# Are smooth BV functions dense in the set of all BV functions?

Consider the set $B$ of functions of bounded variation which are of the form $f: (0,1) \to \mathbb{R}$ and the subset $S$ which contains all the elements of $B$ that are smooth. I'd like to know whether $S$ dense in $B$ ? One argument comes to my mind is , Yes $S$ is dense in $B$ and the reason is the existence of a sequence of smooth functions in the form of Fourier series. But some how this argument seems to be not fully true as I am not sure whether all functions of BV have a Fourier series converging to them. I request you to clarify first of all whether $S$ is dense in $B$ and is the argument using Fourier series sufficient ?

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I don't know what the topology on $B$ is that you have in mind (presumably the total variation norm?) but most likely the answer is no, simply because functions in $B$ are not necessarily continuous, and a uniform limit of continuous functions is continuous. That is, the closure of $S$ in $B$ is going to consist of continuous functions. – Akhil Mathew Jul 28 '11 at 4:28
@Akhil : the norm is $\mathcal{L}^2$. – Rajesh Dachiraju Jul 28 '11 at 4:37
@Jonas Meyer : here the set $S$ is not the set of all smooth functions but only the set of smooth functions that are of Bounded variation. This makes the question non trivial. – Rajesh Dachiraju Jul 28 '11 at 4:57
Rajesh: Sorry, you're right that it wasn't as immediate as I was making it out to be. I'll delete my previous comment because it could be misleading, and properly answer in an answer. – Jonas Meyer Jul 28 '11 at 4:59
To put the $\mathcal{L}^2$ norm on $BV$ strikes me as very strange... – t.b. Jul 28 '11 at 5:11

Let $S'$ denote the set of smooth functions on $(0,1)$ with compact support. Then $S'\subset S$ and $S'$ is dense in $L^2$. Therefore $S'$ is dense in $B$ in the restricted $L^2$ norm.

For example, suppose that $f$ has a jump discontinuity at $a$, with $$\left|\lim_{x\to a^+}f(x)-\lim_{x\to a^-}f(x)\right|=c>0.$$ If $g$ is smooth, then $g$ is continuous, so $$\lim_{x\to a^{\pm}}(f(x)-g(x))=\left(\lim_{x\to a^{\pm}}f(x)\right)-g(a),$$ and therefore $$\left|\lim_{x\to a^+}(f(x)-g(x))-\lim_{x\to a^-}(f(x)-g(x))\right|=c.$$ This implies that the total variation of $f-g$ is at least $c$, which shows that $f$ cannot be approximated in the total variation seminorm by smooth functions.