1322 CHAPTER 38. SOBOLEV SPACES BASED ON L2

Now using what was just shown, it follows that for r small enough, ||ur−u||Ht (Rn) < ε andthis proves the claim.

Now suppose v ∈ Ht (Rn) . By the claim,

||vr− v||Ht (Rn)→ 0

and so by continuity of γ,γvr→ γv in Ht−1/2 (Rn−1) . (38.5.21)

Note vr = 0 a.e. on

Ur ≡{

u ∈ Rn : u′ ∈U ′ and − r < un < φ(u′)− r}

Let φ ∈ C∞c (Ur) and consider φvr. Then it follows φvr = 0 a.e. on Rn. Let w ≡ 0.

Then w ∈S and so γw = 0 = γ (φvr) = γφγvr in Ht−1/2(Rn−1

). It follows that for mn−1

a.e. x′ ∈ [φ ̸= 0]∩Rn−1, γvr (x′) = 0. Now let U ′ = ∪∞k=1Kk where the Kk are compact sets

such that Kk ⊆ Kk+1 and let φ k ∈C∞c (U) such that φ k has values in [0,1] and φ k (x′) = 1 if

x′ ∈ Kk. Then from what was just shown, γvr = 0 for a.e. point of Kk. Therefore, γvr = 0for mn−1 a.e. point in U ′. Therefore, since each γvr = 0, it follows from 38.5.21 that γv = 0also. This proves the lemma.

Theorem 38.5.7 Let t > 1/2 and let U be of the form{u ∈ Rn : u′ ∈U ′ and 0 < un < φ

(u′)}

where U ′ is an open subset of Rn−1 and φ (u′) is a positive function such that φ (u′) ≤ ∞

andinf{

φ(u′)

: u′ ∈U ′}= δ > 0.

Then there exists a unique

γ ∈L(

Ht (U) ,Ht−1/2 (U ′))which has the property that if u= v|U where v is continuous and also a function of H1 (Rn) ,then γu(x′) = u(x′,0) for a.e. x′ ∈U ′.

Proof: Let u ∈ Ht (U) . Then u = v|U for some v ∈ Ht (Rn) . Define

γu≡ γv|U ′

Is this well defined? The answer is yes because if vi|U = u a.e., then γ (v1− v2) = 0 a.e. onU ′ which implies γv1 = γv2 a.e. and so the two different versions of γu differ only on a setof measure zero.

If u = v|U where v is continuous and also a function of H1 (Rn) , then for a.e. x′ ∈Rn−1,it follows from Lemma 38.5.5 on Page 1319 that γv(x′) = v(x′,0) . Hence, it follows thatfor a.e. x′ ∈U ′, γu(x′)≡ u(x′,0).

In particular, γ is determined by γu(x′) = u(x′,0) on S|U and the density of S|U andcontinuity of γ shows γ is unique.