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

and $\displaystyle W^{k, p} (\Omega) \subset \subset L^{q} (\Omega)$ 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 $1 \leqslant q \leqslant p^\star$

and $\displaystyle W^{j+k, p} (\Omega) \subset \subset W^{j,q} (\Omega)$ 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 $|\alpha|\leqslant j$.

Thus $\displaystyle \{u'_i\}$ 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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