There is a very interesting application of the degree to integration [?]. Recall Lemma ??.
I want to generalize this to the case where h :ℝ^{n}→ ℝ^{n} has the property that its
weak partial derivatives and h are in L^{p}
(ℝn;ℝn )
,p > n. This is denoted by
saying
h ∈ W 1,p(ℝn;ℝn ).
In the following proposition, let ϕ_{ε} be a symmetric nonnegative mollifier,
ϕ (x) ≡ 1-ϕ (x) ,sptϕ ⊆ B (0,1).
ε εn ε
Ω will be a bounded open set. By Theorem 24.6.10, h may be considered continuous and
it satisfies
Dh (x) exists a.e., (27.4.48)
(27.4.48)
For any p > n,
p( n n×n)
mlim→∞ D (h∗ψm ) = Dh in L ℝ ;ℝ (27.4.49)
(27.4.49)
where ψ_{m} is a mollifier. Here ℝ^{n×n} denotes the n × n matrices with any norm you
like.
Proposition 27.4.1Let S ⊆ h(∂Ω)^{C}such that
dist(S,h(∂Ω)) > 0
where Ω is a bounded open set and also let h be in W^{1,p}
n n
(ℝ ;ℝ )
. Then whenever ε > 0 issmall enough,
∫
d(h,Ω,y ) = ϕ ε(h (x)− y)detDh (x)dx
Ω
for all y ∈ S.
Proof: Let ε_{0}> 0 be small enough that for all y ∈ S,
B (y,3ε0)∩ h(∂Ω) = ∅.
Now let ψ_{m} be a mollifier as m →∞ with support in B
( −1)
0,m
and let
hm ≡ h∗ψm.
Thus h_{m}∈ C^{∞}
(Ω; ℝn)
and,
||hm − h||L∞(Ω),||Dhm − Dh ||Lp(Ω) → 0 (27.4.50)
(27.4.50)
as m →∞. The first claim above follows from the definition of convolution and the
uniform continuity of h on the compact set Ω and the second follows by 27.4.49. Choose
M such that for m ≥ M,
||hm − h||L∞(Ω) < ε0. (27.4.51)
(27.4.51)
Thus h_{m}∈U_{y}∩ C^{2}
(-- n)
Ω;ℝ
for all y ∈ S.
For y ∈ S, let z ∈ B
(y,ε)
where ε < ε_{0} and suppose x ∈ ∂Ω, and k,m ≥ M. Then for
t ∈
for all k,m ≥ M. By this lemma again, which says that for small enough ε the integral is
constant and the definition of the degree in Definition 21.2.3,
∫
d(y,Ω, h ) = ϕ (h (x)− y)det(Dh (x ))dx (27.4.53)
m Ω ε m m
(27.4.53)
for all ε small enough. For x ∈ ∂Ω, y ∈ S, and t ∈
and so by Theorem 21.3.4, the part about homotopy, for each y ∈ S,
d(y,Ω,h) = d (y,Ω,hm ) =
∫
Ω ϕε(hm (x)− y)det(Dhm (x))dx
whenever ε is small enough. Fix such an ε < ε_{0} and use 27.4.52 to conclude the right side
of the above equation is independent of m > M.
By 27.4.50, there exists a subsequence still denoted by m such that Dh_{m}
(x)
→ Dh
(x )
a.e. Since p > n, det
(Dhm )
is bounded in L^{r}
(Ω)
for some r > 1 and so the integrands in
the following are uniformly integrable. By the Vitali convergence theorem, one can pass
to the limit as follows.
∫
d (y,Ω,h ) = lim ϕ (h (x)− y)det(Dh (x))dx
m∫→∞ Ω ε m m
= Ωϕε(h (x )− y)det(Dh (x ))dx.
This proves the proposition.
Next is an interesting change of variables theorem. Let Ω be a bounded open set and
let h ∈ W^{1,p}
∫ ∫ ∫
f (y)d (y,Ω, h)dy = f (y) ϕε(h(x)− y)detDh (x)dxdy
∫ Ω ∫
= detDh (x) f (y )ϕε(h(x)− y)dydx
∫Ω ∫
= detDh (x) f (h (x)− εu)ϕ(u)dudx
Ω
Using the uniform continuity of f, you can now pass to a limit as ε → 0 and obtain, using
the fact that detDh
(x)
is in L^{r}
(ℝn )
for some r > 1,
∫ ∫
f (y )d(y,Ω,h)dy = f (h(x))detDh (x)dx
Ω
This has proved the following interesting lemma.
