Kenneth Kuttler

37.2. AN EXTENSION THEOREM 1291

≤ 4p (1+Lip(g)p)∫

∣∣∣∣ ∂u∂yi

(ŷ,2g(ŷ)− yn)

∣∣∣∣p+

∣∣∣∣ ∂u∂yn

(ŷ,2g(ŷ)− yn)

∣∣∣∣p dy

= 4p (1+Lip(g)p)∫

∫∞

g(ŷ)

∣∣∣∣ ∂u∂yi

(ŷ,2g(ŷ)− yn)

∣∣∣∣p+

∣∣∣∣ ∂u∂yn

(ŷ,2g(ŷ)− yn)

∣∣∣∣p dyndŷ

= 4p (1+Lip(g)p)∫

∫ g(ŷ)

−∞

∣∣∣∣ ∂u∂yi

(ŷ,zn)

∣∣∣∣p + ∣∣∣∣ ∂u∂yn

(ŷ,zn)

∣∣∣∣p dzndŷ

= 4p (1+Lip(g)p)∫

∫ g(ŷ)

∣∣∣∣ ∂u∂yi

(ŷ,zn)

∣∣∣∣p+

∣∣∣∣ ∂u∂yn

(ŷ,zn)

∣∣∣∣p dzndŷ≤ 4p (1+Lip(g)p) ||u||p1,p,V−

Now by similar reasoning,

||w,n||pLp(U)=

∫U

∣∣∣∣−∂u∂yn

(ŷ,2g(ŷ)− yn)

∣∣∣∣p dy

=∫

∫∞

g(ŷ)

∣∣∣∣−∂u∂yn

(ŷ,2g(ŷ)− yn)

∣∣∣∣p dyndŷ

=∫

∫ g(ŷ)

∣∣∣∣−∂u∂yn

(ŷ,zn)

∣∣∣∣p dzndŷ = ||u,n||p1,p,V− .

It follows

||w||p1,p,Rn = ||w||p1,p,U + ||u||p1,p,V−≤ 4pn(1+Lip(g)p) ||u||p1,p,V− + ||u||

p1,p,V−

and so||w||p1,p,Rn ≤ 4pn(2+Lip(g)p) ||u||p1,p,V−

which implies||w||1,p,Rn ≤ 4n1/p (2+Lip(g)p)

1/p ||u||1,p,V−It is obvious that E0 is a continuous linear mapping. This proves the lemma.

Now recall Definition 37.2.1, listed here for convenience.

Definition 37.2.5 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), (37.2.31)

37.2. AN EXTENSION THEOREM 1291P<4” (1-+Lip(e)”) [ |S" @.28@) yn)Ppdydu.ae-wcsure” [ [° [2+o G.2e@) ym)= 4? (1 +Lip(g mf fe= 4? (1+Lip(g opp(Y, Zn)ou~ ~(¥,28( y)- vol"dyadsP Ou P _=— (Y,zn) a dindyP5>— (Y, zn)OuIYnNow by similar reasoning,+* dend <4? (1+Lip (g yell? pv-Pp—duIallee = 5, 624)- vn) dyoe"dnday “ (¥,28 (¥) —yn)(y)=f LTP Gen) aendd= Ital pyIt followslee = Wl pe HII y-< 4?n(1+Lip(g)’) |lull? .y- + lM pvand so[lvl lt pen <4? (2+ Lip(g)”) [lly v-which implies. 1lIlipen <4n'/? (2+Lip(g)’)'’? |lullyv-It is obvious that Ep is a continuous linear mapping. This proves the lemma.Now recall Definition 37.2.1, listed here for convenience.Definition 37.2.5 An open subset, U, of R" has a Lipschitz boundary if it satisfies thefollowing conditions. For each p € OU =U \U, there exists an open set, Q, containing p,an open interval (a,b), a bounded open box B C R""!, and an orthogonal transformationR such thatRQ =Bx (a,b), (37.2.31)