37.2. AN EXTENSION THEOREM 1287

where q/p+ 1/r = 1. Now φ l (x)→XU (x) and so the integrand in the last integral con-verges to 0 by the dominated convergence theorem. Therefore, k may be chosen largeenough that for all u ∈ S,

||u−φ ku||qLq(U)≤(

ε

3

)q.

Fix such a value of k. Then ∣∣∣∣wq−wp∣∣∣∣

Lq(U)≤∣∣∣∣wq−φ kwq

∣∣∣∣Lq(U)

+∣∣∣∣φ kwq−φ kwp

∣∣∣∣Lq(U)

+∣∣∣∣wp−φ kwp

∣∣∣∣Lq(U)

≤ 2ε

3+∣∣∣∣φ kwq−φ kwp

∣∣∣∣Lq(U)

.

But {φ kwm}∞

m=1 converges in Lq (U) and so the last term in the above is less than ε/3whenever p,q are large enough. Thus {wm}∞

m=1 is a Cauchy sequence and must thereforeconverge in Lq (U). This proves the theorem.

37.2 An Extension TheoremDefinition 37.2.1 An open subset, U, of Rn has a Lipschitz boundary if it satisfies thefollowing conditions. For each p ∈ ∂U ≡U \U, there exists an open set, Q, containing p,an open interval (a,b), a bounded open box B⊆ Rn−1, and an orthogonal transformationR such that

RQ = B× (a,b), (37.2.26)

R(Q∩U) = {y ∈ Rn : ŷ ∈ B, a < yn < g(ŷ)} (37.2.27)

where g is Lipschitz continuous on B,a < min{

g(x) : x ∈ B}, and

ŷ≡ (y1, · · · ,yn−1).

Letting W = Q∩U the following picture describes the situation.

xW

Q

R R(W )

a

bR(Q)

y

The following lemma is important.

Lemma 37.2.2 If U is an open subset ofRn which has a Lipschitz boundary, then it satisfiesthe segment condition and so Xm,p (U) =W m,p (U) .