29.11. INTEGRATION AND THE DEGREE 1047
29.11 Integration and the DegreeThere is a very interesting application of the degree to integration [52]. Recall Lemma23.1.11. I want to generalize this to the case where h : Rn→ Rn is only Lipschitz continu-ous, vanishing outside a bounded set. In the following proposition, let φ ε be a symmetricnonnegative mollifier,
φ ε (x)≡1εn φ
(xε
),sptφ ⊆ B(0,1) .
Ω will be a bounded open set. By Theorem 26.6.7, h satisfies
Dh(x) exists a.e., (29.11.62)
For any p > n,lim
m→∞D(h∗ψm) = Dh in Lp (Rn;Rn×n) (29.11.63)
where ψm is a mollifier.
Proposition 29.11.1 Let S⊆ h(∂Ω)C such that
dist(S,h(∂Ω))> 0
where Ω is a bounded open set and also let h be Lipschitz continuous, vanishing outsidesome bounded set. Then whenever ε > 0 is small enough,
d (h,Ω,y) =∫
Ω
φ ε (h(x)−y)detDh(x)dx
for all y ∈ S.
Proof: Let ε0 > 0 be small enough that for all y ∈ S,
B(y,5ε0)∩h(∂Ω) = /0.
Now let ψm be a mollifier as m→ ∞ with support in B(0,m−1
)and let
hm ≡ h∗ψm.
Thus hm ∈C∞(Ω;Rn
)and for any p > n,
||hm−h||L∞(Ω) , ||Dhm−Dh||Lp(Ω)→ 0 (29.11.64)
as m→∞. The first claim above is obvious and the second follows by 29.11.63. Choose Msuch that for m≥M,
∥hm−h∥∞< ε0. (29.11.65)
Thus hm ∈Uy∩C2(Ω;Rn
)for all y ∈ S.
For y ∈ S, let z ∈ B(y,ε) where ε < ε0 and suppose x ∈ ∂Ω, and k,m ≥M. Then fort ∈ [0,1] ,
|(1− t)hm (x)+hk (x) t− z| ≥ |hm (x)− z|− t |hk (x)−hm (x)|> 2ε0− t2ε0 ≥ 0