1366 CHAPTER 41. ELLIPTIC REGULARITY

Suppose also that u ∈ Hk (Ω) and∫Ω

ai j (x)u,i (x)v, j (x)dx+∫

Ω

hk (x)v,k (x)dx =∫

Ω

f (x)v(x)dx

for all v ∈ Hk (Ω) . Then u ∈ Hk+1 (Ω) and for some C independent of f ,g, and u,

||u||2Hk+1(Ω) ≤C(|| f ||2Hk−1(Ω)+ ||u||

2Hk(Ω)+∑

s||hs||2Hk(Ω)

).

Proof: Let the Wi for i = 1, · · · , l be as described in Definition 41.2.1. Thus ∂Ω ⊆∪l

j=1Wj. Then let C1 ≡ ∂Ω\∪li=2Wi, a closed subset of W1. Let D1 be an open set satisfying

C1 ⊆ D1 ⊆ D1 ⊆W1.

Then D1,W2, · · · ,Wl cover ∂Ω. Let C2 = ∂Ω\(D1∪

(∪l

i=3Wi))

. Then C2 is a closed subsetof W2. Choose an open set, D2 such that

C2 ⊆ D2 ⊆ D2 ⊆W2.

Thus D1,D2,W3 · · · ,Wl covers ∂Ω. Continue in this way to get Di ⊆Wi, and ∂Ω⊆∪li=1Di,

and Di is an open set. Now letD0 ≡Ω\∪l

i=1Di.

Also, let Di ⊆ Vi ⊆ Vi ⊆Wi. Therefore, D0,V1, · · · ,Vl covers Ω. Then the same estimationprocess used above yields

||u||Hk+1(D0)≤C

(|| f ||2Hk−1(Ω)+ ||u||

2Hk(Ω)+∑

k||hk||2Hk(Ω)

).

From Lemma 41.2.5

||u||Hk+1(Vi∩Ω) ≤C

(|| f ||2Hk−1(Ω)+ ||u||

2Hk(Ω)+∑

k||hk||2Hk(Ω)

)

also. This proves the theorem since

||u||Hk+1(Ω) ≤l

∑i=1||u||Hk+1(Vi∩Ω)+ ||u||Hk+1(D0)

.