It turns out that Lipschitz functions on ℝ^{n} can be differentiated a.e. This is called
Rademacher’s theorem. It also can be shown to follow from the Lebesgue theory of
differentiation. We denote D_{v}f
(x)
the directional derivative of f in the direction v. Here v is
a unit vector. In the following lemma, notation is abused slightly. The symbol f
(x+tv)
will
mean t → f
(x+tv )
and
d-
dt
f
(x+tv)
will refer to the derivative of this function of t.
Lemma 17.3.1Let f : ℝ^{n}→ ℝ be a Lipschitz function with constant K and let vbe a unit vector. Then the following hold.
D_{v}f
(x)
exists for a.e. x. Also off a set of measure zero,
|Dvf (x )|
≤ K.
For every x,
∫ t
f (x+tv)− f (x ) = Dvf (x+ sv)ds
0
D_{v}f
(x)
= ∇f
(x)
⋅ v.
For a given x and for σ the measure on S^{n−1}of Section 9.10and for w ∈ S^{n−1},
Dvf (x+tw ) = ∇f (x+tw )⋅w
for a.e. t for σ a.e. w. Thus
∫ ∫ r
tn−1Dwf (x+tw )dtdσ (w)
∫Sn−1∫0
r n−1
= Sn−1 0 t ∇f (x+tw )⋅wdtd σ(w)
= 1, denote the measure of Section 9.10 defined on the unit sphere S^{n−1} as σ. Let
N_{w} be defined as those t ∈ [0,∞) for which D_{w}f
(x+ tw)
≠∇f
(x + tw)
⋅w.
B ≡ {w ∈ Sn −1 : N has positive measure}
w
The set of points of ℝ^{n} for which D_{w}f
(x)
≠∇f
(x)
⋅ w consists of all points x + tw where
t ∈ N_{w} and w ∈ S^{n−1}. Thus from Section 9.10 the measure of this set is
∫ ∫
ρn−1dρdσ (w)
B Nw
This must equal zero from what was just shown and so σ
(B )
= 0. The claimed formula
follows from this. ■
The following lemma gives an interesting inequality due to Morrey.
Lemma 17.3.2Let u be Lipschitz continuous on ℝ^{n}. Then there exists a constant C,depending only on n such that for any x,y ∈ ℝ^{n},
|u(x)− u(y)|
( )
∫ p 1∕p ( (1−n∕p))
≤ C |∇u (z)|dz | x − y| . (17.6)
B(x,2|x−y|)
(17.6)
Here p > n.
Proof: In the argument C will be a generic constant which depends on n. Consider the
following picture.
PICT
This is a picture of two balls of radius r in ℝ^{n}, U and V having centers at x and y
respectively, which intersect in the set W. The center of U is on the boundary of V and the
center of V is on the boundary of U as shown in the picture. There exists a constant, C,
independent of r depending only on n such that
m-(W-) m-(W-) -1
m (U ) = m (V) = C .
You could compute this constant if you desired but it is not important here.
