1318 CHAPTER 38. SOBOLEV SPACES BASED ON L2

Proof: It only remains to verify that γ is onto and that the continuous map, ζ exists.Now define

φ (y)≡ φ(y′,yn

)≡

(1+ |y′|2

)t−1/2

(1+ |y|2

)t .

Then for u ∈S′, letζ u(x)≡CF−1 (φFu)(x) =

C∫Rn

eiy·x

(1+ |y′|2

)t−1/2

(1+ |y|2

)t Fu(y′)

dy (38.5.19)

Here the inside Fourier transform is taken with respect to Rn−1 because u is only definedon Rn−1 and C will be chosen in such a way that γ ◦ ζ = id. First the existence of C suchthat γ ◦ζ = id will be shown. Since u ∈S′ it follows

y→

(1+ |y′|2

)t−1/2

(1+ |y|2

)t Fu(y′)

is in S. Hence the inverse Fourier transform of this function is also in S and so for u ∈S′,it follows ζ u ∈S. Therefore, to check γ ◦ζ = id it suffices to plug in xn = 0. From Lemma38.5.2 this yields

γ (ζ u)(x′,0

)= C

∫Rn

eiy′·x′

(1+ |y′|2

)t−1/2

(1+ |y|2

)t Fu(y′)

dy

= C∫Rn−1

(1+∣∣y′∣∣2)t−1/2

eiy′·x′Fu(y′)∫R

1(1+ |y|2

)t dyndy′

= CCt

∫Rn−1

(1+∣∣y′∣∣2)t−1/2

eiy′·x′Fu(y′)(

1+∣∣y′∣∣2)−2t+1

2dy′

= CCt

∫Rn−1

eiy′·x′Fu(y′)

dy′ =CCt (2π)n/2 F−1 (Fu)(x′)

and so the correct value of C is(

Ct (2π)n/2)−1

to obtain γ ◦ ζ = id. It only remains to

verify that ζ is continuous. From 38.5.19, and Lemma 38.5.2,

||ζ u||2Ht (Rn)

=∫Rn

(1+ |x|2

)t|Fζ u(x)|2 dx

= C2∫Rn

(1+ |x|2

)t ∣∣F (F−1 (φFu)(x))∣∣2 dx