38.5. THE TRACE ON THE BOUNDARY OF A HALF SPACE 1317
and s > 0, ∫Rn−1
(1+∣∣y′∣∣2)s ∣∣Fγu
(y′)∣∣2 dy′
= Cn
∫Rn−1
(1+∣∣y′∣∣2)s
∣∣∣∣∫RFu(y′,yn
)dyn
∣∣∣∣2 dy′
=Cn
∫Rn−1
(1+∣∣y′∣∣2)s
∣∣∣∣∫RFu(y′,yn
)(1+ |y|2
)t/2(1+ |y|2
)−t/2dyn
∣∣∣∣2 dy′
Then by the Cauchy Schwarz inequality,
≤Cn
∫Rn−1
(1+∣∣y′∣∣2)s ∫
R
∣∣Fu(y′,yn
)∣∣2(1+ |y|2)t
dyn
∫R
(1+ |y|2
)−tdyndy′.
(38.5.17)Consider ∫
R
(1+ |y|2
)−tdyn =
∫R
(1+∣∣y′∣∣2 + y2
n
)−tdyn
by Lemma 38.5.2 and taking a =(
1+ |y′|2)1/2
, this equals
Ct
((1+∣∣y′∣∣2)1/2
)−2t+1
=Ct
(1+∣∣y′∣∣2)(−2t+1)/2
.
Now using this in 38.5.17,∫Rn−1
(1+∣∣y′∣∣2)s ∣∣Fγu
(y′)∣∣2 dy′
≤ Cn,t
∫Rn−1
(1+∣∣y′∣∣2)s ∫
R
∣∣Fu(y′,yn
)∣∣2(1+ |y|2)t
dyn ·(1+∣∣y′∣∣2)(−2t+1)/2
dy′
= Cn,t
∫Rn−1
(1+∣∣y′∣∣2)s+(−2t+1)/2 ∫
R
∣∣Fu(y′,yn
)∣∣2(1+ |y|2)t
dyndy′.
What is the correct choice of t so that the above reduces to ∥u∥2Ht (Rn)? It is clearly the one
for whichs+(−2t +1)/2 = 0
which occurs when t = s+ 12 . Then for this choice of t, the following inequality is obtained
for any u ∈S.∥γu∥Ht−1/2(Rn−1) ≤Cn,t ∥u∥Ht (Rn) . (38.5.18)
This has proved part of the following theorem.
Theorem 38.5.4 For each t > 1/2 there exists a unique mapping
γ ∈L(
Ht (Rn) ,Ht−1/2 (Rn−1))which has the property that for u ∈S, γu(x′) = u(x′,0) . In addition to this, γ is onto. Infact, there exists a continuous map, ζ ∈L
(Ht−1/2
(Rn−1
),Ht (Rn)
)such that γ ◦ζ = id.