
again supports a measurable differentiable structure simply by restriction. There is, of
course one small detail to check (you may have missed this in the previous example as
well) - uniqueness of the differential df α . This is an excellent exercise (hint:use density
points and Lipschitzness).
We can also glue structures together. Suppose (X, d, µ) and (Y, ρ, ν) are metric measure
spaces with measurable differentiable structures {(Xα , φα )} and {(Yβ , ψβ )}. Let E ⊆ X
and F ⊆`Y be subsets such that there is an isometry g : E → F , and consider any metric
d0 on X Y /g with the property that d0 |X×X = d and d0Y ×Y = ρ. Then
{(Xα , φα )} ∪ {(Yβ , ψβ )}
`
is a measurable differentiable structure on X Y /g.
3. The Cantor set has no measurable differentiable structure
It’s easy to see that an infinite dimensional normed vector space cannot have a measurable differentiable structure. However, it’s a more subtle to construct a doubling metric
space that has no measurable differentiable structure. To do so, we’ll use the following
convenient terminology:
Definition 3.1. Let (X, d, µ) be a metric measure space and let z0 ∈ X. We say that a
collection of functions {f1 , . . . , fn : X → R} is dependent to first order at z0 if there is a
vector (λ1 , . . . , λn ) ∈ Rn \{0} such that
n
X
λi (fi (z) − fi (z0 )) = o(d(z, z0 )) as z → z0 in X.
i=1
Definition 3.2. Let (X, d, µ) be a metric measure space. We say that the Lipschitz
differentials in X have dimension at most N if every collection of (N + 1) Lipschitz
functions is dependent to first order at µ-almost every point of X.
Proposition 3.3. If (X, d, µ) supports a measurable differentiable structure with with
dimension N , then the Lipschitz differentials in X have dimension at most N .
Proof. Let f1 , . . . , fN +1 : X → R be an (N + 1)-tuple of Lipschitz functions on X. Let
(Xα , φα ) be a chart, and consider that for any z0 ∈ Xα the vectors df1α (z0 ), . . . , df αN+1 (z0 ) ∈
RN(α) are necessarily linearly dependent. Hence there is λ ∈ RN +1 \0 such that
N
+1
X
λi dfiα (z0 ) = 0.
i=1
Moreover, at µ-almost every point z0 ∈ Xα , for every i = 1, . . . , N + 1,
fi (z) − fi (z0 ) = dfiα (z0 ) · (φα (z) − φα (z0 )) + o(d(z, z0 )) as z → z0 .
Hence
N
+1
X
λi (fi (z) − fi (z0 )) =
i=1
N
+1
X
λi (dfiα (z0 ) · (φα (z) − φα (z0 )) + o(d(z, z0 )))
i=1
=
N
+1
X
!
λi dfiα (z0 ) · (φα (z) − φα (z0 ))
+ o(d(z, z0 ))
i=1
= o(d(z, z0 ).
Since this is true at almost every point of an arbitrary chart, it is true almost everywhere.
In fact, the converse of Proposition 3.3 is also true, as we will later show.
We will use Proposition 3.3 to show that the standard 1/3-Cantor set does not support
a measurable differentiable structure.
Recall that the 1/3-Cantor set is constructed as follows. Let S denote the collection
of all finite or countably infinite sequences with entries in {0, 1}. Denote the length of a
sequence σ ∈ S by |σ|. We declare that the empty sequence has length 0.
Define I∅ = [0, 1]. Now, suppose that Iσ has been defined for all sequences σ ∈ S
with |σ| = k ∈ N. We then define Iσ,0 and Iσ,1 to be the left-most and right-most closed
subintervals of Iσ with length 1/3 of that of Iσ .
For each k ∈ N, let
[
Ck =
Iσ .
σ∈S,|σ|=k
The Cantor set C is defined by
C=
\
Ck .
k∈N
Further, for each number k ≥ 1, set
Uk = Ck−1 \Ck .
Then Uk is the union of 2
k−1
open intervals of length 3−k .
