Departments of Mathematics
Montana State University
Fall 2015
Prof. Kevin Wildrick
An introduction to non-smooth analysis and geometry
Lecture 10: The Poincaré Inequality in Euclidean space
1. What is the Poincaré inequality?
In the last section, we gave a reasonable generalization of the norm of the gradient
of a smooth function on Euclidean space. This was motivated by the fact that our
generalization was now a purely metric concept, and didn’t rely on the linear structure of
the domain. This is a pretty weak effort: what good does it do us? In order to understand
the usefulness of this generalization, we need to know something about how the norm of
the gradient of a smooth function on Euclidean space behaves. We’ve seen that the norm
of the gradient controls the behavior of a function on all rectifiable curves, but what can
we say about its behavior on larger sets?
Theorem 1.1 (The Poincaré inequality in Rn .). Let f : Rn → R be a C 1 -smooth mapping,
and let p ≥ 1. Then there is a constant C(n, p) depending only on n and p such that for
any ball B ⊆ Rn ,
Z
p1
Z
− |f − fB | dLn ≤ C(n, p) diam(B) − ||∇f ||p dLn
.
B
B
Let’s examine what this inequality says. At a point x ∈ B, the quantity
|f (x) − fB |
measures the deviation of f from its average on B. We then average this over B to get
the left hand side - so the left hand side of the inequality is a measure of how much, on
average, the function deviates from its average value. One can think of this as a measure
of the average variation of f on B. On the right hand side, we have a o-average of the
value of the norm of the gradient:
p1
Z
p
n
− ||∇f || dL
.
B
If p = 1, this is truly just the usual average, but as p gets larger, the average become
less “spread-out” and more biased to larger values of ||∇f ||. In particular, as p → ∞,
the p-average tends to the essential supremum of ||∇f || divided by the measure of the
ball. We then multiply this p-average by the diameter of the ball - this is just the good
old “total change is rate times distance”. For example, consider f : R → R defined by
f (x) = x. Then
Z
r
−
|f − fB(0,r) | dL1 = ,
2
B(0,r)
while
Z
p1
p
1
− ||∇f || dL
= 1.
B
So, in total, the Poincaré inequality allows us to control the average variation of a function
by the average size of its gradient.
The Poincaré inequality, and variants thereof, are very useful in regularity theory for
PDE’s - if the size of the (weak) gradient can be estimated, then bounds on the variation
of the function can be acheived.
1
The Poincaré inequality is closely related to geometry as well. Let us suppose, for the
moment, that f is a C 1 -smooth function defined not on all of R2 , but instead only on the
slit disk D∗ = B(0, 1)\([0, 1) × {0}). It’s easy to construct such an f that is identically 0
on the set
[1/2, 3/4] × (0, 1/8)
and identically 1 on the set
[1/2, 3/4] × (−1/8, 0).
Then ||∇f || is identically 0 on B((5/8, ), 1/8), but the average variation of f on this ball
is certainly not 0. This is because the curves joining the bottom of the slit to the top of
the slit are too long compared to the distance between the points they join. We will see
that in order for a Poincaré inequality to be valid on some domain, the domain must have
a sufficiently large collection of short curves joining each point.
Let us note that the Poincaré inequality as described above makes sense in a general
metric measure space if we replace ||∇f || with any upper gradient of f .
2. A first proof of the Poincaré inequality in Euclidean space.
The Poincaré inequality is easy to prove on the real line. Fix a ball B ⊆ R, and let x
and y be points of B. By the fundamental theorem of calculus,
Z
Z
0
|f (t)| dt = diam(B)− |f 0 (t)| dt.
|f (x) − f (y)| ≤
B
B
Now, note that since the right hand side above is independent of x and y,
Z
Z Z
Z
− |f( x) − fB | dx ≤ − − |f (x) − f (y)| dxdy ≤ diam(B)− |f 0 (t)| dt,
B
B
B
B
showing the 1-Poincaré inequality. The p-Poincaré inequality now follows by Hölder’s
inequality.
