# Embedding of BV and L^P spaces

An elementary question about Sobolev spaces:

Is there some explicit theorem about embedding relation between spaces BV(Omega) and L^P(Omega)?

e.g. is BV a subset of L^2 (i.e. BV possess regularities of L^2)?

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Edit: The most general imbedding I know of about $L^p$ spaces is that $BV(\Omega) \subset \subset L^{n/n-1}(\Omega)$ where $\Omega \subset \mathbb{R}^n$ and $n > 1$ (replace $n/(n-1)$ with $1$ when $n=1$). This embedding is compact for any $p < n/n-1$. Hence for your $n=2$, $n=3$ interest we have $f \in L^{2}(\Omega)$ for $n=2$ and $f \in L^{3/2}(\Omega)$ for $n=3$. On bounded domains you have all lower $L^p$ norms as well by an application of Holder's inequality. Sorry for my initial mis-understanding of your question.

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Thanks Dorian, It gives some sense, seems in the case one-dimension answer is clear, in general also L1 regularity is evident. But some doubts of mime (sorry i'm beginner): does concept "imbedding" is different from "embedding"? why you continued argument by compact imbedding? also what you wanted to imply by your last paragraph about "distributional derivatives", in fact, its connection to my question? assume my functions are of class BV(r^n) (n=2,3), then in functional analysis i need at least L^2 regularity, then how i should proceed is some assumption/regularization is required? –  Jean-Marie Sep 5 '10 at 6:14
Hi there, I fixed my answer to better answer your question. I perhaps am just misspelling imbedding/embedding; they are the same thing. The compactness is not important to you it appears but still interesting in it's own right. You can ignore all statements about compactness above if you don't understand them and just focus on the statements about embeddings. –  Dorian Sep 5 '10 at 14:44

An $L^1_{loc}$ function on $\mathbb{R}^n$ is in $BV_{loc}$ iff its distributional derivatives $\partial_i f\in\mathcal{M}^1_{loc}$, i.e. they are all locally finite (Radon) measures. If $n=1$, the situation is well-known, and $BV_{loc}\subset L^\infty_{loc}$. So assume $n\geq 2$. Since $W^{s,p}_{c}(\mathbb{R}^n)\subset C^0_{c}(\mathbb{R}^n)$ if $s>n/p$, you have that $$BV_{loc}(\mathbb{R}^n)\subset W^{1-s,p'}_{loc}(\mathbb{R}^n)\subset L^{p'}_{loc}(\mathbb{R}^n)$$ if $s\leq 1$ and $1/p+1/p'=1$, that is if $p'<n/(n-1)$. On the other hand, when $n>1$, $1/r^\alpha$ is in $BV_{loc}(\mathbb{R}^n)$ if $\alpha<n-1$, since partial derivatives are in $L^1_{loc}$, but it is in $L^q$ only for $q<n/\alpha$, so that $BV_{loc}(\mathbb{R}^n)\subset L^q_{loc}(\mathbb{R}^n)$ fails for any $q>n/(n-1)$. I wouldn't bet on the limiting case.

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BS: does your answer implies that in general, for n>1, BV(R^n) does not possess regularity of L^p for p>=2? –  Jean-Marie Sep 5 '10 at 6:27
It implies that the best local regularity to hope for in $BV(R^n)$ is $L^{n/(n-1)}$. The argument above only gives $L^q$ for any $q<n/(n-1)$, but I sort of remember now that this also true for $q=n/(n-1)$ as a consequence of the isoperimetric inequality (the case of characteristic functions in $BV$) and coarea formula en.wikipedia.org/wiki/Coarea_formula, applied to regularized functions in $BV$. So $BV$ is in $L^2$ for $n\leq2$ (and bounded domains), but not for larger $n$. –  BS. Sep 5 '10 at 7:43
Thanks BS: it gives more sense, what about the connection of H^m to BV spaces, e.g. does every H^m (m>=1) already possess the regularities of BV? could you please introduce a minimal text to aid elementary understanding of Sobolev spaces, assuming a little knowledge (not solid) in measure and topology? –  Jean-Marie Sep 5 '10 at 8:22
$L_{loc}^p$ means that $f \in L^p(K)$ for every compact $K$, meaning bounded and closed. In other words, it's integrable on bounded sets. A good example is $f(x) = x^2$ on $\mathbb{R}$. It is in $L_{loc}^p(\mathbb{R})$ for all $p \in [1,\infty]$ but it is not in any space $L^p(\mathbb{R})$. If you are in $H^m(\Omega)$ for any $m \geq 1$ then in fact you can see by Cauchy-Schwartz that you have $\int |Du(x)| dx \leq C||u||_{H^1}$ but this of course is the $W^{1,1}$ norm and $W^{1,1}$ and $BV$ have the same norm on $W^{1,1}$; $BV$ is just a weaker class. –  Dorian Sep 7 '10 at 17:41
Just to be clear, my previous argument shows that if you are in $H^m$ then you are in fact in $W_{loc}^{1,1}$ and consequently in $BV_{loc}$ because $BV \subset W^{1,1}$. –  Dorian Sep 7 '10 at 20:07