→ ℝ is Lipschitz if there is a constant Ksuch that for all x,y,
|f (x)− f (y)| ≤ K |x − y|.
More generally, f is Lipschitzon a subset of ℝ^{n}if for allx,yin this set,
|f (x) − f (y )| ≤ K |x − y|.
In what follows, dt will be used instead of dm in order to make the notation more
familiar.
Lemma 17.2.2Suppose f :
[a,b]
→ ℝ is Lipschitz continuous and increasing. Then f^{′}exists a.e., is in L^{1}
([a,b])
, and
∫ x
f (x) = f (a) + f′(t)dt.
a
In fact, the almost everywhere existence of the derivative holds with only the assumption thatf is increasing. If f : ℝ → ℝis Lipschitz, then f^{′}is in L_{loc}^{1}
(ℝ)
and the above formulaholds.
Proof: The Dini derivates are defined as follows.
+ f (x+ h)− f (x) f (x+ h)− f (x)
D f (x) ≡ lim sup ------h-------, D+f (x) ≡ limhi→n0f+------h------
h→0+
D − f (x) ≡ lim sup f (x)−-f (x-−-h),D − f (x ) ≡ lim inf f (x)−-f-(x-−-h)
h→0+ h h→0+ h
For convenience, just let f equal f
(a)
for x < a and equal f
(b)
for x > b. Let
(a,b)
be an
open interval and let
{ + }
Nab ≡ x ∈ (a,b) : D f (x) > q > p > D+f (x)
Let V ⊆
(a,b)
be an open set containing N_{pq} such that m
(V)
< m
(Npq)
+ ε. By assumption,
if x ∈ N_{pq}, there exist arbitrarily small h such that
f (x-+h-)−-f (x)< p,
h
These intervals
[x,x + h]
are then a Vitali covering of N_{pq}. It follows from Corollary 9.3.6
that there is a disjoint union of countably many,
{[xi,xi + hi]}
_{i=1}^{∞} which cover all of N_{pq}
except for a set of measure zero. Thus also the open intervals
{(xi,xi + hi)}
_{i=1}^{∞} also cover
all of N_{pq} except for a set of measure zero. Now for points x^{′} of N_{pq} so covered, there are
arbitrarily small h such that
f-(x′ +-h′)−-f-(x′)
h′ > q
and
[x′,x′ + h′]
is contained in one of these original open intervals
(xi,xi + hi)
. By the Vitali
covering theorem again, Corollary 9.3.6, it follows that there exists a countable disjoint
sequence
{ [x′,x′+ h′]}
j j j
_{j=1}^{∞} which covers all of N_{pq} except for a set of measure zero, each
of these
[x ′,x′ +h ′]
j j j
being contained in some
(x,x + h )
i i i
. Then it follows that
∑ ′ ∑ (′ ′) ( ′) ∑
qm (Npq) ≤ q hj ≤ f xj + hj − f xj ≤ f (xi + hi) − f (xi)
j∑ j i
≤ p hi ≤ pm (V) ≤ p (m (Npq)+ ε)
i
Since ε > 0 is arbitrary, this shows that qm
(Npq)
≤ pm
(Npq )
and so m
(Npq)
= 0. Now taking
the union of all N_{pq} for p,q ∈ ℚ, it follows that for a.e. x,D^{+}f
(x)
= D_{+}f
(x)
and so the
derivative from the right exists. Similar reasoning shows that off a set of measure zero the
derivative from the left also exists. You just do the same argument using D^{−}f
(x)
and
D_{−}f
(x)
to obtain the existence of a derivative from the left. Next you can use the same
argument to verify that D^{−}f
(x)
= D_{+}f
(x)
off a set of measure zero. This is outlined next.
Define a new N_{pq},
{ − }
Npq ≡ x ∈ (a,b) : D+f (x) > q > p > D f (x)
Let V be an open set containing N_{pq} such that m
(V)
< m
(Npq)
+ ε. For each x ∈ N_{pq} there
are arbitrarily small h such that
f (x )− f (x− h)
-------h------ < p
Then as before, there is a countable disjoint sequence of closed intervals contained in
V,
{[xi − hi,xi]}
_{i=1}^{∞} such that their union includes all of N_{pq} except a set of measure
zero. Thus this is also true of the open intervals
{(xi − hi,xi)}
_{i=1}^{∞}. Then for the
points of N_{pq} covered by these open intervals x^{′}, there are arbitrarily small h^{′} such
that
f-(x-′+-h′)−-f-(x′)
h′ > q.
and each
[x′,x′ +h ′]
is contained in an interval
(xi − hi,xi)
. Then by the Vitali covering
theorem again, Corollary 9.3.6 there are countably many disjoint closed intervals
{[x ′,x′ +h ′]}
j j j
_{j=1}^{∞} whose union includes all of N_{pq} except for a set of measure zero such
that each of these is contained in some
(xi − hi,xi)
described earlier. Then as before,
∑ ∑ ( ) ( ) ∑
qm (Npq) ≤ q h′j ≤ f x′j + h′j − f x′j ≤ f (xi)− f (xi − hi)
j j i
∑
≤ p hi ≤ pm (V) ≤ p (m (Npq)+ ε)
i
Then as before, this shows that qm
(Npq)
≤ pm
(Npq)
and so m
(Npq)
= 0. Then taking the
union of all such for p,q ∈ ℚ yields D_{+}f
(x)
= D^{−}f
(x)
for a.e. x. Taking the union of all
these sets of measure zero and considering points not in this union, it follows that f^{′}
(x )
exists
for a.e. x. Thus this has shown that whenever f is increasing, its derivative exists
a.e.
