38.5. THE TRACE ON THE BOUNDARY OF A HALF SPACE 1319

= C2∫Rn

(1+ |x|2

)t ∣∣φ (x)Fu(x′)∣∣2 dx

= C2∫Rn

(1+ |x|2

)t

∣∣∣∣∣∣∣(

1+ |x′|2)t−1/2

(1+ |x|2

)t Fu(x′)∣∣∣∣∣∣∣

2

dx

= C2∫Rn

(1+ |x|2

)−t∣∣∣∣(1+

∣∣x′∣∣2)t−1/2Fu(x′)∣∣∣∣2 dx

= C2∫Rn−1

(1+∣∣x′∣∣2)2t−1 ∣∣Fu

(x′)∣∣2 ∫

R

(1+ |x|2

)−tdxndx′

= C2Ct

∫Rn−1

(1+∣∣x′∣∣2)2t−1 ∣∣Fu

(x′)∣∣2(1+

∣∣y′∣∣2)−2t+12

dx′

= C2Ct

∫Rn−1

(1+∣∣x′∣∣2)t−1/2 ∣∣Fu

(x′)∣∣2 dx′ =C2Ct ||u||2Ht−1/2(Rn−1) .

This proves the theorem because S is dense in Rn.

Actually, the assertion that γu(x′) = u(x′,0) holds for more functions, u than just u ∈S. I will make no effort to obtain the most general description of such functions but thefollowing is a useful lemma which will be needed when the trace on the boundary of anopen set is considered.

Lemma 38.5.5 Suppose u is continuous and u ∈ H1 (Rn) . Then there exists a set of m1measure zero, N such that if xn /∈ N, then for every φ ∈ L2

(Rn−1

)(γu,φ)H +

∫ xn

0(u,n (·, t) ,φ)H dt = (u(·,xn) ,φ)H

where here

( f ,g)H ≡∫Rn−1

f gdx′,

just the inner product in L2(Rn−1

). Furthermore,

u(·,0) = γu a.e. x′.

Proof: Let {uk} be a sequence of functions from S which converges to u in H1 (Rn)and let {φ k} denote a countable dense subset of L2

(Rn−1

). Then

(γuk,φ j

)H+∫ xn

0

(uk,n (·, t) ,φ j

)H

dt =(

uk (·,xn) ,φ j

)H. (38.5.20)