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Let $[a,b] \subset \mathbb{R}$ be a bounded interval. A function $f: > [a,b] \to \mathbb{R}$ is of bounded variation if $\sum_{i = > 1}^{n}|f(x_i) - f(x_{i+1})| \leq C$ for all partitions $\{x_1,\ldots,x_n\}$ of $[a,b]$. Let \begin{equation} [f]_{BV} := \sup_{\text{partitions}}\sum_{i = 1}^{n}|f(x_i) - f(x_{i+1})| \end{equation} denote the BV seminorm.

What I was wondering is the following:

Do there exist piecewise constant $f_k$ such that $[f_k - f]_{BV} \to 0$?

I know there exists piecewise constant $f_k$ satisfying $[f_k]_{BV} \to [f]_{BV}$, but I was wondering if the stronger result holds too. (Intuitively, it seems like it should hold, but I can imagine there may be some pathological counterexample out there.)

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    $\begingroup$ For a strictly real-variable proof (that this is wrong), perhaps use the fact that $$\int_a^b |g'(t)|dt \leq V(g,[a.b])$$ and that $f_k'=0$ a.e. for your piecewise constant functions? $\endgroup$
    – B. S. Thomson
    Dec 7, 2015 at 23:32
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    $\begingroup$ I like this idea - so if $V(f_k - f, [a,b]) \to 0$, then that would imply \begin{equation} \int_{a}^{b}|g'| = \int_{a}^{b}|f'_k - g'| \leq V(f_k - f, [a,b]) \to 0 \end{equation} which is clearly false unless $g' = 0$ a.e. $\endgroup$
    – Lost in a Maze
    Dec 8, 2015 at 1:21
  • $\begingroup$ Exactly! We have a necessary condition. I won't speculate on what exactly are the functions that are limits in this norm of sequences of piecewise constant functions, but maybe you should. $\endgroup$
    – B. S. Thomson
    Dec 8, 2015 at 2:46
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    $\begingroup$ Perhaps you wanted only a negative answer, but one shouldn't leave it without characterizing this class. David Ullrich's answer should be enough of a clue. (Hint: not all singular functions are the limits in norm of sequences of piecewise constant functions.) $\endgroup$
    – B. S. Thomson
    Dec 9, 2015 at 0:47

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No. If $f$ is BV, right continuous and satisfies $f(-\infty)=0$ then there exists a complex measure $\mu$ so $f(x)=\mu((-\infty,x])$. The BV seminorm is the same as the total variation norm $||\mu||$.

If $f$ is piecewise constant then $\mu$ is a linear combination of point masses. So your question is equivalent to asking whether a complex measure can be approximated in norm by a linear combination of point masses, and the answer to that is no.

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