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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. –  Akhil Mathew Jul 28 '11 at 4:28
    
@Akhil : the norm is $\mathcal{L}^2$. –  Rajesh D 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 D 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
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To put the $\mathcal{L}^2$ norm on $BV$ strikes me as very strange... –  t.b. Jul 28 '11 at 5:11

1 Answer 1

up vote 2 down vote accepted

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. –  Rajesh D Jul 28 '11 at 6:04
    
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. –  Jonas Meyer Jul 28 '11 at 6:18
    
Thank you for the comment and the answer, that was one good thing to know. –  Rajesh D Jul 28 '11 at 6:22

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