Embedding $BV$ into $L^p$

Question

In my study of $BV$ spaces I became aware of the embedding $BV_{\operatorname{loc}}(\mathbb{R}^n)\hookrightarrow L^q_{\operatorname{loc}}(\mathbb{R}^n)$ for $q\leq\frac{n}{n-1}$. I'm really interested in embedding results and I wonder how to prove this. Unfortunately I couldn't find any proof. Does anybody know how to prove this or knows where to find it?

Answer 1

Usually the space $BV_{\mathrm{loc}}(\Bbb R^n)$ is defined as a space of $u \in L^1_{\mathrm{loc}}(\Bbb R^n)$ such that its distributional derivative $Du$ is represented by a Radon measure, which equivalently can be expressed as $$ \sup\left\{ \int_{\Bbb R^n} u \operatorname{div} \varphi \,\mathrm{d}x : \varphi \in C^1_c(\Bbb R^n,\Bbb R^n), \operatorname{supp}(\varphi) \leq K, |\varphi| \leq 1\right\} < \infty$$ for all $K \subset \Bbb R^n$ compact, by the Riesz representation theorem. Under this definition the embedding $BV_{\mathrm{loc}}(\Bbb R^n) \hookrightarrow L^1_{\mathrm{loc}}(\Bbb R^n)$ follows by definition (where the respectively topologies are defined suitably), but one can equivalently define $BV_{\mathrm{loc}}(\Bbb R^n)$ as the space of distributions $u \in \mathcal{D}'(\Bbb R^n)$ whose derivative $Du$ is a measure. I assume this is the definition you are interested in. I am not aware of many references that do this in detail (though Leon Simon's GMT notes does essentially prove this in the context of currents), so I will give a detailed proof.

To see this equivalence, the idea is to mollify such a $u$ and show that it converges in $L^1_{\mathrm{loc}}.$ Given a standard mollifier $\rho_{\varepsilon}$ set $u_{\varepsilon} = u \ast \rho_{\varepsilon},$ which are smooth on $\Bbb R^n$ with $Du_{\varepsilon} = Du \ast \rho_{\varepsilon}.$ Since the latter is a mollification of a measure, for any $U \subset \Bbb R^n$ bounded and $0<\varepsilon<\varepsilon_0$ we have the $L^1$ bound $$ \int_{U} |Du_{\varepsilon}| \,\mathrm{d}x \leq C |Du|(U + B_{\varepsilon_0}) < \infty, $$ so if $U$ is sufficiently regular (say smooth or Lipschitz) combining the Gagliardo-Nirenberg-Sobolev and Poincaré-Wirtinger inequalities (see e.g. Evans Chapter 6 or any text on Sobolev spaces) we can bound $$ \left(\int_U |u_{\varepsilon}-(u_{\varepsilon})_{U}|^q \,\mathrm{d}x\right)^{\frac1q} \leq C_q\int_{U} |Du_{\varepsilon}| \,\mathrm{d}x \leq C_{q,U} $$ for all $1 \leq q \leq \frac{n}{n-1}.$ Here $$ (u_{\varepsilon})_{U} = \def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits} \avint_U u_{\varepsilon} \,\mathrm{d}x = \frac1{|U|} \int_U u_{\varepsilon} \,\mathrm{d} x, $$ which we wish to show converges as $\varepsilon \to 0;$ this would be immediate if we knew a-priori that $u \in L^1_{\mathrm{loc}},$ but in general we need a slightly more involved argument - there may be a simpler argument, but this is what came to mind.

