37.2. AN EXTENSION THEOREM 1289

∣∣u(ŷ1,y1n)−u(ŷ2,y1

n)∣∣+ ∣∣u(ŷ2,y1

n)−u(ŷ2,g(ŷ2))

∣∣+∣∣u(ŷ2,g(ŷ2))−u

(ŷ2,y2

n)∣∣

≤ K |ŷ1− ŷ2|+K[|g(ŷ2)−g(ŷ1)|+g(ŷ1)− y1

n + y2n−g(ŷ2)

]≤

(2K +K2) |ŷ1− ŷ2|+K

∣∣y1n− y2

n∣∣

=(2K +K2)(|ŷ1− ŷ2|+

∣∣y1n− y2

n∣∣)≤ (2K +K2)√2 |y1−y2|

The other cases are similar. Thus u is a Lipschitz continuous function which has compactsupport. By Corollary 35.5.4 on Page 1243 it follows that u,i ∈ L∞ (Rn)∩Lp (Rn) for allp > 1. It remains to verify u,i = (u+),i on V+ and u,i = (u−),i on V−. The last claim isobvious from the definition of weak derivatives.

Lemma 37.2.4 In the situation of Lemma 37.2.3 let u ∈ C1(

V−)∩C1

c (B× (a,b))3 anddefine

w(ŷ,yn)≡

 u(ŷ,yn) if ŷ ∈ B and yn ≤ g(ŷ) ,u(ŷ,2g(ŷ)− yn) , if ŷ ∈ B and yn > g(ŷ)0 if ŷ /∈ B.

Then w∈W 1,p (Rn) and there exists a constant, C depending only on Lip(g) and dimensionsuch that

||w||W 1,p(Rn) ≤C ||u||W 1,p(V−) .

Denote w by E0u. Thus E0 (u)(y) = u(y) for all y ∈ V− but E0u = w is defined on all ofRn. Also, E0 is a linear mapping.

Proof: As in the previous lemma, w is Lipschitz continuous and has compact supportso it is clear w∈W 1,p (Rn) . The main task is to find w,i for ŷ∈ B and yn > g(ŷ) and then toextract an estimate of the right sort. Denote by U the set of points of Rn with the propertythat (ŷ,yn) ∈ U if and only if ŷ /∈ B or ŷ ∈ B and yn > g(ŷ) . Then letting φ ∈ C∞

c (U) ,suppose first that i < n. Then ∫

Uw(ŷ,yn)φ ,i (y)dy

≡ limh→0

∫U

φ (y)u(ŷ−hen−1

i ,2g(ŷ−hen−1

i

)− yn

)−u(ŷ,2g(ŷ)− yn)

hdy (37.2.28)

= limh→0

{−1h

∫U

φ (y)[D1u(ŷ,2g(ŷ)− yn)

(hen−1

i)

+2D2u(ŷ,2g(ŷ)− yn)(g(ŷ−hen−1

i)−g(ŷ)

)]dy

+−1h

∫U

φ (y)[o(g(ŷ−hen−1

i)−g(ŷ)

)+o(h)

]dy}

3This means that spt(u)⊆ B× (a,b) and u ∈C1(

V−).