# Rellich-Kondrachov

I read an article about the Rellich-Kondrachov embedding theorem in Sobolev spaces: https://en.wikipedia.org/wiki/Rellich%E2%80%93Kondrachov_theorem.

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 $k-n/p>l-n/q$?

• 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

• 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 6:57
• 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 7:02