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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!

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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.

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  • $\begingroup$ I am quite unsure why your proposition must be true. While images of totally bounded sets under uniform continuous mappings are always totally bounded (and yes, we do have a uniform continuous map), I cannot see why the converse should be true. Is that a specific property of the Sobolev space? Or does that work for any two complete metric spaces and any uniform continuous mapping? I have tried and failed to prove the converse. Also, how do we know that the image of $\mathcal{F}$ is actually totally bounded? Isn't that basically the same problem? $\endgroup$ – Sellerie Jul 9 '18 at 14:45
  • $\begingroup$ If your proposition were true, wouldn't it mean that for every uniform continuous mapping $W^{k,p}(\mathbb{R}^n)\to \mathbb{R}$ the pre-image of any bounded subset of the reals would be totally bounded? $\endgroup$ – Sellerie Jul 9 '18 at 14:56
  • $\begingroup$ I'm sorry, I only now realised that the main argument here isn't just continuous, but continuous embedding, giving a homeomorphism of the Sobolev space to its image, therefore giving me your proposition. I'm with you up to this point. The only problem left is that from my point of view the question whether the given image is totally bounded or not is essentially the same question, only this time in a different space. My image doesn't include mixed functions as in $(f,g,h,\dots)$ for $g\neq Df$. $\endgroup$ – Sellerie Jul 9 '18 at 16:08
  • $\begingroup$ There are no families of "mixed functions" for which I would have to check total boundedness. The image of a family of Sobolev functions under the above embedding will never be of this "mixed form". I think you misunderstand the "only if" part of your critertion. $\endgroup$ – Peter Wildemann Jul 10 '18 at 11:36
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    $\begingroup$ I see what you mean. However, it's much more fruitful to just think about precompactness (being equivalent to total boundedness in Banach spaces) instead of $\epsilon$-covers. $\endgroup$ – Peter Wildemann Jul 11 '18 at 8:40
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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 $\|\cdot\|_{l,p}$.

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