The Implicit Function Theorem for Lipschitz Functions and Applications
A Masters Thesis
presented to
the Faculty of the Graduate School
University of Missouri
In Partial Fulfillment
of the Requirements for the Degree
Master of Science
by
Michael Wuertz
Dr. Marius Mitrea, Dissertation Supervisor
MAY 2008
The undersigned, appointed by the Dean of the Graduate School, have examined the
thesis entitled
The Implicit Function Theorem for Lipschitz Functions and Applications
presented by Michael Wuertz,
a candidate for the degree of Master of Science in and hereby certify that in their
opinion it is worthy of acceptance.
Professor Marius Mitrea
Professor Fritz Gesztesy
Professor Steve Hofmann
Professor Ioan Kosztin
Thanks to Professor Marius Mitrea, Courtney Bagge, Steven Wuertz, and Susan
Wuertz
ACKNOWLEDGEMENTS
It is a pleasure to thank Simona, Nick, Simon, and James for their special contribution.
ii
Table of Contents
ACKNOWLEDGEMENTS
ii
1 Introduction
1
2 The Implicit and Inverse Function Theorems
2.1 First Proof Using the Contraction Principle . . . . . . . . . . . . . .
2.2 Second Proof Using the Invariance of Domain Theorem . . . . . . . .
11
11
19
3 Lipschitz Functions
24
4 Rademacher’s Theorem
4.1 Almost Everywhere Differentiability of Lipschitz Functions . . . . . .
4.2 Almost Everywhere Differentiability of Absolutely Continuous Functions
4.3 Some Consequences of Rademacher’s Theorem . . . . . . . . . . . . .
31
31
39
43
5 The Implicit Function Theorem for Lipschitz Functions
46
6 Lipschitz Domains
6.1 General Considerations . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Some Toplogical Results . . . . . . . . . . . . . . . . . . . . . . . . .
51
51
53
7 Applications
7.1 First Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2 Second Application . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.3 An Application for the Classical Implicit Function Theorem . . . . .
61
61
68
74
Bibliography
77
iii
Chapter 1
Introduction
A class of domains which plays an important role in the field of partial differential
equations is the collection of all Lipschitz domains in Rn . Recall that a function
f : U → Rm , where U is a subset of Rn , is called Lipschitz if
|f (x) − f (y)|
Lip(f ; U ) := sup
: (x, y) ∈ U × U \ diag < ∞
(1.1)
|x − y|
with Lip (f ; U ) called the Lipschitz constant of f in U . Next, a nonempty, proper
open subset Ω of Rn is called a Lipschitz domain if for every x0 ∈ ∂Ω there exist a
new system of coordinates Rn−1 ×R, which is isometric to the original one, along with
an open upright cylinder C containing x0 , and a Lipschitz function ϕ : Rn−1 → R
whose graph contains x0 and is disjoint from the top and bottom lids of C, such that
C ∩ ∂Ω = C ∩ {(x0 , xn ) ∈ Rn−1 × R : xn = ϕ(x0 )},
(1.2)
C ∩ Ω = C ∩ {(x0 , xn ) ∈ Rn−1 × R : xn > ϕ(x0 )},
(1.3)
C ∩ (Ω)c = C ∩ {(x0 , xn ) ∈ Rn−1 × R : xn < ϕ(x0 )}.
(1.4)
Thus, informally speaking, Ω ⊂ Rn is a Lipschitz domain if in a neighborhood of any
boundary point x0 , the “surface” ∂Ω agrees (in a suitable sense) with the graph of
a Lipschitz function ϕ : Rn−1 → R. Then the outward unit normal has an explicit
formula in terms of ∇0 ϕ namely, in the new system of coordinates,
(∇0 ϕ(x0 ), −1)
ν(x0 , ϕ(x0 )) = p
,
1 + |∇0 ϕ(x0 )|2
if (x0 , ϕ(x0 )) is near x0 ,
(1.5)
where ∇0 denotes the gradient with respect to x0 ∈ Rn−1 . The existence of ∇0 ϕ at
almost every point in Rn−1 is a consequence of the celebrated Rademacher Theorem.
What makes Lipschitz domains both a convenient and a resourceful environment
for doing analysis is:
(i) the fact that they are still retaining some of the most basic geometric properties
of, say, the upper-half space (which, arguably, is the most “standard” type of
domain studied in analysis), and
(ii) the fact that this is a very large class, encompassing many irregular structures
encountered in nature, such as polygonal domains, domains with corners and
edges, etc.
1
To substantiate the claim made in (i), we wish to point to the existence of an
outward unit normal a.e. on the boundary (as discussed above), the fact that the
Divergence Theorem holds in Lipschitz domains, and the fact that the theory of
Sobolev spaces in Lipschitz domains is virtually as rich as the one for the upper-half
space. As regards (ii), it is illuminating to point out that, as opposed to the class of
domains exhibiting corners and edges, the singularities in a Lipschitz domain are not
necessarily isolated, and they can even accumulate.
One of the main issues addressed in this thesis is
determining geometrical criteria for deciding whether a given domain is Lipschitz.(1.6)
As is well-known, if Ω ⊂ Rn is a bounded domain, then Ω is Lipschitz if and only if
it satisfies a uniform cone property. This asserts that there exists an open, circular,
truncated, one-component cone Γ with vertex at 0 ∈ Rn such that for every x0 ∈ ∂Ω
there exist r > 0 and a rotation R about the origin such that
x + R(Γ) ⊆ Ω,
∀ x ∈ B(x0 , r) ∩ Ω.
(1.7)
An open set Ω ⊆ Rn is called starlike with respect to x0 ∈ Ω if I(x, x0 ) ⊆ Ω for
all x ∈ Ω, where I(x, x0 ) := the open segment with endpoint x and x0 . Call an open
set Ω starlike with respect to a ball B ⊆ Ω if Ω is starlike with respect to any point
in B.
One of our main results reads as follows:
Theorem 1.1. Let Ω be a bounded, open, non-empty set which is starlike with respect
to some ball. Then Ω is a Lipschitz domain.
Ω
Q
P
β
2
While this result appears to be folklore, no complete written proof has appeared as
of now in the literature. Here we address this issue thoroughly and establish the
aforementioned result by proceeding in two steps:
Step I. If Ω ⊆ Rn is a bounded, open, non-empty set which is starlike with respect to
a ball B ⊆ Ω centered at the origin, then there exists a Lipschitz function
ϕ : S n−1 −→ (0, ∞),
(1.8)
where S n−1 is the unit sphere in Rn centered at the origin, such that, in polar coordinates,
Ω = rω : ω ∈ S n−1 ,
0 ≤ r < ϕ(ω) .
(1.9)
Step II. If Ω ⊆ Rn is as in (1.9), with ϕ as in (1.8) Lipschitz function, then Ω is a
Lipschitz domain (in the sense of the earlier definition).
Step I is of purely geometrical nature. At each vector ω ∈ S n−1 the function ϕ is
defined
ϕ(ω) := the distance from the point in Lω ∩ ∂Ω to the origin,
(1.10)
where
Lω := {rω : r > 0} .
(1.11)
The Lipschitz constant of ϕ then turns out to be controlled in terms of the radius of
B, and the diameter of Ω. Parenthetically, let us point out that the converse of the
claim in Step I also holds. That is, if Ω is as in (1.9) with ϕ as in (1.8) Lipschitz
function, then Ω is a starlike domain with respect to some ball B ⊂ Ω, centered as
the origin.
In contrast to the proof of Step I, the justification of Step II is analytical in nature.
The goal is to show that, near every boundary point, the surface {ϕ(ω)ω : ω ∈ S n−1 }
agrees with the graph of a real-valued Lipschitz function ψ (after a rigid relocation).
Consider, as example, the point −ϕ(−en )en , corresponding to the south pole on S n−1 .
Careful inspection reveals that the goal just stated amounts to finding
ψ : Rn−1 −→ R
(1.12)
such that, if xn = ψ(x0 ), then
ϕ
(x0 , xn )
|(x0 , xn )|
= |(x0 , xn )| .
(1.13)
3
It is therefore natural to try to find the Lipschitz function ψ by solving (1.13) for xn
as a function of x0 . To this end, it is convenient to rephrase (1.13) as
F (x0 , xn ) = 0,
(x0 , xn ) near (00 , −ϕ(−en )),
(1.14)
where
0
2
F (x , xn ) := ϕ
(x0 , xn )
|(x0 , xn )|
2
− |x0 | − xn 2 ,
(x0 , xn ) near (00 , −ϕ(−en )),
(1.15)
While the above reasoning strongly suggests using the Implicit Function Theorem
in order to solve (7.10) for xn in terms of x0 , with
xn (00 ) = −ϕ(−en ),
(1.16)
there is one major problem in the implementation of this result. Specifically, in the
classical formulation of the Implicit Function Theorem the function in question has
to be of class C 1 . In our case, since ϕ is Lipschitz, F given in (1.15) is only Lipschitz.
This gives us the impetus to reconsider the scope of this basic result in the theory of
functions of several real variables, and prove a suitable extension which is capable of
handling the current situation.
The extension of the Implicit Function Theorem to the class of Lipschitz functions
which we establish in this thesis appears to be new, and will most surely prove useful
in many other problems. To explain the genesis of this new result, let us first recall
the standard form of the Implicit Function Theorem, as presented in virtually any
textbook on Advanced Calculus. Specifically, we have:
Theorem 1.2. The Implicit Function Theorem Let U be an open set in Rm ,
x0 ∈ U , V an open set in Rn , y0 ∈ V . Let F : U × V → Rn be a continuous function
such that:
(α) F (x0 , y0 ) = 0;
(β) F is class C 1 in U × V ;
(γ)
∂F
(x0 , y0 )
∂y
is an invertible n × n matrix.
Then, there exist r > 0 and a function ϕ satisfying
(I) ϕ : B(x0 , r) → Rn ;
(II) ϕ is continuous on B(x0 , r), ϕ(x0 ) = y0 ;
(III) F (x, ϕ(x)) = 0 for all x ∈ B(x0 , r);
(IV) ϕ is unique with (I)-(III).
In addition, ϕ is differentiable and
(Dϕ)(x) = −
−1 ∂F
(x, y)
(x, y).
∂y
∂x
∂F
4
(1.17)
There are several features of Theorem1.2 which stand out. Namely, it is assumed
that:
(i) Smoothness of F : the function F is of class C 1 (i.e., continuously differentiable
in all variables collectively);
(ii) Non-degeneracy of F :
∂F
(x0 , y0 )
∂y
is an invertible n × n matrix;
(iii) Regularity of ϕ: the function ϕ is of class C 1 ;
(iv) Uniqueness of ϕ: the only function satisfying (I)-(III) above.
Our result which extends the scope of Theorem 1.2 then reads:
Theorem 1.3 (The Implicit Function Theorem for Lipschitz Functions). Let Um ⊆
Rm and Un ⊆ Rn open. Next fix a ∈ Um and b ∈ Un where U := Um × Un . Consider
F : Um × Un −→ Rn
(1.18)
a Lipschitz such that
F (a, b) = 0
(1.19)
and with the property that there exists a constant K > 0 for which
|F (x, y1 ) − F (x, y2 )| ≥ K |y1 − y2 |
for all
(x, yj ) ∈ U
where j = 1, (1.20)
2.
Then there exist Vm ⊆ Rm open, such that a ∈ Vm , and a Lipschitz function ϕ : Vm →
Un such that ϕ(a) = b, and
{(x, y) ∈ Vm × Un : F (x, y) = 0} = {(x, ϕ(x)) : x ∈ Vm } .
(1.21)
In particular,
F (x, ϕ(x)) = 0,
for all x ∈ Vm .
(1.22)
It is of interest to draw a parallel between the features (i) − (iv) of the classical
formulation of the Implicit Function Theorem above and our version of this result,
valid for Lipschitz functions. Concretely, properties (i) − (iv) should be contrasted
to the following features of Theorem 1.3:
(i) Smoothness of F : the function F is Lipschitz in all variables collectively;
(ii) Non-degeneracy of F : the function y 7→ F (x, y) is bi-Lipschitz, uniformly in
x;
5
(iii) Regularity of ϕ: the function ϕ is Lipschitz;
(iv) Uniqueness of ϕ: the only function satisfying (1.21).
As this analysis reveals, there are natural parallels between the statements and the
conclusions in Theorem 1.2 and Theorem 1.3; heuristically speaking, the latter is a
version of the former written by decreasing the amount of regularity assumed from C 1
to Lipschitz. Alas, this comparative analysis is misleading since there are fundamental differences in how the two theorems under discussion are proved. Indeed, while
Lipschitz functions are differentiable almost everywhere (by the classical Rademacher
Theorem mentioned earlier), the fact that these first-order partial derivatives are
merely L∞ functions, typically lacking continuity, renders the standard proof of the
classical Implicit Function Theorem virtually useless in the more general context we
are considering here.
We are therefore forced to adopt a new approach, which makes more economical
use of the weaker hypotheses we are adopting. Let us stress that, as already explained,
the current extension of is not purely academical, but rather dictated by natural,
practical considerations.
One ingredient in the proof of Theorem 1.2, which has intrinsic interest, is the
so-called Invariance of Domain Theorem, which is a topological result about homeomorphic subsets of Euclidean space Rn . This states:
Theorem 1.4 (Invariance of Domain Theorem). Let U be an open subset of Rn and
let f : U → Rn be an injective continuous function. Then V = f (U ) is open and
f : U → V is a homeomorphism. In particular f −1 : V → U is a continuous function.
The theorem is due to L.E.J. Brouwer, who published it in 1912; see [1]. The proof
uses tools of algebraic topology, notably Brouwer’s Fixed Point Theorem. This is a
very general and useful result. In fact, we show that it is possible to se it in order to
provide a new proof of the Implicit Function Theorem along the line of
Invariance of Domain Theorem
=⇒
Inverse Function Theorem
=⇒
Implicit Function Theorem.
(1.23)
The above is a summary of part of the arguments which go into the proof of
Step II, stated near the bottom of page 3. Returning to that context, it is possible to
check that the hypotheses of Theorem 1.2 are satisfied and, as a result, given x0 ∈ ∂Ω,
there exists a Lipschitz function ψ : B 0 → R, with B 0 a (n − 1)-dimensional ball,
(1.24)
such that the graph of ψ agrees with ∂Ω near x0 .
While this is certainly in the spirit of condition (1.2), for the purpose of fully matching
the conditions stipulated in the definition of a Lipschitz domain, we must extend the
Lipschitz function ψ, originally defined in B 0 , to the entire Rn−1 . To do so, we rely
on the following abstract extension result (well-known to experts):
6
Theorem 1.5 (Extension of Lipschitz Functions). Let (X, d) be a metric space, and
assume that A ⊆ X is a given set and
f : A −→ R
(1.25)
is a Lipschitz function. That is, f satisfies
Lip(f ; A) := sup
|f (x) − f (y)|
:
d(x, y)
(x, y) ∈ A × A \ diag
< ∞.
(1.26)
Then there exists
F : X −→ R
(1.27)
such that
F = f
and
F is Lipschitz with Lip(F ; X) = Lip(f ; A).
(1.28)
A
Let us momentarily digress for the purpose of explaining an interesting feature of
the above result, which highlights the advantages of having such an extension result
stated in the context of abstract metric spaces. Specifically, there holds:
Corollary 1.6 (Extension of Hölder Functions). Let S ⊆ Rn be an arbitrary set and
fix s ∈ (0, 1). Also, let
f : S −→ R
(1.29)
be a Hölder function of order s. That is,
Hols (f ; S) := sup
|f (x) − f (y)|
:
|x − y|s
(x, y) ∈ S × S \ diag < ∞.
(1.30)
Then there exists
F : Rn −→ R
(1.31)
such that
F = f
S
and
F is Hölder with Hols (F ; Rn ) = Hols (f ; S).
7
(1.32)
Indeed, since for every s ∈ (0, 1),
s
|x − z|s ≤ |x − y| + |y − z| ≤ |x − y|s + |y − z|s ,
n
for all x, y, z ∈ R(1.33)
,
the space (S, d) with d(x, y) := |x − y|s becomes a metric space. Since, in the context of (S, d), the function f satisfying (1.31) becomes Lipschitz, the conclusion in
Corollary 1.6 follows from Theorem 1.5.
Returning now to the mainstream discussion, having to do with the proof of Step II
(on page 3), we can use Theorem 1.5 in order to ensure that (1.24) holds. In turn,
this now guarantees that condition (1.2) in the definition of a Lipschitz domain has
been fully checked. The next step would then, naturally, be to proceed to verify the
remaining conditions in this definition, i.e., (1.3), (1.4).
Instead of following this course, we choose to adopt a different point of view whose
aim is to fully clarify the interrelationships amongst the conditions (1.3), (1.2), (1.4).
The result we prove in this regard which is most directly relevant to the current
discussion is as follows:
Proposition 1.7. For a nonempty open set Ω ⊂ Rn which satisfies
(1.34)
∂Ω = ∂Ω,
there holds
(1.2) =⇒ (1.3) and (1.4).
(1.35)
This is a purely topological result, which is quite useful whenever the task at hand
is to show that a given domain is Lipschitz. In our situation (i.e., in the proof of
Step II), given what we have proved up to this stage, and since for Ω as in (1.9)
∂Ω = ϕ(ω)ω : ω ∈ S n−1 = ∂Ω,
(1.36)
(i.e., (1.34) holds), Proposition 1.7 can be employed to finish the proof of Step II. In
turn, Step I and Step II conclude the proof of Theorem 1.1.
In closing, we wish to point out that a suitable version of Theorem 1.1 can be
used to actually fully characterize the class of bounded Lipschitz domains. More
specifically, the following holds:
Theorem 1.8. Let Ω be a bounded, open, non-empty set. Then Ω is a Lipschitz
domain if and only if Ω is locally starlike with respect to balls.
The latter condition means that for every x ∈ ∂Ω there exists O ⊆ Rn open
neighborhood of x, along with a ball B ⊂ Ω ∩ O such that Ω ∩ O is starlike with
respect to B.
8
Indeed, the implication =⇒ follows from (1.3)-(1.4), whereas the implication ⇐= can
be established, without too much difficulty, using Theorem 1.1.
