38.5. THE TRACE ON THE BOUNDARY OF A HALF SPACE 1321

xn replaced with (xn)k and φ in place of φ j. Now taking k→ ∞, this uniform continuityimplies

(γu,φ)H = (u(·,0) ,φ)H

This implies since Cc(Rn−1

)is dense in L2

(Rn−1

)that γu = u(·,0) a.e. and this proves

the lemma.

Lemma 38.5.6 Suppose U is an open subset of Rnof the form

U ≡{

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

Suppose v ∈ Ht (Rn) such that v = 0 a.e. on U. Then γv = 0 mn−1 a.e. point of U ′. Also, ifv ∈ Ht (Rn) and φ ∈C∞

c (Rn) , then γvγφ = γ (φv) .

Proof: First consider the second claim. Let v∈Ht (Rn) and let vk→ v in Ht (Rn) wherevk ∈S. Then from Lemma 38.3.4 and Theorem 38.5.4

||γ (φv)− γφγv||Ht−1/2(Rn−1) = limk→∞

||γ (φvk)− γφγvk||Ht−1/2(Rn−1) = 0

because each term in the sequence equals zero due to the observation that for vk ∈S andφ ∈C∞

c (U) , γ (φvk) = γvkγφ .Now suppose v = 0 a.e. on U . Define for 0 < r < δ , vr (x)≡ v(x′,xn + r) .Claim: If u ∈ Ht (Rn) , then

limr→0||vr− v||Ht (Rn) = 0.

Proof of claim: First of all, let v ∈S. Then v ∈ Hm (Rn) for all m and so by Lemma38.2.5,

||vr− v||Ht (Rn) ≤ ||vr− v||θHm(Rn) ||vr− v||1−θ

Hm+1(Rn)

where t ∈ [m,m+1] . It follows from continuity of translation in Lp (Rn) that

limr→0||vr− v||θHm(Rn) ||vr− v||1−θ

Hm+1(Rn)= 0

and so the claim is proved if v ∈S. Now suppose u ∈ Ht (Rn) is arbitrary. By density ofS in Ht (Rn) , there exists v ∈S such that

||u− v||Ht (Rn) < ε/3.

Therefore,

||ur−u||Ht (Rn) ≤ ||ur− vr||Ht (Rn)+ ||vr− v||Ht (Rn)+ ||v−u||Ht (Rn)

= 2ε/3+ ||vr− v||Ht (Rn) .