41.1. THE CASE OF A HALF SPACE 1351

=∫

Uα

rs (y+hek)∂(Dh

kw)

∂yr

∂(η2Dh

kw)

∂ys dy+

1h

∫U(αrs (y+hek)−α

rs (y))∂w∂yr

∂(η2Dh

kw)

∂ys dy (41.1.11)

Now∂(η2Dh

kw)

∂ys = 2η∂η

∂ys Dhkw+η

2 ∂(Dh

kw)

∂ys . (41.1.12)

therefore,

=∫

Uη

2α

rs (y+hek)∂(Dh

kw)

∂yr

∂(Dh

kw)

∂ys dy

+

{∫W∩U

αrs (y+hek)

∂(Dh

kw)

∂yr 2η∂η

∂ys Dhkwdy

+1h

∫W∩U

(αrs (y+hek)−αrs (y))

∂w∂yr

∂(η2Dh

kw)

∂ys dy

}≡ A.+{B.} . (41.1.13)

Now consider these two terms. From 41.1.2,

A.≥ δ

∫U

η2∣∣∣∇Dh

kw∣∣∣2 dy. (41.1.14)

Using the Lipschitz continuity of αrs and 41.1.12,

B.≤C (η ,Lip(α) ,α)

{∣∣∣∣∣∣Dhkw∣∣∣∣∣∣

L2(W∩U)

∣∣∣∣∣∣η∇Dhkw∣∣∣∣∣∣

L2(W∩U ;Rn)+

||η∇w||L2(W∩U ;Rn)

∣∣∣∣∣∣η∇Dhkw∣∣∣∣∣∣

L2(W∩U ;Rn)

+ ||η∇w||L2(W∩U ;Rn)

∣∣∣∣∣∣Dhkw∣∣∣∣∣∣

L2(W∩U)

}. (41.1.15)

≤C (η ,Lip(α) ,α)Cε

(∣∣∣∣∣∣Dhkw∣∣∣∣∣∣2

L2(W∩U)+ ||η∇w||2L2(W∩U ;Rn)

)+

εC (η ,Lip(α) ,α)

(∣∣∣∣∣∣η∇Dhkw∣∣∣∣∣∣2

L2(W∩U ;Rn)+∣∣∣∣∣∣Dh

kw∣∣∣∣∣∣2

L2(W∩U)

). (41.1.16)

Now ∣∣∣∣∣∣Dhkw∣∣∣∣∣∣

L2(W )≤ ||∇w||2L2(U ;Rn) . (41.1.17)

To see this, observe that if w is smooth, then(∫W

∣∣∣∣w(y+hek)−w(y)h

∣∣∣∣2 dy

)1/2

≤

(∫W

∣∣∣∣1h∫ h

0∇w(y+ tek) · ekdt

∣∣∣∣2 dy

)1/2