For a function $f:[0,1]\to\mathbb{R}$, define $$ V(f)=\sup_{0=x_0<x_1<\ldots<x_n=1}\sum_{i=1}^{n}|f(x_n)-f(x_{n-1})|. $$ 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 $$ V(f) \leqslant \liminf_{\epsilon \to 0^+} \inf_{g \in C_\epsilon(f)} V(g) .$$

EDIT: I forgot about the converse inequality:

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 $$V(g) \leqslant V(g_+) + V(g_-) \leqslant V(f_+) + V(f_-) = V(f).$$


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