We conclude with a brief outline of the contents of the chapters of this thesis.
• Chapter 2 is devoted to discussing the classical Implicit and Inverse Function Theorems. It is divided into two sections, the first of which contains a proof based on
the Contraction Principle (Proposition 2.1). The actual Implicit Function Theorem
is stated as Theorem 2.2. Subsequently, we addressed the Inverse Function Theorem,
which appears as Theorem 2.3. The second section of Chapter 2 contains a new
proof of the aforementioned theorems, based on the Invariance of Domain Theorem
(Theorem 2.4).
• Chapter 3 contains a number of useful results about Lipschitz functions, including:
(i) the fact that their distributional derivatives are in L∞ (Proposition 3.1) and,
(ii) conversely, if a distribution has a its gradient in L∞ then this is given by a
Lipschitz function (Proposition 3.2; see also Theorem 3.3);
(iii) the fact that real-valued Lipschitz functions on arbitrary sets can be extended
to the entire space with preservation of the Lipschitz constant (Theorem 3.4;
cf. also Theorem 3.5 for a related version).
• Chapter 4 is devoted to presenting a proof of the famous Rademacher Theorem.
It is divided into three sections, the first of which contains the actual statement and
proof of this result. In the second section of this chapter, we discuss the issue of the
almost everywhere differentiability of absolutely continuous functions on the real line
(which is used in the proof of Rademacher’s theorem). Finally, in the third section
we prove the following result:
Theorem 1.9. Let U ⊆ Rn be an open set and let
f : U −→ Rm
(1.37)
be a bi-Lipschitz function. That is, there exist 0 < C1 ≤ C2 < ∞ such that
C1 |x − y| ≤ |f (x) − f (y)| ≤ C2 |x − y|,
(1.38)
for every x, y ∈ U . Then there exists E ⊂ U , of Lebesgue measure zero, such that
Df (x)
exists for every
x ∈ U \ E,
(1.39)
and
C1 |v| ≤ |Df (x)v| ≤ C2 |v|,
for every v ∈ Rn , and x ∈ U \ E.
9
(1.40)
As a corollary,
rank Df (x) = n for every x ∈ U \ E,
(1.41)
which showes that necessarily
m ≥ n.
(1.42)
In addition,
m = n =⇒ f (U ) is an open set.
(1.43)
It is worth pointing out that, as the above theorem entails,
there exist bi-Lipschitz mappings f : Rn → Rm ⇐⇒ m ≥ n.
(1.44)
• Chapter 5 is where we present the main novel technical tool of this thesis, namely
a version of the Implicit Function Theorem for Lipschitz functions. See Theorem 5.1
for the actual statement. The proof of this result occupies the bulk of Chapter 5.
• Chapter 6 consists of two sections. In the first, we collect some definitions and
generalities about bounded Lipschitz domains in Rn . In the second section of this
chapter we then proceed to study in greater detail the degree of independence (or
rather lack thereof) of the conditions appearing in the definition of a Lipschitz domain
(Definition 6.5). See Proposition 6.1 and Proposition 6.2 for precise formulations. The
latter result has been rephrased as Proposition 1.7 earlier in the introduction.
• Chapter 7 is devoted to discussing some applications of our main technical results.
It is partitioned into three sections, and the main results, Theorem 7.1 and Theorem 7.2, have been reformulated as Step I and Step II, respectively, earlier in the
introduction.
Acknowledgments It is a pleasure to thank Simona, Nick, Simon, and James for their
special contribution.
10
Chapter 2
The Implicit and Inverse Function
Theorems
2.1
First Proof Using the Contraction Principle
Proposition 2.1 (The Contraction Principle). Let x0 ∈ Rm , y0 ∈ Rn , r, ρ > 0 and
assume that f : B(x0 , r) × B(y0 , ρ) → Rn is a continuous function for which there
exists α ∈ (0, 1) such that
(i) kf (x, y 0 ) − f (x, y 00 )k ≤ αky 0 − y 00 k for all x ∈ B(x0 , r) and y 0 , y 00 ∈ B(y0 , ρ)
(i.e., for every x ∈ B(x0 , r), f (x, ·) is a contraction on B(y0 , ρ)),
(ii) kf (x, y0 ) − y0 k < ρ(1 − α) for all x ∈ B(x0 , r).
Then, there exists a function ϕ with the following properties:
1. ϕ : B(x0 , r) → B(y0 , ρ),
2. ϕ is continuous,
3. f (x, ϕ(x)) = ϕ(x) for all x ∈ B(x0 , r),
4. ϕ is unique with these properties.
Proof. We will prove this result in a sequence of 6 steps.
Step I: For x ∈ B(x0 , r) arbitrary, define inductively the sequence {ϕj (x)}j∈N0 by
ϕ0 (x) = y0 and ϕj+1 (x) = f (x, ϕj (x)) for j ∈ N0 . Observe that
kϕj+1 (x) − ϕj (x)k = kf (x, ϕj (x)) − f (x, ϕj−1 (x)k ≤ αkϕj (x) − ϕj−1 (x)k ≤ . . .
≤ αj kϕ1 (x) − ϕ0 (x)k = αj kf (x, y0 ) − y0 k
≤ αj ρ(1 − α).
(2.1)
Step II: We claim that for each x ∈ B(x0 , r) the sequence {ϕj (x)}j∈N0 is Cauchy.
Indeed, if j ∈ N0 and l ∈ N, applying the triangle inequality repeatedly and then
using (2.1) we can write
11
kϕj+l (x) − ϕj (x)k ≤ kϕj+l (x) − ϕj+l−1 (x)k + kϕj+l−1 (x) − ϕj+l−2 (x)k
+ · · · + kϕj+1 (x) − ϕj (x)k
≤ αj ρ(1 − α)(αl−1 + αl−2 + ... + α + 1)
= αj ρ(1 − α)
1 − αl
< αj ρ.
1−α
(2.2)
Hence, since lim αj ρ = 0, for ε > 0 there exists j0 ∈ N such that 0 < αj ρ < ε for
j→∞
j ≥ j0 , which in turn, used in (2.2), shows that if ε > 0 there exists j0 ∈ N such that
kϕm (x) − ϕj (x)k < ε for all j, m ≥ j0 . This proves the claim.
Step III: As seen in Step II, for each x ∈ B(x0 , r) the sequence {ϕj (x)}j∈N0 is Cauchy,
thus convergent. Hence, we can define ϕ(x) := lim ϕj (x). If we pass to the limit as
j→∞
j → ∞ in the identity ϕj+1 (x) = f (x, ϕj (x)) (recall f is continuous) it follows that
ϕ(x) = f (x, ϕ(x)), proving (1) and (3) in the statement of the proposition.
Step IV: (the uniqueness of ϕ) Assume that there exist two functions ϕ, ϕ0 satisfying
(1) − (3). Then, for x ∈ B(x0 , r),
kϕ(x) − ϕ0 (x)k = kf (x, ϕ(x)) − f (x, ϕ0 (x))k ≤ αkϕ(x) − ϕ0 (x)k,
(2.3)
which implies that (1 − α)kϕ(x) − ϕ0 (x)k ≤ 0. The latter is only possible if kϕ(x) −
ϕ0 (x)k = 0, which in turn implies ϕ(x) = ϕ0 (x). This completes the proof of (4).
Step V: We will prove by induction that each ϕj is continuous. Clearly the constant
function ϕ(x) = y0 , x ∈ B(x0 , r), is continuous. Suppose ϕj is continuous for some
j ∈ N. Since we know that ϕj+1 = f (x, ϕj (x)), f is continuous, and the composition
of continuous functions is continuous, it follows that ϕj+1 must be continuous. By the
Principle of Mathematical Induction, if follows that ϕj is continuous for any j ∈ N0 .
Step VI: (continuity of ϕ) Fix an arbitrary point x ∈ B(x0 , r) and ε > 0. By the
triangle inequality, if x0 ∈ B(x0 , r), we have for each j ∈ N that
kϕ(x) − ϕ(x0 )k ≤ kϕ(x) − ϕj (x)k + kϕj (x) − ϕj (x0 )k + kϕj (x0 ) − ϕ(x0 )k. (2.4)
Since ϕ(y) = lim ϕj (y) for each y ∈ B(x0 , r), there exists j1 , j2 ∈ N such that
j→∞
kϕ(x) − ϕj (x)k ≤
ε
3
kϕ(x0 ) − ϕj (x0 )k ≤
for j ≥ j1 ,
ε
3
for j ≥ j2 . (2.5)
Select j0 = max{j1 , j2 }. By the continuity of ϕj0 at x, there exists δ > 0 such that
kϕj0 (x) − ϕj0 (x0 )k <
ε
3
whenever kx − x0 k < δ.
12
(2.6)
Now we set j = j0 in (2.4) and use (2.5)-(2.6) to conclude that kϕ(x) − ϕ(x0 )k < whenever kx − x0 k < δ. This proves that ϕ is continuous at x. Since x was arbitrary
in B(x0 , r) we obtain that ϕ is continuous on B(x0 , r). The proof of the proposition
is therefore complete at this point.
2
Theorem 2.2 (The Implicit Function Theorem). Let U be an open set in Rm , x0 ∈ U ,
V an open set in Rn , y0 ∈ V . Let F : U × V 7→ Rn be a continuous function such
that:
(α) F (x0 , y0 ) = 0,
(β)
∂F
(x, y)
∂y
(γ)
∂F
(x0 , y0 )
∂y
exists and is continuous on U × V ,
is an invertible n × n matrix.
Then, there exist r > 0 and a function ϕ satisfying
(I) ϕ : B(x0 , r) 7→ Rn ;
(II) ϕ is continuous on B(x0 , r), ϕ(x0 ) = y0 ;
(III) F (x, ϕ(x)) = 0 for all x ∈ B(x0 , r);
(IV) ϕ is unique with (I)-(III).
Furthermore, if F is differentiable at (x0 , y0 ), then ϕ is differentiable at x0 and
∂F
−1 ∂F
(Dϕ)(x0 ) = −
(x0 , y0 )
(x0 , y0 ).
∂y
∂x
(2.7)
Proof. Introduce the function defined by f (x, y) = y − ( ∂F
(x0 , y0 ))−1 (F (x, y)) for
∂y
each (x, y) ∈ U × V . Observe that f (x0 , y0 ) = y0 and, for (x, y) ∈ U × V , f (x, y) = y
⇔ F (x, y) = 0.
Claim: There exist r, ρ > 0 such that f : B(x0 , r) × B(y0 , ρ) 7→ Rn satisfies the
hypothesis of the Contraction Principle with α = 21 .
First we focus on condition (i) from the Contraction Principle. We have
13
kf (x, y 0 ) − f (x, y 00 )k|
−1
−1
∂F
∂F
(x0 , y0 )
(x0 , y0 )
= y 0 −
F (x, y 0 ) − y 00 +
F (x, y 00 )
∂y
∂y
∂F
−1 h ∂F
i
(x0 , y0 )
(x0 , y0 ) (y 0 − y 00 ) − F (x, y 0 ) + F (x, y 00 ) =
∂y
∂y
∂F
0
00
0
00 (x0 , y0 ) (y − y ) − F (x, y ) + F (x, y )
≤ c
∂y
∂F
∂F
∂F
0
00
0
00
0
00
0
00 = c
(x0 , y0 )(y − y ) −
(x, y0 ) (y − y ) +
(x, y0 ) (y − y ) − F (x, y ) + F (x, y )
∂y
∂y
∂y
h ∂F
∂F
i
∂F
≤ c
(x0 , y0 ) −
(x, y0 ) (y 0 − y 00 ) + c
(x, y0 )(y 0 − y 00 ) − F (x, y 0 ) + F (x, y 00 )
∂y
∂y
∂y
=: A + B,
(2.8)
where c = k( ∂F
(x0 , y0 ))−1 k is a positive constant depending only on F, x0 , y0 , n, m. If
∂y
we use the component-wise notation F = (f1 , f2 , ..., fn ), x = (x1 , x2 , ..., xm ) and y =
∂F
∂fi
(y1 , y2 , ..., yn ), then the entries of the matrix
are of the form
for 1 ≤ i, j ≤ n.
∂y
∂yj
Hence,
∂f
∂fi
i
(x0 , y0 ) −
(x, y0 )ky 0 − y 00 k,
A ≤ ccn max 1≤i,j≤n ∂yj
∂yj
(2.9)
with cn > 0 a constant depending only on n. Since for each 1 ≤ i, j ≤ n the function
∂fi
(·, y0 ) is continuous at x0 , there exists rij > 0 such that
∂yj
∂f
∂fi
1
i
(x0 , y0 ) −
(x, y0 ) <
whenever kx − x0 k < rij .
(2.10)
∂yj
∂yj
4ccn
If we now choose 0 < r < min rij , then from (2.9) and (2.10) it follows that
1≤i,j≤n
1
A ≤ ky 0 − y 00 k
whenever kx − x0 k < r.
(2.11)
4
Now we will proceed with the second norm in the very last expression in (2.8). By the
∂F
Mean Value Theorem, F (x, y 0 ) − F (x, y 00 ) =
(x, z) for some z on the line segment
∂y
joining y 0 and y 00 (we denote this by z ∈ [y 0 , y 00 ]). Hence,
∂F
∂F
(x, y0 ) −
(x, y)
B ≤ cky 0 − y 00 k sup ∂y
z∈[y 0 ,y 00 ] ∂y
∂f
∂fi
i
≤ ccn ky 0 − y 00 k sup max (x, y0 ) −
(x, z)
1≤i,j≤n
0
00
∂y
∂y
j
j
z∈[y ,y ]
≤ ky 0 − y 00 k B1 + B2
14
(2.12)
where
∂f
∂fi
i
B1 := ccn sup max (x, y0 ) −
(x0 , y0 )
∂yj
z∈[y 0 ,y 00 ] 1≤i,j≤n ∂yj
(2.13)
∂f
∂fi
i
(x, z) −
(x0 , y0 ).
B2 := ccn sup max ∂yj
z∈[y 0 ,y 00 ] 1≤i,j≤n ∂yj
(2.14)
and
Recall that we are assuming kx − x0 k < r and observe that z ∈ [y 0 , y 00 ] ⊂ B(y0 , ρ)
∂fi
entails kz − y0 k < ρ. Hence, since each
is continuous at (x0 , y0 ), it follows that
∂yj
we can choose r > 0 and ρ > 0 small enough such that
B1 <
1
1
and B2 <
whenever y 0 , y 00 ∈ B(y0 , ρ) and x ∈ B(x0 , r).
8
8
(2.15)
Combining (2.8), (2.11), (2.12), and (2.15) we can conclude that f satisfies (i) with
the r and ρ selected above and α = 12 .
Next we will show that (ii) in the Contraction Principle holds, that is, kf (x, y0 )−y0 k <
ρ
, ∀ x ∈ B(x0 , r). We know that f (x0 , y0 ) = y0 . Fix ε > 0. Since f is continuous at
2
(x0 , y0 ), there exists δ > 0 such that kf (x, y) − f (x0 , y0 )k < if k(x, y) − (x0 , y0 )k < δ.
In particular, kf (x, y0 ) − y0 k < if kx − x0 k < δ. Now choosing = ρ2 (with ρ > 0 as
selected earlier in the proof) we obtain δ = δ(ρ) > 0 such that kf (x, y0 ) − y0 k < ρ/2
if kx − x0 k < δ. At this point all we have to do is to return to the r determined so far
and selected so that r ≤ δ and we see that, with this new r, f verifies the hypotheses
of the Contraction Principle.
Applying the Contraction Principle, we obtain the existence of a function ϕ satisfying (1) − (4) in the conclusion of the Contraction Principle.
So far we have that f (x0 , ϕ(x0 )) = ϕ(x0 ) and f (x0 , y0 ) = y0 . To conclude that
y0 = ϕ(x0 ), observe that the function ψ : B(x0 , r) → B(y0 , ρ) defined by ψ(x) = ϕ(x)
for x 6= x0 and ψ(x) = y0 for x = x0 is continuous and satisfies f (x, ψ(x)) = ψ(x)
for all x ∈ B(x0 , r). By the uniqueness property 4. in the Contraction Principle,
we obtain that ψ = ϕ in B(x0 , r), and in particular, y0 = ϕ(x0 ). Hence, conditions
(I) − (II) in the statement of the Implicit Function Theorem are satisfied.
To conclude that F (x, ϕ(x)) = 0 for all x ∈ B(x0 , r) (condition (III) in the
statement of the Implicit Function Theorem) combine the equivalence f (x, y) = y ⇔
F (x, y) = 0 with (3) in the Contraction Principle. By a similar reasoning we see that
the uniqueness statement in (IV ) follows.
We are left with the proof of the statement regarding the differentiability of ϕ
under the additional condition that F is differentiable at (x0 , y0 ). Assume first that
we have proved that if F is differentiable at (x0 , y0 ) then ϕ is differentiable at x0 (this
15
fact will be proved latter), and introduce the function G : B(x0 , r) 7→ Rm ×Rn , defined
by G(x) := (x, ϕ(x)) for x ∈ B(x0 , r). Then by (III), F ◦ G = 0 on B(x0 , r), and by
(II) G(x0 ) = (x0 , y0 ). Under the assumption that ϕ is differentiable at x0 , we obtain
that G is differentiable at x0 and, since F is differentiable at G(x0 ), by the Chain Rule,
F ◦ G is differentiable at x0 and D(F ◦ G)(x0 ) = (DF )(G(x0 )) · (DG)(x0 ). Hence,
Im×m
0 = (DF )(x0 , y0 )(DG)(x0 ). A direct computation reveals that DG(x0 ) =
,
Dϕ(x0 )
while DF (x0 , y0 ) = ( ∂F
(x0 , y0 ) ∂F
(x0 , y0 )). Multiplying the two we obtain
∂x
∂y
0 =
=
∂F
∂F
Im×m
(x0 , y0 )
(x0 , y0 ) ·
Dϕ(x)
∂x
∂y
∂F
∂F
(x0 , y0 ) +
(x0 , y0 ) · Dϕ(x0 )
∂x
∂y
and the formula (Dϕ)(x0 ) = −( ∂F
(x0 , y0 ))−1 ( ∂F
(x0 , y0 )) now follows.
