1506 CHAPTER 45. TRACES OF SOBOLEV SPACES

here this is centering at 0 and stretching B since λ will be large. Let h be the name of thismapping. Thus

h(x̂) ≡ λ

(x̂− b̂0

),

k(x̂) ≡ h−1 (x̂) =1λ

x̂+ b̂0

These mappings are defined on all of Rn. Now let u′ be defined on U ′ as follows.

u′ (x̂,xn)≡ k∗u(x̂,xn) .

Also letg′ (x̂)≡ k∗g(x̂) .

Thus g′ (x̂) ≡ g(

1λ

x̂+ b̂0

)= g(k(x̂)) . Then choosing λ large enough the Lipschitz con-

dition for g′ is as small as desired. Always assume λ has been chosen this large and alsoλ ≥ 1. Furthermore, g′

(x̂′)= x′n describes the boundary in the same way as xn = g(x̂).

Now I need to consider whether u′ ∈ ˜W θ ,p (U ′). Consider

∫U ′

∫U ′

∣∣∣u′(x̂′,xn

)−u′

(ŷ′,yn

)∣∣∣p(∣∣∣x̂′− ŷ′∣∣∣2 +(xn− yn)

2)p+nθ

dx′dy′

=∫

U ′

∫U ′

∣∣∣k∗u(x̂′,xn

)−k∗u

(ŷ′,yn

)∣∣∣p(∣∣∣x̂′− ŷ′∣∣∣2 +(xn− yn)

2)p+nθ

dx′dy′

Then change the variables x̂′ = λ

(x̂− b̂0

)= h(x̂) with a similar change for ŷ′, the above

expression equals (λ

n−1)2 ∫

U

∫U

|u(x̂,xn)−u(ŷ,yn)|p(λ

2 |x̂− ŷ|2 +(xn− yn)2)p+nθ

dxdy

Thus ∣∣∣∣u′∣∣∣∣ ˜W θ ,p(U ′)≤ λ

n−1 ||u|| ˜W θ ,p(U)< ∞ (45.3.21)

and k∗ : ˜W θ ,p (U) to ˜W θ ,p (U ′)is continuous and linear. Similar reasoning shows that h∗ is

continuous and linear mapping ˜W θ ,p (Rn) to ˜W θ ,p (Rn). By the first part of the argumentthere exists a continuous linear map

E ′ : ˜W θ ,p (U ′)→ ˜W θ ,p (Rn)