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This post has been on MathOverflow for couple of days but receive no response. So I put it here hoping for more attentions.

Thank you guys!


Let $\Omega\subset \mathbb R^N$ be open bounded with smooth boundary. Let $0<s<1$, $1\leq p<\infty$, and $1\leq \theta\leq\infty$. We denote by $B^{s;p,\theta}(\Omega)$ the Besov space. For definition of Besov space we refer to Leoni's book, Chapter 14, section 14.1. (Also this book by Adam, page 230, section 7.32.)

Theorem 14.29 in Leoni's book states the continuous imbedding theorem for Besov space. (For simplification, let's assume $p=1$.) We have $B^{s;1,\theta}(\Omega)$ continuous imbedded in $L^{\frac{N}{N-s}}(\Omega)$ for $1\leq \theta\leq \frac{N}{N-s}$.

We now take $r<\frac{N}{N-s}$.

My question is: do we have $B^{s;1,\theta}(\Omega)$ is COMPACT imbedded in $L^{r}$? I think the answer is yes because according to this post, exercise 15, that

sequences bounded in a high regularity space, and constrained to lie in a compact domain, will tend to have convergent subsequences in low regularity spaces.

So I would think my conjecture is true. However, I did a deep search over the internet but has no lucky to find such result.

If there is no such result, please let me know (and maybe a counterexample?). If there is, please direct me to a reference.

Thank you!

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  • $\begingroup$ For what it's worth, I would also believe your conjecture. But if it's not already in Adams and Fournier, then it may be hard to find the exact statement you're looking for. But if you want a rigorous proof perhaps you can continuously embed your Besov space into a Sobolev space for which compact embeddings are more standard. Alternatively it may be possible to compactly embed your Besov space into one with a bit less regularity, for which you already know the embedding into $L^r$. $\endgroup$ – felipeh Aug 4 '15 at 7:24
  • $\begingroup$ @felipeh Thank you for your suggestions! However, for what I know that we have $BV\subset B^{s;1,\theta}\subset L^r$ so it is hard to find a space between $B^{s;1,\theta}$ and $L^r$, which has some compactly embedding properties since $BV$ is already very weak... $\endgroup$ – spatially Aug 4 '15 at 12:02
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Your conjecture is true. Since $\Omega$ is smooth, another way to see Besov spaces is to use interpolation between $L^p(\Omega)$ and $W^{1,p}(\Omega)$, that is, $(L^p(\Omega),W^{1,p}(\Omega))_{\theta,q}=B^{s,p,\theta}(\Omega)$. Now the injection $T:W^{1,p}(\Omega))\to L^{p}(\Omega))$ given by $T(u)=u$ is compact (by Rellich-Kondrachov theorem) while $T:L^{p}(\Omega))\to L^{p}(\Omega))$ is continuous. You can now apply standard theorems in interpolation to conclude that $T:B^{s,p,\theta}(\Omega)\to L^{p}(\Omega)$ is compact. Hence, if you have a sequence bounded in $B^{s,p,\theta}(\Omega)$ you know that it has a subsequence converging in $L^p(\Omega)$, but since the sequence is bounded in $L^{N/(N-s)}(\Omega)$, playing with Holder's inequality you get that it converges in $L^r(\Omega)$ for all $1\le r<N/(N-s)$. Classical books on interpolation are Bergh and Lofstrom or Bennet and Sharpley.

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