I'd like to use something like Aubin-Lions lemma for the following spaces: $$ C^{0, \alpha}(B) \subset L^1(B) \subset W^{-1, q}(B),$$ with $B \subset \mathbb{R}^n$ being a compact, say a closed ball and $q$ is some constant s.t. $q < 3/2$ but I guess we will need $q \leq 1$. The first inclusion is trivial, since all continuous functions are integrable. I also need it to be compact, but that's ok too, since we know that $C^{0,\alpha} \subset C^{0,\beta}$ is compact for $\beta < \alpha$ so I can just use that I guess - would appreciate a correction if this doesn't work. Nevertheless my main problem is the second injection. Negative Sobolev means that my function $f \in L^1$ should be in $L^q$ itself so I guess $q$ should be smaller/equal to $1$. then again $f$ also has to be a weak derivative of something from $L^q$. If $q < 1$ then since my domain is bounded then $f \in L^q$ so there's no problem with that part, the only issue is finding a function $g \in L^q$ s.t. $$\sum_i^n a_i \partial_{x_i} g = f$$ can anyone provide some insight? does this even hold? if so, what are the necessary/sufficient constraints on $q$? a reference would be perfectly fine answer
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Sobolev embedding theorem says $W^{1,n}(B) \subset L^\infty(B)$, so taking duals, $L^1(B) \subset L^\infty(B)^* \subset W^{-1,n'}(B)$ where $1/n + 1/n' = 1$. I believe these exponents are sharp so you should only get your inclusion for $q \ge n'$. – Nate Eldredge Jun 16 '14 at 01:21
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1@Nate Eldredge: The famous counterexample $$u(x)=\begin{cases} \log{\Bigl(\log{\bigl(\frac{1}{|x|}\bigr)}\Bigr)},\quad |x|<e^{-1},\ 0,\quad |x|\geqslant e^{-1},\end{cases}$$ shows that $W^{1,n}(B)\not\subset L^{\infty}(B)$ for a unit ball $B\subset\mathbb{R}^n$, $n\geqslant 2$. – mkl314 Jun 16 '14 at 08:16
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@mm-aops: For $L^q$ with $q<1$, you will need something very far different from the Aubin-Lions lemma. A dual to the non locally convex metric space $L^q(B)$ with $q<1$ is trivial, for which reason $W^{-1,q}(B)$ with $q<1$ looks most intriguing. How do you define it? – mkl314 Jun 16 '14 at 09:10
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right, I just thought $q \leq 1$ might be good cause then we don't have a problem with $f \in L^q$ but if we could get it somehow else I'd be happy to stick with $1 \leq q < 3/2$. What I want to do is to use a version of Aubin-Lions that doesn't need reflexivity (i.e. not the one from wikipedia, rather from Simon's paper 'compact sets in the space $L^p(0,T;B)$'). I want to use negative Sobolev because I know that the sequence to which I want to apply Aubin-Lions is a combination of weak derivatives of functions in $L^q$ for $q < 3/2$ but frankly I never had much to do with negative Sobolev – mm-aops Jun 16 '14 at 10:13
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which is why I'm puzzled. then again I think that @NateEldredge 's idea with Sobolev embedding is nice and might work - I'll try to figure it out today and if I do I'll post my findings. in the meantime all insights are still most welcome, thanks! – mm-aops Jun 16 '14 at 10:15
1 Answers
The negative-order Sobolev spaces are usually defined as dual spaces: $W^{-1,q}=(W_0^{1,p})^*$ where $1/p+1/q=1$. This works only for $q>1$. Another definition, which makes sense also for $q=1$, is that $W^{-1,q}=\{\operatorname{div}\vec F: \vec F\in L^q\}$ where $\vec F$ is a vector field. The derivatives forming the divergence of $\vec F$ are understood in the sense of distributions. The norm of a distribution $T\in W^{-1,q}$ is the infimum of $\|\vec F\|_q$ over all $\vec F$ such that $T=\operatorname{div}\vec F$. See Sobolev spaces by Adams. Note that neither definition makes sense for $q<1$.
However, you don't need $q<1$. Nate Eldredge gave a hint in the right direction: the embedding $L^1\subset W^{-1,q}$ holds for $q<\frac{n}{n-1}$. This statement simply means that we can integrate $L^1$ functions against functions in $W^{1,q'}$. Which is true because $W^{1,q'}\subset L^\infty$ by Morrey-Sobolev embedding.
