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 ?
Are smooth BV functions dense in the set of all BV functions?
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0A related question: http://math.stackexchange.com/questions/8504/are-the-smooth-functions-dense-in-either-l2-or-l1 – 2011-07-28
1 Answers
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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0Thank you for the comment and the answer, that was one good thing to know. – 2011-07-28