Departments of Mathematics
Montana State University
Fall 2015
Prof. Kevin Wildrick
An introduction to non-smooth analysis and geometry
Lecture 15: The Proof of Cheeger’s Theorem
1. The Lip-lip condition and finite dimensionality
In this section, we will show
Theorem 1.1. Suppose that (X, d, µ) is a complete doubling metric space that satisfies
the K-Lip-lip condition. Then there is a constant N0 such that for almost every x ∈ X,
every (N0 + 1)-tuple of Lipschitz functions is dependent to first order at x.
As we have seen in the past, if (X, d, µ) has a measurable differentiable structure, then
the conclusion of Theorem 1.1 is valid. We will see that in fact the two are essentially
equivalent.
Let f1 , . . . fN : X → R be a collection of Lipschitz functions so that the set of points
ind where they are not dependent has positive µ measure. We must bound the number
N from above.
We consider the collection of all rational linear combinations of f1 , . . . , fn :
F := {
N
X
λi fi : X → R : λ1 , . . . , λN ∈ Q}.
i=1
Note that F is a countable collection.
Definition 1.2. Let > 0. A set Y ⊆ X is -good for F if for every f ∈ F, there is a
scale rf > 0 such that if 0 < r < rf and x ∈ Y , the n
Lip f (x)
− ≤ lip f (x) − ≤ var f (x, r) ≤ Lip f (x) + .
K
We say that Y is good for F if it is -good for all > 0.
Notice that the first inequality in the above definition is valid for µ-almost every x ∈ X
by the K-Lip-lip condition.
Lemma 1.3. Let Y0 be any measurable subset of X of finite (and positive) measure. For
every δ > 0, there is a subset Y ⊆ Y0 so that µ(Y0 \Y ) < δ and Y is good for F.
Proof. For f ∈ F, recall that
lip f (x) = lim inf var f (x, r) ≤ lim sup var f (x, r) = Lip f (x).
r→0
r→0
By a version of Egoroff’s theorem for lim inf and lim sup, for any δ > 0, there is a set
Y0 (f, δ) ⊆ Y0 so that the convergence of the lim inf and lim sup above is uniform on
Y0 (f, δ) and µ(Y0 \Y0 (f, δ)) < δ. In other words, for > 0, there is r such that if r < r ,
then for all x ∈ Y0 (f, δ)
lip f (x) − ≤ var f (x, r) ≤ Lip f (x) + .
This is just another way of saying that Y0 (f, δ) is good for f .
Define
i+1
[ 2−(i+1)
Yi+1 =
Yi f j , δ
.
i
+
1
j=1
1
Then Y0 ⊇ Y1 ⊇ Y2 . . . , and so we set
Y =
∞
\
Yi
i=1
and see that
µ (Y0 \Y 0 ) ≤ δ.
Then Y is good for each function in F.
Let Y be a subset of ind satisfying the conclusions of Lemma 1.3. By Lusin’s theorem
we may also assume that for each f ∈ F, the function lip f is continuous on Y .
Lemma 1.4. Let x ∈ Y be a point of density of Y . Let (X∞ , d∞ , x∞ ) be a GromovHausdorff tangent of X at x such that each f ∈ F has a tangent function f∞ : X∞ → R.
Then, for every f ∈ F, every p ∈ X∞ and every r > 0,
Lip f (x)
≤ var f∞ (p, r) ≤ Lip f (x).
K
In particular, the Lipschitz constant of f∞ is no greater than Lip f (x).
Proof. The novelty of this lemma, compared to what we did in the last lecture, is that we
estimate var f∞ at every point of X∞ , not just at the basepoint. Let {φi : (X∞ , d∞ , x∞ ) →
(X, d/ri , x)} be a Hausdorff approximation. We begin with an exercise: since x is a density
point of Y and µ is doubling, we may find another Hausdorff approximation
{φ0i : (X∞ , d∞ , x∞ ) → (Y, d/ri , x)}.
To complete this exercise, you may want to use the lemma involved in the proof of
Stepanov’s theorem.
We have seen that
f ◦ φ0i − f (x)
f∞ = lim
i→∞
ri
uniformly on compact subsets of X∞ . Fix p, q ∈ X∞ , and set pi = φ0i (p) and qi = φ0i (q).
Then for each > 0, for all sufficiently large i ∈ N, we use the fact that Y is good for F
to see that
|f∞ (p) − f∞ (q)|
|f (pi ) − f (qi )|
≤
+
d∞ (p, q)
d(pi , qi )
≤ (1 + ) var f (pi , (1 + )d(pi , qi )) + ≤ Lip f (pi ) + 2.
