38.1. FOURIER TRANSFORM TECHNIQUES 1303

Now let ψε be a mollifier and pick εk small enough that∣∣∣∣∣∣uk ∗ψεk−uk

∣∣∣∣∣∣L2(µ)

<12k .

Then uk ∗ψεk∈C∞

c (Rn)⊆S. Therefore, there exists wk ∈ G such that Fwk = uk ∗ψεk. It

follows||Fwk−Fu||L2(µ) ≤ ||Fwk−uk||L2(µ)+ ||uk−Fu||L2(µ)

and these last two terms converge to 0 as k→ ∞. Therefore, wk → u in Hm (Rn) and thisproves the first part of this lemma.

Now let u ∈ Hm (Rn) . By what was just shown, there exists a sequence, uk → u inHm (Rn) where uk ∈S. It follows from 38.1.2 that

||uk−ul ||Hm ≥ ||uk−ul ||m,2,Rn

and so {uk} is a Cauchy sequence in W m,2 (Rn) . Therefore, there exists w∈W m,2 (Rn) suchthat

||uk−w||m,2,Rn → 0.

But this implies0 = lim

k→∞

||uk−w||0,2,Rn = limk→∞

||uk−u||0,2,Rn

showing u = w which verifies Hm (Rn)⊆W 2,m (Rn) . The opposite inclusion is proved thesame way, using density of S and the fact that the norms in both spaces are larger thanthe norms in L2 (Rn). The equivalence of the norms follows from the density of S and theequivalence of the norms on S. This proves the lemma.

The conclusion of this lemma with the density of S and 38.1.2 implies you can useeither norm, ||u||Hm(Rn) or ||u||m,2,Rn when working with these Sobolev spaces.

What of open sets satisfying Assumption 38.1.2? How does W m,2 (U) relate to theFourier transform?

Definition 38.1.6 Let U be an open set in Rn. Then

Hm (U)≡ {u : u = v|U for some v ∈ Hm (Rn)} (38.1.4)

Here the notation, v|U means v restricted to U. Define the norm in this space by

||u||Hm(U) ≡ inf{||v||Hm(Rn) : v|U = u

}. (38.1.5)

Lemma 38.1.7 Hm (U) is a Banach space.

Proof: First it is necessary to verify that the given norm really is a norm. Suppose thenthat u = 0. Is ||u||Hm(U) = 0? Of course it is. Just take v≡ 0. Then v|U = u and ||v||Hm = 0.Next suppose ||u||Hm(U) = 0. Does it follow that u = 0? Letting ε > 0 be given, there existsv ∈ Hm (Rn) such that v|U = u and ||v||Hm(Rn) < ε. Therefore,

||u||0,U ≤ ||v||0,Rn ≤ ||v||Hm(U) < ε.

38.1. FOURIER TRANSFORM TECHNIQUES 1303Now let y, be a mollifier and pick €, small enough that1Uk * We, — Uk 22(n) < 5kThen uz * We, © CP (R") CG. Therefore, there exists wz, € Y such that Fw, = uz * We,- Itfollows||Fwe — Ful|p2(yy < \|Fwe — uel |r) + |e — Full p2(u)and these last two terms converge to 0 as k > ce. Therefore, w, > u in H” (R") and thisproves the first part of this lemma.Now let u € H”(R”"). By what was just shown, there exists a sequence, uz — u inH™ (R") where uz € G. It follows from 38.1.2 that|e — ui | ppm 2 |[Ue — Ui ||n.2 90and so {uz} is a Cauchy sequence in W”? (R"). Therefore, there exists w € W’"? (IR”) suchthat|| — w| [2,8 > 0.But this implies0= jim ||ux —Wlo2.R" = dim ||uu —Ulo2.R0showing u = w which verifies H” (IR") C W?”" (IR"). The opposite inclusion is proved thesame way, using density of G and the fact that the norms in both spaces are larger thanthe norms in L? (IR”). The equivalence of the norms follows from the density of G and theequivalence of the norms on G. This proves the lemma.The conclusion of this lemma with the density of G and 38.1.2 implies you can useeither norm, ||i||;mcgn) OF ||u4||,,,227 When working with these Sobolev spaces.What of open sets satisfying Assumption 38.1.2? How does W”? (U) relate to theFourier transform?Definition 38.1.6 Let U be an open set in R". ThenH"” (U) = {u: u=vly for some v € H™ (R")} (38.1.4)Here the notation, v|y means v restricted to U. Define the norm in this space byllellam) = inf { [loll jam a :vly = ub. (38.1.5)Lemma 38.1.7 H” (U) is a Banach space.Proof: First it is necessary to verify that the given norm really is a norm. Suppose thenthat u = 0. Is ||u|| ym) = 0? Of course it is. Just take v= 0. Then vly = wand ||v||;7m = 0.Next suppose ||u||;m(z7) = 0. Does it follow that u = 0? Letting € > 0 be given, there existsv € H™(R") such that vly = wu and ||v||;7m¢gn) < €. Therefore,ll4Hlo.0 <[Mllowe S Ivlliemuy < €: