I have two questions about weak derivatives and bounded variation, as you can see they are very similar, help with either would be much appreciated. I will start with one confusion I have
- In the definition of bounded variation functions https://en.wikipedia.org/wiki/Bounded_variation test functions are taken from $C^1_c(\mathbb{R}^d,\mathbb{R}^d)$. I see some authors take functions from $C^\infty_c(\mathbb{R}^d,\mathbb{R}^d)$ e.g Proposition 3.13 of https://arxiv.org/abs/1512.02783 . Maybe one implies the other though some density argument.
$\underline{Questions :}$
$\textbf{Question 1 :}$
Assume I have a $L^1(\mathbb{R}^d)$ function $f$, where we know
$$ \int_{\mathbb{R}^d}f\text{div}(\phi)dx=\int_{\mathbb{R}^d} \langle g , \phi \rangle dx ~~\forall \phi\in C^{\infty}_c(\mathbb{R}^d,\mathbb{R}^d)$$
and $\sup_{\phi\in C^{\infty}_c(\mathbb{R}^d,\mathbb{R}^d)}| \int_{\mathbb{R}^d}f\text{div}(\phi)dx|<c$. can I conclude $f \in W^{1,1}(\mathbb{R}^d)$ and $\nabla f=g$ $\textbf{?} $ Basically asking does begin in $BV(\mathbb{R}^d)$ imply weak derivatives are in $L^1(\mathbb{R}^d)$ ?
$\textbf{Question 2 :}$
Assume I have an sequence of uniformly bounded $L^1(\mathbb{R}^d)$ functions $\|f_k\|_{L^1(\mathbb{R}^d)}<C~ \forall k $. Furthermore I know that they are of uniformly bounded variation $L^1(\mathbb{R}^d)$ functions $\|f_k\|_{BV(\mathbb{R}^d)}<C~ \forall k $. What can I say about the $\|\nabla f_k\|_{L^1(\mathbb{R}^d)}$ weak derivatives of $f_k$ $\textbf{?}$
Best Answer
Answer to Question 1.
For the first part of question 1 the answer is yes i.e. $f\in W^{1,1}(\Bbb R^d)$, while for the second part the answer is no.
For the first part: we assume that $$\DeclareMathOperator{\dvg}{\mathrm{div}} \sup_{\substack{\phi\in C^{\infty}_c(\mathbb{R}^d,\mathbb{R}^d) \\ |\phi(x)|\le 1}}\Bigg| \,\int\limits_{\mathbb{R}^d}f(x)\dvg\phi(x)\,\mathrm{d}x\,\Bigg|=TV(f)<+\infty\label{1}\tag{1} $$ and that there exists a $\Bbb R^d$-valued function $g$ such that $$ \int\limits_{\mathbb{R}^d}f(x)\dvg\phi(x)\,\mathrm{d}x = \int\limits_{\mathbb{R}^d} \langle g(x) , \phi(x) \rangle \mathrm{d}x \qquad\forall \phi\in C^{\infty}_c(\mathbb{R}^d,\mathbb{R}^d).\label{2}\tag{2} $$ which is thus at least locally integrable (i.e. $g\in L^1_\text{loc}(\Bbb R^d)$). Now consider the positive and negative parts of the components of $g$, defined as $$ g^+_i(x)= \begin{cases} g_i(x) & g_i(x)> 0\\ 0 & g_i(x)\le 0 \end{cases}\qquad g^-_i(x)= \begin{cases} 0 & g_i(x)\ge 0\\ -g_i(x) & g_i(x)< 0 \end{cases}\qquad\forall x\in\Bbb R^d, i=1, \ldots, d $$ and their support sets defined as $$ \begin{align} G^+_i & \triangleq \{x\in\Bbb R^d : g_i(x) > 0\} \\ G^-_i & \triangleq \{x\in\Bbb R^d : g_i(x) < 0\} \end{align} \qquad i=1, \ldots, d. $$ From \eqref{2} we have that, for each $i=1, \ldots, d$ $$ \int\limits_{\mathbb{R}^d} \langle g(x) , \phi(x) \rangle \mathrm{d}x = \sum_{i=1}^d \Bigg[\,\int\limits_{\mathbb{R}^d} g^+_i(x) \phi_i(x)\, \mathrm{d}x - \int\limits_{\mathbb{R}^d} g^-_i(x) \phi_i(x) \, \mathrm{d}x \Bigg]\qquad\forall \phi\in C^{\infty}_c(\mathbb{R}^d,\mathbb{R}^d). $$ Now, for each compact set $K_i^\pm\Subset G^\pm_i$ (possibly one or more of these sets, but not all, can be empty) it is possible construct compactly supported functions $\varphi^\pm_i(x)\in C^\infty_c(\mathbb{R}^d,\mathbb{R})$ by using partitions of unity in such a way that $$ \begin{split}\DeclareMathOperator{\supp}{\mathrm{supp}} 0\le\varphi^\pm_i &(x) \le 1\quad \forall x\in G^\pm_i\\ \varphi^\pm_i &(x) = 1 \quad \forall x\in K^\pm_i\\ &\supp \varphi^\pm_i \Subset G^\pm_i \end{split} $$ By defining $\varphi_i(x)=\varphi_i^+(x)-\varphi^-_i(x)$ and $\varphi(x)= \big(\varphi_1(x), \ldots, \varphi_d(x)\big)$ we have $$ \int\limits_{\mathbb{R}^d} \langle g(x) , \varphi(x) \rangle \mathrm{d}x = \sum_{i=1}^d \Bigg[\,\int\limits_{\mathbb{R}^d} g^+_i(x) \varphi_i^+(x)\, \mathrm{d}x + \int\limits_{\mathbb{R}^d} g^-_i(x) \varphi_i^-(x) \, \mathrm{d}x \Bigg], $$ and then $$ \begin{split} TV(f) & \ge \sup_{\varphi(x)} \sum_{i=1}^d \Bigg[\,\int\limits_{\mathbb{R}^d} g^+_i(x) \varphi_i^+(x)\, \mathrm{d}x + \int\limits_{\mathbb{R}^d} g^-_i(x) \varphi_i^-(x) \, \mathrm{d}x \Bigg]\\ & = \sum_{i=1}^d \int\limits_{\mathbb{R}^d} |g_i(x)| \, \mathrm{d}x \ge 0 \iff g_i\in L^1(\Bbb R^d)\quad i=1, \ldots,d \end{split} $$ where the supremum on the right side is taken on the subset of $C^{\infty}_c(\mathbb{R}^d,\mathbb{R}^d)$ consisting of functions $\varphi$ constructed as shown above: thus $f\in W^{1,1}(\Bbb R^d)$.
For the second part, as David C. Ulrich also noted in his comment on the $d=1$ case, weak derivatives of functions $f\in L^1(\Bbb R^d)$ do not need to be functions: they can be measures or even more general distributions, and precisely $BV$ functions are those integrable function whose weak derivative are possibly singular (i.e. supported on sets of dimensions $<d$) measures.
Answer to Question 2.
The answer to question 2 is no: as a simple counterexample you may consider the following sequence of monotonically decreasing functions $$ f_n(x) = \begin{cases} 1 & x\in [-1,1]\\ {n(1+x) +1 } & x\in\big]-\frac{1}{n}-1, -1\big[\\ n(1-x)+1 & x\in \big] 1,1+\frac{1}{n}\big[\\ 0 & \text{elsewere} \end{cases} $$ The variation of all these function is constant and equal to $+2$ but the limit function does not belong to $W^{1,1}(\Bbb R)$.
A final Note.
A more rigorous way to define the sets $G_i^\pm$, $i=1,\ldots, d$ is the following:$\DeclareMathOperator{\intr}{\operatorname{int}} \DeclareMathOperator{\closr}{\operatorname{clo}}$ $$ \begin{align} G^+_i & \triangleq \intr\closr\{x\in\Bbb R^d : g_i(x) > 0\} \\ G^-_i & \triangleq \intr\closr\{x\in\Bbb R^d : g_i(x) < 0\} \end{align} \qquad i=1, \ldots, d. $$ where $\closr$ and $\intr$ are respectively the closure and the interior operators. In this way, the partitions of unity can be constructed since $G_i^\pm$ is open and non empty (at least not for all $i$) and thus contains non empty compact sets $K_i^\pm$ for all $i=1, \ldots, d$.