Let $[a,b]$ be a real bounded interval and let $\mathfrak{P}([a,b])$ be the set of all partitions of $[a,b]$, i.e. $\mathfrak{P}([a,b])=\{\{t_0,\dotsc,t_n\}:a=t_0<t_1<\dotso<t_n=b\}$. Define:
$$\mathrm{Var}_f[a,b]:=\sup_{\{t_0,\dotsc,t_n\}\in\mathfrak{P}([a,b])}\sum_{i=1}^n|f(t_i)-f(t_{i-1})|,$$
for an arbitrary $f:[a,b]\to\mathbb{R}$, to be the total variation of $f$ over $[a,b]$.
Definition
We call a function $f:[a,b]\to\mathbb{C}$ BV if $\operatorname{Var}_f[a,b]<\infty$.
Definition 2
We call $f:[a,b]\to\mathbb{C}$ BVD if the induced distribution has a distributional derivative $D_f'$ which is a finite signed measure.
Since $L^\infty$ has the space of finitely additive signed measures as its dual, BVD is equivalent to:
$$\left|\int f\phi'\right|\leq C\|\phi\|_{L^\infty}. \tag{$\ast$}$$
$\chi_{\mathbb{Q}}$ is not BV since one needs only select $n$ points of $[0,1]$, starting with 0 and ending with 1, such that if $x_i$ is rational $x_{i+1}$ is irrational and viceversa to get a partition giving a sum of $n$, thus proving $\operatorname{Var}_f[a,b]\geq n$ for any $n$, since the above procedure is possible for any $n$. However, as a distribution, it is the zero distribution, meaning the derivative is also zero, which is clearly a finite signed measure. So BVD does not imply BV.
Now I was wondering: does the other implication hold? Otherwise known as: (how) can I find $C$ such that $(\ast)$ holds, knowing only that $\operatorname{Var}_f[a,b]<\infty$?