Lemma 21.2.1Let g : U → ℝ^{p}be C^{2}where U is an open subset of ℝ^{p}. Then
∑p
cof (Dg)ij,j = 0,
j=1
where here
(Dg )
_{ij}≡ g_{i,j}≡
∂∂gxij
. Also,cof
(Dg )
_{ij} =
∂de∂t(giD,jg)
.
Proof: From the cofactor expansion theorem,
∑p
det(Dg ) = gi,j cof (Dg)ij
i=1
and so
∂det(Dg )
---------= cof (Dg )ij (21.2.4)
∂gi,j
(21.2.4)
which shows the last claim of the lemma. Also
∑
δkjdet(Dg ) = gi,k(cof (Dg ))ij (21.2.5)
i
(21.2.5)
because if k≠j this is just the cofactor expansion of the determinant of a matrix in which
the k^{th} and j^{th} columns are equal. Differentiate 21.2.5 with respect to x_{j} and sum on j.
This yields
∑ ∑ ∑
δkj∂-(detDg-)gr,sj = gi,kj(cof (Dg ))ij + gi,kcof (Dg )ij,j.
r,s,j ∂gr,s ij ij
Another simple lemma which will be used whenever convenient is the following
lemma.
Lemma 21.2.2Let K be a compact set and C a closed set in ℝ^{p}such that K ∩C = ∅.Then
dist(K, C) ≡ inf{∥k − c∥ : k ∈ K, c ∈ C } > 0.
Proof:Let
d ≡ inf{||k− c|| : k ∈ K, c ∈ C }
Let
{ki}
,
{ci}
be such that
1
d + i > ||ki − ci||.
Since K is compact, there is a subsequence still denoted by
{ki}
such that k_{i}→ k ∈ K.
Then also
||c − c || ≤ ||c − k ||+ ||k − k ||+||c − k ||
i m i i i m m m
If d = 0, then as m,i →∞ it follows
||ci − cm ||
→ 0 and so
{ci}
is a Cauchy sequence
which must converge to some c ∈ C. But then
||c − k||
= lim_{i→∞}
||ci − ki||
= 0 and so
c = k ∈ C ∩ K, a contradiction to these sets being disjoint. ■
In particular the distance between a point and a closed set is always positive if the
point is not in the closed set. Of course this is obvious even without the above lemma.
The above lemmas will be used now to prove technical lemmas which are the basis for
everything.
Definition 21.2.3Let g ∈ C^{∞}
(-- )
Ω;ℝp
where Ω is a bounded open set. Also let ϕ_{ε}be amollifier.
∫
ϕε ∈ C ∞c (B (0,ε)),ϕε ≥ 0, ϕεdx = 1.
First, here is a technical lemma which will end up being the reason one can define the
degree in terms of an integral. It is a result on homotopy invariance for functions which
are C^{∞}.
Lemma 21.2.4If h : ℝ^{p}×ℝ → ℝ^{p}is in C_{c}^{∞}
p p
(ℝ × ℝ, ℝ )
, and0
∈∕
h
(∂Ω × [α,β])
thenfor 0 < ε <dist
(0,h(∂Ω × [α,β ]))
,
∫
t → ϕε(h(x,t))detD1h (x,t)dx
Ω
is constant for t ∈
(a,b)
, an open set which contains
[α,β]
.
Proof:By continuity of h, h
(∂Ω × [α,β])
is compact and so is at a positive distance
from 0. Therefore, there exists an open interval,
(a,b)
such that
(a,b)
contains
[α, β]
and
0
∕∈
h
(∂ Ω× [a,b])
. Let ε > 0 be such that for all t ∈
[a,b]
,
B(0,ε)∩ h(∂Ω × [a,b]) = ∅ (21.2.6)
(21.2.6)
Define for t ∈
(a,b)
,
∫
H (t) ≡ ϕε(h (x,t))detD1h (x,t)dx
Ω
Then if t ∈
(a,b)
,
∫ ∑
H ′(t) = ϕε,α (h(x,t))hα,t(x,t)detD1h (x,t)dx
Ω α
In this formula, the function det is considered as a function of the n^{2} entries in the n×n
matrix and the ,αj represents the derivative with respect to the αj^{th} entry h_{α,j}. Now as
in the proof of Lemma 21.2.1 on Page 2372,
Now the sum on j is the dot product of the β^{th} row with the α^{th} row of the cofactor
matrix which equals zero unless β = α because it would be a cofactor expansion of a
matrix with two equal rows. When β = α, the sum on j reduces to det
which is the same thing with opposite sign. Hence these sum to 0. Therefore, H^{′}
(t)
= 0
and so H is a constant on
(a,b)
⊇
[α,β ]
. ■
The following is a situation in which one only has continuity in t. Of course the
difficulty is that it is not yet clear whether the constant depends on ε.