Are smooth bounded-variation functions in the Sobolev Space $W_{1,1}$? Let $\Omega \subseteq \mathbb{R}^n$ open and $f \in BV(\Omega) \cap C^{\infty}(\Omega)$. Now I would like to prove that $f \in W_1^1(\Omega)$. I know that the distributional derivates of $f$ are the classical derivates, but I don't find a way to show that $\partial_if \in L_1(\Omega)$ for $i=1,...,n$. Can maybe someone help me? :)  
 A: EDIT. In 2019 I left unfinished this answer. In 2020, just after Christmas I answered a related question, and today, after a few months of lock down, I decided to finish it by using the same method employed for the latter answer, just for the sake of completeness, and in the hope that this will be useful for someone.

In order to correctly answer the question, let's state explicitly the two definitions lying at the roots of the theory of BV functions (see for example [1], chapter 1, pp. 3-4):
Definition 1. Let $\Omega\subseteq \Bbb R^n$ be an open set and let $f\in L^1(\Omega)$. The total variation of f (in $\Omega$) is the following non negative quantity
$$\DeclareMathOperator{\grd}{\operatorname{grad}} \DeclareMathOperator{\dvg}{\operatorname{div}}
\DeclareMathOperator{\supp}{\mathrm{supp}}
V(f,\Omega)=\sup\Bigg\{\int\limits_\Omega\! f(x)\dvg{\mathbf{g}(x)}\,\mathrm{d}x\,\Big|\,\mathbf{g}=(g_1,\ldots,g_n)\in C_0^1(\Omega,\Bbb R^n),\;|\mathbf{g}(x)|\le1\,\forall x\in\Omega\Bigg\}
$$
Definition 2. A function $f\in L^1(\Omega)$ is said to have bounded  variation if $V(f,\Omega)<\infty$. $BV(\Omega)$ is defined as the space of all functions in $L^1(\Omega)$ of bounded variation.
Now consider $f\in BV(\Omega)\cap C^1(\Omega)$: then, for every $\mathbf{g}(x) \in C_0^1(\Omega,\Bbb R^n)$ such that $\;|\mathbf{g}(x)|\le1$ on $\Omega$, we have that
$$
\begin{split}
\int\limits_\Omega\! f(x)\dvg{\mathbf{g}(x)}\,\mathrm{d}x 
&= - \int\limits_\Omega \sum_{i=1}^n \frac{\partial f(x)}{\partial x_i}\cdot{ g_i(x)}\,\mathrm{d}x\\
&= -\int\limits_\Omega\langle\grd f(x),{\mathbf{g}(x)}\rangle\,\mathrm{d}x.\\
\end{split}\label{1}\tag{1}
$$
A first consequence of this formula is the following result, obtained by using the Cauchy-Schwartz inequality,
$$
\begin{split}
-\langle\grd\!f(x) ,{\mathbf{g}(x)}\rangle
&\le |\langle\grd\!f(x) ,{\mathbf{g}(x)}\rangle| \\
& = \left| \sum_{i=1}^n \frac{\partial f(x)}{\partial x_i}\cdot{ g_i(x)}\right|\\
&\le \left( \sum_{i=1}^n \left|\frac{\partial f(x)}{\partial x_i}\right|^2\right)^\frac{1}{2}\left( \sum_{i=1}^n g_i(x)^2\right)^\frac{1}{2} \\
\\
& = \vert\grd\!f(x)\vert|\mathbf{g}(x)|\le\vert\grd\!f(x)\vert
\end{split}
$$
which in turn implies
$$
-\int\limits_{\supp\mathbf{g}}\langle\grd\!f(x),{\mathbf{g}(x)}\rangle\,\mathrm{d}x\le \int\limits_{\supp\mathbf{g}}\vert\grd\!f(x)\vert\,\mathrm{d}x \label{2}\tag{2}
$$

Define the positive and negative parts of the components of $\grd f$ as
$$
\partial_i f^+(x)=
\begin{cases}
\dfrac{\partial f(x)}{\partial x_i} & \dfrac{\partial f(x)}{\partial x_i}> 0\\
0 & \dfrac{\partial f(x)}{\partial x_i}\le 0
\end{cases}\qquad
\partial_if^-(x)=
\begin{cases}
0 & \dfrac{\partial f(x)}{\partial x_i}\ge 0\\
-\dfrac{\partial f(x)}{\partial x_i} & \dfrac{\partial f(x)}{\partial x_i}< 0
\end{cases}\\ \forall x\in\Bbb R^n,\: i=1, \ldots, n,
$$
and their support sets as
$$\DeclareMathOperator{\intr}{\operatorname{int}} \DeclareMathOperator{\closr}{\operatorname{clo}}
\begin{align}
\Omega^+_i & \triangleq \intr\closr\Big\{x\in\Omega : \tfrac{\partial f(x)}{\partial x_i} > 0\Big\} \\
\Omega^-_i & \triangleq \intr\closr\Big\{x\in\Omega : \tfrac{\partial f(x)}{\partial x_i} < 0\Big\}
\end{align} \qquad i=1, \ldots, n.
