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I read an article about the Rellich-Kondrachov embedding theorem in Sobolev spaces. Nevertheless, when I checked the refererence in Evans' PDE book, I only find the proof of the special case $W^{1,p}(\Omega)\subset\subset L^q(\Omega)$ where $\Omega\subset \mathbb{R}^n$ with $\partial \Omega \in C^1$, $1\le p<n$, and $1\le q<\frac{np}{n-p}$.

Do you know the proof (or references) for general result $W^{k,p}(\Omega)\subset\subset W^{l,q}(\Omega)$ whenever $k-\frac{n}{p} > l-\frac{n}{q}$?

ViktorStein
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tes tes
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    You can simply iterate the argument to get a sequence of inclusions: $W^{k,p} \subset W^{k-1,p_2} \subset \ldots \subset W^{l+1,\hat q} \subset W^{l,q}$. – gerw Jun 29 '16 at 17:42
  • @gerw How do I get the condition $k-n/p>l-n/p$ by iterating? – avati91 Jan 31 '17 at 13:50
  • You have $W^{1,p} \subset L^q$ for $n / q > n/p - 1$. Hence, you gain $1$ in this inequality for each level of differentiability. From $W^{k,p}$ to $W^{l,q}$ you spend $k - l$ levels of differentiability. Hence, $k-l$ comes into play. – gerw Jan 31 '17 at 21:15

2 Answers2

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You can find the general statement and proof in Chapter 6 of the book "Sobolev Spaces" by Robert A. Adams and John. J. F. Fournier.

levap
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  • Thanks. I have checked but the formulation is different, the author use the assumption of cone condition and the second space $W^{l,q}(\Omega)$ is replaced by $W^{l,q}(\Omega_0^k)$ where $\Omega_0^k$ denote the intersection of $k$-dimensional plane with a subset $\Omega_0$ – tes tes Jun 28 '16 at 06:57
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    This version is more general than Evans' version. If $\Omega$ is bounded (as it is in Evans' case), you can take $\Omega = \Omega_0$ and if you take $k = n$ you get $\Omega_0^k = \Omega$. The cone condition is much weaker than requiring smoothness of the boundary. In any case, I assume you can mimic his argument of moving from the $k = 1$ case to general $k$ and it will work the same way with Evans's assumptions. – levap Jun 28 '16 at 07:02
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(Rellich–Kondrachov I) \label{(Rellich-Kondrachov)}
Let $1 \leq p < n$ and let $\Omega \subset \mathbb{R}^{n}$ be an extension domain for $\mathcal{W}^{1,p}(\Omega)$ of finite measure. Let $(u_{n}) \subset \mathcal{W}^{1,p}(\Omega)$ be a bounded sequence. Then there exists a subsequence $(u_{n_{k}})$ of $(u_{n})$ and a function $u \in L^{q}(\Omega)$ such that $u_{n_{k}} \rightarrow u$ in $L^{q}(\Omega)$ for every $1\leq p \leq q < p^{*}$. \cite[p.322]{Leoni}

\cite[2.33]{Aubin1}(Rellich–Kondrachov II)
Let $\Omega \subset \mathbb{R}^{n}$ be a bounded domain with Lipschitz boundary. Then:
$i)$ If $m \in \mathbb{N}$, $1 \leq p < \infty$, and $0 \leq \ell < m$, the inclusion \begin{equation} \mathcal{W}^{m,p}(\Omega) \hookrightarrow \mathcal{W}^{\ell,q}(\Omega) \end{equation} is compact if $$ \frac{1}{q} \geq \frac{1}{p} - \frac{m - \ell}{n}, \quad q < \frac{n p}{n - (m - \ell)p}. $$ If $m p < n$, $1\geq \frac{1}{q} > \frac{1}{p} - \frac{k}{n}$ and $q \in [1, \frac{n p}{n - m p})$ then $$ \mathcal{W}^{m,p}(\Omega) \hookrightarrow L^q(\Omega). $$

$ii)$ If $s \in \mathbb{R}$, $1 \leq p < \infty$, and $0 \leq \ell < s$, the inclusion \begin{equation} \mathcal{W}^{s,p}(\Omega) \hookrightarrow \mathcal{W}^{\ell,q}(\Omega) \end{equation} is compact if $$ \frac{1}{q} \geq \frac{1}{p} - \frac{s - \ell}{n}, \quad q < \frac{n p}{n - (s - \ell)p}. $$ In particular, if $s p < n$, and $q \in \left[1, \frac{n p}{n - s p}\right)$ then $$ \mathcal{W}^{s,p}(\Omega) \hookrightarrow L^q(\Omega). $$ If $m p = n$ (resp.\ $s p = n$), the inclusion $\mathcal{W}^{m,p}(\Omega) \hookrightarrow L^q(\Omega)$ is not compact for any $q \geq 1$.

\label{Rellich-Kondrachov III} \cite[t.3.6, c.3.7]{Hebey}(Rellich–Kondrachov III).
Let $(M,g)$ be a compact boundaryless Riemannian $n$-manifold. Then:
i) For integers $j \geq 0$ and $m \geq 1$, and any real $q \geq 1$ and $p \in \mathbb{R}$ such that $1 \leq p < \tfrac{n q}{n - m q}$, the embedding \begin{equation} \mathcal{W}^{j+m,q}(M) \hookrightarrow \mathcal{W}^{j,p}(M) \end{equation} is compact. Equivalently, if $k,\ell \in \mathbb{N}$ and $k-\tfrac{n}{p} > \ell-\tfrac{n}{q}$, then the map \begin{equation} \mathcal{W}^{k,p}(M) \hookrightarrow \mathcal{W}^{\ell,q}(M) \end{equation} is compact. For $1 \leq q < n$ and any $p \geq 1$ such that $\tfrac{1}{p} > \tfrac{1}{q} - \tfrac{1}{n}$, the embedding $\mathcal{W}^{1,q}(M) \hookrightarrow L^{p}(M)$ is compact.

ii) If $(M,g)$ is compact without boundary, for $s \in \mathbb{R}$ with $s p < n$, the embedding \begin{equation} \mathcal{W}^{s,p}(M) \hookrightarrow L^{q}(M) \end{equation} is continuous for $1<p \leq q \leq p^{*}$ and is compact when $q < p^{*}$. The above embedding is also compact for $q=\infty$ if and only if $s p > n$ and $p < \infty$, namely \begin{equation} \mathcal{W}^{s,p}(M) \hookrightarrow L^{\infty}(M). \end{equation}

Santiago
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