Lemma 11.2.1Let g : U → ℝ^{p}be C^{2}where U is an open subset of ℝ^{p}. Then
p
∑ cof (Dg) = 0,
j=1 ij,j
where here
(Dg )
_{ij}≡ g_{i,j}≡
∂gi
∂xj
. Also,cof
(Dg )
_{ij} =
∂det(Dg)
∂gi,j
.
Proof: From the cofactor expansion theorem,
p
det(Dg ) = ∑ g cof (Dg )
i=1 i,j ij
and so
∂det(Dg )
---------= cof (Dg )ij (11.3)
∂gi,j
(11.3)
which shows the last claim of the lemma. Also
∑
δkjdet(Dg ) = gi,k(cof (Dg ))ij (11.4)
i
(11.4)
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 11.4 with respect to x_{j} and sum on j. This yields
∑ ∂ (det Dg) ∑ ∑
δkj---∂g----gr,sj = gi,kj(cof (Dg ))ij + gi,kcof (Dg )ij,j.
r,s,j r,s ij ij
Another simple lemma which will be used whenever convenient is the following lemma.
Lemma 11.2.2Let 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
{k }
i
such that k_{i}→ k ∈ K. Then
also
||ci − cm|| ≤ ||ci − ki||+ ||ki − km ||+ ||cm − km ||
If d = 0, then as m,i →∞ it follows
||c − c ||
i m
→ 0 and so
{c }
i
is a Cauchy sequence which must
converge to some c ∈ C. But then
||c − k||
= lim_{i→∞}
||c − k ||
i i
= 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 11.2.3Let g ∈ C^{∞}
(-- )
Ω;ℝp
where Ω is a bounded open set. Also let ϕ_{ε}be a mollifier.
∫
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 11.2.4If h : ℝ^{p}×ℝ → ℝ^{p}is in C_{c}^{∞}
(ℝp × ℝ,ℝp)
, and0
∕∈
h
(∂Ω× [α,β])
then for
0 < ε <dist
(0,h(∂Ω × [α,β ]))
,
∫
t → Ω ϕε(h(x,t))detD1h (x,t)dx
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]) = ∅ (11.5)
(11.5)
Define for t ∈
(a,b)
,
∫
H (t) ≡ Ωϕε(h (x,t))detD1h (x,t)dx
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 11.2.1 on
Page 873,
detD1 (h (x,t)),αj = (cof D1 (h (x,t)))αj
and so
∫
B = ∑ ∑ ϕ (h (x,t))(cof D (h (x,t))) h dx.
Ω α j ε 1 αj α,jt
By hypothesis
x → ϕ (h(x,t))(cof D (h (x,t)))
ε 1 αj
is in C_{c}^{∞}
(Ω )
because if x ∈ ∂Ω, it follows that for all t ∈
[a,b]
h(x,t) ∕∈ B (0,ε)
and so ϕ_{ε}
(h (x,t))
= 0. Thus it equals 0 on ∂Ω. Therefore, integrate by parts and write
∫
B = − ∑ ∑ -∂-(ϕ (h(x,t)))(cof D (h (x,t))) h dx+
Ω α j ∂xj ε 1 αj α,t
∫
− ∑ ∑ ϕ (h (x,t))(cof D (h(x,t))) h dx
Ω α j ε αj,j α,t
The second term equals zero by Lemma 11.2.1. Simplifying the first term yields
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.