Bounded Variation and Sobolev Norms. 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

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*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 $f_k$ $\textbf{?}$
 A: 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 $i=1, \ldots, d$.
