1312 CHAPTER 38. SOBOLEV SPACES BASED ON L2

In the second term, let t = x−y. Then this term is of the form∫ ∣∣∣h∗(Dβ u)(y)∣∣∣2 ∫ ∣∣gβ (y+ t)−gβ (y)

∣∣2 |t|−n−2s dtdy (38.3.15)

≤ C∫ ∣∣∣h∗(Dβ u

)(y)∣∣∣2 dy≤C ||u||2Hm(Rn) . (38.3.16)

because the inside integral equals a constant which depends on the Lipschitz constants andbounds of the function, gβ and these things depend only on h. The reason this integral isfinite is that for |t| ≤ 1,∣∣gβ (y+ t)−gβ (y)

∣∣2 |t|−n−2s ≤ K |t|2 |t|−n−2s

and using polar coordinates, you see∫[|t|≤1]

∣∣gβ (y+ t)−gβ (y)∣∣2 |t|−n−2s dt < ∞.

Now for |t| > 1, the integrand in 38.3.15 is dominated by 4 |t|−n−2s and using polar coor-dinates, this yields∫

[|t|>1]

∣∣gβ (y+ t)−gβ (y)∣∣2 |t|−n−2s dt ≤ 4

∫[|t|>1]

|t|−n−2s dt < ∞.

It follows 38.3.14 is dominated by an expression of the form

C (h)∫|F (u)(z)|2

∣∣∣zβ

∣∣∣2 |z|2s dz+C ||u||2Hm(Rn)

and so the sum in 38.3.13 is dominated by

C (m,h)∫|F (u)(z)|2 |z|2s

∑|β |≤m

∣∣∣zβ

∣∣∣2 dz+C ||u||2Hm(Rn)

≤ C (m,h)∫|F (u)(z)|2

(1+ |z|2

)s(1+ |z|2

)mdz+C ||u||2Hm(Rn)

≤ C ||u||2Hm+s(Rn) .

This proves the theorem because the assertion about h−1 is obvious. Just replace h withh−1 in the above argument.

Next consider the case where U is an open set.

Lemma 38.3.3 Let h(U) ⊆ V where U and V are open subsets of Rn and suppose thath,h−1 :Rn→Rn are both functions in Cm,1 (Rn) . Recall this means Dα h and Dα h−1 existand are Lipschitz continuous for all |α| ≤ m. Then h∗ ∈L (Hm+s (V ) ,Hm+s (U)).

Proof: Let u∈Hm+s (V ) and let v∈Hm+s (Rn) such that v|V = u. Then from the above,h∗v ∈ Hm+s (Rn) and so h∗u ∈ Hm+s (U) because h∗u = h∗v|U . Then by Lemma 38.3.2,

||h∗u||Hm+s(U) ≤ ||h∗v||Hm+s(Rn) ≤C ||v||Hm+s(Rn)