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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.2011-07-28
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    @Akhil : the norm is $\mathcal{L}^2$.2011-07-28
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    @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.2011-07-28
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    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.2011-07-28
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    To put the $\mathcal{L}^2$ norm on $BV$ strikes me as very strange...2011-07-28
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    @Rajesh: In general, smooth functions will be dense in any space like $L^2$ because of the existence of approximations to the identity (a standard trick). The idea is you approximate the "delta function" by smooth functions $\delta_\epsilon, \epsilon \simeq 0$, and approximate $f = f \ast \delta$ by $f \ast \delta_\epsilon$. Note that the convolution of something with something smooth is smooth.2011-07-28
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    A related question: http://math.stackexchange.com/questions/8504/are-the-smooth-functions-dense-in-either-l2-or-l12011-07-28

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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.

(Please ask for elaboration if any of these claims is unclear.)


If instead you were considering some norm like the one in the Wikipedia article, then the smooth functions would not be dense.

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.

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    From the initial part of your answer, it seems that one thing which i did not know was that all smooth functions with compact support are of bounded variation. Please clarify whether its correct.2011-07-28
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    Rajesh, yes, they are. This follows from the fact that their derivatives are continuous with compact support, hence bounded. From here you can apply the mean value theorem or the fundamental theorem of calculus to see that smooth functions with compact support are Lipschitz, from which BV easily follows.2011-07-28
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    Thank you for the comment and the answer, that was one good thing to know.2011-07-28