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 24.5.1Let f : ℝ^{n}→ ℝ be a Lipschitz functionwith constant K and letv be 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) = 0 Dvf (x+ sv)ds
D_{v}f
(x)
= ∇f
(x)
⋅ v.
For a given x and for σ the measure on S^{n−1}of Section 11.9and forw ∈ S^{n−1},
Dvf (x+tw ) = ∇f (x+tw )⋅w
for a.e. t for σ a.e. w. Thus
∫ ∫
r n−1
Sn−1 0 t Dwf (x+tw )dtdσ(w)
∫ ∫ r n−1
= n−1 t ∇f (x+tw )⋅wdtdσ(w )
S 0
= 1, denote the measure of Section 11.9 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 : Nw has positive measure
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 11.9 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 24.5.2Let 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 )|
(∫ )1∕p( )
≤ C |∇u (z)|pdz | x− y|(1−n∕p) . (24.5.10)
B (x,2|x−y|)
(24.5.10)
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
∫
|u(x)− u(y)| = --1--- |u (x)− u(y)|dz
m (W )W∫ ∫
≤ --1--- |u (x)− u(z)|dz + --1--- |u (z) − u(y)|dz
m (W )W m (W )W
--C-- [∫ ∫ ]
= m (U) W |u (x )− u(z)|dz + W |u (z)− u (y)|dz
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 11.9, and letting C denote a generic constant
which depends on n, and also using Lemma 24.5.1
C ∫ r∫ ∫ r
≤ -- |∇u (x + tw )|dtdσ (w )dρ
r∫ 0 S∫n−r1 0
= C |∇u (x+ tw)|tn−1-1--dtdσ(w)
∫Sn−1 0 tn−1
|∇u-(x-+z)|
= C U0 |z|n−1 dz
(∫ )1∕p (∫ ′ ′)1∕p′
≤ C |∇u (x+ z)|p dz |z|p−np
U0 U
( ∫ p )1∕p( ∫ ∫ r ′ ′ )(p−1)∕p
= C |∇u (z)| dz n−1 ρp−np ρn−1dρdσ
U ( S 0 )(p− 1)∕p
( ∫ p )1∕p ∫ ∫ r--1--
= C U |∇u (z)| dz Sn−1 0 np−−11 dρdσ
ρ
( )(p−1)∕p(∫ )1∕p
= C p-− 1 |∇u (z)|pdz r1− np
p− n U
( p − 1 )(p−1)∕p(∫ p )1∕p 1− n
= C p−-n- |∇u (z)| dz |x − y| p
U
Similarly,
--1---∫ ( p−-1-)(p− 1)∕p(∫ p )1∕p 1− np
m (V )U |u(y)− u (z)|dz ≤ C p− n V |∇u (z)| dz |x − y|
Therefore,
( )
( p− 1) (p−1)∕p ∫ p 1∕p 1− n
|u(x)− u(y)| ≤ C p−-n- |∇u (z)| dz |x − y| p
B(x,2|x− y|)
because B
(x,2|x− y|)
⊇ V ∪ U. ■
Here is Rademacher’s theorem.
Theorem 24.5.3Suppose 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|.
(24.5.11)
(24.5.11)
Also u is differentiable at a.e. x and also
∫
t
u(x+tv )− u(x) = 0 Dvu (x +sv )ds (24.5.12)
(24.5.12)
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 24.5.2,
|u(y)− u (x) − ∇u (x) ⋅(y − x)|
= |g(y)| = |g (y )− g(x)|
( ∫ )1∕p
≤ C |∇u (z)− ∇u (x)|pdz |x− y |1− np
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− 1 )1∕p
≤ C ---------------- |∇u (z)− ∇u (x)|2 nK dz | 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 24.5.12, this follows from Lemma 24.5.2.
■
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 24.5.4If 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
|h (x)− h(y)| ≤ K |x− y|. (24.5.13)
(24.5.13)
Define
--
h(x) ≡ inf{h (w)+ K |x− w | : w ∈ Ω}. (24.5.14)