Bounded function of variation

Question

Show that if $f$ is continuous at $[a,b]$ and $\left|f\right|\in BV\left[a,b\right]$

then $f\in BV\left[a,b\right]$.

I don't have a solution for this excersise. Any help or hint would be appreciated.

Answer 1

Hint:

You can't always be sure that $ |f(x_k) - f(x_{k-1})| \leqslant |\,|f(x_k)| - |f(x_{k-1})|\,|$ unless the signs of $f(x_k)$ and $f(x_{k-1})$ are the same, but if $f(x_k)$ and $f(x_{k-1})$ have opposite signs then by continuity there is a point $c_k \in (x_{k-1},x_k)$ where $f(c_k) = 0$ and

$$|f(x_k) - f(x_{k-1})| \leqslant |\,|f(x_k)| - |f(c_{k})|\,|+ |\,|f(c_k)| - |f(x_{k-1})|\,| $$

Answer 2

This is false for complex-valued functions. For example if $\theta:[a,b]\to\mathbb R$ is any function whatever and $f=e^{i\theta}$ then $|f|\in BV$.