1298 CHAPTER 37. BASIC THEORY OF SOBOLEV SPACES

Proof: Define

Eu(x′,xn

)≡{

u(x′,xn) if xn < 0∑

kj=1 λ ju(x′,− jxn) if xn ≥ 0

where the λ j are chosen in such a way that for l ≤ k−1,

Dlenu(x′,0

)−Dlen

(k

∑j=1

λ ju

)(x′,0

)= 0

so that Lemma 37.4.1 may be applied. Do there exist such λ j? It is necessary to have thefollowing hold for each r = 0,1, · · · ,k−1.

k

∑j=1

(− j)rλ jDrenu

(x′,0

)= Drenu

(x′,0

).

This is satisfied ifk

∑j=1

(− j)rλ j = 1

for r = 0,1, · · · ,k−1. This is a system of k equations for the k variables, the λ j. The matrixof coefficients is of the form

1 1 1 · · · 1−1 −2 −3 · · · −k1 4 9 · · · k2

......

......

(−1)k (−2)k (−3)k · · · (−k)k

This matrix has an inverse because its determinant is nonzero.

Now from Lemma 37.4.1, it follows from the above description of E that for |α| ≤ k,

Dα (Eu)(x′,xn

)≡{

Dα u(x′,xn) if xn < 0∑

kj=1 λ j (− j)αn (Dα u)(x′,− jxn) if xn ≥ 0

It follows that E is linear and there exists a constant, C independent of u such that 37.4.36holds. This proves the lemma.

Corollary 37.4.3 Let H− be the half space of 37.4.35. There exists E with the propertythat E : W l,p (H−)→W l,p (Rn) and is linear and continuous for each l ≤ k.

Proof: This immediate from the density of C∞c

(H−)

in W k,p(

H−)

and Lemma 37.4.2.There is nothing sacred about a half space or this particular half space. It is clear that

everything works as well for a half space of the form

H−k ≡ {x : xk < 0} .

Thus the half space featured in the above discussion is H−n .

1298 CHAPTER 37. BASIC THEORY OF SOBOLEV SPACESProof: Define1 f u(x xp) ifn <0Eu(x',xn) = { A A ju(x!, jin) if.xn > 0where the A ; are chosen in such a way that for / < k—1,Du (x’,0) —pD' (Jk2) (x’,0) =0=1so that Lemma 37.4.1 may be applied. Do there exist such A ;? It is necessary to have thefollowing hold for each r=0,1,---,k-—1.k(—j)'A,D"*"u (x’,0) = Du (x’,0).j=1JThis is satisfied ifkY (AA; =1j=lfor r=0,1,--- ,k—1. This is a system of k equations for the k variables, the A ;. The matrixof coefficients is of the formWk yk ak yk(—1)" (-2)" (-3)" +++ (Fk)This matrix has an inverse because its determinant is nonzero.Now from Lemma 37.4.1, it follows from the above description of E that for |a| < k,Du (x! Xx, ) if x, <0a / = um ”D4 (Eu) (x.%9) = { EE Ay (—1)% (Du) (x!,— jn) if.xy > 0It follows that E is linear and there exists a constant, C independent of u such that 37.4.36holds. This proves the lemma.Corollary 37.4.3 Let H~ be the half space of 37.4.35. There exists E with the propertythat E : W''? (H~) — W"? (IR") and is linear and continuous for each | < k.Proof: This immediate from the density of C> (7) in WP (7) and Lemma 37.4.2.There is nothing sacred about a half space or this particular half space. It is clear thateverything works as well for a half space of the formH, = {x:x% <0}.Thus the half space featured in the above discussion is H,, .