If f is Lipschitz continuous, then all of the difference quotients are bounded by the
Lipschitz constant of f. Thus f^{′}
(t)
≥ 0 and is a limit of measurable even continuous
functions for a.e. x so f^{′} is clearly measurable. This only uses an assumption that f is
continuous. In fact, it does not even need this, because for f increasing, the difference
quotients are clearly measurable because f^{−1}
([a,∞ ))
is an interval.
The remaining issue is whether f
(y)
− f
(x)
= ∫X_{[x,y]
}
(t)
f^{′}
(t)
dm. Let h > 0.
∫ x f (t)− f (t− h) 1 ∫ x 1 ∫ x
-------------dt = -- f (t)dt− -- f (t − h )dt
a h h ∫ax h ∫ax−h
= 1- f (t)dt− 1- f (t)dt
h a h a−h
1 ∫ x 1∫ a
= h- f (t)dt − h- f (t)dt
∫x−xh a−h
= 1- f (t)dt − f (a)
h x−h
Therefore, by continuity of f it follows from Fatou’s lemma that
∫ ∫ ∫
xD f (t)dt = xf′(t)dt ≤ lim inf x f (t)−-f (t−-h)dt = f (x)− f (a)
a − a h→0+ a h
and this shows that f^{′} is in L_{loc}^{1}. This part only used the fact that f is increasing and
continuous. That f is Lipschitz has not been used.
If it were known that there is a dominating function for t →
f(t)−f(t−h)
h
, then you could
simply apply the dominated convergence theorem in the above inequality instead
of Fatou’s lemma and get the desired result. But from Lipschitz continuity, you
have
| |
||f-(t)−-f-(t-− h-)||
| h | ≤ K
and so one can indeed apply the dominated convergence theorem and conclude
that
∫
x ′
a f (t)dt = f (x )− f (a)
The last claim follows right away from consideration of intervals since the restriction of a
Lipschitz function is Lipschitz. ■
Note that the above has shown that every increasing function is differentiable off a set of
measure zero. It says more in the case that the function is Lipschitz in that it is possible to
recover the function as an integral of its derivative.
With the above lemmas, the following is the main theorem about absolutely continuous
functions.
The following simple corollary is a case of Rademacher’s theorem.
Corollary 17.2.3Suppose f :
[a,b]
→ ℝ is Lipschitz continuous,
|f (x)− f (y)| ≤ K |x − y|.
Then f^{′}
(x)
exists a.e. and
∫ x ′
f (x) = f (a) + a f (t)dt.
Proof: If f were increasing, this would follow from the above lemma. Let
g
(x)
= 2Kx − f
(x)
. Then g is Lipschitz with a different Lipschitz constant and also if
x < y,
g(y)− g(x) = 2Ky − f (y)− (2Kx − f (x))
≥ 2K (y− x)− K |y− x| = k |y − x| ≥ 0
and so Lemma 17.2.2 applies to g and this shows that f^{′}
(t)
exists for a.e. t and
g^{′}
(x)
= 2K − f^{′}
(x)
. Also
2K (x− a)− (f (x)− f (a))
∫ x
= g(x)− g(a) = 2Kx − f (x)− (2Ka − f (a)) = (2K − f′(t))
∫ x a
= 2K (x− a)− f ′(t)dt
a
showing that f
(x)
− f
(a)
= ∫_{a}^{x}f^{′}
(t)
dt. ■
The following is a basic lemma which will be used in what follows. First recall the
following definition.
Definition 17.2.4For Ω an open set in ℝ^{n}, C_{c}^{∞}
(Ω )
denotes those functionsϕ which are infinitely differentiable and have compact support in Ω. This is a nonemptyset of functions by Lemma 10.1.2.
With this definition, the fundamental lemma is as follows.
Lemma 17.2.5Suppose f ∈ L_{loc}^{1}
(ℝn)
and suppose
∫
fϕdx = 0
for all ϕ ∈ C_{c}^{∞}(ℝ^{n}). Then f
(x)
= 0 a.e. x.
Proof:Without loss of generality f is real-valued.
E ≡ { x : f (x) > ε}
and let
Em ≡ E ∩ B(0,m ).
We show that m
(Em )
= 0. If not, there exists an open set, V , and a compact set K
satisfying
K ⊆ Em ⊆ V ⊆ B (0,m ),m (V ∖K ) < 4−1m (Em ),
∫
|f|dx < ε4−1m (Em ).
V ∖K
Let H and W be open sets satisfying
-- ---
K ⊆ H ⊆ H ⊆ W ⊆ W ⊆ V
and let
--
H ≺ g ≺ W
where the symbol, ≺, has the same meaning as it does in 9.4. That is, g equals 1 on H and
has compact support contained in W. Then let ϕ_{δ} be a mollifier and let h ≡ g ∗ϕ_{δ} for δsmall
enough that
K ≺ h ≺ V.
Thus
∫ ∫ ∫
0 = f hdx = fdx + f hdx
K V∖K
≥ εm (K )− ε4− 1m (Em )
≥ ε(m (E )− 4−1m (E ))− ε4−1m (E )
−1 m m m
≥ 2 εm (Em ).