41.1. THE CASE OF A HALF SPACE 1357

+ ∑s

∣∣∣∣∣∣∣∣∣∣∑τ<α

C (τ)Dα−τ (αrs)∂ (Dτ w)

∂yr +Dα (hs)

∣∣∣∣∣∣∣∣∣∣H1(Û1)

and by 41.1.28, this implies

||Dα w||H2(U1)≤C

(|| f ||Hk−1(U)+ ||w||Hk(U)+∑

s||hs||Hk(U)

)which proves the Claim.

To establish 41.1.24 it only remains to verify that if |α| ≤ k+1, then

||Dα w||L2(U1)≤C

(|| f ||Hk−1(U)+ ||w||Hk(U)+∑

s||hs||Hk(U)

). (41.1.29)

If |α| < k + 1, there is nothing to show because it is given that w ∈ Hk (U) . Therefore,assume |α| = k+ 1. If αn equals 0 the conclusion follows from the claim because in thiscase, you can subtract 1 from a pair of positive α i and obtain a new multi index, β suchthat |β |= k−1 and β n = 0 and then from the claim,

||Dα w||L2(U1)≤∣∣∣∣∣∣Dβ w

∣∣∣∣∣∣H2(U1)

≤C(|| f ||Hk−1(U)+ ||w||Hk(U)+∑

s||hs||Hk(U)

).

If αn = 1, then subtract 1 from some positive α i and consider

β = (α1, · · · ,α i−1,α i+1, · · · ,αn−1,0)

Then from the claim,

||Dα w||L2(U1)≤∣∣∣∣∣∣Dβ w

∣∣∣∣∣∣H2(U1)

≤C(|| f ||Hk−1(U)+ ||w||Hk(U)+∑

s||hs||Hk(U)

).

Suppose 41.1.29 holds for αn ≤ j−1 where j−1≥ 1 and consider α for which |α|= k+1and αn = j. Let

β ≡ (α1, · · · ,αn−1,αn−2) .

Thus Dα = Dβ D2n. Restricting 41.1.23 to z ∈ C∞

c (U1) and using the density of this set offunctions in L2 (U1) , it follows that

− ∂

∂ys

(α

rs (y)∂w∂yr

)− ∂hs

∂ys = f .

Therefore, from the product rule,

∂αrs

∂ys∂w∂yr +α

rs ∂ 2w∂ys∂yr +

∂hs

∂ys =− f

and so

αnnD2

nw = −

(∂αrs

∂ys∂w∂yr + ∑

r≤n−1∑

s≤n−1α

rs ∂ 2w∂ys∂yr +

∑s

αns ∂ 2w

∂ys∂yn +∑r

αrn ∂ 2w

∂yn∂yr +∂hs

∂ys + f).