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 16.2.3. 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 16.3.1Lety
∈∕
g
(∂Ω)
for g ∈ C^{∞}
(- p)
Ω, ℝ
. 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
(y,ε)
⊆∩_{i=1}^{m}g
(Ui)
and let
V_{i}≡ g^{−1}
(B (y,ε))
∩ U_{i}.
PICT
Therefore, for any ε this small,
∫ ∑m ∫
Ω ϕε(g(x)− y)detDg (x)dx = V ϕε(g (x) − y )detDg (x )dx
i=1 i
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