By weak compactness in $L^p$ for $1 < p \leq \frac{n}{n-1},$ we can find a subsequence $\varepsilon_k \to 0$ such that $u_{\varepsilon_k}-(u_{\varepsilon_k})_{U} \rightharpoonup v$ weakly in $L^p(U)$ and hence in $\mathcal{D}'(U).$ However we also know that $u_{\varepsilon} \rightharpoonup u$ weakly in $\mathcal{D}'(U),$ so it follows that $(u_{\varepsilon_k})_{U} \rightharpoonup w$ in $\mathcal{D}'(U).$ We claim $w$ is constant, which (as we will see later) would imply $(u_{\varepsilon_k})_{U}$ converges as a sequence of reals and hence $u_{\varepsilon}$ converges to $u$ weakly in $L^p(U).$ For this note that for all $\varphi_1,\varphi_2 \in C^{\infty}_c(U)$ such that $\varphi_1,\varphi_2$ are non-negative and non-zero, we have $$ \frac{\langle w, \varphi_1\rangle}{\int_U \varphi_1 \,\mathrm{d}x} = \frac{\langle w, \varphi_2\rangle}{\int_U \varphi_2 \,\mathrm{d}x}; $$ this follows since it holds for each $(u_{\varepsilon_k})_{U}$ and we can pass to the limit in $\mathcal{D}'(U).$ From here one can argue that $w \equiv a$ is constant and equals the common value (noting the distributional pairing $\langle \cdot,\cdot\rangle : \mathcal{D}(U) \times \mathcal{D}'(U) \to \Bbb R$ is exact). Hence we have shown that \begin{align*} u_{\varepsilon_k} - (u_{\varepsilon_k})_{U} \rightharpoonup v &\qquad\text{ weakly in } L^p(U), \\ (u_{\varepsilon_k})_{U} \to a &\qquad\text{ in } \Bbb R, \end{align*} so it follows that $u_{\varepsilon_k}$ converges weakly to in $L^p(U)$ for each $1 < p \leq \frac{n}{n-1}.$ Since the distributional limit is unique, we conclude that it coincides with $u|_U \in L^p(U).$

Remark: It is also possible to show that $u_{\varepsilon_k}$ converges strongly in $L^p$ for all $1 \leq p < \frac{n}{n-1},$ but this is not necessary for our argument. Also the argument simplifies in we know a-priori that $u \in L^1_{\mathrm{loc}},$ but otherwise justifying the sequence of averages converges will need to follow from the assumption that $u$ is a distribution.

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Added later: It was pointed out in the comments that I've not fully discussed the embedding part. We equip $L^p_{\mathrm{loc}}(\Bbb R^n)$ with the seminorms $u \mapsto \lVert u \rVert_{L^p(K)}$ for each $K \subset \Bbb R^n,$ which gives it the structure of a locally convex space.

Note from the above that we have the estimate $$ \left(\int_U |u - (u)_{U}|^q \,\mathrm{d}x\right)^{\frac1q} \leq C |Du|(\overline U), $$ which holds for all $1 \leq q \leq \frac{p}{p-1}$ and $U \subset \Bbb R^n$ bounded Lipschitz. Note this estimate follows from the work above by sending $\varepsilon \to 0$ in the mollified estimate, applying Fatou's lemma on the LHS. Given this, if we define the topology on $BV_{\mathrm{loc}}(\Bbb R^n)$ by the seminorms $u \mapsto \lVert u \rVert_{L^1(K)}$ and $u \mapsto |Du|(K)$ for each $K \subset \Bbb R^n$ compact, then the embedding is a straightforward consequence of the above estimate.

However we would like to prove this without assuming $BV_{\mathrm{loc}}(\Bbb R^)n \hookrightarrow L^1_{\mathrm{loc}}(\Bbb R^n).$ This however raises the question of how to define the topology on $BV_{\mathrm{loc}};$ the seminorms $u \mapsto |Du|(K)$ alone are not sufficient, since it cannot distinguish constant functions (the associated topology is not Hausdorff). Instead we have define it to inherit the topology of $\cal D'(\Bbb R^n)$ and further require the seminorms $u \mapsto |Du|(K)$ to be continuous.

This does make the question of continuity a bit hairy however, as the topology on $\cal D'(\Bbb R^n)$ is rather complicated. The easiest way is probably to take a sequence $u_n \in BV_{\mathrm{loc}}(\Bbb R^n)$ such that $u_n \to u$ in $\cal D'(\Bbb R^n)$ and $Du_n \to Du$ locally in $\mathcal M(\Bbb R^n)$ and show that $u_n \to u$ in $L^p_{\mathrm{loc}}(\Bbb R^n).$ This does use the fact that the topology is sequential, which it essentially inherits from $\mathcal D'(\Bbb R^n).$ To show this let $U \subset \Bbb R^n$ be bounded Lipschitz and note we have $$ \lVert u_n - u \rVert_{L^p(U)} \leq |D(u_n-u)|(\overline U) + |(u_n-u)_U| |U|^{\frac1p}. $$ We know from our work above that $u_n,u \in L^p_{\mathrm{loc}}(\Bbb R^n)$ so we can take averages. As $n \to \infty$ the $D(u_n-u)$ term vanishes by assumption, so it remains to argue that $$ \lim_{n \to \infty} |(u_n-u)_U| = 0. $$ This involves a simlar argument to above, showing $(u_n-u)_U$ converges in $\mathcal D'(U)$ to a constant function and that said constant is zero by uniqueness of limits.