1324 CHAPTER 38. SOBOLEV SPACES BASED ON L2

Are there functions which are in Hm+s (Γ)? The answer is yes. Just take the restrictionto Γ of any function, u ∈C∞

c (Rm) . Then each h∗i (uψ i) ∈ Hm+s (Ui) and the sum is finitebecause sptu has nonempty intersection with only finitely many Wi.

It is not at all obvious this norm is well defined. What if {W ′i ,ψ ′i,Γ′i,Ui,h′i,g′i}∞

i=1 isas described above? Would the two norms be equivalent? If they aren’t, then this is not agood way to define Hm+s (Γ) because it would depend on the choice of partition of unityand functions, hi and choice of the open sets, Ui. To begin with pick a particular choice for{Wi,ψ i,Γi,Ui,hi,gi}∞

i=1 .

Lemma 38.6.4 Hm+s (Γ) as just described, is a Banach space.

Proof: Let{

u j}∞

j=1 be a Cauchy sequence in Hm+s (Γ) . Then{

h∗i (u jψ i)}∞

j=1 is aCauchy sequence in Hm+s (Ui) for each i. Therefore, for each i, there exists wi ∈Hm+s (Ui)such that

limj→∞

h∗i (u jψ i) = wi in Hm+s (Ui) . (38.6.22)

It is required to show there exists u ∈ Hm+s (Γ) such that wi = h∗i (uψ i) for each i.Now from Corollary 36.2.5 it follows easily by approximating with simple functions

that for ever nonnegative µ measurable function, f ,∫Γ

f dµ =∞

∑r=1

∫grΓr

ψr f (hr (u))Jr (u)du.

Therefore, ∫Γ

∣∣u j−uk∣∣2 dµ =

∞

∑r=1

∫grΓr

ψr

∣∣u j−uk∣∣2 (hr (u))Jr (u)du

≤ C∞

∑r=1

∫grΓr

ψr

∣∣u j−uk∣∣2 (hr (u))du

= C∞

∑r=1

∣∣∣∣h∗r (ψr

∣∣u j−uk∣∣)∣∣∣∣2

0,2,Ur

≤ C∣∣∣∣u j−uk

∣∣∣∣Hm+s(Γ)

and it follows there exists u ∈ L2 (Γ) such that∣∣∣∣u j−u∣∣∣∣

0,2,Γ→ 0.

and a subsequence, still denoted by u j such that u j (x)→ u(x) for µ a.e. x∈Γ. It is requiredto show that u ∈ Hm+s (Γ) such that wi = h∗i (uψ i) for each i. First of all, u is measurablebecause it is the limit of measurable functions. The pointwise convergence just establishedand the fact that sets of measure zero on Γi correspond to sets of measure zero on Ui whichwas discussed in the claim found in the proof of Theorem 36.2.4 on Page 1259 shows that

h∗i (u jψ i)(x)→ h∗i (uψ i)(x)