∂y
∂x
Lastly, we will show that ϕ is differentiable at x0 if F is differentiable at (x0 , y0 ).
Fix an arbitrary > 0. We want to find δ > 0 such that
∂F
−1 ∂F
(x0 , y0 )
(
(x0 , y0 )) h ≤ khk
ϕ(x0 + h) − ϕ(x0 ) +
∂y
∂x
(2.16)
if h ∈ Rm is such that khk < δ. To this end, we write
∂F
−1 ∂F
(x0 , y0 )
(
(x0 , y0 )) h
ϕ(x0 + h) − ϕ(x0 ) +
∂y
∂x
∂F
−1 h ∂F
∂F
i
=
(x0 , y0 )
(x0 , y0 ) ϕ(x0 + h) − ϕ(x0 ) +
(x0 , y0 ) h ∂y
∂y
∂x
∂F
∂F
≤ c
(x0 , y0 ) ϕ(x0 + h) − ϕ(x0 ) +
(x0 , y0 ) h
∂y
∂x
= ckDF (x0 , y0 )(h, ϕ(x0 + h) − ϕ(x0 ))k,
(2.17)
∂F
∂F
(x0 , y0 )
(x0 , y0 ) (an n × (n + m) matrix).
since DF (x0 , y0 ) =
∂x
∂y
Note that F ((x0 , y0 ) + (h, ϕ(x0 + h) − ϕ(x0 ))) = F (x0 + h, ϕ(x0 + h)) = 0 whenever
x0 + h ∈ B(x0 , r), or equivalently, whenever khk < r. Since F (x0 , y0 ) = 0, under the
assumption that khk < r, we can write
c(DF (x0 , y0 )) (h, ϕ(x0 + h) − ϕ(x0 )))
(2.18)
= cF (x0 , y0 ) + (h, ϕ(x0 + h) − ϕ(x0 )) − F (x0 , y0 ) − DF (x0 , y0 ) (h, ϕ(x0 + h) − ϕ(x0 )).
16
Because F is differentiable at (x0 , y0 ), for each ε1 > 0, there exists δ1 > 0 such that
if k(h, ϕ(x0 + h) − ϕ(x0 ))k < δ1 then
cF (x0 , y0 ) + (h, ϕ(x0 + h) − ϕ(x0 )) − F (x0 , y0 ) − DF (x0 , y0 )(h, ϕ(x0 + h) − ϕ(x0 ))
≤ c ε1 k(h, ϕ(x0 + h) − ϕ(x0 ))k
≤ c ε1 (khk + kϕ(x0 + h) − ϕ(x0 )k).
(2.19)
If we denote the norm in the left hand-side of the estimate in (2.16) as LHS, if
k(h, ϕ(x0 + h) − ϕ(x0 ))k < δ1 we can write
∂F
−1 ∂F
kϕ(x0 + h) − ϕ(x0 )k ≤ LHS + (x0 , y0 )
(
(x0 , y0 ) h)
∂y
∂x
≤ c ε1 (khk + kϕ(x0 + h) − ϕ(x0 )k) + c0 khk,
(2.20)
−1
∂F
where c0 = k ∂F
(x
,
y
)
(x0 , y0 )k. For the first inequality in (2.20) we have
0 0
∂y
∂x
applied the triangle inequality, while the estimate of LHS for the second inequality
in (2.20) is due to a combination of (2.17), (2.18), and (2.19). Now we require that
ε1 > 0 is such that c ε1 < 21 . Then (2.20) becomes
1
1
kϕ(x0 + h) − ϕ(x0 )k ≤ khk + kϕ(x0 + h) − ϕ(x0 )k + c0 khk,
2
2
which further implies that
kϕ(x0 + h) − ϕ(x0 )k ≤ (1 + 2c0 )khk,
provided khk + kϕ(x0 + h) − ϕ(x0 )k < δ(2.21)
1.
Observe that if we impose the conditions khk < δ21 and kϕ(x0 + h) − ϕ(x0 )k <
δ1
, then (2.21) holds. Since ϕ is continuous at x0 , there exists δ2 > 0 such that
2
kϕ(x0 + h) − ϕ(x0 )k < δ21 whenever khk < δ2 . Returning with (2.21) in (2.19) and
also recalling (2.17)-(2.18), we see that (with appropriate restrictions on the size of
khk)
LHS ≤ c ε1 (2 + 2c0 )khk.
(2.22)
To arrive at the desired conclusion we impose one more condition on ε1 to ensure
that c ε1 (2 + 2c0 ) < ε. The final conclusion is as follows: for each ε > 0, take
1
, c(2+2c
0 < 1 < min{ 2c
0 ) }, which yields a corresponding value for δ1 , which further
provides a δ2 , and if we take δ := min{r, δ21 , δ2 } then (2.16) holds for all h ∈ Rm is
such that khk < δ. This concludes the proof of the fact that ϕ is differentiable at x0 .
2
17
Theorem 2.3 ( The Inverse Function Theorem). Let U be an open set in Rn , x0 ∈ U ,
f : U → Rn be a continuously differentiable function (equivalently f is C 1 in U ), and
(Df )(x0 ) is a n × n invertible matrix. Then there exists an open set V ⊂ U with
x0 ∈ V , such that f (V ) is open in Rn and f |V : V → f (V ) is a bijective function
with a continuous inverse (f |V )−1 . Moreover (f |V )−1 is differentiable at f (x0 ) and
D(f |V )−1 (f (x0 )) = ((Df )(x0 ))−1 .
(2.23)
Proof. Let us define the function F : U × Rn 7→ Rn by F (x, y) := y − f (x) for x ∈ U ,
y ∈ Rn . Then F is C 1 on U × Rn and if we set y0 := f (x0 ) then F (x0 , y0 ) = 0.
In addition, from the expression of F we see that ∂F
(x, y) = −(Df )(x). Hence,
∂x
∂F
(x0 , y0 ) = −(Df )(x0 ) is an invertible matrix. We can apply the Implicit Function
∂x
Theorem to obtain r > 0 and a function ϕ satisfying
(i) ϕ : B(y0 , r) −→ Rn ;
(ii) ϕ is continuous on B(y0 , r), ϕ(y0 ) = x0 ;
(iii) (ϕ(y), y) ∈ U × Rn for all y ∈ B(y0 , r);
(iv) F (ϕ(y), y) = 0 for all y ∈ B(y0 , r);
(v) ϕ is unique with (i)-(iv);
(vi) ϕ is C 1 at y0 and (Dϕ)(y0 ) = −( ∂F
(x0 , y0 ))−1 ( ∂F
(x0 , y0 )).
∂x
∂y
Now set V := ϕ(B(y0 , r)). By (iii) we have that V ⊂ U and (ii) ensures that
x0 ∈ V . Making use of (iv) we see that f (ϕ(y)) = y for all y ∈ B(y0 , r). Thus,
f (V ) = f (ϕ(B(y0 , r))) = B(y0 , r), which is open.
So far we can conclude that the restriction mapping f |V : V 7→ f (V ) is well defined
and onto. To prove that this mapping is one-to-one, let x1 , x2 ∈ V be such that
f (x1 ) = f (x2 ). From the definition of V it follows that there exist y1 , y2 ∈ B(y0 , r)
such that x1 = ϕ(y1 ) and x2 = ϕ(y2 ). Consequently, f (ϕ(y1 )) = f (ϕ(y2 )). However,
f (ϕ(y1 )) = y1 and f (ϕ(y2 )) = y2 , thus y1 = y2 . Applying ϕ to this last equality it
follows that ϕ(y1 ) = ϕ(y2 ), which is the same as x1 = x2 . This completes the proof
of the fact that f |V : V → f (V ) is one-to-one.
The mapping f |V : V 7→ f (V ) being bijective is invertible. We claim that
(f |V )−1 = ϕ, when we consider ϕ : f (V ) = B(y0 , r) 7→ V = ϕ(B(x0 , r)). To prove
this claim, let y ∈ f (V ). Then ϕ(y) ∈ V and f (ϕ(y)) = y. We already This means
that ϕ(y) is the unique z ∈ V for which f (z) = y, i.e., (f |V )−1 (y) = ϕ(y).
We have already seen that ∂F
(x0 , y0 ) = −(Df )(x0 ). Moreover, by a direct com∂x
∂F
putation, ∂y0 (x0 , y0 ) = I. Hence,
∂F
−1 ∂F
D(f |V )−1 (f (x0 )) = Dϕ(y0 ) = −
(x0 , y0 )
(x0 , y0 ) = ((Df )(x0 ))−1
(2.24)
.
∂x
∂y0
We are left with proving that V is open. To do so, fix x∗ ∈ V . We want to find
ρ > 0 such that B(x∗ , ρ) ⊆ V . From the definition of V , we see that there exists y ∗ ∈
18
B(y0 , r) such that x∗ = ϕ(y ∗ ). This implies that f (x∗ ) = f (ϕ(y ∗ )) = y ∗ ∈ B(y0 , r).
Hence, x∗ ∈ f −1 (B(y0 , r)) (where f −1 is the preimage of f ) and since f is continuous,
f −1 (B(y0 , r)) is open. Thus there exists ρ > 0 such that B(x∗ , ρ) ⊆ f −1 (B(y0 , r)),
which implies that f (B(x∗ , ρ)) ⊆ B(y0 , r). Now apply ϕ to the last inclusion to obtain
that ϕ(f (B(x∗ , ρ))) ⊆ ϕ(B(y0 , r)) = V . If we can prove that
ϕ(f (x̂)) = x̂ for all x̂ ∈ B(x∗ , ρ),
(2.25)
then the conclusion will be that B(x∗ , ρ) ⊆ V as desired. Finally, to see that (2.25)
holds, define ψ : B(y0 , r) → Rn , by ψ(y) = ϕ(y) if y ∈ B(y0 , r), y 6= f (x̂), and
ψ(y) = x̂ if y = f (x̂) ∈ B(y0 , r). Then this function verifies (i)-(iv). By (v), we must
have ψ = ϕ, hence ϕ(f (x̂)) = ψ(f (x̂)) = x̂ for all x̂ ∈ B(x∗ , ρ) which gives (2.25).
2.2
Second Proof Using the Invariance of Domain
Theorem
In this subsection we present a second proof of the Implicit and Inverse Function
Theorems. As a preamble to this we isolate a useful result, called the Invariance of
Domain theorem, which is a theorem from topology about homeomorphic subsets of
Euclidean space Rn . It states:
Theorem 2.4 (Invariance of Domain). Let U be an open subset of Rn and let f :
U → Rn be an injective continuous function. Then V = f (U ) is open and f : U → V
is a homeomorphism. In particular f −1 : V → U is a continuous function.
The theorem and its proof are due to L.E.J. Brouwer, published in 1912; see [1]. The
proof uses tools of algebraic topology, notably the Brouwer fixed point theorem.
We will now prove the Inverse Function Theorem stated in Section 1.
Proof of Theorem 2.3. We make the following claim. There exists r > 0 such that f
is injective when restricted to B(x0 , r).
Proof of Claim. Reasoning by contradiction, we may assume that for all j ∈ N, f is
not injective when restricted to B(x0 , 1j ). This means that for all j ∈ N there exists
xj , yj ∈ B(x0 , 1j ), xj 6= yj and such that f (xj ) = f (yj ). Define
wj :=
y j − xj
∈ S n−1 , j ∈ N.
|yj − xj |
(2.26)
Also, consider the path
γj (t) := xj + t(yj − xj ), 0 ≤ t ≤ 1
(2.27)
and fix an arbitrary vector v ∈ Rn . Then the function
19
[0, 1] 3 t 7→ f (γj (t)) · v ∈ R
(2.28)
is of class C 1 and takes the same values at t = 0 and t = 1, i.e.
[f (γj (t)) · v]
t=0
= f (xj ) · v
= f (yj ) · v
= [f (γj (t)) · v]
.
(2.29)
t=1
Rolle’s Theorem then implies that there exists a number
cj ∈ (0, 1)
(2.30)
such that
d
[f (γj (t)) · v]
= 0.
dt
t=cj
(2.31)
Using Chain Rule, we see that
d
[f (γj (t)) · v] = (Df )(γj (t)) · γj (t) · v
dt
(2.32)
so that, keeping in mind that γj (t) = yj − xj , we arrive at
(Df )(γj (cj ))(yj − xj ) · v = 0,
(2.33)
(Df )(γj (cj ))wj · v = 0.
(2.34)
or
Now |xj − x0 | < 1j , |yj − x0 | <
1
j
implies that xj → x0 , yj → y0 as j → ∞. Hence
γ(xj ) = x0 + xj (yj − xj ) −→ x0
(2.35)
Since wj ∈ S n−1 , by passing the subsequence it can be assumed that wj → w0 ∈
S
as j → ∞ for some w0 ∈ S n−1 . Passing the limit j → ∞ in (2.33) we then
obtain
n−1
(Df )(x0 )w0 · v = 0 for all v ∈ Rn .
20
(2.36)
Choosing v := (Df )(x0 )w0 , this implies
k(Df )(x0 )w0 k2 = 0
(2.37)
and ultimately
(Df )(x0 )w0 = 0.
(2.38)
Since (Df )(x0 ) is, by assumption, an invertible matrix, this forces
w0 = 0,
(2.39)
contradicting the fact that w ∈ S n−1 . This contradiction finishes the proof of the
claim made at the beginning of the proof of the theorem.
Once this claim has been established, the Invariance of Domain Theorem implies
that
W := f (B(x0 , r))
(2.40)
is an open set in Rn and, if V := B(x0 , r) then
f : V −→ W
(2.41)
is a topological homeomorphism (i.e., continuous, bijective, and the inverse function
is continuous as well). There remains to prove that
f −1 : W −→ V is differentiable at f (x0 )
(2.42)
(Df )−1 (f (x0 )) = ((Df )(x0 ))−1 .
(2.43)
and
To this end, consider
f −1 (f (x0 ) + h) − f −1 (f (x0 )) − ((Df )(x0 ))−1 h
.
khk
(2.44)
Our goal is to show that this expression goes to zero as h → 0. Pick xh ∈ B(x0 , r)
such that
21
f (xh ) = f (x0 ) + h,
i.e., xh = f −1 (f (x0 ) + h).
(2.45)
Then f (xh ) → f (x0 ) as h → 0. Thus since f − 1 is continuous, then xh = f −1 (f (xh )) →
f −1 (f (x0 )) = x0 as h → ∞. Thus,
xh −→ x0
as h → ∞.
(2.46)
Then
f −1 (f (x0 ) + h) − f −1 (f (x0 ) − ((Df −1 (f (x0 ))h
khk
=
(2.47)
f −1 (f (xh )) − f −1 (f (x0 ) − ((Df −1 (f (x0 ))h
khk
xh − x0 − ((Df −1 (f (x0 ))h
khk
−1 (Df )(x )(x − x ) − h 0
h
0
= (Df )(x0 )
.
khk
=
(2.48)
Hence, for our goals, it suffices to prove that
(Df )(x0 )(xh − x0 ) − h
khk
−→ 0,
as h → ∞.
(2.49)
Recall that f is differentiable at x0 means that there exists
η : [0, ∞) −→ [0, ∞)
(2.50)
lim η(t) = 0
(2.51)
with
t→0+
such that
kf (x0 + k) − f (x0 ) − (Df )(x0 )kk
≤ η(kkk)
kkk
for every k ∈ Rn \ {0}. In our case, take
22
(2.52)
k := xh − x0
(2.53)
so that (2.52) becomes (applying (2.53) as well)
kh − (Df )(x0 )(xh − x0 )k ≤ kxh − x0 k · ηkxh − x0 k.
(2.54)
This implies
kxh − x0 k = k((Df )(x0 ))−1 (Df )(x0 )(xh − x0 )k
≤ Ck(Df )(x0 )(xh − x0 )k
≤ Ckh − (Df )(x0 )(xh − x0 )k + Ckhk
≤ Cη(kxh − x0 k) kxh − x0 k + Ckhk,
(2.55)
by (2.54). Choose h close to 0 so that
1
2C
1
then absorb 2 kxh − x0 k in the left-hand side to obtain
η(kxh − x0 k) <
(2.56)
kxh − x0 k ≤ 2Ckhk
(2.57)
if h is sufficiently close to 0. Finally, using our previous results we can write the
following
k(Df )(x0 )(xh − x0 ) − hk
kxh − x0 k η(kxh − x0 k)
≤
khk
khk
≤
Ckhk η(kxh − x0 k)
khk
(applying (2.54))
(applying (2.57))
≤ Cη(kxh − x0 k) → 0 as h → 0,
(2.58)
since
xh −→ x0 as h → 0.
(2.59)
This completes the proof of Theorem 2.3.
2
Finally, observe that
Inverse Function Theorem =⇒ Implicit Function Theorem.
(2.60)
Indeed, the Implicit Function Theorem (as stated in Theorem 2.2) follows immediately
from the Inverse Function Theorem (as stated in Theorem 2.3) applied to the function
f (x, y) := (x, F (x, y)).
(2.61)
23
Chapter 3
Lipschitz Functions
We begin by recalling the class of Lipschitz functions.
Definition 3.0.1 (Definition of a Lipschitz Function). Let f : U → Rm where U is
an open set in Rn . Then call f Lipschitz if
Lip(f ; U ) := sup
(x, y) ∈ U × U \ diag < ∞
|f (x) − f (y)|
:
|x − y|
(3.1)
where Lip(f ; U ) is the Lipschitz constant.
It follows from the above definition that a function f : U −→ R is Lipschitz if and
only if there exists M ≥ 0 such that
|f (x) − f (y)| 6 M |x − y|
for every x, y ∈ U.
(3.2)
The Lipschitz constant Lip(f ; U ) is, in fact, the smallest M for which (3.2) holds.
Remark 3.0.1. If the function f is Lipschitz then f is continuous (in fact, uniformly
continuous). In turn, any continuous function f belongs L1loc . Also, any function f
0
in L1loc induces a regular distribution (i.e., f ∈ D (Rn )).
Proposition 3.1. If f : Rn → R is Lipschitz then the components of ∇f , computed
in the sense of distributions, belong to L∞ (Rn ), and k∇f kL∞ (Rn ) 6 M .
