Examples of $f \in C^2[a,b]$ where the total variation of $f$ on $[a,x]$ is not in $C^2[a,b]$

Question

Suppose $f:[a,b]\to\mathbb{R}$ is of bounded variation; define $V(x) = V[f;a,x]$ (the total variation of $f$ on $[a,x]$.

I want to show that $V \in C^1[a,b]$. Since $f'$ is continuous, hence bounded on $[a,b]$ I have that $f$ is absolutely continuous. From that I have $V' = |f'|$, and since $f'$ is continuous, $|f'| = V'$ must be as well, correct?

If instead $f \in C^2[a,b]$, I'm looking for an example $f$ where $V \not\in C^2[a,b]$. From the above result, I think that means I should be looking for an $f$ where $f''$ is continuous but $|f'|'$ is not.

Thanks for any examples and/or suggestions on how to construct such an $f$.

Answer 1

How about $x\mapsto x^2$ on $[-1,1]$?

Answer 2

I think $f(x) = \frac{1}{2}x^2 - x$ on $[0,2]$ would do the trick. Then $f'(x) = x-1$, $f''(x) = 1$ so $f \in C^2$. But $|f'|' = \pm 1$ on $[0,1)$ and $(1,2]$ respectively, so $|f'|' = V''$ is not continuous.