According to Rellich compactness theorem, for a bounded set $\Omega \subset \mathbb{R}^n$ and $\frac{1}{p^*}=\frac1p-\frac1n$ the following compact embedding holds:
- If $p<n$ then $W^{1,p}(\Omega)$ is compactly embedded in $L^q(\Omega)$ for $q\in [1,p^*)$
- If $p=n$ then $W^{1,p}(\Omega)$ is compactly embedded in $L^q(\Omega)$ for $q\in [1,\infty)$
Now I have the following question:
If $\Omega=\mathbb{R}^n$ do we have such embedding results for $W^{1,p}(\mathbb{R}^n)$ into $L^q$ for some $q?$
If not what are the counter examples.
Essentially, the compactness results of $W^{1,p}$ is proved using the Kolmogorov compactness theorem, which says $\mathcal{F} \subset L^q(\Omega)$ is compact in $L^q(\Omega)$ iff it is uniformly bounded, equicontinuous and equi-tight ($\Omega$ can be unbounded set as well).
Even for $\Omega =\mathbb{R}^n$ uniform bounded ness of $\mathcal{F}$ in $W^{1,p}$ implies the uniform boundedness of $L^q$ at least for some suitable $q$ which depends on $p$ and $n.$ So, the non existence of the compact embedding must be due to the lack of either uniform bound or equi-tightness. Can someone throw light on these aspects. Why no compact embedding for $\Omega$ unbounded and $\Omega=\mathbb{R}$ in particular.
Best Answer
Yes, the lack of equitightness is the key.
For example, let $\phi \in C^\infty(\mathbb{R};\mathbb{R})$ be a compactly supported function (say in $[-1,1]$), and consider the family of its translations $\phi_n := \phi(\cdot+n)$ for $n \in \mathbb{Z}$: $$ \mathcal{F} := \big\{ \phi_n ; ~n \in \mathbb{Z} \big\}. $$ For any $p \in [1,+\infty]$, the family $\mathcal{F}$ is uniformly bounded in $W^{1,p}(\mathbb{R})$.
Nevertheless, you can check that, you cannot extract any converging subsequence in $W^{1,p}$ from the sequence $(\phi_n)_{n\in \mathbb{N}}$. Indeed, the only possible limit would be the identically null function, but for every $n \in \mathbb{N}$, $\|\phi_n-0\|=\|\phi_n\|=\|\phi\|$ does not tend to $0$.