R
Proof. Let θ ∈ Cc∞ (Rn )be such that θ(x)dx = 1 and θ > 0. For each number ε > 0,
define θε (x) := ε−n θ xε , and let fε := f ∗ θε . Then the following are true:
(1) fε −→ f as ε → 0+ uniformly on compact sets;
(2) |fε (x) − fε (y)| 6 M |x − y|;
(3) fε ∈ C ∞ (Rn ) and k∇fε kL∞ (Rn ) ≤ M .
24
Let us justify these claims. Claim (1) follows from the fact that f is continuous,
as known results. As far as (2) is concerned, we write
Z
|fε (x) − fε (y)| 6
|f (x − z) − f (y − z)| |θε (z)| dz 6 M |x − y| ,
(3.3)
Rn
as desired. Turning our attention to (3), we first note that
fε ∈ C ∞ (Rn ) since by
R
differentiating under the integral sign we obtain ∂ α fε (x) = Rn f (y)∂ α θε (x − y) dy for
every multiindex α ∈ Nn0 . In turn, this implies that, for every j ∈ {1, ..., n},
∂j fε (x) = lim
h→0
fε (x + hej ) − fε (x)
h
(3.4)
exists. Keeping this in mind and invoking (2) we also obtain k∂j fε kL∞ (Rn ) ≤ M ,
finishing the proof of the claim made in (3) above.
Moving on, for any test function ψ ∈ Cc∞ (Rn ), we write
Z
h∂j f, ψi = −hf, ∂j ψi = − lim+ hfε , ∂j ψi = lim+ h∂j fε , ψi = lim+
ε→0
ε→0
ε→0
(∂j fε )(x)ψ(x) dx.(3.5)
Rn
Using (3) we get
Z
≤ k∂j fε kL∞ (Rn ) kψkL1 (Rn ) ≤ M kψkL1 (Rn ) .
(∂
f
)(x)ψ(x)
dx
j
ε
n
(3.6)
R
Hence, from (3.6) and (3.5) we obtain that
Z
(∂j fε )(x)ψ(x) dx ≤ M kψkL1 (Rn ) .
h∂j f, ψi ≤ lim sup ε→0+
(3.7)
Rn
Consequently, |h∂j f, ψi| 6 M kψkL1 (Rn ) so that the linear assignment
Cc∞ (Rn ) 3 ψ 7→ h∂j f, ψi ∈ R
(3.8)
is continuous in the L1 -norm, and has norm less or equal to M . Since Cc∞ (Rn )
embeds densely into L1 (Rn ), the linear functional in (3.8) extends to a linear, bounded
functional
Λ : L1 (Rn ) −→ R, with norm less or equal to M.
(3.9)
∗
Thus, Λ ∈ L1 (Rn ) has norm less or equal to M , and since the dual space of
L1 (Rn ) is L∞ (Rn ), we obtain that there exists a unique
25
gj ∈ L∞ (Rn )
with kgj kL∞ (Rn ) ≤ M,
(3.10)
such that
Z
hΛ, ηi =
∀ η ∈ L1 (Rn ) .
gj (x)η(x) dx,
(3.11)
Rn
Granted this and keeping in mind that Λ is an extension of the linear assignment in
(3.8), we arrive at the conclusion that
Z
h∂j f, ψi =
gj (x)ψ(x) dx,
Rn
∀ ψ ∈ Cc∞ (Rn ) .
(3.12)
In other words,
∂j f = gj
in the sense of distributions in Rn .
(3.13)
In concert, (3.13) and (3.10) give that ∂j f ∈ L∞ (Rn ) for every j, and k∂j f kL∞ (Rn ) 6
M . This completes the proof of the proposition.
2
Proposition 3.2. If f is a distribution in Rn with the property that ∇f ∈ L∞ (Rn ),
then f is a Lipschitz function.
0
Proof. Let f ∈ D (Rn ) be as in the statement of the proposition, and define
fε (x) := hf, θε (x − ·)i,
x ∈ Rn ,
(3.14)
where θε is as before. It can then be checked that
fε ∈ C ∞ (Rn )
and fε −→ f
in the sense of distributions.
(3.15)
Furthermore,
∇fε (x) = hf, ∇θε (x − ·)i = −hf, ∇. (θε (x − ·))i = h∇f, θε (x − ·)i = (∇f )ε (x)
(3.16)
and we claim that
(∇f )ε ∈ L∞ (Rn )
satisfies
k∇fε kL∞ (Rn ) 6 k∇f kL∞ (Rn ) .
To see this, we estimate
26
(3.17)
Z
6 k∇f k ∞ n ,
(∇f
)(x
−
y)θ
(y)
dy
ε
L (R )
n
(3.18)
R
which justifies (3.17).
Going further, fix x0 ∈ Rn and consider the sequence of C ∞ functions
gε (x) := fε (x) − fε (x0 ),
for x ∈ Rn .
(3.19)
Then
|gε (x)| 6 k∇f kL∞ (Rn ) |x − x0 | and k∇gε kL∞ (Rn ) 6 k∇fε kL∞ (Rn ) 6 k∇f kL∞ (R(3.20)
n)
By the Arzela-Ascoli Theorem, for any compact subset of Rn there exists a subsequence {gε }ε which converges uniformly on it. By a diagonal argument we can assume
there exists a subsequence of {gε }ε which converges uniformly on any compact subset
of Rn to some function g ∈ C 0 (Rn ). On the other hand, gε = fε − fε (x0 ) −→ f − c
in the sense of distributions (where c is a constant). Then we have f = g + c.
Combining the fact that f (x)−f (y) = g(x)−g(y) and fε (x)−fε (y) −→ g(x)−g(y)
with
| fε (x) − fε (y)| 6 k∇fε kL∞ (Rn ) |x − y| 6 k∇f kL∞ (Rn ) |x − y|
(3.21)
we may finally conclude that |f (x) − f (y)| 6 k∇f kL∞ (Rn ) |x − y|. This shows that f
is a Lipschitz function, as desired.
2
At this stage we would like to summarize the conclusions in Proposition (3.1) and
Proposition (3.2) in the form of the following theorem.
Theorem 3.3. For a function f : Rn → R the following two statements are equivalent,
(i) f is a Lipschitz function;
(ii) in the sense of distributions ∇f ∈ L∞ (Rn ).
Next we consider the issue of extending a given Lipschitz function originally defined on a set A to the entire space, denoted by X. The most general and natural
context in which this task can be carried out is that of metric spaces. For the convenience of the reader we first summarize the definition of an abstract metric space.
Specifically we have the following:
Definition 3.0.2. Call (X, d) a metric space if X is an arbitrary set and
d : X × X −→ [0, ∞)
(3.22)
is a function satisfying the following properties:
27
(i) non-degeneracy: d(x, y) = 0 ⇐⇒ x = y;
(ii) triangle inequality: d(x, y) ≤ d(x, z) + d(z, y) for every x, y, z ∈ X;
(iii) symmetry: d(x, y) = d(y, x) for every x, y ∈ X.
In this setting, the Extension result alluded to before reads as follows.
Theorem 3.4 (Extension of Lipschitz Functions). Let (X, d) be a metric space such
that
f : A −→ R is Lipschitz, for
A⊆X
and
A 6= X
(3.23)
then there exists
F : X −→ R
(3.24)
such that:
F = f
and
F is Lipschitz with Lip(F ; X) = Lip(f ; A).
(3.25)
A
Proof. Let us first define the formula
F (x) := inf {f (a) + M d(a, x) : a ∈ A}
x ∈ X,
(3.26)
where
M := Lip(f ; A).
(3.27)
Claim 1. If x ∈ A then F (x) = f (x).
Proof of Claim 1. Fix three arbitrary points a1 , a ∈ A and x ∈ X. We may then
write the following:
f (a1 ) = f (a) + [f (a1 ) − f (a)]
(3.28)
≤ f (a) + M d(a1 , a)
(3.29)
≤ f (a) + M [d(x, a1 ) + d(a, x)]
(3.30)
= f (a) + M d(x, a1 ) + M d(a, x).
(3.31)
Hence combining (3.28) and (3.31) we get the following
f (a1 ) − M d(x, a1 ) ≤ f (a) + M d(a, x).
28
(3.32)
A few justifications are in order for the previous statements. In (3.28) we make use of
the additive inverse property. Statement (3.30) makes used of the triangle inequality
on metric spaces (X, d). The last two statements are algebraic simplifications.
Since a ∈ A this implies that f (a1 ) − M d(x, a1 ) is an arbitrary lower bound for
{f (a) + M d(a, x) : a ∈ A}. Given the definition of F in (3.26), this gives:
f (a1 ) − M d(x, a1 ) ≤ F (x) ≤ f (a) + M d(a, x)
for all a, a1 ∈ X and x ∈ X.
(3.33)
If x ∈ A we may specialize (3.33) to the case when a1 := x and a := x. In this
scenario, (3.33), becomes
f (x) − 0 ≤ F (x) ≤ f (x) + 0
i.e. F (x) = f (x) if x ∈ A.
(3.34)
Thus this concludes the proof of Claim 1.
Claim 2. F : X → R is Lipschitz with Lip (F ; X) = Lip (f ; A).
Proof of Claim 2. Fix two points x0 , x1 ∈ X. We want |F (x1 ) − F (x0 )| ≤ M d(x0 , x1 ).
Hence it is enough to show that for all x0 , x1 ∈ X we have F (x1 )−F (x0 ) ≤ M d(x0 , x1 ).
Since we can interchange the roles of x0 and x1 , we obtain F (x0 ) − F (x1 ) ≤
M d(x0 , x1 ). Thus fix an arbitrary ε > 0. Then there exists a0 ∈ A such that
f (a0 ) + M d(a0 , x0 ) − ε < F (x0 ).
(3.35)
and thus
−F (x0 ) < −f (a0 ) − M d(a0 , x0 ) + ε
(3.36)
F (x1 ) ≤ f (a0 ) + M d(a0 , x1 ).
(3.37)
also
Combining (3.36) and (3.37) we get
F (x1 ) − F (x0 ) ≤ M [d(x0 , x)] < ε.
(3.38)
Since ε is arbitrary this finishes the proof of Claim 2. Thus (3.32) follows from the
proof of Claim 1 and 2.
2
29
Theorem 3.5 (Extension of Lipschitz Functions on Rn ). Let (X, d) be a metric space
such that:
f : A −→ Rm
is Lipschitz, for
A⊆X
and
A 6= X
(3.39)
then there exists
F : X −→ Rm
(3.40)
such that
F = f
and
F is Lipschitz with Lip(F ; X) ≤
√
mLip(f ; A).
(3.41)
A
Proof. Applying Theorem 2.4 we know that there exists an Fj : X → R such that
F j = fj
A
and Lip(Fj ; X) = Lip(fj ; A).
(3.42)
Now take F := (F1 , · · · , Fm ) then
F : X → Rm
and F = f.
(3.43)
A
Also, for every x1 , x2 ∈ X we have
|F (x1 ) − F (x2 )| =
m
X
! 21
(Fj (x1 ) − Fj (x2 ))2
j=1
√
=
m · M d(x1 , x2 ).
This proves that F : X → Rm is a Lipschitz function with Lip(F ; X) ≤
finishing the proof of the theorem.
30
(3.44)
√
m · M,
2
Chapter 4
Rademacher’s Theorem
Here we present a proof of Rademacher’s Theorem on the a.e. differentiability of
Lipschitz functions, following [2]
4.1
Almost Everywhere Differentiability of Lipschitz Functions
Theorem 4.1 (Rademacher). Let f : Rn → Rm be locally Lipschitz. Then f is
differentiable Ln -a.e., ie gradf (x) = (D1 f (x), ....., Dn f (x)) exists and
f (y) − f (x) − grad f (x) · (y − x)
=0
y→x
|y − x|
lim
(4.1)
for Ln -a.e. x ∈ Rn .
Proof. We assume m = 1 and f is Lipschitz. Fix v ∈ Rn with |v| = 1, and define
Dv f (x) =: lim
t→0
f (x + tv) − f (x)
t
for x ∈ Rn
(4.2)
if this limit exists.
Claim 1. Dv f (x) exsists for Ln -a.e. x ∈ Rn .
Proof of Claim 1. In fact, since f is continuous,
Dv f (x) = lim sup
t→0
f (x + tv) − f (x)
t
(4.3)
is Borel measurable. Why? Def: f is Borel measurable if f −1 (I) is a Borel set for
every I ⊂ R open interval. Fact 1: If fk are Borel measurable k ∈ N, then so is
f (x) := limk→∞ fk (x) if the limit exists a.e. Fact 2: If fk are Borel measurable for
every k, then f (x) = supk∈N fk (x) is Borel measurable. Fact 3: Every continuious
function is Borel measurable.
31
Dv f (x) = lim inf
t→0
f (x + tv) − f (x)
.
t
(4.4)
And by using the definition of Borel measurable sets on the inverse of the function
lim inf
t→0
f (x + tv) − f (x)
f (x + tv) − f (x)
− lim sup
t
t
t→0
on the interval (0, ∞).
(4.5)
Thus
Bv =: {x ∈ Rn ; Dv f (x) does not exsist } = {x ∈ Rn ; Dv f (x) < Dv f (x)}
(4.6)
is Borel measurable.
Now for each x, v ∈ Rn , with |v|=1, define φ : R → R by
φ(t) := f (x + tv),
t ∈ R.
(4.7)
Then φ is Lipschitz, thus absolutely continous, and thus differentiable L1 -a.e. Hence
for each line L parallel to v,
L1 (Bv ∩ L) = 0.
(4.8)
Then Fubini’s Theorem implies
Ln (Bv ) = 0.
(4.9)
As a consequence, we see that
grad f (x) =: (D1 f (x), ..., Dn f (x))
(4.10)
2
exsists for Ln -a.e. x ∈ Rn . This finishes the proof of Claim 1.
Claim 2. Show that Dv f (x) = v · grad f (x) for Ln -a.e x ∈ Rn .
Proof of Claim 2. Assume that η ∈ C0∞ (Rn ) (i.e. η is compactly supported) then
Z
Rn
f (x + tv) − f (x)
η(x)dx = −
t
Z
f (x)
Rn
To see this, split the integrals
32
η(x) − η(x − tv)
dx.
t
(4.11)
1
t
Z
Z
f (x + tv)η(x)dx −
f (x)η(x)dx .
Rn
(4.12)
Rn
Applying a change of variables to the first integral above we get
1
t
Z
Z
f (y)η(y − tv)dy −
Rn
f (x)η(x)dx .
(4.13)
Rn
Changing x to y in the first integral and further simplifications yield
Z
Rn
f (x + tv) − f (x)
η(x)dx = −
t
Z
Rn
η(x) − η(x − tv)
f (x)dx,
t
finishing the proof of (4.11). Next take t =
the following
f (x + k1 v) − f (x) Lip(f )( k1 )|v|
≤
1
1
k
1
k
(4.14)
where k = 1, 2, .... in (4.14) and note
because f is Lipschitz.
(4.15)
k
Hence
f (x + k1 v) − f (x) ≤ Lip(f )
1
because |v| = 1.
(4.16)
k
Since f was assumed to be Lipschitz then f is continuous and hence locally bounded
(ie. f is bounded on every compact set). Furthermore since f is locally bounded and
η(x) ∈ C0∞ (Rn ) (ie. η(x) is compactly supported) then
Z
Rn
f (x + tv) − f (x)
η(x)dx < ∞.
t
(4.17)
Now in claim 1 we have shown that Dv f (x) exists for Ln a.e., hence we have established pointwise convergence for Dv f (x). Furthermore since the sequence of functions
Dv f (x) for t = k1 is dominated by the Lipschitz constant and since constants are integrable on bounded sets then
Z
Lipf |η(x)| dx < ∞.
(4.18)
Rn
thus we have fulfilled the crtiteria for the Dominated Covergence Theorem. Hence
Z
lim
t→0
Rn
f (x + tv) − f (x)
η(x)dx =
t
f (x + tv) − f (x)
η(x)dx.
t
Rn t→0
Z
lim
33
(4.19)
Appyling our change of variables from (4.14) we get
Z
Z
Dv f (x)η(x)dx = −
Rn
f (x)Dv η(x)dx.
(4.20)
Rn
Applying the definition of Dv f (x) from (4.2), we see that
η(x + tv) − η(x)
d
= [η(x + tv)] .
t→0
t
dt
t=0
Dv η(x) =: lim
(4.21)
With this in mind we can further simplify (4.20) through the definition of the Chain
Rule.
Let r(t) = x + tv
and r0 (t) = v.
(4.22)
Hence
d
[η(x + tv)] = ∇η(r(t)) · r0 (t).
dt
(4.23)
Thus
∇η(x + tv)v =
n
X
∂i η(x + tv)vi .
(4.24)
i=1
Replacing Dv η(x) in (4.20) with (4.24) we get
Z
−
f (x)Dv η(x) dx = −
Rn
n
X
i=1
Z
vi
f (x)
Rn
∂η
(x) dx.
∂xi
(4.25)
Now since f is absolutely continuous and η ∈ C0∞ (Rn ) we can apply integration by
parts to (4.25) through the use of a result which will be proved later in Section 4.2.
Theorem 4.2. Let f : [a, b] → R be an absolutely continuous function. Then f is
differentiable almost everywhere with integrable derivative such that
Z
f (b) − f (a) =
b
f 0 (x) dx
(4.26)
a
34
Thus, we get the following statements.
−
n
X
Z
vi
Rn−1
i=1
·
=−
n
X
=−
Rn−1
Z
−
vi
n
X
Z
=
Z
vi
Rn
i=1
Rn i=1
vi
f (x1 , · · · , xi−1 , xi , · · · , xn )
R
∂η
(x1 , · · · , xi−1 , xi , · · · , xn ) dxi dx1 · · · dxn
∂xi
Z
i=1
n
X
Z
∂f
−
(x)η(x) dxi
∂xi
R
(4.27)
dx1 · · · dxi−1 · · · dxn
∂f
(x)η(x) dx
∂xi
(4.28)
(4.29)
∂f
(x)η(x) dx
∂xi
(4.30)
Z
[∇f (x) · v]η(x) dx.
