Characterization of precompactness/totally boundedness in Sobolev Spaces

Question

within the paper about the Kolmogorov Riesz compactness theorem by Hanche-Olsen and Holden it is stated that for Sobolev Spaces any subset $\mathcal{F}$ of $W^{k,p}(\mathbb{R}^n)$ is totally bounded if, and only if the images of the canonical embeddings $D^{\alpha}[\mathcal{F}]:=\{D^{\alpha}f\,,\, f\in\mathcal{F}\}$ are totally bounded with respect to the $L^p$-Norm for every $|\alpha|\le k$. With that in mind the Kolmogorov Riesz Theorem for Sobolev Spaces becomes obvious.

I can easily see why a totally bounded subset $\mathcal{F}$ would have totally bounded embeddings in $L^p$ considering that the Sobolev-Norm features every $L^p$-Norm of the function within the sum. Proving the other implication though has been a problem I cannot solve. If I choose a particular $\varepsilon>0$ and try to find a finite $\varepsilon$-cover, I notice that while for every weak derivative of an arbitraty $f$ I can find a function $h_{\alpha}$ within a finite subset, I cannot guaranteee that one function $h$ satisfies the condition $\|D^{\alpha}f-D^{\alpha}h\|<\varepsilon$ for every $\alpha$.

I am stuck and would greatly appreciate if someone could point me in the right direction. I would gladly continue trying to prove it myself, but I have run out of ideas due to my rather limited knowledge of Sobolev Spaces. Every inequality I have found so far has some sort of condition for $k,n$ and $p$, which I cannot guarantee.

Thanks for all your help!

Answer 1

Using a different approach to totally boundedness I think I found a way to prove my proposition using Cauchy sequences. Since a subset of any metric space is pre-compact if and only if for every sequence there exists a Cauchy-subsequence, I considered the sequences $(D^\alpha f_n)_{n\in\mathbb{N}}$, which all have a Cauchy-subsequence with respect to the $L^p$-norm. Passing to a diagonal subsequence $(f_{n_k})_{k\in\mathbb{N}}$ we have a sequence whose projetions $(D^\alpha f_{n_k})_{k\in\mathbb{N}}$ are all Cauchy-sequences in $L^p$. Seeing that the Sobolev-norm is given by $\|\cdot\|_{l,p}=\left(\sum_{|\alpha|\le l}\|D^\alpha\cdot\|_p^p\right)^{1/p}$, we can easily show that $(f_{n_k})_{k\in\mathbb{N}}$ must be Cauchy with respect to $\|\cdot\|_{l,p}$.

Answer 2

The mapping \begin{align} W^{k,p}(\mathbb{R}^n) &\to L^p(\mathbb{R}^n,\mathbb{R}\times\mathbb{R}^n\times\cdots\times\mathbb{R}^{n^k})\\ f &\mapsto (f,\mathrm{D}f,\ldots,\mathrm{D}^kf) \end{align} is a continuous embedding of the Sobolev space $W^{k,p}(\mathbb{R}^n)$ into another Banach space. Therefore, a subset of $W^{k,p}(\mathbb{R}^n)$ is totally bounded if and only if its image under the above mapping is totally bounded.