Show that if $f$ is continuous and has finite number of maxima and minima on $[a,b]$, then $f \in BV([a,b]).$

Question

I am not entirely sure how to approach this problem so I would very much appreciate some help on getting me on the right track...

I know that $\, f \in BV([a,b]) \iff V_{a}^{b} f < \infty \iff$ sup{$T(f,P) | P \in \mathscr{P}[a,b] $} $< \infty$.

We also have that since $f$ is a continuous function on a compact set $[a,b]$ that $f$ must attain its maximum and minimum values on $[a,b]$. In other words, $f$ is bounded.

So $m := \inf_{x \in [a,b]} f(x)$ and $M := \sup_{x \in [a,b]} f(x)$ and that $m \leq f(x) \leq M$ for all $x \in [a,b]$.

But I'm not sure how to proceed.. I started off doing a proof by contradiction:

Suppose that $f$ were continuous and has finitely many minima and maxima on $[a,b]$, but $f \notin BV([a,b]).$ Then it must be the case that $\sup${$T(f,P) | P \in \mathscr{P}[a,b] $} $\rightarrow \infty$. I'm not sure where to go next...

Thanks in advance for the help!

Answer 1

List the local extrema: $\left \{ x_1,\cdots, x_n \right \}.$

Now make the following observations:

$1).$ Since $f$ is continuous, between any two successive (local) maxima, there is exactly one (local) minimum, and vice versa. (why?). Wlog the first extremum is a maximum. It either occurs at $x=a,\ $ or at a point interior to $[a,b].$ In the first case,

$2). f$ is strictly decreasing on $(a,x_1),\ $ strictly increasing on $(x_2,x_3)\ $ (why?), and so on until the process stops at $(x_{n-1},x_n).$ In the second case $f$ is strictly increasing on $(a,x_1),\ $ strcitly decreasing on $(x_1,x_2)\ $ and so on until the process stops at $(x_{n-1},x_n).$ In either case, $f$ is either strictly increasing or strictly decreasing on each $(x_{i-1},x_i).$ This implies that we now have

$3).$ $V_f[a,b]=V_f[a,x_1]+V_f[x_1,x_2]+\cdots +V_f[x_{n},b]=|f(x_1)-f(a)|+\cdots +|f(b)-f(x_n)|,\ $ and this sum is clearly finite.