We will now give the classical proof of the Poincaré inequality in the case that n ≥ 2,
and along the way we will also prove some other important inequalities from classical
function theory. The approach we will take is based on harmonic analysis. There are two
important parts to this approach - the boundedness of certain operators on Lp (which is
very general), and the foliated structure of Euclidean space (which is not very general).
2.1. Maximal functions. We have already met, in a way, one of the main tools from
harmonic analysis that will play a central role in the proof of the Poincaré inequality: the
maximal function.
Definition 2.1. Let (X, d, µ) be a metric measure space, and let f be an integrable
function on X. The maximal function M f : X → R is defined by
Z
M f (x) = sup −
|f | dµ.
r>0
B(x,r)
Many variants of the maximal function are occasionally useful (for example, the restricted maximal function, where the supremum is only taken over radii less than some
fixed R > 0. Notice that by the Lebesgue differentiation theorem, |f | < M f at µ-almost
every point.
Note that M is not a linear operator, but rather only sublinear:
M (f + g) ≤ M f + M g.
The key property of the maximal function is its boundedness on certain function spaces.
Theorem 2.2. Let (X, d, µ) be a doubling metric measure space. Then there is a constant
C1 , depending only on the doubling constant of µ, such that if f ∈ L1 (X, µ), then for all
t > 0,
Z
C1
µ({x ∈ X : M f (x) > t}) <
|f | dµ.
t X
Moreover, for each p > 1, then there is a constant Cp , depending only on the doubling
constant of µ, such that if f ∈ Lp (X, µ), then
||M f ||p ≤ Cp ||f ||p .
In otherwords, M is a bounded sublinear operator L1 (X, µ) → weak − L1 (X, µ) and
Lp (X, µ) → Lp (X, µ).
Lemma 2.3. Let f be a measurable function on (X, µ). Then for p ≥ 1,
Z
Z ∞
p
|f | dµ = p
tp−1 µ({x ∈ X : |f (x)| > t}) dt.
X
0
Proof. Use the fact that
Z
p
|f (x)|
|f (x)| = p
tp−1 dt
0
and Fubini’s Theorem.
Proof. For R > 0, let us denote
Z
MR f = sup −
|f | dµ.
0<r<R B(x,r)
We will prove the results for MR f , and then let R → ∞.
We start with the weak-type estimate. For each x ∈ X such that MR f (x) > t, choose
a ball B(x, r) with r < R such that
Z
−
|f | dµ > t.
B(x,r)
Using the 5B covering lemma, extract a countable collection G of such balls so that 5-times
dialted balls cover. Then
X
X
µ({x ∈ X : MR f (x) > t}) ≤
µ(5B) ≤ C
µ(B)
B∈G
B∈G
Z
Z
CX
C
≤
|f | dµ ≤
|f | dµ.
t B∈G B
t X
Letting R → ∞ now shows the first statement. The second statement will follow only
from the first inequality, the sublinearity of M , and the fact that M f (x) ≤ ||f ||∞ . This
is an example of an interpolation argument; we have shown that M is bounded from L1
to weak-L1 , and it is clearly bounded from L∞ to L∞ , and now we everything in between
for free.
To this end, let f ∈ Lp (X, µ), and fix t > 0. Let’s write f as the sum of a good part g
and a bad part b:
f = f χ{|f |≤t/2} + f χ{|f |>t/2} = g + b.
Since M g ≤ ||g||∞ , the sublinearity of M yields
t
M f ≤ + M b,
2
implying that
{x ∈ X : M f (x) > t} ⊆ {x ∈ X : M b(x) > t/2}.
