Ngô Quốc Anh

May 5, 2010

An extension of the Rellich-Kondrachov theorem

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 14:00

In mathematics, the Rellich-Kondrachov theorem is a compact embedding theorem concerning Sobolev spaces. It is named after the Italian-Austrian mathematician Franz Rellich.

Theorem (Rellich-Kondrachov). Let \Omega \subset \mathbb R^n be an open, bounded Lipschitz domain, and let 1 \leqslant p \leqslant kn. Set

\displaystyle p^\star := \frac{np}{n - kp}.

Then the Sobolev space W^{k,p}(\Omega) is continuously embedded in the L^q(\Omega) space for every 1 \leqslant q \leqslant p^\star and is compactly embedded in L^q(\Omega) for every 1 \leqslant q < p^\star. In symbols,

\displaystyle W^{k, p} (\Omega) \hookrightarrow L^{p^\star} (\Omega)


\displaystyle W^{k, p} (\Omega) \subset \subset L^{q} (\Omega) for 1 \leqslant q < p^\star.

It is worth noticing from the theory of Sobolev spaces that

\displaystyle W^{0, p} (\Omega) \equiv L^p(\Omega).

Therefore, we have the following extension

Theorem (Extension of Rellich-Kondrachov). Let \Omega  \subset \mathbb R^n be an open, bounded Lipschitz domain, and let 1 \leqslant p \leqslant kn. Set

\displaystyle p^\star := \frac{np}{n - kp}.

Then we have

\displaystyle W^{j+k, p} (\Omega) \hookrightarrow W^{j, q} (\Omega) for 1 \leqslant q \leqslant p^\star


\displaystyle W^{j+k, p} (\Omega) \subset \subset W^{j,q} (\Omega) for 1 \leqslant q < p^\star.

Proof. We first place here the proof of the compactness. In fact, its proof comes from the Rellich-Kondrachov theorem (the case j=0). Indeed, assume j \geqslant 1 and \{u_i\} a bounded sequence in W^{j+k, p} (\Omega). Our aim is to prove \{u_i\} is precompact (its closure is compact or there exists a convergent subsequence) in W^{j,q} (\Omega).

Clearly from the definition of W^{j+k, p} (\Omega) one has

\displaystyle {\left\| {{D^\alpha }{u_i}} \right\|_{{W^{j,p}}(\Omega )}} \leqslant {\left\| {{u_i}} \right\|_{{W^{j + k,p}}(\Omega )}}

for every i and multi-index \alpha satisfying |\alpha| \leqslant j. In other words, \{D^\alpha u_i\} a bounded sequence in W^{k, p} (\Omega) for each multi-index \alpha satisfying |\alpha| \leqslant j.

By the Rellich-Kondrachov theorem, there exists a convergent subsequence in L^q(\Omega) (still denoted by \{u_i\}), that means

\displaystyle \{D^\alpha{u_i}\} converges in L^q(\Omega).

Since there are finite number of multi-indexes \alpha satisfying |\alpha|\leqslant j, by finite induction it is possible to select a subsequence \{u'_i\} \subset \{u_i\} for which

\displaystyle \{D^\alpha {u'_i}\} converges in L^q(\Omega) for any fixed |\alpha|\leqslant j.


\displaystyle \{u'_i\} converges in W^{j,q}(\Omega).

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 C such that

\displaystyle {\left\| {{u_i}} \right\|_{{W^{j,q}}(\Omega )}} \leqslant C{\left\| {{u_i}} \right\|_{{W^{j + k,p}}(\Omega )}}.

Again, by the definition of norms in Sobolev spaces, the Rellich-Kondrachov theorem and a very fundamental inequality, one has

\displaystyle\begin{gathered} {\left\| {{u_i}} \right\|_{{W^{j,q}}(\Omega )}} = {\left( {\sum\limits_{|\alpha | \leqslant j} {\left\| {{D^\alpha }{u_i}} \right\|_{{L^q}(\Omega )}^q} } \right)^{\frac{1}{q}}} \hfill \\ \qquad\leqslant {\left( {\sum\limits_{|\alpha | \leqslant j} {{C_\alpha }\left\| {{D^\alpha }{u_i}} \right\|_{{W^{m,p}}(\Omega )}^p} } \right)^{\frac{1}{p}}} \hfill \\ \qquad\leqslant C{\left( {\sum\limits_{|\alpha | \leqslant j + k} {\left\| {{D^\alpha }{u_i}} \right\|_{{L^p}(\Omega )}^p} } \right)^{\frac{1}{p}}}= C{\left\| {{u_i}} \right\|_{{W^{j + k,p}}(\Omega )}}. \hfill \\ \end{gathered}

The proof now follows.

Corollary. We always have the following compact embedding

\displaystyle W^{j+k, p} (\Omega) \subset \subset W^{j,p}  (\Omega).


  1. In the statements of both theorems (Rellich-Kondrachov and its extension), it must be written “kp < n” instead of “p ≤ kn”, for p* being correctly defined through the subsequent formula.

    Comment by Mărgărit Baubec — March 8, 2019 @ 10:11

    • Thanks! You are right.

      Comment by Ngô Quốc Anh — March 8, 2019 @ 23:50

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