Bounded variation on $\mathbb{R}$ We define for $v \in L^1(\mathbb{R})$
$$
Var_{\mathbb{R}}(v) := \mathrm{sup}\left \lbrace\int_{\mathbb{R}} u(x)\zeta'(x)~\mathrm{d}x: \zeta \in C_0^\infty(\mathbb{R}),~ \lvert \zeta(x) \rvert \leq 1\text{ for all }x \in \mathbb{R} \right \rbrace.
$$
If $u \in L^1(\mathbb{R}) \cap C^1(\mathbb{R})$ and $Var_{\mathbb{R}}(u) < \infty$, is it true that
$$
\int_{\mathbb{R}} \lvert v'(x) \rvert~\mathrm{d}x < \infty \quad ?
$$
I am sure that it is, but I can not quite find a rigorous proof. I also know that this holds for bounded domains. I also know that
$$
\int_{\mathbb{R}} v(x) \zeta'(x)~\mathrm{d}x = - \int_{\mathbb{R}} v'(x) \zeta(x)~\mathrm{d}x.
$$
Maybe approximating $-\mathrm{sign}(u')$ could help, but I do not see how. Mabye Riesz Representation Theorem can be helpful...
I am grateful for help or a reference. I have found some posts on SE that could not really help me. Maybe it is not even true...
 A: One can use the duality between $L^1(\mathbb R)$ and $L^\infty(\mathbb R)$ to give an elegant proof. First observe that by duality one has
\begin{align*}
\lVert f \rVert_1 &= \sup \bigg \lbrace \int_{\mathbb R} f(x) g(x) \, \mathrm d x  : g \in L^\infty(\mathbb R), \, \lVert g \rVert_\infty = 1 \bigg\rbrace \\
&= \sup \bigg \lbrace \int_{\mathbb R} f(x) g(x) \, \mathrm d x : g \in C^\infty_0(\mathbb R), \, \lVert g \rVert_\infty = 1 \bigg\rbrace
\end{align*}
for all $f \in L^1(\mathbb R)$. The second equality one can get as follows: "$\geq$" follows from the fact that $C^\infty_0(\mathbb R) \subseteq L^\infty(\mathbb R)$ and  "$\leq$" holds as $C^\infty_0(\mathbb R)$ is weak$^\ast$-dense in $L^\infty(\mathbb R)$ (see this post).
Now let $f \in L^1(\mathbb R) \cap C^1(\mathbb R)$ such that $\mathrm{TV}_{\mathbb R}(f) < \infty$. Then it follows from the above identity and integration by parts that
\begin{align*}
\lVert f' \rVert_1 &= \sup \bigg \lbrace \int_{\mathbb R} f'(x) g(x) \, \mathrm d x : g \in C^\infty_0(\mathbb R), \, \lVert g \rVert_\infty = 1 \bigg\rbrace \\
&= \sup \bigg \lbrace \int_{\mathbb R} f(x) g'(x) \, \mathrm d x : g \in C^\infty_0(\mathbb R), \, \lVert g \rVert_\infty = 1 \bigg\rbrace = \mathrm{TV}_{\mathbb R}(f) < \infty.
\end{align*}
Thus, we have $f' \in L^1(\mathbb R)$ with $\lVert f' \rVert_1 = \mathrm{TV}_{\mathbb R}(f)$.
