Are $C^1$, $BV$ functions dense in set of all continuous $BV$ functions with respect to $BV$ norm? I was wondering whether any $BV$ function can be approximated arbitrarily well with a piecewise linear function in sense of $BV$ norm.
Let:
$A$ - be the set of all continuous functions $[0,1]\rightarrow\mathbb{R}^n$ of bounded variation which map $0$ to $0$,$B$ - be the set of all continuous piecewise linear functions $[0,1]\rightarrow\mathbb{R}^n$ of bounded variation which map $0$ to $0$.
Is $B$ dense in $A$ with respect to the $BV$ norm: $\left\|f-g\right\|=\text{var}(f-g)$ ($\text{var}$ is like the variation...)?
More generally, is something like the opposite of the next statement true?
The statement that I know is true is:
$f_n\xrightarrow[BV]{}f\Rightarrow f_n\xrightarrow[\left\|\cdot\right\|_\infty]{}f$, when $n\rightarrow\infty$,
but can it be that the opposite is also true? Something like:
$f_n\xrightarrow[\left\|\cdot\right\|_\infty]{}f\wedge\text{var}(f_n)\rightarrow\text{var}(f)\Rightarrow f_n\xrightarrow[BV]{}f$, when $n\rightarrow\infty$.
I ask the second question because if it was true, then I would be able to answer the first question with a construction.
 A: No, the type of convergence you're considering is too strong.  (In the answer below, I'm assuming $B$ is the set of continuous, piecewise linear functions on $[0,1]$, hence a subset of $A$.)
On the definition of $A$:  The first thing to check is that $A$ is actually a Banach space as defined.  This isn't completely obvious, not only because of the requirement that the functions vanish at zero but also because of the continuity requirement.
It is worth recalling that if $f : [0,1] \to \mathbb{R}$ has bounded variation, then there is a finite (signed) Borel measure $\mu_{f}$ on $[0,1]$ acting as the weak derivative of $f$:
\begin{equation*}
\int_{0}^{1} f(x) g'(x) \, dx = f(1^{-}) g(1) - f(0^{+}) g(0) -\int_{0}^{1} g(\xi) \, \mu_{f}(d\xi) \quad \text{if} \, \, g \in C^{\infty}((0,1)).
\end{equation*}
It is possible to show that if $[a,b] \subseteq [0,1]$, then
\begin{equation*}
TV(f; [a,b]) = \|\mu_{f}\|([a,b]),
\end{equation*}
where $\|\mu_{f}\|$ is the total variation measure of $\mu$ (i.e. if $\mu_{f} = \mu^{+}_{f} - \mu^{-}_{f}$ is the Hahn decomposition of $\mu$, then $\|\mu_{f}\| = \mu^{+}_{f} + \mu^{-}_{f}$).  In particular,
\begin{equation*}
\text{var}(f) = \|\mu_{f}\|([0,1]).
\end{equation*}
After recalling the previous preliminaries, it becomes clear that $A$ is a Banach space.  Indeed, if $(f_{m})_{m \in \mathbb{N}}$ is a Cauchy sequence in $A$ as defined above, then $(\mu_{f_{m}})_{m \in \mathbb{N}}$ is Cauchy with respect to the total variation norm $\|\cdot\|_{TV}$ defined by $\|\nu\|_{TV} = \|\nu\|([0,1])$.  In particular, it converges to some measure $\mu$.  Further, since the continuity of the functions $(f_{m})_{m \in \mathbb{N}}$ implies $\|\mu_{f_{m}}\|(\{x\}) = 0$ independently of the point $x \in [0,1]$ or the index $m \in \mathbb{N}$, the total variation convergence implies $\|\mu(\{x\})\| = 0$ independently of $x \in [0,1]$.  Finally, using the following identity (which follows by choosing an appropriate sequence of $g$ in the definition of weak derivative above)
\begin{equation*}
f_{m}(x) = f_{m}(0) + \mu_{f_{m}}([0,x]) = \mu_{f_{m}}([0,x]) \quad \text{if} \, \, x \in [0,1], \, \, m \in \mathbb{N},
\end{equation*}
we see that we can define a function $f : [0,1] \to \mathbb{R}$ by
\begin{equation*}
f(x) = \mu([0,x]) = \lim_{m \to \infty} \mu_{f_{m}}([0,x]) = \lim_{m \to \infty} f_{m}(x).
\end{equation*}
In fact, the convergence $f_{m} \to f$ is uniform and $\mu(\{0\}) = 0$ so $f(0) = 0$.  This proves $A$ is a Banach space.
$B$ is not dense:  In the previous discussion, we actually saw that the norm $\|\cdot\| = \text{var}(\cdot)$ is slightly stronger than the supremum norm.  It's not enough to make $B$ (or even the $C^{1}$ functions) dense in $A$.
The key here is that $B$ is a subset of $W^{1,1}([0,1])$, but many functions in $A$ are not in $W^{1,1}([0,1])$.
The Sobolev space $W^{1,1}([0,1])$ is the space of (absolutely) continuous functions $f : [0,1] \to \mathbb{R}$ with a weak derivative $\mu_{f}$ satisfying $\mu_{f} \ll dx$.  (Here $dx$ is the Lebesgue measure on $[0,1]$.)  Usually we define the measurable function $f'$ by $f' = \frac{d \mu}{dx}$.  It is possible to show that $f$ is differentiable at almost every point in $[0,1]$ and its derivative equals $f'$ at those points.
$W^{1,1}([0,1])$ is made a Banach space with the following norm:
\begin{equation*}
\|f\|_{W^{1,1}([0,1])} = \int_{0}^{1} \left( |f(x)| + |f'(x)| \right) \, dx.
\end{equation*}
If $f \in W^{1,1}([0,1])$, then an exercise shows that $TV(f; \cdot)$ can be computed as
\begin{equation*}
TV(f; [a,b]) = \int_{a}^{b} |f'(x)| \, dx.
\end{equation*}
Hence the norm $\|\cdot\|$ defined above in $A$ induces the same topology on $B$ as the norm $\|\cdot\|_{W^{1,1}([0,1])}$.  Thus, $\overline{B} \subseteq A \cap W^{1,1}([0,1])$.  (In fact, equality holds.)
On the other hand, $A \setminus W^{1,1}([0,1])$ is not empty.  If $\mu$ is any singularly continuous measure in $[0,1]$ (that is, a measure that is singular with respect to $dx$, but has no atoms), then the function $f(x) = \mu([0,x])$ is in $A$, but not in $W^{1,1}([0,1])$.  (An example of such a function is the "Devil's Staircase.")