Moreover, if we apply Lemma 2.3 to b, we see that
Z ∞
Z
µ({x ∈ X : f χ|f |>t/2 (x) > s}) ds
|b| dµ =
0
X
Z ∞
t
µ({x ∈ X : f (x) > s}) ds
= µ({x ∈ X : |f | > t/2}) +
2
t/2
Now, by Lemma 2.3, the weak type estimate proved above applied to b, the above estimate,
and Fubini’s theorem
Z ∞
p
tp−1 µ({x ∈ X : M f (x) > t}) dt
||M f ||p = p
Z0 ∞
tp−1 µ({x ∈ X : M b(x) > t/2}) dt
≤p
0
Z
Z ∞
p−2
t
|b| dµ dt
≤ 2Cp
0
X
Z ∞ p−1
Z ∞
t
p−2
≤ 2Cp
µ({x ∈ X : f (x) > s}) ds dt
µ({x ∈ X : |f | > t/2}) + t
2
0
t/2
Z ∞Z ∞
p
p
tp−2 µ({x ∈ X : f (x) > s})χ{t<2s} dt ds
≤ 2 C||f ||p + 2Cp
0
0
2p C
≤ 2p C||f ||pp +
||f ||pp .
p−1
This completes the proof.
We now move towards the Poincaré inequality.
Theorem 2.4 (The Riesz Potential Estimate). For each n ∈ N, there is a constant
C = C(n) such that if f : Rn → R is a C 1 -smooth function, B is a ball in Rn , and x ∈ B,
then
Z
|∇f (z)|
dLn (z).
|f (x) − fB | ≤ C
n−1
B |z − x|
Proof. We prove the theorem only when n ≥ 2; the case that n = 1 is mostly the same,
but easier.
Think of fixing x and letting y vary, and consider the parameterized line l : R → Rn
between them, defined by
ly (t) = x + t(y − x).
Then, by the fundamental theorem of calculus and the chain rule
Z 1
Z 1
0
f (x) − f (y) = −
(f ◦ ly ) (t) dt = −
∇f (ly (t)) · (y − x) dt
0
0
n
Now, fix a ball B in R of radius R, and let x and y be points of B. Integrating with
respect to y ∈ B and applying Fubini’s theorem yields
Z Z
n
L (B)|f (x) − fB | ≤
|∇f (ly (t))||(y − x)| dL1 (t) dLn (y).
B [0,1]
Z
Z
≤
|∇f (ly (t))||(y − x)| dLn (y) dL1 (t).
[0,1]
B
We now condsider the change of variables ψt : B → ψt (B) defined by
ψt (y) = ly (t).
Note that
det ψt = tn
ψt (B) ⊆ B ∩ B(x, 2tR)
|y − x| =
|ψt (y) − x|
.
t
Thus,
n
Z
Z
L (B)|f (x) − fB | ≤
|∇f (z)|
[0,1]
B∩B(x,2tR)
Z
Z
|∇f (z)|t−n dLn (z) dL1 (t)
≤ 2R
[0,1]
|z − x| −n
t dLn (z) dL1 (t)
t
B∩B(x,2tR)
Note that if t ∈ [0, 1] and z ∈ B satisfies |z − x| < 2tR, then t satisfies
|z − x|
< t < 1.
2R
Hence, Fubini’s theorem and an integration imply
Z Z 1
n
|∇f (z)|t−n dLn (z) dL1 (t)
L (B)|f (x) − fB | ≤ 2R
B |z−x|(2R)−1
n Z
≤
2R
n−1
B
|∇f (z)|
dLn (z).
|z − x|n−1
Given this result, we make the following definition.
Definition 2.5 (The Riesz Transform). Let g ∈ L1loc (Rn ) and let B be a ball. The Riesz
transform of g on B is defined by
Z
g(z)
I1 (g)(x) =
dLn (z).
n−1
B |z − x|
Hence, we have shown that for a C 1 -smooth function f , a ball B ⊆ Rn , and x ∈ B,
|f (x) − fB | ≤ C(n)I1 (∇f )(x).
Theorem 2.6 (The boundedness of the Riesz transform). Let B be a ball in Rn , n ≥ 2.
Then there are constants C1 and Cp , 1 < p < n, such that if g ∈ L1 (B), then
n
||g||1 n−1
n
L ({I1 (g) > t}) ≤ C1
,
t
and if g ∈ Lp (B), 1 < p < n, then
||I1 (g)||np (n − p) ≤ Cp ||g||p .