Then
∫
--1---
|u (x )− u(y)| = m (W ) W |u(x) − u (y)|dz
1 ∫ 1 ∫
≤ m-(W-) |u(x) − u (z)|dz + m-(W-) |u(z)− u (y)|dz
[∫ W ∫ W ]
= --C-- |u(x)− u (z)|dz + |u(z)− u (y )|dz
m (U)[∫W W∫ ]
≤ --C-- |u (x)− u(z)|dz + |u(y)− u (z)|dz
m (U) U V
Now consider these two terms. Using spherical coordinates and letting U_{0} denote the ball of
the same radius as U but with center at 0,
1 ∫
----- |u (x)− u(z)|dz
m (U) ∫U
= ---1-- |u(x)− u(z +x )|dz
m (U0) U0
Now using spherical coordinates, Section 9.10, and letting C denote a generic constant which
depends on n, and also using Lemma 17.3.1
∫ r ∫
= --1--- ρn−1 |u(x)− u (ρw +x )|dσ (w )dρ
m (U0)∫0 ∫Sn−1∫
--1--- r n−1 ρ
≤ m (U0) 0 ρ Sn−1 0 |Dwu (x + tw )|dtdσ (w )dρ
1 ∫ r ∫ ∫ ρ
= m-(U-) ρn−1 n−1 |∇u (x+ tw )⋅w|dtdσ(w) dρ
0 0 S 0
C ∫ r ∫ ∫ r
≤ -r n−1 |∇u (x +tw )|dtdσ(w )dρ
∫ 0 S∫ r 0
= C |∇u (x + tw)|tn− 1n1−-1dtdσ (w )
∫Sn−1 0 t
= C |∇u-(x+-z)|dz
U0 |z|n−1
(∫ )1∕p(∫ ′ ′)1∕p′
≤ C |∇u (x +z)|pdz |z|p− np
U0 U
( ∫ )1∕p(∫ ∫ ) (p−1)∕p
= C |∇u (z)|pdz r ρp′−np′ρn− 1dρdσ
U Sn−1 0
( ∫ )1∕p( ∫ ∫ r ) (p−1)∕p
= C |∇u (z)|pdz -1n−1dρdσ
U Sn−1 0 ρp−1
( p− 1) (p−1)∕p(∫ p )1∕p n
= C p−-n- |∇u (z)| dz r1− p
( ) (p−1)∕p(∫U )1∕p
= C p−-1- |∇u (z)|p dz |x − y|1− np
p− n U
Similarly,
∫ ( ) (p−1)∕p (∫ )1∕p
--1--- |u (y)− u(z)|dz ≤ C p−-1- |∇u (z)|pdz |x − y |1− np
m (V ) U p− n V
Therefore,
( )1∕p
( p−-1-)(p− 1)∕p ∫ p 1− np
|u(x)− u (y )| ≤ C p− n B (x,2|x−y|)|∇u (z)| dz |x− y|
because B
(x,2|x− y|)
⊇ V ∪ U. ■
Here is Rademacher’s theorem.
Theorem 17.3.3Suppose u is Lipschitzwith constant K then if x is a point where∇u
(x)
exists,
|u(y)− u (x )− ∇u (x )⋅(y − x)|
( ∫ )1 ∕p
--------1------- p
≤ C m (B (x,2|x− y|)) B(x,2|x−y|)|∇u (z)− ∇u (x)|dz | x− y|. (17.7)
(17.7)
Also u is differentiable at a.e. x and also
∫ t
u(x+tv) − u (x) = Dvu (x+ sv)ds (17.8)
0
(17.8)
Proof: This follows easily from letting g
(y)
≡ u
(y)
− u
(x)
−∇u
(x)
⋅
(y − x)
. As
explained above,
|∇u (x)|
≤
√n--
K at every point where ∇u exists, the exceptional points
being in a set of measure zero. Then g
(x)
= 0, and ∇g
(y)
= ∇u
(y)
−∇u
(x)
at the points y
where the gradient of g exists. From Lemma 17.3.2,
|u (y)− u(x)− ∇u (x)⋅(y − x)|
= |g (y )| = |g(y)− g (x )|
(∫ )1∕p n
≤ C |∇u (z) − ∇u(x)|pdz |x− y|1−p
B (x,2|x−y|)
( ∫ )1 ∕p
= C --------1------- |∇u (z)− ∇u (x)|pdz | x− y|.
m (B (x,2|x− y|)) B(x,2|x−y|)
Now this is no larger than
( ∫ ( √ -- ) )1∕p
≤ C --------1------- |∇u (z)− ∇u (x)|2 nK p− 1dz | x − y|
m (B (x,2|x− y|)) B(x,2|x−y|)
It follows that at Lebesgue points of ∇u, the above expression is o
(|x− y|)
and so
at all such points u is differentiable. As to 17.8, this follows from Lemma 17.3.2.
■
In the above major theorem, the function u is defined on all of ℝ^{n}. However, it is always
the case that Lipschitz functions can be extended off a given set. Thus if a Lipschitz function
is defined on some set Ω, then it can always be considered the restriction to Ω of a Lipschitz
map defined on all of ℝ^{n}.
Theorem 17.3.4If h : Ω → ℝ^{m}is Lipschitz, then there exists h : ℝ^{n}→ ℝ^{m}which extends h and is also Lipschitz.
Proof:It suffices to assume m = 1 because if this is shown, it may be applied to the
components of h to get the desired result. Suppose