The closure of $C^1$ in the  functions of bounded variation Consider the space $(BV[0,1];||.||)$ with the norm
$$||f||=|f(0)|+V_{f}[0,1]$$
Where $V_{f}[0,1]$ is the variation of $f$. My questions
what is the closure of $C^1[0,1]$ with respect to this norm?
Another question is how to prove that this norm is Banach?
 A: I think the space is $W^{1,1}[0, 1]$. We clearly have that the closure (say $B$) is in $W^{1, 1}$. Furthermore, $W^{1, 1}$ is a proper subset of $\text{BV}$.
So, take a function $f$ in $W^{1, 1}$ and take an approximating sequence $f_n$ consisting of $C^\infty$ functions in the $W^{1, 1}$ norm.
So, we have $\|f_n - f\|_{\text{BV}} \lesssim \|f_n - f\|_{W^{1, 1}} \to 0$.
As $f_n$ are all in $C^1$ we also have that $f$ is in $B$.
Now we have 
$$f(x) = f(0) + \int_0^x f'(t) \, \textrm{d}t.$$
So, $\|f\|_{W^{1, 1}} = \|f\|_{L^1} + \|f'\|_{L^1}$.
And $$\|f\|_{L^1} \leqslant |f(0)| + \int_0^1 \left | \int_0^x f'(t) \, \textrm{d}t \right | \leqslant |f(0)| + \int_0^1 |f'(t)| \, \textrm{d}t.$$
A: sIt is the space $W^{1,1}[0, 1]$. 
Note initially that $W^{1,1}[0, 1]$ is a subspace of $BV$ moreover in there the norms are equivalent. In fact if $f\in W^{1,1}[0, 1]$, for its continuous representative 
$$f(x)=f(0)+\int_{0}^{x}f'(t)dt\tag{1}$$
Then
$$|f(x)|\leq|f(0)|+|\int_{0}^{x}f'(t)dt|\leq |f(0)|+\int_{0}^{1}|f'(t)|dt.$$
But $V_f[0,1]=\int_{0}^{1}|f'(t)|dt$. So 
$||f||_L^1\leq||f||_{\infty}\leq||f||_{BV}$
since
$||f'||_L^1\leq||f||_{BV}$ then
$||f||_{W^{1,1}}\leq 2||f||_{BV}.$
Using $(1)$ again we get
$$f(0)=-f(x)+\int_{0}^{x}f'(t)dt$$
then
$|f(0)|\leq |f(x)|+\int_{0}^{1}|f'(t)|dt$,
integrating from $0$ to $1$
$|f(0)|\leq ||f||_{W^{1,1}}$ and analogously
$||f||_{BV}\leq 2||f||_{W^{1,1}}$ ergo both norms are equivalent.
If $f_n \to f$ in  $BV$ with $(f_n)\subset C^{1}$   then $(f_n)$ is a Cawchy 
sequence in $W^{1,1}$, since it is a complete space $f_n\to g$ in $W^{1,1}$.
Then 
$f_n\to g$ in $BV$ because the norms are equivalents and by the unity of the limit  $g=f\in BV$.
This proves that the closure of ${C^1}$ in the $BV$ norm is a subspace of $W^{1,1}$. 
The other inclusion is given because any function $f\in W^{1,1}$ 
can be approximated by a 
sequence $f_n\to f $ in $W^{1,1}$ what is the same $f_n\to f $ in $BV$ $\blacksquare$
PS: That is my answer it may contains some English problems but it is mathematically correct! And all details are included! It is of course inspired on @Jonas answer! And he deserved to gain the bounty, and I thanks to him! But how he was not so sure ( he wrote I think it is W1,1) I had to put a complete answer. My first thought were to edit his answer but I realized it was a complete rewriting. I decided to put a new answer.