We will not prove this theorem, but mention that it is a corollary of the HardyLittlewood maximal function theorem. As a relatively quick corollary, we achieve
Corollary 2.7 (The Sobolev-Poincaré Inequality). Let 1 ≤ p < n, and let f : Rn → R be
a C 1 -smooth function. Then for any ball B,
||f − fB ||np/(n−p) ≤ C(n, p)||∇f ||p .
This is an extraordinary fact - it says that the p-norm of the gradient of f controls its
the np/(n − p)-norm of its variation. This fact has profound impacts in partial differential
equations. Tracing through the argument, we see that the following is true: let f : Rn → R
be any function that has an upper gradient in Lp (Rn ), for 1 < p < n. Then f − fB is in
Lnp/(n−p) for any ball B.
The Sobolev-Poincaré inequality, after two applications of Hölder’s inequality, leads
to a proof of the Poincaré inequality. However, we can proceed directly from the Riesz
potential estimate to give a quick proof.
Proof of the Poincaré inequality. By the Riesz potential estimate and Fubini’s inequality
Z
Z Z
|∇f (z)|
n
dLn (z) dLn (x)
− |f (x) − fB | dL (x) ≤ C−
n−1
|z
−
x|
B
ZB B
Z
1
≤ C− |∇f (z)|
dLn (x) dLn (z).
n−1
|z
−
x|
B
B
Polar coordinates shows that
Z
Z
1−n
n
|z − x|
dL (x) ≤
B
|z − x|1−n dLn (x) ≤ C(n) diam B.
B(z,diam B))
This shows the 1-Poincaré inequality. The general case now follows from Hölder’s inequality.
If f : Rn → R is a C 1 -smooth function, B is a ball in Rn , and x and y are in B, then the
triangle inequality and the Riesz potential estimate proven before imply the symmetric
estimate
Z
Z
|∇f (z)|
|∇f (z)|
n
n
dL (z) +
dL (z) .
|f (x) − f (y)| ≤ C
n−1
n−1
B |z − y|
B |z − x|
This inequality can also be used to prove the Poincaré inequality, since
Z
Z Z
Z Z
n
n
n
− |f (x)−fB | dL (x) = − − f (x) − f (y) dL (y) dL (x) ≤ − − |f (x)−f (y)| dLn (y) dLn (x).
B
B
B
B
B
Plugging in the symmetric estimate and arguing as before now yields the desired result.
3. Pencils of Curves
We now give a geometric way of verifying the Poincaré inequality - one that does not
rely so heavily on the structure of Rn .
Let f : Rn → R be a C 1 -function, and let x and y be points of a ball B in Rn . Suppose
that there is a family of rectifiable curves Γ and a probability measure P on Γ. Then we
may form another measure µ on B by
Z Z
Z
µ(A) :=
ds dP (γ) = length(γ ∩ A) dP (γ).
Γ
A∩γ
γ
If one can show that for each Borel set A ⊆ B,
Z
Z
|∇f (z)|
|∇f (z)|
n
n
µ(A) ≤ C
dL (z) +
dL (z) ,
n−1
n−1
A |z − x|
A |z − y|
then a similar result follows for Borel functions:
Z
Z Z
Z
ρ(z)
ρ(z)
n
n
ρ ds dP (γ) ≤ C
dL (z) +
dL (z) .
n−1
n−1
Γ A∩γ
A |z − x|
A |z − y|
Thus integrating the upper gradient inequality yields
Z Z
Z
|∇f | ds dP
|f (x) − f (y)| = |f (x) − f (y)|dP ≤
Γ
Γ γ
Z
Z
|∇f (z)|
|∇f (z)|
n
n
≤C
dL (z) +
dL (z) ,
n−1
n−1
B |z − x|
B |z − y|
and the Poincaré inequality follows. Thus we have reduced the Poincaré inequality to a
question about finding sufficiently large path families.