=
(4.31)
Rn
A few comments are in order for the previous collection of statements. The first
equality above, (4.27), uses Fubini’s Theorem to split the domain of integration. In
the second equality, (4.28), we have used integration by parts for the function
u(xi ) := f (x1 , · · · , xi−1 , xi , xi+1 , · · · , xn ),
(4.32)
v(xi ) := η(x1 , · · · , xi−1 , xi , xi+1 , · · · , xn ).
(4.33)
With x1 , · · · , xi−1 , xi+1 , · · · , xn fixed and xi variable. Here we have also used that
du =
∂f
(x1 , · · · , xi , · · · , xn )
∂xi
(4.34)
dv =
∂η
(x1 , · · · , xi , · · · , xn )
∂xi
(4.35)
and that
35
xi =∞
= 0 − 0 = 0,
f (x1 , · · · , xi , · · · , xn )η(x1 , · · · , xi , · · · , xn ) (4.36)
xi =−∞
due to the face that η ∈ C0∞ (Rn ). Furthermore in the third equality, (4.28) we use
Fubini’s Theorem to reconstruct the domain of integration back to Rn . In the fourth
inequality, (4.30) we use the additivity and homogeneity of the integral to bring the
summation inside. Lastly, (4.31), is deduced from the definition of the directional
derivative. Altogether
Z
Z
(v · gradf (x))η(x)dx,
Dv f (x)η(x)dx =
Rn
Rn
for all η ∈ C0∞ (Rn ).
(4.37)
Since η is an arbitrary test function, we may conclude from the above equality that
Dv f (x) = v · grad f (x)
(4.38)
for Ln a.e. x ∈ Rn . This finishes the proof of claim 2.
2
Before finishing the proof we will introduce a subset A of Rn as follows. First, choose
a countable set of numbers
n−1
{vk }∞
k=1 ⊆ S
(4.39)
such that
n−1
{vk }∞
.
k=1 is dense in S
(4.40)
Next define:
Ak := {x ∈ Rn : ∃ Dvk f (x), ∃ grad f (x) and Dvk f (x) = vk · grad f (x)}.(4.41)
Finally define:
A :=
∞
\
Ak .
(4.42)
k=1
Now we can finish the proof of Rademacher’s Theorem by establishing the following.
Claim 3. f is differentiable for all x ∈ A.
Proof of Claim 3. This will involve several steps.
Step 1: Show that Ln (Ac ) = 0. From claim 1 we know that for every v ∈ S n−1
36
Dvk f (x) exists a.e.
(4.43)
hence, in particular
grad f (x) exists a.e.
(4.44)
Claim 2 also gives us
Dvk f (x) = vk · grad f (x) a.e.
(4.45)
This implies that Ln (Ack ) = 0. On the other hand,
Ln (Ac ) ≤
X
Ln (Ack ).
(4.46)
k∈N
The right hand side of (4.46) is the sum of terms of which all are equal to zero. Thus
the left hand side of (4.46) is equal to zero. Thus the claim in step 1 has been proved.
Step 2: Now setup a function Q, which will be defined on A × S n−1 × R \ {0} → R,
according to
Q(x, v, t) :=
f (x + tv) − f (x)
− v · grad f (x).
t
(4.47)
Then if x ∈ A and v, w ∈ S n−1 the following result is obtained:
f (x + tv) − f (x + tw) + |(v − w) · grad f (x)|
|Q(x, v, t) − Q(x, w, t)| ≤ (4.48)
t
≤ Lip(f )|v − w| + |grad f (x)| · |v − w|
(4.49)
√
≤ ( n + 1) Lip(f )|v − w|
(4.50)
A few comments are in order for the previous statements. The first equality above,
(4.48), is obtained by the triangle inequality. The next equality, (4.49), is gained
through the Lipschitzianity of f and the Cauchy-Schwartz inequality. The last line,
(4.50), is not entirely obvious; it requires us to examine the limits of the partial
derivatives. Thus let ej denote the j th basis vector of Rn . Furthermore
37
f (x + ej h − f (x) |Dej f (x)| = lim h→0
h
≤ lim Lip(f )
h→0
= Lip(f ).
(4.51)
The previous statement, (4.51), is due to the Lipschitz property of the function f .
Step 3: Here we present the last details in the proof of claim 3. Now fix an ε > 0.
By the compactness of S n−1 and the density of {vk }k in S n−1 it is possible to choose
N ∈ N so large that for all v ∈ S n−1 there exists a k ∈ N, k < N such that
|v − vk | ≤
ε
√
2( n + 1) Lip(f )
(4.52)
where vk was from the set in (4.40). Hence
lim Q(x, vk , t) = 0
(4.53)
t→0
simply by the definition A in (4.41) and (4.42). Thus there exists a δ > 0 such that
for every x ∈ A,
ε
for all 0 < |t| < δ.
2
Thus for all x ∈ S n−1 , there exists k ∈ {1, 2, 3, ...., N − 1} such that:
|Q(x, vt , t)| <
(4.54)
|Q(x, v, t)| ≤ |Q(x, vk , t)| + | Q(x, v, t) − Q(x, vk , t)|
< ε,
if 0 < |t| < δ.
(4.55)
Notice that the same δ works for all v ∈ S n−1 . As a result of (4.55), for all x ∈ A, v ∈
S n−1
lim Q(x, v, t) = 0.
(4.56)
t→0
Next choose any y ∈ Rn , y 6= x ∈ A and define
v :=
y−x
.
|y − x|
(4.57)
Clearly v ∈ S n−1 and y = x + tv where t := |y − x|. Thus the following is derived
f (y) − f (x) − grad f (x) · (y − x)
y−x
lim
= lim Q x,
, |y − x| = 0(4.58)
y→x
y→x
|x − y|
|y − x|
by (4.56). This proves that the function f is differentiable at every x ∈ A, hence a.e.
in Rn . This finishes claims 3 and completes the proof of the theorem.
2
38
4.2
Almost Everywhere Differentiability of Absolutely Continuous Functions
Definition 4.2.1. A function F : R → R is called absolutely continuous on [a, b] if
and only if, for every ε > 0 there exists a δ > 0 such that for any finite set of disjoint
intervals (a1 , b1 ), . . . , (aN , bN ) contained in [a, b],
N
X
(bj − aj ) < δ ⇒
N
X
1
|F (bj ) − F (aj )| < ε.
(4.59)
1
As pointed out in [5], Lipschitz functions are important examples of absolutely
continuous functions . . . To produce an absolutely continuous function that is not
Lipschitz, we pick an arbitrary integrable function g on [a, b] and define
Z
x
g(t) dt.
f (x) =
(4.60)
a
Then f is absolutely continuous, but Lipschitz only if g is bounded. In fact, up
to an additive constant, all absolutely continuous functions are of the form (4.60)
with g = f 0 almost everywhere. Lipschitz functions are precisely those for which g is
bounded.
It was also observed in [5] that the sum and product of two absolutely continuous
functions on [a, b] are again absolutely continuous and that the usual rules of differentiation apply (in the almost everywhere sense). Most notably, the integration by
parts formula holds: If f, g : [a, b] → R are absolutely continuous, then
Z
b
Z
0
f (x)g(x) dx = −
a
b
f (x)g 0 (x) dx + f (b)g(b) − f (a)g(a).
(4.61)
a
Thus, absolutely continuous functions are precisely the functions of one variable
that allow for calculus. These functions need not be smooth; their graphs can have
corners or other bad behavior that prevents pointwise differentiability at many points,
as the representation in (4.60) shows. For example, for any given set E of measure zero
in [a, b], there exists a Lipschitz function f : [a, b] → R that fails to be differentiable
at every point in E.”
Definition 4.2.2. If F : R → R and x ∈ R, define the total variation function of F
by
TF (x) = sup
( n
X
)
|F (xj ) − F (xj−1 )| : n ∈ N, −∞ < x0 < · · · < xn = x (4.62)
.
1
39
Note that if a < b, then
TF (b) − TF (a) = sup
( n
X
)
|F (xj ) − F (xj−1 )| : n ∈ N, a = x0 < · · · < xn = b .
1
(4.63)
If limx→∞ TF (x) is finite, say that F is of bounded variation on R, and denote
the space of all such F by BV . Denote by N BV (N for “normalized”) the space of
functions
N BV := {F ∈ BV : F is right continuous and lim F (x) = 0}.
x→−∞
(4.64)
The right hand side of (4.63) is called the total variation of F on [a, b]. Denote by
BV ([a, b]) the set of all functions on [a, b] whose total variation on [a, b] is finite.
Quoting [5],
Functions of bounded variation are precisely the functions that can be written
as a difference of two increasing functions. Absolutely continuous functions are of
bounded variation, but not conversely. The Cantor function is of bounded variation
but not absolutely continuous.
This lemma appears as Lemma 3.34 in [3].
Lemma 4.3. If F is absolutely continuous on [a, b], then F ∈ BV ([a, b]).
Proof. Let δ be as in the definition of absolute continuity, corresponding to ε = 1.
Let N be a natural number such that N > b−a
. Suppose a = x0 < · · · < xn = b is a
δ
: 0 ≤ i ≤ N } are included in the
partition of [a, b]. Assume that the points {a + i b−a
N
partition.
This
assumption
can
only
increase
the
quantity
we want to bound above.
Pn
Then 1 |F (xj ) − F (xj−1 )| < N · 1 = N , so the total variation of F on [a, b] is at
most N .
2
The next theorem is Theorem 3.29 in [3].
Theorem 4.4. If µ is a complex Borel measure on R and F (x) = µ((−∞, x]), then
F ∈ N BV . Conversely, if F ∈ N BV , there is a unique complex Borel measure µF
such that F (x) = µF ((−∞, x]); moreover, |µF | = µTF .
We denote Lebesgue measure on Rn by m.
Definition 4.2.3. A family {Er }r>0 of Borel subsets of Rn is said to shrink nicely
to x in Rn if
(i) Er ⊂ B(x, r) for each r;
(ii) there is a constant α > 0, independent of r, such that m(Er ) > αm(B(x, r)).
40
The sets Er need not contain x itself.
Let (X, M) be a measure space, and recall the following facts about measures. If
µ is a signed measure on (X, M), then a set E ∈ M is called null for µ if µ(F ) = 0
for all F ∈ M such that F ⊂ E. Two signed measures µ and ν on (X, M) are
called mutually singular precisely when there exist A, B ∈ M such that X = A ∪ B,
A ∩ B = ∅, A is null for µ and B is null for ν. In this case we write µ ⊥ ν. If ν is
a signed measure and µ a positive measure on (X, M), we say that ν is absolutely
continuous with respect to µ and write ν µ if ν(E) = 0 for every E ∈ M for which
µ(E) = 0.
The Lebesgue-Radon-Nikodym theorem for signed measures:
Theorem 4.5. If ν is a σ-finite signed measure and µ a σ-finite positive measure on
(X, M), then there exist unique σ-finite signed measures λ, ρ on (X, M) such that
λ ⊥ µ,
ρ µ,
and
ν = λ + ρ.
(4.65)
Moreover, there is an extended µ-integrable function f : X → R such that dρ = f dµ,
and any two such functions are equal µ-a.e.
The next theorem appears as Theorem 3.22 in [3]. It is a consequence of the
so-called “Lebesgue Differentiation Theorem.”
Theorem 4.6. Let ν be a regular signed or complex Borel measure on Rn , and let
dν = dλ + f dm be its Lebesgue-Radon-Nikodym representation. Then for m-almost
every x in Rn ,
lim+
r→0
ν(Er )
= f (x)
m(Er )
(4.66)
for every family {Er }r>0 that shrinks nicely to x.
The main result in this document is
Theorem 4.7. If F : [a, b] → R is absolutely continuous, then F is differentiable
almost everywhere on [a, b].
Proof. Replacing F by F − F (a) if necessary, assume that F (a) = 0. Extend F from
[a, b] to R by setting F (x) = 0 for x < a and F (x) = F (b) for x > b. By Lemma 4.3,
F |[a,b] is in BV ([a, b]), and therefore F is in N BV . By Theorem 4.4, there is a
unique Borel measure µF on R such that for all x ∈ R, µF ((−∞, x]) = F (x). By
Theorem 1.18 in [3], µF is regular.
Following Proposition 3.30 in [3], observe that for r > 0,
F (x + r) − F (x)
µF ((−∞, x + r]) − µF ((−∞, x])
=
r
r
µF ((x, x + r])
=
,
m((x, x + r])
41
and
F (x − r) − F (x)
µF ((−∞, x − r]) − µF ((−∞, x])
=
−r
−r
=
−µF ((x − r, x])
−r
=
µF ((x − r, x])
.
m((x − r, x])
Therefore
F 0 (x) =
=
F (x + s) − F (x)
s→0, s6=0
s
lim
(4.67)
µF (Er )
,
m(Er )
lim+
r→0
where Er = (x, x + r] or (x − r, x].
Note that {(x, x + r]}r>0 and {(x − r, x]}r>0 are both families shrinking nicely to
x. Then by Theorem 4.6, the limit in (4.67) exists for almost every x in R. Therefore
F |[a,b] is differentiable at almost every point in [a, b].
2
We also have the following, which appears as Theorem 3.32 in [3]:
Theorem 4.8. If F ∈ N BV , then F is absolutely continuous if and only if µF m.
In light of this, and since the shifted and extended function F in the proof of
Theorem 4.7 is absolutely continuous and in N BV , we have that µF m. Therefore
the unique Lebesgue-Radon-Nikodym decomposition of µF is µF = 0 + µF , and
moreover dµF = F 0 dm and
Z
F (x) = µF ((−∞, x]) =
Z
dµF =
(−∞,x]
F 0 (t) dm(t),
(4.68)
(−∞,x]
which is a part of the Fundamental Theorem of Calculus for Lebesgue integrals. See
Theorem 3.35 in [3], and also Theorem 3.1 in [5], which is:
Theorem 4.9. Let f : [a, b] → R be a continuous function. Then f is differentiable
almost everywhere with integrable derivative such that
Z
f (b) − f (a) =
b
f 0 (x) dx
(4.69)
a
if and only if f is absolutely continuous.
42
The Cantor function c : [0, 1] → [0, 1] is continuous, and differentiable almost
everywhere with c0 = 0 (hence c0 is integrable,) and c is of bounded variation, but as
noted, c is not absolutely
continuous.
Sure enough, c fails (4.69), since c(1) − c(0) =
R1
R1 0
1 − 0 = 1 6= 0 = 0 0 dt = 0 c (t)dt. Quoting [5],
There is an important criterion which identifies absolutely continuous functions
among general functions of bounded variation: a continuous function is absolutely
continuous if and only if it maps sets of measure zero to sets of measure zero. That
the Cantor function fails to be absolutely continuous can be seen using this criterion
as well; it maps the Cantor set onto [0, 1].
There is one more important way to think about absolutely continuous functions
of one variable. It follows from (4.61) that if f : [a, b] → R is absolutely continuous,
and if ϕ : [a, b] → R is a smooth function that vanishes at the end points of the
interval, then
Z
b
0
b
Z
f (x)ϕ0 (x) dx.
f (x)ϕ(x) dx = −
(4.70)
a
a
The converse is true as well, in the following improved form, where continuity need
not be assumed: If f : [a, b] → R is an integrable function such that
Z
b
Z
g(x)ϕ(x) dx = −
a
b
f (x)ϕ0 (x) dx
(4.71)
a
for some integrable g : [a, b] → R and for all infinitely many times differentiable,
or smooth, functions ϕ : [a, b] → R such that ϕ(a) = 0 = ϕ(b), then f agrees
almost everywhere with an absolutely continuous function; moreover, f 0 = g almost
everywhere in this case.
Using the language of the theory of distributions, the preceding fact can be stated
as follows: a (continuous) function is absolutely continuous if and only if its distributional derivative is represented by integration against an integrable function.
4.3
Some Consequences of Rademacher’s Theorem
The aim of this section is to prove the following:
Theorem 4.10. Let U ⊆ Rn be an open set and let
f : U −→ Rm
(4.72)
be a bi-Lipschitz function. That is, there exist 0 < C1 ≤ C2 < ∞ such that
C1 |x − y| ≤ |f (x) − f (y)| ≤ C2 |x − y|,
(4.73)
for every x, y ∈ U . Then there exists E ⊂ U , of Lebesgue measure zero, such that
43
Df (x)
exists for every
x ∈ U \ E,
(4.74)
and
C1 |v| ≤ |Df (x)v| ≤ C2 |v|,
for every v ∈ Rn , and x ∈ U \ E.
(4.75)
As a corollary,
rank Df (x) = n for every x ∈ U \ E,
(4.76)
which showes that necessarily
m ≥ n.
(4.77)
In addition,
m = n =⇒ f (U ) is an open set.
(4.78)
Proof. If f = (f1 , ..., fm ), it follows that each component fj : U → R is Lipschitz.
Then Theorem ?? gives that each fj is differentiable in U \ Ej for some set Ej of zero
Lebesgue measure. Taking E := ∪1≤j≤m Ej , (4.74) follows.
Next, for each x ∈ U \ E we have
f (x + h) − f (x) − (Df )(x)h
= 0.
h→0
|h|
lim
(4.79)
Thus, for every ε > 0 there exists δ > 0 such that
|f (x + h) − f (x) − (Df )(x)h| ≤ ε|h|
if |h| ≤ δ.
(4.80)
Using this, given v ∈ Rn \ {0} arbitrary, set
w := δ
v
∈ Rn ,
|v|
|w| = δ.
(4.81)
Then, using (4.73), we may write
C1 δ = C1 |w| ≤ |f (x + w) − f (x)| ≤ |f (x + w) − f (x) − (Df )(x)w| + |(Df )(x)w|
≤ ε|w| + |(Df )(x)w| = εδ + δ
|(Df )(x)v|
,
|v|
44
(4.82)
by (4.80). Dividing by δ then gives
C1 ≤ ε +
|(Df )(x)v|
,
|v|
(4.83)
and further, uppon letting ε go to zero,
C1 ≤
|(Df )(x)v|
,
|v|
for every v ∈ Rn \ {0}.
(4.84)
This readily implies
C1 |v| ≤ |(Df )(x)v| for every v ∈ Rn ,
(4.85)
which proves the first inequality in (4.75).