The Cantor set has a natual “uniformly distributed measure” µ. It can described in
many ways:
• Define a sequence of probability measures µk on [0, 1] that restrict to Lebesgue
measure on each Iσ with |σ| = k. These measures converge weakly to a measure
µ on [0, 1] that is supported on C,
• normalized log3 2-dimensional Hausdorff measure,
• consider C as the compact topological group of infinite binary sequences; then let
µ be the Haar measure on this group,
• consider the Cantor function c : C → [0, 1], a continuous surjection, and consider
a measure µ on C satisfying µ(E) = L(c(E)).
Of course, we endow C with the restriction of the Euclidean metric on R. Let’s check
that µ is a doubling metric measure space. Let x ∈ C, and let r > 0; without loss of
generality we may assume that r ≤ 1. Find k ∈ N such that 3−(k+1) < r ≤ 3−k . By
definition x ∈ Iσ for some σ ∈ S with |σ| = k + 1. Hence, B(x, r) ⊇ Iσ ∩ C and so
µ(B(x, r)) ≥ 2−k+1 ≥ rlog3 2 /2.
On the otherhand, the 3−k -neighborhood of Iσ in C consists of the parent I of Iσ ; hence
µ(B(x, r)) ≤ 2−k ≤ 2rlog3 2 .
Thus
µ(B(x, 2r)) ≤ 21+log3 2 rlog3 2 ≤ 22+log3 2 µ(B(x, r)).
Theorem 3.4. There is no N ∈ N so that the Lipschitz differentials in (C, | · |, µ) have
dimension at most N .
Proof. For an infinite sequence σ ∈ S denote the kth entry of σ by σ(k), and define
uσ : [0, 1] → [0, 1] by
X
uσ =
σ(k)χUk .
k∈N
Then define vσ : [0, 1] → [0, 1] by
Z
vσ (x) =
x
uσ (t) dL1 (t).
0
We may of course consider the restriction of vσ to C. Suppose that σ1 , . . . , σN are infinite
sequences in S, and suppose that the corresponding functions vσ1 , . . . , vσN are dependent
to first order at almost every point z0 ∈ C. Let z0 be such a point, and let (λ1 , . . . , λN ) ∈
RN \{0} be such that
N
X
λi (vσi (z) − vσi (z0 )) = o(|z − z0 |) as z → z0 in C.
i=1
We claim that
(3.1)
lim
k→∞
N
X
λi σi (k) = 0.
i=1
Towards this end, let > 0, and choose δ > 0 so that if |z − z0 | < δ and z ∈ C, then
N
X
λi (vσi (z) − vσi (z0 )) ≤ |z − z0 |.
i=1
Choose k0 to be the smallest natural number so that 2 · 3−k0 < δ. Let k ≥ k0 . Then
z0 ∈ Iσ for some σ ∈ S with |σ| = k. Let U be the interval in Uk adjacent to Iσ that is
further away from z0 . Denote the endpoints of U by y and y 0 where y 0 is between y and
z0 . Then,
N
N
N
X
X
X
0 0 λi (vσi (z0 ) − vσi (y ))
λi (vσi (y) − vσi (z0 )) + λi (vσi (y) − vσi (y )) ≤ i=1
i=1
i=1
0
≤ (|y − z0 | + |y − z0 |)
≤ 33−k .
On the otherhand,
vσi (y) − vσi (y 0 ) = σi (k)3−k ,
so that
N
N
X
X
0 −k λi σi (k) ,
λi (vσi (y) − vσi (y )) = 3 i=1
i=1
showing that
N
X
λi σi (k) ≤ 3
i=1
and proving the claim.
It remains to show the following: for any N , there is a collection σ1 , . . . , σN of infinite
sequences in S so that for every vector (λ1 , . . . , λN ) ∈ RN \{0}, the limit (3.1) fails to
hold. Let
(
1 k = i mod N,
σi (k) =
0 k 6= i mod N.
Then
N
X
i=1
(
λi
λi σi (k) =
0
k=i
k 6= i
mod N,
mod N.
Since (λ1 , . . . , λN ) is not the zero vector, at least one of the limits
N
X
lim λi σi (ik) = lim |λi | =
6 0.
k→∞ k→∞
i=1