Theorem (Rellich-Kondrachov). Let be an open, bounded Lipschitz domain, and let . Set
Then the Sobolev space is continuously embedded in the space for every and is compactly embedded in for every . In symbols,
It is worth noticing from the theory of Sobolev spaces that
Therefore, we have the following extension
Theorem (Extension of Rellich-Kondrachov). Let be an open, bounded Lipschitz domain, and let . Set
Then we have
Proof. We first place here the proof of the compactness. In fact, its proof comes from the Rellich-Kondrachov theorem (the case ). Indeed, assume and a bounded sequence in . Our aim is to prove is precompact (its closure is compact or there exists a convergent subsequence) in .
Clearly from the definition of one has
for every and multi-index satisfying . In other words, a bounded sequence in for each multi-index satisfying .
By the Rellich-Kondrachov theorem, there exists a convergent subsequence in (still denoted by ), that means
converges in .
Since there are finite number of multi-indexes satisfying , by finite induction it is possible to select a subsequence for which
converges in for any fixed .
converges in .
The proof of continuity part is almost the same. What we need here is to derive an estimate to ensure that there is a constant such that
Again, by the definition of norms in Sobolev spaces, the Rellich-Kondrachov theorem and a very fundamental inequality, one has
The proof now follows.
Corollary. We always have the following compact embedding