If function $f : [a,b] \to R$ is of bounded variation and differentiable, the total variation function $V(x) = V(f, [a:x])$ is also differentiable?

Question

I'm new to analytics and having some trouble with total variation.

I can prove that if above function $f$ is continuous, then $V$ is also continuous.

But I cannot thoroughly prove if $f$ is differentiable, $V$ is also differentiable.

Is there any thorough proof of this?

Answer 1

Let us consider the simpler case $f\in C^1([a,b])$, i.e., $f$ is differentiable in $[a,b]$ with continuous derivative. In particular, the map $x\mapsto |f'(x)|$ is continuous in $[a,b]$.

Then $$ V(x) := \text{Tot.Var.}(f, [a,x]) = \int_a^x |f'(t)|\, dt $$ is a $C^1$ function, and $V'(x) = |f'(x)|$ for every $x\in [a,b]$.

Answer 2

The answer is yes. The distributional derivative of a BV function is a signed measure $\mu=\mu_+-\mu_-$, and $V(x)=(\mu_++\mu_-)([a,x])$ is the total variation of this measure. Here we also know that $\mu_+-\mu_-$ is a continuous function, and the only way this can happen is that $\mu_{\pm}=f_{\pm}\, dx$, $f_{\pm}\in C$, are continuous separately. This follows because $\mu_{\pm}$ have disjoint supports, so there can't be any cancellations between singularities. So $V(x)=\int_a^x(f_+(t)+f_-(t))\, dt$ is in $C^1$.