29.12. THE CASE OF W 1,p 1051
Ω will be a bounded open set. By Theorem 26.6.10, h may be considered continuous andit satisfies
Dh(x) exists a.e., (29.12.69)
For any p > n,lim
m→∞D(h∗ψm) = Dh in Lp (Rn;Rn×n) (29.12.70)
where ψm is a mollifier. Here Rn×n denotes the n×n matrices with any norm you like.
Proposition 29.12.1 Let S⊆ h(∂Ω)C such that
dist(S,h(∂Ω))> 0
where Ω is a bounded open set and also let h be in W 1,p (Rn;Rn). Then whenever ε > 0 issmall 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,3ε0)∩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,
||hm−h||L∞(Ω) , ||Dhm−Dh||Lp(Ω)→ 0 (29.12.71)
as m→∞. The first claim above follows from the definition of convolution and the uniformcontinuity of h on the compact set Ω and the second follows by 29.12.70. Choose M suchthat for m≥M,
||hm−h||L∞(Ω) < ε0. (29.12.72)
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
showing that for each y ∈ S, B(y,ε)∩ ((1− t)hm + thk)(∂Ω) = /0. By Lemma 23.1.11, forall y ∈ S, ∫
Ω
φ ε (hm (x)−y)det(Dhm (x))dx =∫Ω
φ ε (hk (x)−y)det(Dhk (x))dx (29.12.73)