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.)
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.