In this section is an important theorem which can be used to verify that d
(f,Ω,y)
≠0. This is significant
because when this is known, it follows from Theorem 11.3.4 that f^{−1}
(y)
≠∅. In other words there exists
x ∈ Ω such that f
(x)
= y.
Definition 11.4.1A bounded open set, Ω is symmetric if −Ω = Ω. Acontinuous function,f : Ω→ ℝ^{p}is odd if f
(− x)
= −f
(x)
.
Suppose Ω is symmetric and g ∈ C^{∞}
( -- p)
Ω;ℝ
is an odd map for which 0 is a regular value. Then the
chain rule implies Dg
(− x)
= Dg
(x)
and so d
(g,Ω,0)
must equal an odd integer because if x ∈ g^{−1}
(0)
, it
follows that −x∈ g^{−1}
(0)
also and since Dg
(− x)
= Dg
(x)
, it follows the overall contribution to the
degree from x and −x must be an even integer. Also 0 ∈ g^{−1}
(0)
and so the degree equals an
even integer added to sgn
(detDg (0 ))
, an odd integer, either −1 or 1. It seems reasonable to
expect that something like this would hold for an arbitrary continuous odd function defined on
symmetric Ω. In fact this is the case and this is next. The following lemma is the key result
used. This approach is due to Gromes [56]. See also Deimling [37] which is where I found this
argument.
The idea is to start with a smooth odd map and approximate it with a smooth odd map which also has
0 a regular value.
Lemma 11.4.2Let g ∈ C^{∞}
(Ω;ℝp )
be an odd map. Then for every ε > 0, there exists h∈ C^{∞}
( Ω;ℝp )
such that h is also an odd map,
||h− g||
_{∞}< ε, and 0 is a regular value ofh.Here
Ω is a symmetric bounded open set. In addition, d
(g,Ω,0)
is an odd integer.
Proof:In this argument η > 0 will be a small positive number. Let h_{0}
(x)
= g
(x)
+ ηx where η is
chosen such that detDh_{0}
(0)
≠0. Note that h_{0} is odd and 0 is a value of h_{0} thanks to h_{0}
(0)
= 0. This has
taken care of 0. Now we need to take care of the other points. Let
∑p j 3 ∑0
h (x) ≡ h0 (x) − y xj, ≡ 0.
j=1 j=1
The y^{j} will each be small. Note that h
(0)
= 0 and det
(Dh (0))
= det
(Dh0 (0))
≠0. We want to choose
small y^{j},
∥ ∥
∥yj∥
< η in such a way that 0 is a regular value for h for each x≠0 such that x ∈ h^{−1}
(0)
.
Let
p p
Ωi ≡ {x ∈ Ω : xi ⁄= 0}, so ∪ j=1 Ωj = {x ∈ ℝ : x ⁄= 0}.
Let Ω_{0}≡
{0}
. Then the theorem will be proved if for each k ≤ p, there exist y^{i},
∥ ∥
∥yi∥
< η and 0 is a
regular value of
k
h (x) ≡ h (x )− ∑ yjx3for x ∈ ∪k Ω
k 0 j=1 j i=0 i
For k = 0 this is done. Suppose then that this holds for k − 1, k ≥ 1. Thus 0 is a regular value
for
k−∑ 1
hk−1(x) ≡ h0 (x) − yjx3j on ∪k− 1Ωi
j=1 i=0
What should y^{k} be? Keeping y^{1},
⋅⋅⋅
,y^{k−1}, it follows that if x_{k} = 0, h_{k}
(x)
= h_{k−1}
(x)
so for
x_{k}∈ Ω_{k},
∑k j3
hk (x ) ≡ h0(x)− y xj = 0
j=1
if and only if
∑k −1 j 3
h0-(x)−---j=1 y-xj = yk
x3k
So let y^{k} be a regular value of h_{k−1}
(x)
≡
h0(x)−∑kj−=11 yjx3j
x3k
on Ω_{k} and also
∥∥ k∥∥
y
< η. Then for such x, and
using the quotient rule,
Ω_{k}, then x_{k} = 0 and so, because of the power of 3 in
x_{k}^{3}, Dh_{k}
(x)
= Dh_{k−1}
(x )
which has nonzero determinant by induction. Thus for each k ≤ n, there is a
function h_{k}of the form described above, such that for x ∈∪_{j=1}^{k}Ω_{k}, with h_{k}
Proof:Let ψ_{n} be a mollifier which is symmetric, ψ
(− x)
= ψ
(x)
. Also recall that f is the restriction to
Ω of a continuous function, still denoted as f which is defined on all of ℝ^{p}. Let g be the odd part of this
function. That is,
1
g (x ) ≡- (f (x) − f (− x))
2
Since f is odd, g = f on Ω. Then
∫
gn(− x) ≡ g∗ ψn(− x) = Ω g(− x − y)ψn (y)dy
∫ ∫
= − g(x+ y )ψn(y)dy = − g(x− (− y ))ψn (− y)dy = − gn (x)
Ω Ω
Thus g_{n} is odd and is infinitely differentiable. Let 3δ = dist