# Characterization of bounded variation

For a function $$f:[0,1]\to\mathbb{R}$$, define $$V(f)=\sup_{0=x_0 For $$f$$ with integrable derivative, the definition coincides with $$V(f)=\int_0^1|f'(x)|dx$$.

Now every continuous $$f:[0,1]\to\mathbb{R}$$ is uniformly approximable by a $$C^\infty$$ function (Weierstrass); denote by $$C_{\epsilon}(f)$$ the collection of all $$g\in C^\infty[0,1]$$ such that $$\sup_{x\in[0,1]}|f(x)-g(x)|\le\epsilon$$.

I conjecture that for every continuous $$f:[0,1]\to\mathbb{R}$$, we have $$V(f) = \limsup_{\epsilon\to0}\inf_{g\in C_\epsilon(f)}V(g).$$

Is this true? Known?

This is quite simple: if $$\|f - g\|_\infty \leqslant \epsilon$$, then clearly $$\sum_{i = 1}^n |f(x_i) - f(x_{i-1})| \leqslant \sum_{i = 1}^n |g(x_i) - g(x_{i-1})| + 2 n \epsilon \leqslant V(g) + 2 n \epsilon.$$ Therefore, $$\sum_{i = 1}^n |f(x_i) - f(x_{i-1})| \leqslant \inf_{g \in C_\epsilon(f)} V(g) + 2 n \epsilon,$$ and consequently $$\sum_{i = 1}^n |f(x_i) - f(x_{i-1})| \leqslant \liminf_{\epsilon \to 0^+} \inf_{g \in C_\epsilon(f)} V(g) .$$ Take the supremum over all possible partitions to get $$V(f) \leqslant \liminf_{\epsilon \to 0^+} \inf_{g \in C_\epsilon(f)} V(g) .$$
On the other hand, $$f$$ can be approximated by functions from $$C_\epsilon(f)$$ in the total variation norm (see below), so that $$V(f) \geqslant \inf_{g \in C_\epsilon(f)} V(g) ,$$ and hence $$V(f) \geqslant \limsup_{\epsilon \to 0^+} \inf_{g \in C_\epsilon(f)} V(g) .$$
Regarding approximation: if $$f$$ is non-decreasing, then there is a smooth function $$g \in C_\epsilon(f)$$ such that $$V(g) \leqslant V(f)$$ (just mollify $$f$$ appropriately). In the general case, if $$V(f) < \infty$$, then $$f = f_+ - f_-$$ with $$f_+$$ and $$f_-$$ non-decreasing and $$V(f) = V(f_+) + V(f_-)$$. Find $$g_+$$ and $$g_-$$ as above; then $$g = g_+ - g_-$$ is in $$C_{2\epsilon}(f)$$, and $$V(g) \leqslant V(g_+) + V(g_-) \leqslant V(f_+) + V(f_-) = V(f).$$