To prove the second inequality in (4.75), we estimate
|(Df )(x)w| ≤ |f (x + w) − f (x) − (Df )(x)w| + |f (x + w) − f (x)| ≤ ε|w| + C2 |w|
= δ(C2 + ε).
(4.86)
Thus,
δ
|Df (x)v|
≤ δ(C2 + ε),
|v|
(4.87)
which implies
|Df (x)v| ≤ (C2 + ε)|v|,
for every v ∈ Rn \ {0}.
(4.88)
Passing to limit ε → 0+ then gives
|Df (x)v| ≤ C2 |v|for every v ∈ Rn ,
(4.89)
completing the proof of (4.75).
Having established (4.75), it follows from this that the matrix Df (x) sends nonzero vectors into non-zero vectors, for each x ∈ U \ E. From this, (4.76) follows. Now,
(4.77) is a consequence of (4.76).
Finally, (4.78) follows from the Invariance of Domain Theorem (Theorem 2.4). 2
45
Chapter 5
The Implicit Function Theorem for
Lipschitz Functions
In this chapter we present a version of the classical Implicit Function Theorem in
a more general setting than classical framework of functions of class C 1 , namely
functions which are only Lipschitz (among other things). This result, presented in
Theorem 5.1, appears to be new and will play a significant role later on, in Chapter 5,
when we discuss various applications. Before proceeding with the statement and proof
of the main result the reader is advised to revisit the Invariance of Domain Theorem
stated in an earlier chapter.
Theorem 5.1 (The Implicit Function Theorem for Lipschitz Functions). Let Um ⊆
Rm and Un ⊆ Rn open. Next fix a ∈ Um and b ∈ Un where U := Um × Un . Consider
F : Um × Un −→ Rn
(5.1)
a Lipschitz such that
F (a, b) = 0
(5.2)
and with the property that there exists a constant K > 0 for which
|F (x, y1 ) − F (x, y2 )| ≥ K |y1 − y2 |
for all
(x, yj ) ∈ U
where j = 1, 2.(5.3)
Then there exist Vm ⊆ Rm open, such that a ∈ Vm , and a Lipschitz function ϕ : Vm →
Un such that ϕ(a) = b, and
{(x, y) ∈ Vm × Un : F (x, y) = 0} = {(x, ϕ(x)) : x ∈ Vm } .
(5.4)
In particular,
F (x, ϕ(x)) = 0,
for all x ∈ Vm .
46
(5.5)
Proof. For some ε > 0 small (to be specified later) consider the function
f : U → Rm+n
defined by f (x, y) := (x, εF (x, y)).
(5.6)
We will first show that f is Lipschitz. To this end, for every (x1 , y1 ), (x2 , y2 ) in U we
estimate
|f (x1 , y1 ) − f (x2 , y2 )|2 = |(x1 − x2 , ε(F (x1 , y1 ) − F (x2 , y2 ))|2
= |x1 − x2 |2 + ε2 |F (x1 , y1 ) − F (x2 , y2 )|2 .
(5.7)
By the Lipschitzianity of F we get the following:
|f (x1 , y1 ) − f (x2 , y2 )|2 ≤ |x1 − x2 |2 + ε2 M 2 |(x1 , y1 ) − (x2 , y2 )|2
= (1 + ε2 M 2 ) |x1 − x2 |2 + ε2 M 2 |y1 − y2 |2
≤ C12 |(x1 , y1 ) − (x2 , y2 )|2 ,
where we have set C1 :=
(x2 , y2 ) in U ,
(5.8)
p
max {1 + ε2 M 2 , ε2 M 2 }. Thus, for every pairs (x1 , y1 ),
|f (x1 , y1 ) − f (x2 , y2 )|2 ≤ C12 |(x1 , y1 ) − (x2 , y2 )|2 ,
(5.9)
or equivalently,
|f (x1 , y1 ) − f (x2 , y2 )| ≤ C1 |(x1 , y1 ) − (x2 , y2 )| .
(5.10)
Hence f is Lipschitz.
Next, we establish a bound from below which will show that f is bi-Lipschitz.
Concretely, given two points of the form (x1 , y1 ), (x1 , y2 ) in U , using (5.3) we write
|y1 − y2 |2 ≤ K −2 |F (x1 , y1 ) − F (x1 , y2 )|2
≤ K −2 (|F (x1 , y1 ) − F (x2 , y2 )| + |F (x2 , y2 ) − F (x1 , y2 )|)2
≤ 2K −2 (|F (x1 , y1 ) − F (x2 , y2 )|2 + |F (x2 , x2 ) − F (x1 , y2 )|2 ). (5.11)
Since F is Lipschitz, estimate (5.11) implies
K 2 |y1 − y2 |2 ≤ 2 |F (x1 , x1 ) − F (x2 , y2 )|2 + 2M 2 |x2 − x1 |2
and further
47
(5.12)
K 2 ε2
|y1 − y2 |2 − M 2 ε2 |x2 − x1 |2 ≤ ε2 |F (x1 , y1 ) − F (x2 , y2 )|2 .
2
(5.13)
Recall now from (5.7) that
|f (x1 , y1 ) − f (x2 , y2 )|2 = |x1 − x2 |2 + ε2 |F (x1 , y1 ) − F (x2 , y2 )|2 .
(5.14)
Combining this with the above estimate we get
K 2 ε2
|y1 − y2 |2 − M 2 ε2 |x2 − x1 |2 ≤ |f (x1 , y1 ) − f (x2 , y2 )|2 − |x1 − x2 |2(5.15)
.
2
In other words, for every (x1 , y1 ), (x2 , y2 ) in U ,
K 2 ε2
|y1 − y2 |2 + (1 − M 2 ε2 ) |x2 − x1 |2 ≤ |f (x1 , y1 ) − f (x2 , y2 )|2 .
(5.16)
2
q
22
At this point we fix ε such that 0 < ε < M −1 and take C0 := min K 2ε , 1 − ε2 M 2 .
Thus, C0 > 0 is well-defined and
C02 (|x1 − x2 |2 + |y1 − y2 |2 ) ≤ |f (x1 , y1 ) − f (x2 , y2 )|2 .
(5.17)
Taking square-roots we obtain that for every points (x1 , y1 ), (x2 , y2 )
C0 |(x1 , y1 ) − (x2 , y2 )| ≤ |f (x1 , y1 ) − f (x2 , y2 )| .
(5.18)
Thus by (5.10) and (5.18) f is bi-Lipschitz.
Since f is bi-Lipschitz, it follows that f is continuous and injective (as mentioned
earlier). Hence by the Invariance of Domains Theorem the following hold:
V := f (U ) is an open subset of Rm+n ,
(5.19)
f : U → V is bijective (i.e., one-to-one and onto),
(5.20)
the inverse function f −1 : V → U is continuous.
(5.21)
In addition, (5.2) and (5.6) ensure that
f −1 (a, 0) = (a, b).
(5.22)
Next, consider the coordinate projection functions
48
1
: Rm × Rn −→ Rm ,
πm
πn2 : Rm × Rn −→ Rn ,
1
(x, y) := x,
πm
πn2 (x, y) := y,
(5.23)
(5.24)
and define
Vm := B(a, r) ⊂ Rm ,
(5.25)
where r > 0 is so small that
B(a, r) ⊆ Um and πn2 (f −1 (x, 0)) ∈ Un for every x ∈ B(a, r).
(5.26)
Since πn2 (f −1 (a, 0)) = b ∈ Un and since both πn2 and f −1 are continuous functions,
this is certainly possible. After this preamble, introduce
ϕ : B(a, r) −→ Un ,
ϕ(x) := πn2 (f −1 (x, 0)).
(5.27)
It follows that ϕ is well-defined, and since f −1 (a, 0) = (a, b) we have ϕ(a) = b.
Going further, we may invoke (5.20) in order to write
(x, 0) = f (f −1 (x, 0))
1
= f (πm
(f −1 (x, 0)), πn2 (f −1 (x, 0))),
(5.28)
which is valid for every x ∈ Vm . Recalling our previous definition f (x, y) := (x, εF (x, y)),
from the above statement we then get
1
1
(x, 0) = (πm
(f −1 (x, 0)), ε F (πm
(f −1 (x, 0)), ϕ(x)))
(5.29)
which shows that
1
1
πm
(f −1 (x, 0)) = x and F (πm
(f −1 (x, 0)), ϕ(x)) = 0
(5.30)
for every x ∈ Vm . Note that this implis that F (x, ϕ(x)) = 0 for x ∈ Vm , proving (5.5).
In turn, this justifies the inclusion
{(x, ϕ(x)) : x ∈ Vm } ⊆ {(x, y) ∈ Vm × Un : F (x, y) = 0} .
Thus, as far as (5.4) is concerned, there remains to prove that
49
(5.31)
{(x, y) ∈ Vm × Un : F (x, y) = 0} ⊆ {(x, ϕ(x)) : x ∈ Vm } .
(5.32)
However, if (x, y) ∈ Vm × Un is such that F (x, y) = 0 then f (x, y) = (x, 0) and, hence,
(x, y) = f −1 (x, 0) which implies 0 = πn2 (f −1 (x, 0)). Thus, by definition we obtain
ϕ(x) = 0, proving (5.32).
Finally, let us check that the function ϕ in (5.27) is Lipschitz. For this, given any
x1 , x2 ∈ Vm we write
f (xj , ϕ(xj )) = (xj , εF (xj , ϕ(xj ))) = (xj , 0),
j = 1, 2.
(5.33)
Based on this and the bi-Lipschitzianity of f we then estimate
C0 |ϕ(x1 ) − ϕ(x2 )| ≤ C0 |(x1 , ϕ(x1 )) − (x2 , ϕ(x2 ))|
≤ |f (x1 , ϕ(x1 )) − f (x2 , ϕ(x2 ))|
= |(x1 , 0) − (x2 , 0)| = |x1 − x2 |.
(5.34)
Thus,
|ϕ(x1 ) − ϕ(x2 )| ≤ C0−1 |x1 − x2 |
(5.35)
for every x1 , x2 ∈ Vm , proving that ϕ in (5.27) is indeed Lipschitz. This concludes
the proof of the theorem.
2
50
Chapter 6
Lipschitz Domains
6.1
General Considerations
A class of domains which plays an important role in the field of partial differential
equations is the collection of all Lipschitz domains in Rn . For the clarity of the
exposition we record here a formal definition (recall that the superscript c is the
operation of taking the complement of a set, relative to Rn ).
Definition 6.1.1. Let Ω be a nonempty, proper open subset of Rn . Also, fix x0 ∈
∂Ω. Call Ω a Lipschitz domain near x0 if there exist b, c > 0 with the following
significance. There exist an (n − 1)-plane H ⊂ Rn passing through x0 , a choice N of
the unit normal to H, and an open cylinder Cb,c := {x0 + tN : x0 ∈ H, |x0 − x0 | <
b, |t| < c} (called coordinate cylinder near x0 ) such that
Cb,c ∩ Ω = Cb,c ∩ {x0 + tN : x0 ∈ H, t > ϕ(x0 )},
(6.1)
Cb,c ∩ ∂Ω = Cb,c ∩ {x0 + tN : x0 ∈ H, t = ϕ(x0 )},
(6.2)
c
Cb,c ∩ Ω = Cb,c ∩ {x0 + tN : x0 ∈ H, t < ϕ(x0 )},
(6.3)
for some Lipschitz function ϕ : H → R satisfying
ϕ(x0 ) = 0
and
|ϕ(x0 )| < d if |x0 − x0 | ≤ b.
(6.4)
Finally, call Ω a Lipschitz domain if it is a Lipschitz domain near every point
x ∈ ∂Ω.
A few comments are appropriate at this stage.
Remark 1. Definition 6.1.1 can be rephrased as follows.
Call a nonempty, proper open subset Ω of Rn a Lipschitz domain if for every
x0 ∈ ∂Ω there exist a new system of coordinates Rn−1 × R, which is isometric to the
original one, along with an open upright cylinder C containing x0 , and a Lipschitz
function ϕ : Rn−1 → R whose graph conatins x0 and is disjoint from the top and
bottom lids of C, such that
51
C ∩ Ω = C ∩ {(x0 , xn ) ∈ Rn−1 × R : xn > ϕ(x0 )},
(6.5)
C ∩ ∂Ω = C ∩ {(x0 , xn ) ∈ Rn−1 × R : xn = ϕ(x0 )},
(6.6)
C ∩ (Ω)c = C ∩ {(x0 , xn ) ∈ Rn−1 × R : xn < ϕ(x0 )}.
(6.7)
Remark 2. Given a bounded Lipschitz domain Ω ⊂ Rn , the number and size of
coordinate cylinders in a finite covering of ∂Ω, along with the quantity max k∇ϕkL∞
(called the Lipschitz constant of Ω), where the supremum is taken over all Lipschitz
functions ϕ associated with these coordinate cylinders) make up what is called the
Lipschitz character of Ω.
Remark 3. The classes of bounded C k domains, k ∈ N, is defined analogously requiring that the defining functions ϕ is of class C k .
Remark 4. Definition 6.1.1 shows that if Ω ⊂ Rn is a Lipschitz domain near a boundary point x0 then, in a neighborhood of x0 , ∂Ω agrees with the graph of a Lipschitz
function ϕ : Rn−1 → R, considered in a suitably chosen system of coordinates (which
is isometric with the original one). Then the outward unit normal has an explicit
formula in terms of ∇0 ϕ namely, in the new system of coordinates,
(∇0 ϕ(x0 ), −1)
ν(x0 , ϕ(x0 )) = p
,
1 + |∇0 ϕ(x0 )|2
if (x0 , ϕ(x0 )) is near x0 ,
(6.8)
where ∇0 denotes the gradient with respect to x0 ∈ Rn−1 . The existence of ∇0 ϕ at
almost every point in Rn−1 is a consequence of the Rademacher Theorem proved in
an earlier section.
Remark 5. As is well-known, if Ω ⊂ Rn is a Lipschitz domain with compact boundary
then Ω satisfies a uniform cone property. This asserts that there exists an open,
circular, truncated, one-component cone Γ with vertex at 0 ∈ Rn such that for every
x0 ∈ ∂Ω there exist r > 0 and a rotation R about the origin such that
x + R(Γ) ⊆ Ω,
∀ x ∈ B(x0 , r) ∩ Ω.
(6.9)
Let us point out that if Ω satisfies a uniform cone property, as described above, then
also
x0 ∈ ∂Ω =⇒ x0 − R(Γ) ⊆ Ωc ,
(6.10)
at least if the height of Γ is sufficiently small relative to r (appearing in (6.9)). Indeed,
the existence of a point y ∈ (x0 −R(Γ))∩Ω would entail x0 ∈ y +R(Γ). Since y ∈ Ω is
also close to x0 (assuming that Γ has small height, relative to r), (6.9) further implies
that x0 belongs to the interior of Ω, in contradiction with x0 ∈ ∂Ω.
Granted (6.10), it is not difficult to see that the converse statement regarding
Lipschitzianity implying a uniform cone condition is also true. That is, a bounded
open set Ω ⊂ Rn satisfying a uniform cone property is, necessarily, Lipschitz. See,
e.g., Theorem 1.2.2.2 on p. 12 in [4] for a proof.
52
6.2
Some Toplogical Results
It should be noted that the conditions (6.5)-(6.7) in the definition of a Lipschitz
domain are not independent, since (6.5) implies (6.6)-(6.7). This follows directly
from the toplogical result presented below.
Proposition 6.1. Assume that Ω ⊂ Rn is an open set, and fix x0 ∈ ∂Ω. Also,
consider
ϕ : Rn−1 −→ R continuous,
Σ := graph of ϕ ⊆ Rn ,
(6.11)
D± := connected components of Rn \ Σ.
Finally, assume that B 0 ⊂ Rn−1 is an open ball, I ⊆ R is an open interval, with the
property that
ϕ(B 0 ) ⊆ I,
(6.12)
and such that the cylinder
C := B 0 × I ⊂ Rn−1 × R = Rn
(6.13)
contains x0 and satisfies
C ∩ D+ = C ∩ Ω.
(6.14)
Then also
(Ω)c ∩ C = D− ∩ C
(6.15)
Σ ∩ C = ∂Ω ∩ C.
(6.16)
and
Proof. Use double inclusion to prove (6.16). In one direction, it suffices to establish
that
∂Ω ∩ C ⊆ Σ.
(6.17)
To justify this inclusion, pick an arbitrary point x ∈ ∂Ω ∩ C. Then there exist
xj ∈ Ω ∩ C = D+ ∩ C, j ∈ N, such that xj → x as j → ∞. Thus, x ∈ D+ = D+ ∪ Σ,
53
i.e. either x ∈ Σ (which suits our purposes), or x ∈ D+ . However, in the latter case,
we obtain x ∈ D+ ∩ C = Ω ∩ C which forces x ∈ Ω, in contradiction with x ∈ ∂Ω.
Hence, x ∈ D+ never reallu happens. This proves (6.17).
In the opposite direction, it suffices to establish that
Σ ∩ C ⊆ ∂Ω.
(6.18)
To see this, pick an arbitrary point x ∈ Σ ∩ C and, reasoning by contradiction, assume
that x ∈
/ ∂Ω. By (6.34), it follows that either x ∈ Ω, which cannot really happen since
otherwise x ∈ Ω ∩ C = D+ ∩ C contradicting x ∈ Σ, or x ∈
/ Ω. Nonetheless, this last
eventuality does not materialize either, since this would otherwise imply the existence
of some r > 0 such that B(x, r) ∩ Ω = ∅. By further decreasing r if necessary, we can
in fact assume that B(x, r) ⊆ C, so that
∅ = B(x, r) ∩ Ω ∩ C = B(x, r) ∩ D+ ∩ C = B(x, r) ∩ D+ .
(6.19)
Thus, x ∈
/ D+ , contradicting x ∈ Σ. This finishes the proof of (6.18). Together,
(6.17)-(6.18) readily give (6.16).
At this point, there remains to prove (6.15). Staring with the decompositions
[
[
(Ω)c ∂Ω,
[
[
Rn = D+ D− Σ,
Rn = Ω
disjoint unions,
(6.20)
disjoint unions,
(6.21)
we obtain (by taking the intersection with C)
[
[
C = Ω∩C
(Ω)c ∩ C
∂Ω ∩ C ,
[
[
+
−
C = D ∩C
D ∩C
Σ∩C ,
disjoint unions,
(6.22)
disjoint unions.
