40.1. A FUNDAMENTAL INEQUALITY 1343
+2∫
U−Diφ
(x,
32
g(x)− yn
2
)f (x,yn)dyndx. (40.1.3)
Now let
ψ1 (x,yn)≡ φ (x,2g(x)− yn) , ψ2 (x,yn)≡ φ
(x,
32
g(x)− yn
2
).
Then∂ψ1∂xi
= Diφ (x,2g(x)− yn)+Dnφ (x,2g(x)− yn)2Dig(x) ,
∂ψ2∂xi
= Diφ
(x,
32
g(x)− yn
2
)+Dnφ
(x,
32
g(x)− yn
2
)32
Dig(x) .
Also∂ψ1∂yn
(x,yn) =−Dnφ (x,2g(x)− yn) ,
∂ψ2∂yn
(x,yn) =
(−12
)Dnφ
(x,
32
g(x)− yn
2
).
Therefore,∂ψ1∂xi
(x,yn) = Diφ (x,2g(x)− yn)−2∂ψ1∂yn
(x,yn)Dig(x) ,
∂ψ2∂xi
(x,yn) = Diφ
(x,
32
g(x)− yn
2
)−3
∂ψ2∂yn
(x,yn)Dig(x) .
Using this in 40.1.3, the integrals in this expression equal
−3∫
U−
[∂ψ1∂xi
(x,yn)+2∂ψ1∂yn
(x,yn)Dig(x)]
f (x,yn)dyndx+
2∫
U−
[∂ψ2∂xi
(x,yn)+3∂ψ2∂yn
(x,yn)Dig(x)]
f (x,yn)dyndx
=∫
U−
[−3
∂ψ1 (x,y)∂xi
+2∂ψ2 (x,yn)
∂xi
]f (x,yn)dyndx.
Therefore, ∫Rn
∂φ
∂xif dx =
∫U−
[∂φ
∂xi−3
∂ψ1∂xi
+2∂ψ2∂xi
]f dxdyn
and alsoφ (x,g(x))−3ψ1 (x,g(x))+2ψ2 (x,g(x)) =
φ (x,g(x))−3φ (x,g(x))+2φ (x,g(x)) = 0
and so φ −3ψ1 +2ψ2 ∈ H10 (U
−) . It also follows from the definition of the functions, ψ iand the assumption that g is Lipschitz, that
||ψ i||1,2,U− ≤Cg ||φ ||1,2,U− ≤Cg ||φ ||1,2,Rn . (40.1.4)