37.3. GENERAL EMBEDDING THEOREMS 1295

Corollary 37.3.3 Suppose m ≥ 1 and j is a nonnegative integer satisfying jp < n. Alsosuppose U has a (1, p) extension operator for all p≥ 1 and satisfies the segment condition.Then

id ∈L(W m+ j,p (U) ,W m,q (U)

)where

q≡ npn− jp

. (37.3.34)

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

id ∈L(W m+ j,p (U) ,W m,r (U)

)and is compact.

Proof: If |α| ≤ m, then Dα u ∈W j,p (U) and so by Corollary 37.3.2, Dα u ∈ Lq (U)where q is given above. Since U has the segment property, this means u ∈W m,q (U). Itremains to verify the assertion about compactness of id.

Let S be bounded in W m+ j,p (U) . Then S is bounded in W m,q (U) by the first part. Nowlet {uk}∞

k=1 be any sequence in S. The corollary will be proved if it is shown that anysuch sequence has a convergent subsequence in W m,r (U). Let {α1,α2, · · · ,αh} denote theindices satisfying |α| ≤ m. Then for each of these indices, α,

supu∈S

{||Dα u||1,1,U + ||Dα u||Lq(U)

}< ∞

and so for each such α, satisfying |α| ≤ m, it follows from Lemma 37.1.16 on Page 1284that {Dα u : u ∈ S} is precompact in Lr (U) . Therefore, there exists a subsequence, still de-noted by uk such that Dα1uk converges in Lr (U) . Applying the same lemma, there exists asubsequence of this subsequence such that both Dα1 uk and Dα2uk converge in Lr (U) . Con-tinue taking subsequences until you obtain a subsequence, {uk}∞

k=1 for which {Dα uk}∞

k=1converges in Lr (U) for all |α| ≤m. But this must be a convergent subsequence in W m,r (U)and this proves the corollary.

Theorem 37.3.4 Let U be a bounded open set having a (1, p) extension operator and letp > n. Then id : W 1,p (U)→C

(U)

is continuous and compact.

Proof: Theorem 37.1.3 on Page 37.1.3 implies rU : W 1,p (Rn)→C(U)

is continuousand compact. Thus

||u||∞,U = ||Eu||

∞,U ≤C ||Eu||1,p,Rn ≤C ||u||1,p,U .

This proves continuity. If S is a bounded set in W 1,p (U) , then define S1 ≡ {Eu : u ∈ S} .Then S1 is a bounded set in W 1,p (Rn) and so by Theorem 37.1.3 the set of restrictions toU, is precompact. However, the restrictions to U are just the functions of S. Therefore, idis compact as well as continuous.

Corollary 37.3.5 Let p> n, let U be a bounded open set having a (1, p) extension operatorwhich also satisfies the segment condition, and let m be a nonnegative integer. Then id :W m+1,p (U)→Cm,λ

(U)

is continuous for all λ ∈ [0,1− np ] and id is compact if λ < 1− n

p .