First is a simple lemma which links an integral to something which is an integer. It will
always be the case that ϕ_{ε} will denote a mollifier as in Definition 21.2.3. Also, there are
various conditions which hold if something is small enough. Numbers like 1/5 and 1/10
are used to express suitable smallness assumptions. There is nothing particularly special
about these numbers. I just chose some which were sufficiently small to work out. I am
not trying to give anything precise in expressing these estimates. The thing that is
interesting is the degree for continuous functions and anything which will get you this is
what is wanted.
Lemma 21.3.1Lety
∕∈
g
(∂ Ω)
for g ∈ C^{∞}
(-- )
Ω,ℝp
. Also suppose y is a regular value ofg. Then for all ε small enough,
Proof: First note that the sum is finite. Indeed, if there were a sequence of x_{k} such
that x_{k}∈ g^{−1}
(y)
, then it would have a subsequence converging to some x ∈Ω . Thus
g
(x )
= y and by assumption, y
∕∈
∂Ω. Since y is a regular value, Dg
(x)
is invertible and
so by the inverse function theorem, there is an open set containing x on which g is one to
one, contrary to x being a limit of a sequence of points in g^{−1}
(y)
. It only remains to
verify the equation.
I need to show the left side of this equation is constant for ε small enough and
equals the right side. By what was just shown, there are finitely many points,
{xi}
_{i=1}^{m} = g^{−1}
(y )
. By the inverse function theorem, there exist disjoint open sets U_{i}
with x_{i}∈ U_{i}, such that g is one to one on U_{i} with det
(Dg (x))
having constant sign on
U_{i} and g
(Ui)
is an open set containing y. Then let ε be small enough that
B
The reason for this is as follows. The integrand on the left is nonzero only if
g
(x )
− y ∈ B
(0,ε)
which occurs only if g
(x)
∈ B
(y,ε)
which is the same as
x ∈ g^{−1}
(B (y,ε))
. Therefore, the integrand is nonzero only if x is contained in
exactly one of the disjoint sets, V_{i}. Now using the change of variables theorem,
(z = g