(6.23)
With this in hand, and recalling (6.14) (which is given) and (6.16) (which is has
already been proved), we see that (6.15) follows.
2
Going further, we wish to mention that (6.6) implies (6.5), (6.7) (up to reversing
the sense on the vertical axis in |mathbbRn−1 × R) if, for example, x0 ∈
/ (Ω)◦ (where,
◦
n
generally speaking, E stands for the interior of the set E ⊆ R ). The latter condition
is guaranteed if it is known a priori that
∂Ω = ∂Ω.
(6.24)
This follows from the proposition below.
54
Proposition 6.2. Assume that Ω ⊂ Rn is an open set, and fix x0 ∈ ∂Ω. Also,
consider
ϕ : Rn−1 −→ R continuous,
Σ := graph of ϕ ⊆ Rn ,
(6.25)
D± := connected components of Rn \ Σ.
Finally, assume that B 0 ⊂ Rn−1 is an open ball, I ⊆ R is an open interval, with the
property that
ϕ(B 0 ) ⊆ I,
(6.26)
and such that the cylinder
C := B 0 × I ⊂ Rn−1 × R = Rn
(6.27)
contains x0 and satisfies
C ∩ Σ = C ∩ ∂Ω.
(6.28)
Then one of the following three alternatives holds:
Ω ∩ C = D+ ∩ C
and
(Ω)c ∩ C = D− ∩ C,
(6.29)
Ω ∩ C = D− ∩ C
and
(Ω)c ∩ C = D+ ∩ C,
(6.30)
or
or
x0 ∈ (Ω)◦ .
(6.31)
A few comments are in order.
Remark 1. Condition (6.26) simply means that the surface Σ does not escape through
the top or bottom lids of C. In particular, Σ is disjoint from the top and bottom lids
of C, indeed disjoint from certain open neighborhoods of them.
Remark 2. Conditions (6.29) and (6.30) transform into one another by reversing the
sense on the vertical axis in Rn−1 × R (a transformation which changes D+ into D− ).
Remark 3. If the point x0 ∈ ∂Ω does not belong to (Ω)◦ , then conclusion (6.31) is
apriori excluded. Note that if the open set Ω is such that
55
∂Ω = ∂Ω
(6.32)
then (6.31) never happens, since (Ω)◦ ∩ ∂Ω = (Ω)◦ ∩ ∂Ω = ∅.
Let us now turn to the
Proof of Proposition 6.2. We proceed in a series of steps, starting with
Step I. We claim that
C ∩ D+
and C ∩ D−
are connected sets.
(6.33)
Proof of Step I. Consider C ∩ D+ , as the argument for C ∩ D− is similar. It suffices
to show that the set in question is path-connected, i.e. for every x, y ∈ C ∩ D+ there
exists a continuous curve γ contained in C ∩ D+ and which joins x and y. Such
a curve, however, can always be taken to consist of three line segments, L1 , L2 , L3 ,
where L1 and L2 are vertical line segments contained in C ∩ D+ which emerge from x
and y, respectively, and where L3 is a horizontal line segment making the transition
between L1 and L2 near the very top of C (the observation in Remark 1 above then
ensures that L2 can be taken in C ∩ D+ ). This finishes the proof of Step I.
Step II. We claim that
Rn = Ω
[
[
(Ω)c ∂Ω,
(6.34)
with disjoint unions.
Proof of Step II. This follows from the fact that Ω is open which, in turn, implies that
∂Ω = Ω ∩ Ωc = Ω ∩ Ωc
(6.35)
since Ωc is closed. Taking complements in (6.35) then gives
Rn \ ∂Ω = Ω
[
(Ω)c
(6.36)
from which (6.34) follows.
Step III. We claim that one of the following situations necessarily happens:
(i) D+ ∩ C ⊆ Ω and D− ∩ C ⊆ (Ω)c ,
or
(ii) D− ∩ C ⊆ Ω and D+ ∩ C ⊆ (Ω)c ,
(iii) D+ ∩ C ⊆ Ω and D− ∩ C ⊆ Ω,
(iv) D− ∩ C ⊆ (Ω)c
or
(6.37)
or
and D+ ∩ C ⊆ (Ω)c .
56
Proof of Step III. Note that
±
Σ ∩ D ∩ C = ∅,
(6.38)
by the definition of D± . This implies that D± ∩ C are disjoint from Σ and, hence,
from Σ ∩ C = ∂Ω ∩ C. In turn, this further entails that
D± ∩ C are disjoint from ∂Ω.
(6.39)
Based on this and (6.34), we then conclude that
D± ∩ C ⊆ Ω ∪ (Ω)c .
(6.40)
Recall from Step I that D± ∩ C are connected sets, and observe that Ω are (Ω)c open,
disjoint sets. In concert with the definition of connectivity, (6.40) then implies that
each of the two sets D+ ∩ C, D− ∩ C is contained in either Ω or ∪(Ω)c . Unraveling
the various possibilities now proves that one of the four scenarios in (6.37) must hold.
This concludes the proof of Step III.
Step IV. If conditions (i) in (6.37) happen, then conditions (6.29) happen as well. In
particular,
(Ω)c ∩ C = D− ∩ C,
Ω ∩ C = D+ ∩ C,
∂Ω ∩ C = Σ ∩ C.
(6.41)
Proof of Step IV. Assume (i) happens, i.e., we have
D+ ∩ C ⊆ Ω,
(6.42)
D− ∩ C ⊆ (Ω)c ,
(6.43)
Σ ∩ C = ∂Ω ∩ C.
(6.44)
We split the remainder of the proof of Step IV into two claims.
Claim 1. There holds D+ ∩ C = Ω ∩ C.
Proof of Claim 1. The left-to-right inclusion is immediate from (6.42). To prove the
oposite one we reason by contradiction and assume that there exists x ∈ Ω ∩ C such
that x 6∈ D+ ∩ C, thus x ∈ Ω, x ∈ C, x 6∈ D+ . Note that
C = (C ∩ D+ )
[
(C ∩ D− )
[
(C ∩ Σ)
[
[
= (C ∩ D+ ) (C ∩ D− ) (C ∩ ∂Ω) disjoint unions,
since
57
(6.45)
Rn = D+ ∪ D− ∪ Σ,
disjoint unions.
(6.46)
From the assumptions on x we have that x 6∈ C ∩ D+ , x 6∈ C ∩ ∂Ω (since x ∈ Ω and
Ω ∩ ∂Ω = ∅) and consequently, using also (6.45), x ∈ C ∩ D− ⊆ (Ω)c . This yields that
x 6∈ Ω, contradicting the assumption x ∈ Ω. This completes the proof of Claim 1.
Claim 2. There holds D− ∩ C = (Ω)c ∩ C.
Proof of Claim 2. the proof is similar to the proof of Claim 1. More specifically, the
inclusion ⊆ is clear, while for the other inclusion reason by contradiction and assume
that there exists x such that x ∈ C, x 6∈ Ω and x 6∈ D− . Then it follows that x 6∈ ∂Ω
since ∂Ω ⊆ Ω. Combining this with (6.45) it follows that x ∈ D+ and in turn, based
on (6.42), x ∈ Ω ⊆ Ω leading to a contradiction.
Step V. If conditions (ii) in (6.37) happen, then conditions (6.30) happen as well. In
particular, if (ii) happens, then
Ω ∩ C = D− ∩ C,
(Ω)c ∩ C = D+ ∩ C,
∂Ω ∩ C = Σ ∩ C.
(6.47)
Proof of Step V. Assume (ii) happens, i.e., we have
D+ ∩ C ⊆ (Ω)c ,
(6.48)
D− ∩ C ⊆ Ω,
(6.49)
Σ ∩ C = ∂Ω ∩ C.
(6.50)
We split the remainder of the proof of Step V into two claims.
Claim 1. There holds D+ ∩ C = (Ω)c ∩ C.
Proof of Claim 1. The inclusion “⊆” is clear. Assume that ∃ x ∈ C such that x 6∈ Ω
(thus x 6∈ ∂Ω) and x 6∈ D+ . These facts together with (6.45) we conclude that x ∈ D−
so x ∈ D− ∩ C ⊆ Ω ⊆ Ω. which is in contradiction with our assumptions.
Claim 2. There holds D− ∩ C = Ω ∩ C.
Proof of Claim 2. The inclusion “⊆” is clear. Assume that ∃ x ∈ C such that x ∈ Ω
(so x 6∈ ∂Ω) yet x 6∈ D− . Invoking (6.45) it follows that x ∈ C ∩ D+ ⊆ (Ω)c hence
x 6∈ Ω contradicting the assumption on x.
Step VI. If conditions (iii) in (6.37) happen, then x0 ∈ (Ω)◦ ∩ ∂Ω, i.e. (6.31) happens.
Proof of Step VI. Assume (iii) happens, i.e., we have
58
D+ ∩ C ⊆ Ω,
(6.51)
D− ∩ C ⊆ Ω,
(6.52)
Σ ∩ C = ∂Ω ∩ C.
(6.53)
Informally speaking, in this case Ω sits on both sides of ∂Ω. Let x∗ ∈ ∂Ω ∩ C. Then
∃ r > 0 such that
B(x∗ , r) ⊆ C.
(6.54)
We claim that
B(x∗ , r) ⊆ Ω
(6.55)
Indeed, by (6.46), we have
B(x∗ , r) = B(x∗ , r) ∩ D+ ∪ B(x∗ , r) ∩ D− ∪ B(x∗ , r) ∩ Σ
(6.56)
making use of (6.54), we have
B(x∗ , r) ∩ D+ ⊆ C ∩ D+ ⊆ Ω
B(x∗ , r) ∩ D− ⊆ C ∩ D− ⊆ Ω
(6.57)
B(x∗ , r) ∩ Σ ⊆ C ∩ Σ = C ∩ ∂Ω ⊆ ∂Ω.
Combining all these with (6.56), it follows that B(x∗ , r) ⊆ Ω ∩ ∂Ω = Ω, proving
(6.55).
Going further, the above claim implies that x∗ ∈ (Ω)◦ ∩ ∂Ω so that, ultimately,
C ∩ ∂Ω ⊆ (Ω)◦ ∩ ∂Ω.
(6.58)
Since x0 ∈ C ∩ ∂Ω, this forces x0 ∈ (Ω)◦ ∩ ∂Ω, as claimed. This finishes the proof of
Step VI.
Step VII. Conditions (vi) in (6.37) never happen.
Proof of Step VII. Reasoning by contradiction, assume that conditions (iv) in (6.37)
actually do happen. Thus, we have
D− ∩ C ⊆ (Ω)c ,
D+ ∩ C ⊆ (Ω)c
and Σ ∩ C = ∂Ω ∩ C.
Taking the union of the first two inclusions above yields
59
(6.59)
(D+ ∪ D− ) ∩ C ⊆ (Ω)c =⇒ (Rn \ Σ) ∩ C ⊆ (Ω)c =⇒ C \ Σ ⊆ (Ω)c
=⇒ C ∩ (Σc ) ⊆ (Ω)c =⇒ Ω ⊆ Σ ∪ C c ,
(6.60)
where the last implication follows by taking complements. Taking the intersection
with C, this further gives
C ∩ Ω ⊆ C ∩ Σ = C ∩ ∂Ω,
(6.61)
by (6.28). Thus,
C ∩ Ω ⊆ C ∩ Ω ⊆ C ∩ ∂Ω ⊆ ∂Ω,
(6.62)
C ∩ Ω ⊆ ∂Ω.
(6.63)
i.e.,
Since, by assumption, C is an open neighborhood of the point x0 ∈ ∂Ω, the definition
of the boundary implies that C ∩ Ω 6= ∅. Therefore, there exists x∗ ∈ C ∩ Ω. From
(6.63) it follows that x∗ ∈ ∂Ω, which forces us to conclude that the open set Ω
contains some of its own boundary points. This is a contradition which shows that
(iv) in (6.37) never happens.
The proof of Proposition 6.2 is therefore complete.
2
60
Chapter 7
Applications
7.1
First Application
Define S n−1 to be the unit sphere in Rn centered at the origin.
Theorem 7.1. Assume that
ϕ : S n−1 −→ (0, ∞) is Lipschitz
(7.1)
and define Ω ⊆ Rn by
Ω := rω : ω ∈ S n−1 ,
0 ≤ r < ϕ(ω) .
(7.2)
Then Ω is a Lipschitz domain.
Proof. Fix a point
x0 ∈ ∂Ω = rω : ω ∈ S n−1 ,
r = ϕ(ω) .
(7.3)
We want a new system of coordinates which is a rigid motion of the original and along
with a function
ψ : Rn−1 → R Lipschitz
(7.4)
and a cylinder, Ca,b (in this new system of coordinates) such that
61
R
Ω
ζ
Rn−1
R
x0
Ca,b = {X = (x0 , xn ) : |x0 | < a,
|xn | < b}
(7.5)
with the property that
x0 ∈ Ca,b
(7.6)
∂Ω ∩ Ca,b = (x0 , ψ(x0 ) : x0 ∈ Rn−1 ∩ Ca,b
(7.7)
and
Without loss of generality, assume that x0 lies at the intersection of the vertical axis
with ∂Ω.
62
R
Rn−1
en
X=(x’,xn )
−ϕ(−e n )e n = −λe n
That is,
x0 = (00 , −λ),
where λ := ϕ(−en ).
(7.8)
Condition (7.7) requires that if xn = ψ(x0 )
ϕ
(x0 , xn )
|(x0 , xn )|
= |(x0 , xn )|
(7.9)
and the idea is to find the Lipschitz function ψ by solving (7.9) for xn as a function
of x0 . To this end, it is convenient to rephrase (7.9) as
F (x0 , xn ) = 0,
(x0 , xn ) near (00 , −λ),
(7.10)
where
0
2
F (x , xn ) := ϕ
(x0 , xn )
|(x0 , xn )|
2
− |x0 | − xn 2 ,
(x0 , xn ) near (00 , −λ),
(7.11)
Note that
F (00 , −λ) = ϕ2 (−en ) − λ2 = 0.
(7.12)
63
We want to use the Implicit Function Theorem for Lipschitz Functions in order to
solve (7.10) for xn in terms of x0 such that
xn (00 ) = −λ.
(7.13)
At this stage, we need to check the hypothesis of the Implicit Function Theorem for
F given in (7.11). First, let us check that F is a Lipschitz function (in x0 and xn ).
To this end, we make use of (7.11) and the fact that ϕ is Lipschitz in order to write
0
0
2 (x , xn )
(y , yn ) 2
0
0
−ϕ
|F (x , xn ) − F (y , yn )| ≤ ϕ
|(x0 , xn )|
|(y 0 , yn )| 2
2
+ |x0 | − |y 0 | + xn 2 − yn 2 =: I + II.
(7.14)
Note that
0
0
0
0
(x
,
x
)
(y
,
y
)
(y
,
y
)
(x
,
x
)
n
n
n
n
· ϕ
I = ϕ
−
ϕ
+
ϕ
|(x0 , xn )|
|(y 0 , yn )| |(x0 , xn )|
|(y 0 , yn )| 0
(y 0 , yn ) n−1 (x , xn )
−
.
(7.15)
≤ 2kϕkL∞ (S n−1 ) Lip (ϕ; S ) 0
|(x , xn )| |(y 0 , yn )| To continue, bring in the general idenity
|ω1 − ω2 |2 = r1−1 r2−1 |r1 ω1 − r2 ω2 |2 − r1−1 r2−1 |r1 − r2 |2 ,
(7.16)
valid for any ω1 , ω2 ∈ S n−1 , r1 , r2 > 0. Incidentally, (7.16) is easily established by
expanding
|r1 ω1 − r2 ω2 |2 = r12 + r22 − 2r1 r2 ω1 · ω2
|ω1 − ω2 |2 = 2 − 2ω1 · ω2 ,
(7.17)
|r1 − r2 |2 = r12 + r22 − 2r1 r2 ,
and then naturally combining these expressions.
Returing to the mainstream discussion, let us use (7.16) with
ω1 :=
(x0 , xn )
,
|(x0 , xn )|
r1 := |(x0 , xn )| ,
ω2 :=
This gives
64
(y 0 , yn )
|(y 0 , yn )|
r2 := |(y 0 , yn )| .
(7.18)
0
2
0
(x , xn )
(y
,
y
)
n
−
|(x0 , xn )| |(y 0 , yn )| −1
0
= |(x , xn )|
(7.19)
0
−1
|(y , yn )|
2 0
2
0
0
0
|(x , xn ) − (y , yn )| − |(x , xn )| − |(y , yn )| .
Using the triangle inequality, it is then easy to bound
0
0
(x , xn )
(y
,
y
)
n
0
0
|(x0 , xn )| − |(y 0 , yn )| ≤ C |(x , xn ) − (y , yn )|
(7.20)
for (x0 , xn ), (y 0 , yn ) near (00 , −λ), where C > 0 is a fixed contant (which depends on
λ). Plugging this back in (7.15) then gives
I ≤ C |(x0 , xn ) − (y 0 , yn )|
for (x0 , xn ), (y 0 , yn ) near (00 , −λ).
(7.21)
In a more direct fashion (using just the triangle inequality), we also obtain
II ≤ C |(x0 , xn ) − (y 0 , yn )|
for (x0 , xn ), (y 0 , yn ) near (00 , −λ).
(7.22)
All together, (7.14) and (7.22)-(7.23) give that
|F (x0 , xn ) − F (y 0 , yn )| ≤ C |(x0 , xn ) − (y 0 , yn )|
for (x0 , xn ), (y 0 , yn ) near (00 , −λ).
(7.23)
This proves that F is Lipschitz in a neighborhood of (00 , −λ).
Thus, for Theorem 5.1 to apply we need to check that F is bi-Lipschitz in the
second variable. That is, we seek an estimate of the form
C |y1 − y2 | ≤ |F (x0 , y1 ) − F (x0 , y2 )| for x0 near 00 and y1 , y2 near −λ.
