Departments of Mathematics
Montana State University
Fall 2015
Prof. Kevin Wildrick
An introduction to non-smooth analysis and geometry
Lecture 14: The Lip-lip condition
1. The Lip-lip condition
Let (X, d, µ) be a doubling metric measure space, and suppose that f0 : X0 → R is a
tangent function of a Lipschitz function f : X → R at p. We have shown last time that
lip f (p) ≤ var f0 (p0 , r) ≤ Lip f (p).
If we wish to have the limit function f0 play the role of a “linear function”, we will need
to have more precise control.
Definition 1.1. A metric measure space (X, d, µ) has the Lip-lip condition if there is a
constant K ≥ 1 such that for every Lipschitz function f : X → R,
Lip f (x) ≤ K lip f (x)
for µ-almost every x ∈ X.
If (X, d, µ) is a Lip-lip space, and f0 is a tangent function of a Lipschitz function f on
X at p, then for almost every choice of p ∈ X,
Lip f (p)
≤ var f0 (p0 , r) ≤ Lip f (p).
K
This indicates that f0 is similar to a linear function.
2. The Poincaré Inequality and the Lip-lip condition
Theorem 2.1. Suppose that a complete doubling metric measure space (X, d, µ) admits
a p-Poincaré inequality for some p ≥ 1. Then there is a constant K ≥ 1, depending only
on the doubling and Poincaré inquality constants, so that (X, d, µ) is a K-Lip-lip space.
We will need two lemmas:
Lemma 2.2. A doubling metric measure space that supports a p-Poincaré inequalit for
some p ≥ 1 is quasiconvex, i.e., there is a constant C0 ≥ 1 such that for any pair of
points x, y ∈ X, there is a path γ from x to y with
length(γ)Cd(x, y).
d
Proof. By iteration, it suffices to show the following: there is a constant C0 ≥ 1such that
for any pair of points x, y ∈ X, there is a path γ connecting a point of Ax = B x, d(x,y)
4
d(x,y)
to a point of Ay = B y, 4
.
A convenient notion in this proof is that of an -path: for > 0, a sequence x0 , . . . xk is
called
P an -path if consecutive points have distance less than . The length of an -path
is
d(xi , xi+1 ).
By the Arzela-Ascoli theorem (again!), it suffices to show that for any > 0, there
is an -path connecting a point of Ax to a point of Ay . To this end, for K ≥ 1, define
fK : X → [0, K] by
fK (z) = min{K, the minimal length of an -path connecting z and Ay }.
1
Note that Then fK is Iocally 1-Lipschitz and vanishes on Ay . Let
5d(x, y)
M = inf fK (z), and B = B y,
.
z∈Ax
4
Since µ is doubling,
µ(Ax ) ' µ(Ay ) ' µ(B).
Either 0 or M is at least M/2 away from the average value (fK )B , so
Z
− |fK − (fK )B | dµ & M/2.
B
However, the Poincaré inequality implies
Z
− |fK − (fK )B | dµ . Cd(x, y).
B
Thus M . d(x, y). Choosing K large enough now yields the result.
We note that the Lip-lip condition is invariant under bi-Lipschitz changes of metric:
if Φ : (X, dX ) → (Y, dY ) is bi-Lipschitz, and (X, dX ) has the Lip-lip condition, then so
does (Y, dY ). The same is true for the completeness and the Poincaré inequality (with the
pushed forward measure). Furthermore, a quasiconvex metric space (X, d) is bi-Lipschitz
equivalent to its induced path-metric space (X, dlength ). It is a nice exercise to show that
a locally compact and complete path metric space is geodesic, i.e., any two points can be
connected by a path whose length is the distance between the points.
The upshot of the previous paragraph is that we may assume, when proving Theorem
2.1, that (X, d, µ) is a geodesic metric measure space.
Recall that by Lusin’s theorem, given a measurable function g : (X, d, µ) → R and a
set A ⊆ X of finite measure, for any > 0 there is a subset E ⊆ A with µ(A\E) < and
a continuous function ge : X → R of compact support in A with ge|E = g|E .
Let us say that a measurable function g : X → R is approximately continuous at a
point x ∈ X if there is a set E for which g|E is continuous and x is a density point of E.
By Lusin’s theorem and the Lebesgue differentiation theorem, every measureable function
is approximately continuous almost everywhere!. We will apply this to lip.
Proof of Theorem 2.1. As discussed above, we may assume without loss of generality that
(X, d, µ) is a geodesic space.
Let f : X → R be an L-Lipschitz function, and let x ∈ X, r > 0, and y ∈ B(x, r). Let
λ ≥ 1. Then we may find a r/λ-chain of points x = p0 , p1 , . . . , pn = y with n . λ. Let
Bi = B(pi , r/λ). Then
Z
n−1 Z
Z
Z
X
|f (x) − f (y)| ≤ f (x) − − f dµ + − f dµ − − f dµ + − f dµ − f (y) .
B0
Bi
Bi+1
B0
i=0
First, note that
Z
Z
f (x) − − f dµ ≤ − L|x − z| dµ(z) ≤ Lr .
λ
B0
B0
A similar estimate holds for the correspodning expression involving y.
Now, let Bi0 = B(pi , 3r/λ), so that Bi ∪ Bi+1 ⊆ Bi0 . Then, by the doublign assumption,
Z
Z
|fBi − fBi0 | ≤ − |f − fBi0 | dµ . − |f − fBi0 | dµ.
Bi
Bi0
Thus, the Poincaré inequality (and the fact that lip f is an upper gradient) show that,
!1/p
Z
r
− (lip f )p dµ
|fBi − fBi0 | .
.
λ
CBi0
Now, suppose x is an approximate continuity point of (lip f )p , and let > 0. We may
find a set E ⊆ B(x, r) and a continuous function g ≥ 0 so that (lip f )p = g on E, and
µ(B(x, r)\E) < . Thus, if λ ≥ 3C,
Z
Z
p
− (lip f ) dµ ≤ − g dµ + Lp .
CBi0
CBi0
By the Lebesgue differentiation theorem, if r < r(, λ), then
Z
− g dµ ≤ (lip f (x))p + .
CBi0
The upshot is that for r < r(, λ),
!1/p
Z
. lip f (x) + o().
− (lip f )p dµ
CBi0
Thus
|fBi − fBi0 | .
r
(lip f (x) + o()).
λ
Since
|fBi − fBi+1 | ≤ |fBi − fBi0 | + |fBi+1 − fBi0 |,
we conclude that for r < r(, λ),
lip f (x) + o()
|fBi − fBi+1 | . r
.
λ
Hence, since n . λ, when r < r(, λ),
L
|f (x) − f (y)|
. + lip f (x) + o().
r
λ
This impiles that
Lip f (x) . lip f (x).