38.2. FRACTIONAL ORDER SPACES 1305

Next take u ∈W m,2 (U) . Then Eu ∈W m,2 (Rn) = Hm (Rn) and this shows u ∈Hm (U) .This has shown the two spaces are the same. It remains to verify their norms are equivalent.Let u ∈ Hm (U) and let v|U = u where v ∈ Hm (Rn) and

||u||Hm(U)+ ε > ||v||Hm(Rn) .

Then recalling that ||·||Hm(Rn) and ||·||m,2,Rn are equivalent norms for Hm (Rn) , there existsa constant, C such that

||u||Hm(U)+ ε > ||v||Hm(Rn) ≥C ||v||m,2,Rn ≥C ||u||m,2,U

Now consider the two Banach spaces,(Hm (U) , ||·||Hm(U)

),(

W m,2 (U) , ||·||m,2,U

).

The above inequality shows since ε > 0 is arbitrary that

id :(

Hm (U) , ||·||Hm(U)

)→(

W m,2 (U) , ||·||m,2,U

)is continuous. By the open mapping theorem, it follows id is continuous in the other di-rection. Thus there exists a constant, K such that ||u||Hm(U) ≤ K ||u||k,2,U . Hence the twonorms are equivalent as claimed.

Specializing Corollary 37.3.3 and Theorem 37.3.6 starting on Page 1295 to the case ofp = 2 while also assuming more on U yields the following embedding theorems.

Theorem 38.1.9 Suppose m≥ 0 and j is a nonnegative integer satisfying 2 j < n. Also letU bean open set which satisfies Assumption 38.1.2. Then id ∈ L

(Hm+ j (U) ,W m,q (U)

)where

q≡ 2nn−2 j

. (38.1.6)

If, in addition to the above, U is bounded and 1≤ r < q, then

id ∈L(Hm+ j (U) ,W m,r (U)

)and is compact.

Theorem 38.1.10 Suppose for j a nonnegative integer, 2 j < n < 2( j+1) and let m be apositive integer. Let U be any bounded open set in Rn which satisfies Assumption 38.1.2.Then id ∈L

(Hm+ j (U) ,Cm−1,λ

(U))

for every λ ≤ λ 0 ≡ ( j+1)− n2 and if λ < ( j+1)−

n2 , id is compact.

38.2 Fractional Order SpacesWhat has been gained by all this? The main thing is that Hm+s (U) makes sense for anys ∈ (0,1) and m an integer. You simply replace m with m+ s in the above for s ∈ (0,1).This gives what is meant by Hm+s (Rn)