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I am trying to understand the proof of a theorem, but I just can't seem to wrap my head around this one step:

We have a sequence of functions $(f_i)_{i \in \mathbb{N}}$ that are bounded in the Sobolev space $W^{k,p}$, where $k \in \mathbb{N}$, $p \geq 2$, i.e. $\|f_i\|_{W^{k,p}} \leq C$ for some constant $C$ and all $i$. We also know that $f_i$ converges to some $g$ in $L^p(\mathbb{R})$ and this $g$ is also an element of $W^{k,p}$. The proof now just claims, that using standard properties of Sobolev spaces, this bound together with the convergence in $L^p$ implies the convergence $f_i \to g$ in all intermediate spaces $W^{k',p}$, where $k' \in (0,k)$.

I am rather new to Sobolev spaces, so I am quite confused as to what these standard properties might be. I tried to read up in the book on Sobolev spaces by Adams, but I could not really find anything helpful. To be honest, I have yet to understand what norms these intermediate spaces use. I also tried to first understand the statement asuming $k'$ is an integer, because in this case I am more familiar with the spaces, but I was also unsuccessful.

Do you have ideas on how to approach this task? Any help is appreciated

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2 Answers 2

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Here is the rough idea: typically, you have a compact embedding $W^{k,p} \hookrightarrow W^{k',p}$ for all $k' \in (0,k)$ [and I would say that in most cases, $k'$ is an integer].

Since $f_i$ is bounded in $W^{k,p}$, there is a subsequence $f_{k_i}$ with $f_{k_i} \to h$ in $W^{k',p}$ for some $h \in W^{k',p}$. However, this gives $f_{k_i} \to g$ in $L^p$, i.e., $g = h$.

Finally, you can use a subsequence-subsequence argument to show that the entire sequence $f_i$ converges in $W^{k',p}$.

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  • $\begingroup$ I have read up a bit on the compact embeddings, and have the impression that they usually require the domain to be at least bounded. In my case the domain is $\mathbb{R}$. To be very precise, the functions I am working with are all of the form $\varphi^{\frac{1}{p}}$, where $\varphi: \mathbb{R} \to \mathbb{R}$ is some density. Will the compact embeddings still work in this case? Furthermore, can you recommend me some literature where I can read up on these embeddings? $\endgroup$
    – Andreas132
    Jun 3, 2021 at 9:39
  • $\begingroup$ On $\mathbb R$, compactness seems to be difficult. In particular, you can consider a sequence of translates, i.e., for each $n \in \mathbb N$ set $u_n(x) = u(x - n)$. Then, $u_n$ will be bounded but will not contain a convergent subsequnce in typical function spaces. $\endgroup$
    – gerw
    Jun 3, 2021 at 9:46
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The most direct way to justify this would be to use the Gagliardo-Nirenberg interpolation inequality. There are certain technical assumptions (on the exponents and the domain), but for simplicity, let us fix $0 \le k' < k$ and take for granted the following: $$ \| u \|_{W^{k',p}} \le A \| u \|^{\alpha}_{L^p} \cdot \| u \|^{1-\alpha}_{W^{k,p}} \qquad \text{for all } u \in W^{k,p}, $$ where the constants $0 < \alpha \le 1$ and $A > 0$ may depend on everything (in particular the choice of $k'$) except $u$.

If $\| f_i \|_{W^{k,p}} \le C$ for all $i$, then $$ \| f_i-f_j \|_{W^{k',p}} \le A \| f_i-f_j \|^\alpha_{L^p} \cdot \| f_i-f_j \|^{1-\alpha}_{W^{k,p}} \le 2C^{1-\alpha}A \| f_i-f_j \|^\alpha_{L^p}. $$ If additionally $f_i$ converges in $L^p$, then by the above, it's a Cauchy sequence in $W^{k',p}$, which implies convergence in this space.

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  • $\begingroup$ From the Wikipedia article you linked I can understand how you reach your first equation in the case that $k'$ is an integer. But will this also work, if $k'$ is an arbitrary positive real number? In that case, the norm of $W^{k', p}$ has the additional term of the Slobodeckij seminorm, and I dont understand what happens to this term. $\endgroup$
    – Andreas132
    Jun 3, 2021 at 11:41

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