Let $f\in BV([a,b])$ then $|f|\in BV([a,b])$

Question

Let $f\in BV([a,b])$ then $|f|\in BV([a,b])$

My proof:

As $f\in BV([a,b])$ by hypothesis then $\sup\{Var(f,P;[a,b]):P\in\mathscr{P}_{[a,b]}\}<\infty$

Let $P=\{t_0,...,t_m\}$ a partition of $[a,b]$ then

$$Var(f,P;[a,b])=\sum\limits_{i=1}^{m}|f(t_i)-f(t_{i-1)}|<\infty$$

Moreover,

$$Var(|f|,P;[a,b])=\sum\limits_{i=1}^{m}||f(t_i)|-|f({t_{i-1}})||\leq\sum\limits_{i=1}^{m}|f(t_i)-f(t_{i-1)}|<\infty$$

Then

$$Var(|f|,P;[a,b])=\sum\limits_{i=1}^{m}||f(t_i)|-|f({t_{i-1}})||<\infty$$

In consequence:

$|f| \in BV([a,b])$

Is correct this?

Answer 1

The variation of a finite-valued function with respect to a particular partition is always finite, so on its face you haven't shown anything here. To fix it, go to $$Var(|f|,P;[a,b])=\sum\limits_{i=1}^{m}||f(t_i)|-|f({t_{i-1}})||\leq\sum\limits_{i=1}^{m}|f(t_i)-f(t_{i-1})|\leq \,? <\infty$$ and put some value that doesn't depend on $P$ in place of the question mark.

Answer 2

Yes, it is correct and as you can expect with these kinds of things more or less a direct consequence of the reverse triangle inequality.

Answer 3

It's correct. All you need to do now is take supremums over all partitions of $[a,b]$ as in the definition of total variation