Notice that as the sequence of scales ri is tending to 0, it holds that d(pi , x) → 0. Hence,
the continuity of Lip f on Y implies that for sufficiently large i,
Lip f (pi ) ≤ Lip f (x) + .
An easy argument now shows that
var f∞ (p, r) ≤ Lip f (x)
as well as the final statement of the lemma.
For the other inequality, for > 0, we may choose ai ∈ B(pi , (r − )ri )) so that
var f (pi , (r − )ri )) ≤
f (pi ) − f (ai )
+
(r − )ri
Since pi tends to x, and x is a density point of Y , an argument as we have seen in the
proof of Stepanov’s theorem implies that we may assume that ai is a point of Y , and
hence ai = φ0i (vi ) for some vi ∈ B∞ (p, r). Thus
var f (pi , (r − )ri )) ≤
|fi (p) − fi (vi )|
+ .
(r − )
Since fi converges uniformly to f∞ on compact sets, we see that for sufficiently large
i ∈ N,
var f (pi , (r − )ri )) ≤
|f∞ (p) − f∞ (vi )|
r
+ 2 ≤
var f∞ (p, r) + 2.
r−
r−
Since Lip f is continuous on Y ,
Lip f (x) = lim Lip f (pi ).
i→∞
Since Y is good for f , for sufficiently large i,
Lip f (pi ) ≤ K(var f (pi , (r − )ri )) + ).
Thus, combining and sending → 0, we see that
Lip f (x) ≤ K var f∞ (p, r).
Let F∞ be a collection of tangent functions as in the statement of Lemma 1.4 (we
know such a collection exists by Gromov’s compactness theorem, as in the last lecture).
Note that F∞ is the Q-span of {(f1 )∞ , . . . , (fN )∞ ). Since each element of F∞ satisfies the
uniform quasilinearity condition (i.e., the conclusion of Lemma 1.4), so does each element
of the R-span of {(f1 )∞ , . . . , (fN )∞ }. Denote this R-span by F ∞ .
Now, recall that (X∞ , d∞ ) is a doubling metric space. Consider a maximal 1/(4K)separated set T in B∞ (x∞ , 1). Then card T is finite, and depends only on K and the
doubling constant.
Consider that the dimension of L∞ (T ) is card T , and that there is a natural restriction
mapping from F ∞ to L∞ (T ) given by f∞ 7→ (f∞ )|T . Suppose that (f∞ )|T is zero, so
that f∞ is in the kernel of this mapping. Let z ∈ B := B∞ (x∞ , 1), and let z 0 ∈ T with
d∞ (z, z 0 ) ≤ 1/(4K). Then
|f∞ (z)| = |f∞ (z) − f∞ (z 0 )| ≤
Lip f (x)
var f∞ (x∞ , 1)
≤
.
4K
4
By the triangle inequality,
var f∞ (x∞ , 1) ≤ 2||f∞ |B ||L∞ ,
and so the above argument shows that ||f∞ |B ||L∞ = 0. This implies that f∞ |B = 0, and
so var f∞ (x∞ , 1) = 0. By the quasilinearity condition, this implies that the variation on
any ball is 0, and so f∞ is constantly 0. This means that F ∞ is the injective image of
L∞ (T ), and so its dimension is at most card T .
Suppose that N > card T . Then there are b1 , . . . , bN ∈ R such that
n
X
bi (fi )∞ = 0.
i=1
We may write bi = limk→∞ ai,k where each ai,k is rational. Then
n
X
i=1
ai,k (fi )∞
converges uniformly on compact subsets of X∞ to the zero function as k → ∞. This
implies that for any p ∈ X∞ and r > 0,
!
n
X
lim var
ai,k (fi )∞ (p, r) = 0.
k→∞
Since each of the functions
Pn
i=1
n
X
i=1
ai,k (fi )∞ is uniformly quasilinear, and
!
n
X
ai,k (fi )∞ =
ai,k fi
,
i=1
i=1
∞
this implies that
lim Lip
k→∞
n
X
!
ai,k fi
(x) = 0.
i=1
Finally,
Lip
n
X
i=1
!
bi f i
(x) ≤ lim sup Lip
k→∞
n
X
i=1
!
ai,k fi
(x) + Lip
n
X
!
(bi − ai,k )fi
(x) = 0.
i=1
This is exactly the statement that the fi are dependent to first order at x, as desired.