$$
where $\closr$ and $\intr$ are respectively the closure and the interior operators. Then,
from \eqref{1}, we have
$$
-\int\limits_{\Omega} \langle\grd f(x) , \mathbf{g}(x) \rangle \mathrm{d}x =
 \sum_{i=1}^n \Bigg[\,\int\limits_{\Omega} \partial_i f^-(x)  g_i(x)\,  \mathrm{d}x -
\int\limits_{\Omega} \partial_if^+(x) g_i(x) \, \mathrm{d}x \Bigg]\qquad\forall \mathbf g\in C^{1}_0(\Omega,\mathbb{R}^n).
$$
Since the sets $\Omega_i^\pm$ are open and non empty (at least not for all $i$) and thus contain non empty compact subsetsets $K_i^\pm$, $i=1, \ldots, n$, for each of them, by using partitions of unity, it is possible construct a compactly supported family of functions ${\varphi}^\pm_i(x)\in C^\infty_0(\Omega,\mathbb{R})$ in such a way that
$$
\begin{split}
0\le\varphi^\pm_i &(x) \le 1\quad \forall x\in \Omega^\pm_i\\
\varphi^\pm_i &(x) = 1 \quad  \forall x\in K^\pm_i\\
&\supp \varphi^\pm_i \Subset \Omega^\pm_i
\end{split}\label{3}\tag{K}
$$
By defining $\varphi_i(x)=\varphi_i^-(x)-\varphi^+_i(x)$ and $\boldsymbol{\varphi}(x)= \big(\varphi_1(x), \ldots, \varphi_n(x)\big)$ we have
$$
-\int\limits_{\Omega} \langle \grd f(x) , \boldsymbol{\varphi}(x) \rangle \mathrm{d}x =
 \sum_{i=1}^n \Bigg[\,\int\limits_{\Omega} g^+_i(x)  \varphi_i^+(x)\,  \mathrm{d}x +
\int\limits_{\Omega} g^-_i(x) \varphi_i^-(x) \, \mathrm{d}x \Bigg],
$$
and then
$$
\begin{split}
V(f,\Omega) & \ge \sup_{\boldsymbol{\varphi}(x)\in K_0(\Omega, \Bbb R^n)} \sum_{i=1}^n \Bigg[\,\int\limits_{\Omega} \partial_i f^+(x)  \varphi_i^+(x)\,  \mathrm{d}x +
\int\limits_{\Omega} \partial_i f^-(x) \varphi_i^-(x) \, \mathrm{d}x \Bigg]\\
& = \sum_{i=1}^n \int\limits_{\Omega} \bigg|\frac{\partial f(x)}{\partial x_i}\bigg| \,  \mathrm{d}x \ge  \int\limits_{\Omega} |\grd f(x)| \,  \mathrm{d}x \ge 0.
\end{split}
$$
where $K_0(\Omega, \Bbb R^n)\subsetneq C^1_0(\Omega, \Bbb R^n)$ is the set of compactly supported functions constructed by using relations \eqref{3}.
This inequality, since $f\in BV(\Omega)$, implies
$$ 
\int\limits_\Omega\vert\operatorname{grad}\!f(x)\vert\,\mathrm{d}x<\infty\iff \operatorname{grad}\!f(x)\in L^1(\Omega,\Bbb R^n)
$$
Having proved the $L^1$ integrability of the gradient of $f$, by applying \eqref{2} we have that
$$
\begin{split}
\int\limits_\Omega\vert\operatorname{grad}\!f(x)\vert\,\mathrm{d}x 
& \ge \sup\Bigg\{ \int\limits_{\supp\mathbf{g}}\vert\grd\!f(x)\vert\,\mathrm{d}x
\,\Big|\,\mathbf{g}\in C_0^1(\Omega,\Bbb R^n),\;|\mathbf{g}(x)|\le1\,\forall x\in\Omega \Bigg\} \\
&\ge \sup\Bigg\{\!\!-\!\!\int\limits_{\supp\mathbf{g}}\langle\grd\!f(x),{\mathbf{g}(x)}\rangle\,\mathrm{d}x \,\Big|\,\mathbf{g}\in C_0^1(\Omega,\Bbb R^n),\;|\mathbf{g}(x)|\le1\,\forall x\in\Omega\Bigg\} =V(f,\Omega)
\end{split}
$$
and thus
$$ 
V(f,\Omega)=\int\limits_\Omega\vert\grd f(x)\vert\,\mathrm{d}x\iff f\in W^{1,1}(\Omega)
$$
The reasoning is exactly the same if we consider $f\in BV(\Omega)\cap C^\infty(\Omega)$.
Notes
As stated above, this answer closely follows this one: the only point of departure is possibly the use of the inequality
$$
\sum_{i=1}^n \int\limits_{\Omega} \bigg|\frac{\partial f(x)}{\partial x_i}\bigg| \,  \mathrm{d}x \ge  \int\limits_{\Omega} |\grd f(x)| \,  \mathrm{d}x 
$$
Reference.
[1] Enrico Giusti (1984), Minimal surfaces and functions of bounded variations, Monographs in Mathematics, 80, Basel–Boston–Stuttgart: Birkhäuser Verlag, pp. XII+240, ISBN 978-0-8176-3153-6, MR 0775682, Zbl 0545.49018