(7.24)
To justify this, we first write
0
0
2
F (x , xn ) − F (y , yn ) = ϕ
(x0 , y1 )
|(x0 , y1 )|
2
−ϕ
(x0 , y2 )
|(x0 , y2 )|
+y2 2 − y1 2
=: III + IV,
(7.25)
and note that
|IV | = |y1 − y2 ||y1 + y2 | ≥ λ|y1 − y2 | for y1 , y2 sufficiently close to −λ. (7.26)
65
As for III in (7.25), we may recycle (7.15) in the form
|III| ≤ 2kϕkL∞ (S n−1 ) Lip (ϕ; S
n−1
0
0
(x , y1 )
(x
,
y
)
2
.
) 0
−
|(x , y1 )| |(x0 , y2 )| (7.27)
Using the Mean Value Theorem we then estimate
0
(x , y1 )
(x0 , y2 ) |(x0 , y1 )| − |(x0 , y2 )| ≤ |y1 − y2 |
!
0
(x
,
x
)
n
max sup ∂n
.
0 , x )|
1≤j≤n xn ∈[y ,y ] |(x
n
1 2
j
(7.28)
Now, observe that for each j ∈ {1, ..., n},
∂n
(x0 , xn )
|(x0 , xn )|
=
j
δjn
xj xn
−
,
0
|(x , xn )| |(x0 , xn )|3
(7.29)
and that, since xn ∈ [y1 , y2 ] and x0 is assumed to be near 00 ,
y1 , y2 close to −λ =⇒ xn close to −λ =⇒
(x0 , xn ) close to (00 , −λ) =⇒
δjn
xj xn
−
close to 0.
|(x0 , xn )| |(x0 , xn )|3
(7.30)
Consequently,
|III| ≤
λ
|y1 − y2 |
2
if (x0 , y1 ), (x0 , y2 ) sufficiently close to (00 , −λ).
(7.31)
Relying on this and recalling (7.26), we may now deduce from (7.25) that,
|F (x0 , y1 ) − F (x0 , y2 )| = |III + IV | ≥ |IV | − |III|
≥ λ|y1 − y2 | −
λ
λ
|y1 − y2 | = |y1 − y2 |,
2
2
(7.32)
whenever (x0 y1 ) and (x0 , y2 ) are sufficiently close to (00 , −λ). This finishes the justification of (7.24).
At this point, Theorem 5.1 applies and gives the existence of a neighborhood V
of 00 in Rn−1 along with a Lipschitz function
ψ : V −→ R,
with ψ(00 ) = −λ,
and such that
66
(7.33)
F (x0 , ψ(x0 )) = 0
for every x0 ∈ V.
(7.34)
Given the definion of F in (7.11), this entails
ϕ
(x0 , ψ(x0 ))
|(x0 , ψ(x0 ))|
= |(x0 , ψ(x0 ))|
for every x0 ∈ V.
(7.35)
With this in mind and observing that
∂Ω = {ϕ(ω)ω : ω ∈ S n−1 },
it follows (by specializing ω :=
0
0
(x , ψ(x )) = ϕ
(7.36)
(x0 ,ψ(x0 ))
|(x0 ,ψ(x0 ))|
(x0 , ψ(x0 ))
|(x0 , ψ(x0 ))|
∈ S n−1 ) that
(x0 , ψ(x0 ))
∈ ∂Ω
|(x0 , ψ(x0 ))|
for every x0 ∈ V.
(7.37)
Thus,
the graph of ψ is included in ∂Ω.
(7.38)
On the other hand, if x = (x0 , xn ) ∈ ∂Ω is near −λen = (00 , −λ), then there exists
ω ∈ S n−1 near −en such that
ϕ(ω)ω = x
(7.39)
which forces
|x| = ϕ(ω) and ω =
x
.
|x|
(7.40)
Thus,
2
F (x) = ϕ
x
|x|
− |x|2 = |x|2 − |x|2 = 0
(7.41)
and since x is near −λen , (5.4) tells us that x belongs to the graph of ψ. Consequently,
∂Ω near x0 = −λen is contained in the graph of ψ.
From (7.38) and (7.42)
67
(7.42)
∂Ω and the graph of ψ coincide near x0 .
(7.43)
At this stage, we may invoke Theorem 3.4 in order to extend ψ : V → R to a function
(for which, by slightly abusing notation, we retain the same symbol ψ)
ψ : Rn−1 −→ R,
with Lip (ψ; Rn−1 ) ≤ Lip (ψ; V ).
(7.44)
Of course, for this new function, (7.43) continues to hold. Consequently, there exists
an open, upright, doubly truncated circular cylinder C in Rn−1 × R such that
C ∩ ∂Ω = C ∩ {(x0 , xn ) ∈ Rn−1 × R : xn = ψ(x0 )},
(7.45)
in agreement with (6.6).
Upon observing that
Ω = {rω : 0 ≤ r ≤ ϕ(ω), ω ∈ S n−1 }
(7.46)
we may conclude (using also (7.36)) that
∂Ω = {ϕ(ω)ω : ω ∈ S n−1 } = ∂Ω.
(7.47)
Hence, ∂Ω = ∂Ω and, having already established (6.6), conditions (6.5), (6.7) are now
guaranteed by Proposition 6.2.
Thus, (6.5)-(6.7) hold for the Lipschitz function (7.44), which proves that Ω is a
Lipschitz domain.
2
7.2
Second Application
Definition 7.2.1.
1. Call an open set Ω starlike with respect to x0 ∈ Ω if I(x, x0 ) ⊆ Ω for all x ∈ Ω,
where I(x, x0 ) := the open segment with endpoint x and x0 .
2. Call an open set Ω starlike with respect to a ball B ⊆ Ω if I(x, x0 ) ⊆ Ω for all
x ∈ Ω and y ∈ B. That is, Ω is starlike with respect to any point in B.
Theorem 7.2. Let Ω be a bounded, open, non-empty set which is starlike with respect
to some ball. Then Ω is a Lipschitz domain.
68
Proof. Without loss of generality, assume that Ω is starlike with respect to B :=
B(0, ρ), for some ρ > 0.
Step I. Assume that for all x ∈ Ω, x0 ∈ B =⇒ I(x, x0 ) ⊆ Ω.
Proof of Step I. Since x ∈ Ω, there exists xj ∈ Ω such that xj → x. Fix y ∈
I(x, xo ), then there exists t ∈ (0, 1) such that y = x0 + t(x − x0 ) and consider
yj := x0 + t(xj − x0 ) ∈ I(x, x0 ) ⊆ Ω. Then
lim yj =
j→∞
lim [x0 + t(xj − x0 )]
j→∞
= x0 + t(x − x0 )
= y
(7.48)
Since yj ∈ Ω for every j, this entails y ∈ Ω. Since y was an arbitrary point in I(x, x0 ),
this implies that
I(x, x0 ) ⊆ Ω.
(7.49)
Thus so far, we have that
I(x, x0 ) ⊆ Ω, for all x ∈ Ω and x0 ∈ B.
(7.50)
We now claim that
I(x, x0 ) ∩ ∂Ω = ∅.
(7.51)
Reasoning by contradiction, assume that there exists z ∈ I(x, x0 ) such that z ∈ ∂Ω.
Define the cone-like region
C :=
[
I(z, y).
(7.52)
y∈B
Note that C is open. By (7.50) we know that C ⊆ Ω. Also, I(x, z) ⊆ I(x, x0 ) ⊆ Ω
and I(x, z) ∩ Ω = ∅. Thus, if w ∈ I(x, z) ∩ Ω then z ∈ I(w, x0 ) ⊆ Ω which is a
contradiction. Hence
I(x, z) ⊆ ∂Ω.
(7.53)
Let z0 ∈ I(x, z) then there exists B0 ball centered at z0 such that B0 ⊆ C ⊆ Ω. Then
there exists three collinear points a, b, c (mutually different) with a ∈ B, b ∈ I(x, z),
and c ∈ B.
69
B
B0
x0
a
z0
z
b
c
x
ζ
This means that a ∈
/ Ω, otherwise I(a, c) ⊆ Ω, which implies that b ∈ Ω. However,
b ∈ I(x, z) ⊆ ∂Ω. Since, nonetheless, a ∈ B0 ⊆ C ⊆ Ω implies a ∈ ∂Ω. Perturbing a,
implies there exists r > 0 such that B(a, r) ⊆ B0 and such that for all a ∈ B(a, r) ⊆
∂Ω implies that a0 ∈ ∂Ω (i.e. B(a, r) ⊆ ∂Ω. However, a ∈ ∂Ω implies B(a, r) ∩ Ω 6= ∅,
in contradiction with
B(a, r) ∩ Ω ⊆ ∂Ω ∩ Ω = ∅.
(7.54)
Hence this proves (7.52). From (7.50) and (7.52), the conclusion in Step I follows.
Step II. Let us assume that for all ω ∈ S n−1 , the open half line
Lw := {rω : r > 0}
(7.55)
must intersect ∂Ω.
Proof of Step II. Note that Lw ⊆ Rn = Ω ∪ ∂Ω ∪ (Ω)c , disjoint union. Thus, if
Lw ∩ ∂Ω = ∅ then this implies that Ω and (Ω)c are and open, disjoint, cover of the
connected set Lw . Consequently, either
(i) Lw ∩ Ω = ∅, or
(ii) Lw ∩ (Ω)c = ∅.
Since ω is open and O ∈ Ω then this implies (i) cannot occur. Now, (ii) implies
that Lw ⊆ Ω which implies is impossible since Ω is bounded. Thus this contradiction
proves Step II.
Step III. Assume that if ω ∈ S n−1 then Lw ∩ ∂Ω contains just one point.
Proof of Step III. By applying the results of Step II we know that Lw ∩ ∂Ω 6= ∅. If
x1 6= x2 ∈ Lw ∩ ∂Ω then either,
(i) x2 ∈ I(x1 , 0) or
(ii) x1 ∈ I(x2 , 0).
70
By Step I, if (i) happens then, since x1 ∈ ∂Ω and ∂Ω 3 x2 ∈ I(x1 , 0) ⊆ Ω, which is a
contradiction. By Step I, if (ii) occurs, then (since x2 ∈ ∂Ω) ∂Ω 3 x1 ∈ I(x2 , 0) ⊆ Ω,
which is a contradiction. Hence, this finishes the proof of Step III.
Step IV. If for all ω ∈ S n−1 we set ϕ(ω) := the distance from the point in Lw ∩ ∂Ω to
the origin then
ϕ : S n−1 −→ (0, ∞)
(7.56)
is well-defined and
∂Ω = ϕ(ω)ω : ω ∈ S n−1 .
(7.57)
Proof of Step IV. This follows directly from the fact that O ∈ Ω and Step III.
Step V. Assume that there exists a C > 0 such that for all
ω1 , ω2 ∈ S n−1 and |ω1 − ω2 | < 1
(7.58)
|ϕ(ω1 ) − ϕ(ω2 )| ≤ C |ω1 − ω2 | .
(7.59)
then
Proof of Step V. Let xj := ϕ(ωj )ωj ∈ ∂Ω where, by Step IV, ϕ(ωj )ωj ∈ ∂Ω for
j = 1, 2. This implies that
|ϕ(ω1 ) − ϕ(ω2 )| ≤ |x1 − x2 | .
(7.60)
Let us also note that if θ :=<
) (ω1 , ω2 ) then
θ
1 > |ω1 − ω2 | = 2sin
2
(7.61)
π
θ ∈ 0,
.
3
(7.62)
then
Recall B = B(0, ρ).
Claim 1. The line Lx1 ,x2 passing through x1 , x2 does not intersect B(0, ρ2 ).
Proof of Claim 1. Otherwise, there exists z ∈ B(0, ρ2 ) ∩ Lx1 ,x2 . Since z ∈ B, then by
Step I, I(xj , z) ⊆ Ω, j = 1, 2. Since xj ∈ ∂Ω, this implies that
z ∈ I(x1 , x2 ).
(7.63)
71
Ω
x1
α2
ρ
θ
O
z
α1
Β
x2
Now x1 , x2 ∈ ∂Ω and B(0, ρ) ⊆ Ω implies that |x1 | , |x2 | ≥ ρ. Thus the following
implications occur
|xj − z| ≥ |xj | − |z| ≥ ρ −
ρ
ρ
= ,
2
2
(7.64)
|z| is the shortest side in ∆Ozxj , αj is the smallest angle in ∆Ozxj , and
α≤
π
, j = 1, 2.
3
(7.65)
Recall that θ < π3 . This makes it impossible to have α1 + α2 + θ = π. Thus this
proves Claim 1 made above.
Claim 2. It holds that
distance (0, Lx1 ,x2 ) =
|x1 | |x2 | sinθ
.
|x1 − x2 |
(7.66)
This is just elementary geometry
72
x1
b
c
θ
Ο
h
a
x2
b · a · sinθ
h·c
= area ∆ =
.
s
2
(7.67)
For us,
h = dist(0, Lx1 ,x2 )
a := |x2 | , b = |x1 | , and
c = |x1 − x2 | .
(7.68)
From (7.67), the desired conclusion readily follows.
Next by Claim 1 and Claim 2,
|x1 | |x2 | sin θ
ρ
≥
|x1 − x2 |
2
(7.69)
and
|x1 − x2 | ≤ 2ρ−1 |x1 | |x2 | sinθ
θ
θ
−1
cos
= 4ρ |x1 | |x2 | sin
2
2
≤ 4ρ−1 (diam(Ω))2 |ω1 − ω2 | .
(7.70)
Hence, all in all,
|ϕ(ω1 ) − ϕ(ω2 )| ≤ |x1 − x2 | ≤ C |ω1 − ω2 | ,
proving Step V.
73
(7.71)
Step VI. Show that
Ω is a Lipschitz domain.
(7.72)
Proof of Step VI. By applying our results from Step V we know that ϕ is locally
Lipschitz. Since S n−1 is compact, then ϕ is Lipschitz. Thus,
Ω = rω : 0 ≤ r < ϕ(ω), ω ∈ S n−1
(7.73)
ϕ : S n−1 −→ (0, ∞) Lipschitz,
(7.74)
with
and it is known that this implies that Ω is a Lipschitz domain.
7.3
An Application for the Classical Implicit Function Theorem
Definition 7.3.1. Let Ω be an open set in Rn . Furthermore define Ω to be a C 1
domain if for all x0 ∈ ∂Ω. Then there exist:
U = open neighborhood of x0
(7.75)
and R > 0 along with a C 1 diffeomorphism
ψ : U −→ B(0, R), ψ = (ψ1 , ...., ψn ) with ψ(x0 ) = 0.
(7.76)
Theorem 7.3. Let Ω be an open set in Rn . Then Ω is a C 1 domain if and only if the
0
following holds. For every x0 = (x0 , t0 ) ∈ ∂Ω and a system of coordinates (isometric
with the original one) and
ϕ : Rn−1 → R
(7.77)
a C 1 function and, in addition, a cylinder
C = B0 × I
(7.78)
0
such that ϕ(x0 ) = t0 , B 0 is a (n − 1)-dimensional ball centered at x0 ∈ Rn , and I is
an open interval containing t0 , and the following are valid:
Ω ∩ C = {(x0 , xn ) ∈ C : xn > ϕ(x0 )}
(7.79)
∂Ω ∩ C = {(x0 , xn ) ∈ C : xn = ϕ(x0 )}.
(7.80)
74
Proof of theorem. Assume Ω is a C 1 domain in the sense of the previous definition.
Next choose j ∈ {1, ...., n} such that
(∂j ψn )(x0 ) 6= 0.
(7.81)
This is possible since otherwise ∇ψ(x0 ) = 0, which implies Dψ(x0 ) has a zero line
and thus
Dψ(x0 )
is not invertible.
(7.82)
Thus yielding a contradiction since ψ ∈ C 1 diffeo at x0 . To simplify the explanation
assume that j = n. In other words that
(∂n ψn )(x0 ) 6= 0.
(7.83)
Using the Implicit Function theorem in the classical sense we can solve
ψ(x) = 0 for xn = xn (x0 ).
(7.84)
Thus by the Implicit Function theorem there exist a (n − 1)-dimensional ball B 0
centered at x0 and an open interval I containing t0 such that
ϕ : B 0 −→ I
0
C 1 function, with ϕ(x0 ) = t0 .
(7.85)
In addition if C := B 0 × I then
{x ∈ C : ψn (x) = 0} = {(x0 , ϕ(x0 )) : x0 ∈ B 0 }.
(7.86)
Without loss of generality we can assume that C ⊆ U then
∂Ω ∩ C
=
(∂Ω ∩ U ) ∩ C
= {x ∈ U : ψn (x) = 0} ∩ C
= {x ∈ C : ψn (x) = 0}
= C ∩ graph of ϕ̃
(7.87)
where ϕ is some C 1 -extension of ϕ to Rn−1 . Thus, if Σ denotes graph ϕ̃ then
∂Ω ∩ C = C ∩ Σ.
(7.88)
75
Also x0 ∈ Σ. Furthermore Σ does not escape through the top or bottom part of C.
This and the results in Section 6.2, plus the fact that x0 ∈
/ (Ω)◦ , imply that
Ω ∩ C = {(x0 , xn ) ∈ C : xn > ϕ(x0 )},
(7.89)
so this direction is proved. For the converse implication, just take
ψ(x0 , xn ) := (x0 , xn − ϕ(x0 )).
(7.90)
Thus ψ is a C 1 -diffeomorphism which does the job (after adjusting it so that it maps
onto B(0, R) for some R > 0).
2
76
Bibliography
[1] L. Brouwer, Zur Invarianz des n-dimensionalen Gebiets, Mathematische Annalen, 72
(1912), 55 – 56.
[2] L. Fanghua and Y. Xiaoping, Geometric Measure Theory. An Introduction, International Press, Boston, 2002.
[3] G.B. Folland, Real Analysis - Modern Techniques and their Applications, 2dn edition,
Wiley, 1999.
[4] P. Grisvard, Elliptic problems in nonsmooth domains, Monographs and Studies in
Mathematics, Vol. 24 Pitman Boston, MA, 1985.
[5] J. Heinonen, Nonsmooth Calculus, Bull. Amer. Math. Soc., Vol. 44, No. 2 (2007), 163–
232.
Michael Wuertz
Department of Mathematics
University of Missouri at Columbia
Columbia, MO 65211, USA
e-mail: [email protected]
77