Lemma 27.4.2Let f ∈ C_{c}
(h (∂Ω)C)
and let h ∈ W^{1,p}
(ℝn;ℝn )
,p > n, h
(∂Ω)
hasmeasure zero for Ω a bounded open set. Then everything is measurable which needs to beand
∫ ∫
f (y)d(y,Ω,h)dy = Ωdet(Dh (x))f (h (x))dx.
Note that h is not necessarily one to one. The difficult issue is handling d
(y,Ω, h)
which has integer values constant on each component of h
(∂Ω)
^{C} and the difficulty
arrises in not knowing how many components there are. What if there are infinitely
many, for example, and what if the degree changes sign. If this happens, it is hard to
exploit convergence theorems to get generalizations of f ∈ C_{c}
(h (∂Ω )C)
. One way
around this is to insist h be one to one and that Ω be connected having a boundary
which separates ℝ^{n} into two components, three if n = 1. That way, you can use the
Jordan separation theorem and assert h
(∂Ω)
also separates ℝ^{n} into the same
number of components with h
(Ω)
being the only one on which the degree is
nonzero.
First recall the following proposition.
Proposition 27.4.3Let Ω be an open connected bounded set in ℝ^{n},n ≥ 1 suchthat ℝ^{n}∖∂Ω consists of two, three if n = 1, connected components. Let f ∈ C
(-- n)
Ω;ℝ
be continuous and one to one. Then f
(Ω)
is the bounded component of ℝ^{n}∖ f
(∂Ω)
and for y ∈ f
(Ω )
, d
(f,Ω,y)
either equals 1 or −1.
Proof: First suppose n ≥ 2. By the Jordan separation theorem, ℝ^{n}∖ f
(∂Ω)
consists
of two components, a bounded component B and an unbounded component U. Using the
Tietze extention theorem, there exists g defined on ℝ^{n} such that g = f^{−1} on
f
(--)
Ω
. Thus on ∂Ω,g ∘ f = id. It follows from this and the product formula that
1 = d (id,Ω,g(y)) = d (g ∘f,Ω,g (y))
= d (g,B, g(y))d(f,Ω,B) +d (f,Ω,U )d(g,U,g (y ))
= d (g,B, g(y))d(f,Ω,B)
Therefore, d
(f,Ω,B )
≠0 and so for every z ∈ B, it follows z ∈ f
(Ω)
. Thus B ⊆ f
(Ω )
. On
the other hand, f
(Ω)
cannot have points in both U and B because it is a connected set.
Therefore f
(Ω)
⊆ B and this shows B = f
(Ω)
. Thus d
(f,Ω, B)
= d
(f,Ω,y )
for each
y ∈ B and the above formula shows this equals either 1 or −1 because the degree is an
integer. In the case where n = 1, the argument is similar but here you have 3 components
in ℝ^{1}∖ f
(∂Ω )
so there are more terms in the above sum although two of them give 0.
This proves the proposition.
The following is a version of the area formula.
Lemma 27.4.4Let h ∈ W^{1,p}
(ℝn;ℝn )
,p > n where h is one to one, h
(∂ Ω)
,∂Ω havemeasure zero for Ω a bounded open connected set in ℝ^{n}. Then h
(∂Ω )
^{C}has twocomponents, three if n = 1, and for y ∈ h
(Ω)
, and f ∈ C_{c}
(ℝn )
.
∫ ∫
f (y )dy = |det(Dh (x))|f (h(x))dx
h(Ω) Ω
If O is an open set, it is also true that
∫ ∫
h(Ω)XO (y)dy = Ω|det(Dh (x))|XO (h (x))dx
Also if f is any nonnegative Borel measurable function
∫ ∫
f (y )dy = |det(Dh (x))|f (h(x))dx
h(Ω) Ω
Proof: Consider the first claim. Let δ be such that B
(x1,δ)
⊆ Ω and let
{fj(y)}
_{j=1}^{∞}
be nonnegative, increasing in j and converging pointwise to X_{h(B (x1,δ))
}
(y )
. This can be
done because h
(B (x1,δ))
is an open bounded set thanks to invariance of domain,
Theorem 21.5.3. By Proposition 27.4.3, d
(y,Ω,h)
either equals 1 or −1. Suppose it
equals −1. Then from Lemma 27.3.2
∫ ∫
h(Ω)fj(y)dy = − Ωdet(Dh (x))fj(h(x))dx
The integrand on the right is uniformly integrable thanks to the fact the f_{j} are bounded
and det
(Dh (x))
is in L^{r}
(Ω )
for some r > 1. Therefore, by the Vitali convergence
theorem and the monotone convergence theorem,