# Approximating BV Function by Piecewise Constant Functions

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 $$[f]_{BV} := \sup_{\text{partitions}}\sum_{i = 1}^{n}|f(x_i) - f(x_{i+1})|$$ 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.)

• 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? Dec 7, 2015 at 23:32
• I like this idea - so if $V(f_k - f, [a,b]) \to 0$, then that would imply $$\int_{a}^{b}|g'| = \int_{a}^{b}|f'_k - g'| \leq V(f_k - f, [a,b]) \to 0$$ which is clearly false unless $g' = 0$ a.e. Dec 8, 2015 at 1:21
• 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. Dec 8, 2015 at 2:46
• 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.) Dec 9, 2015 at 0:47

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.