Is Of Bounded Variation, only for the finite partition?

Question

*prior to the body, note that title might be insufficient or inappropriate. Please edit it if it's needed.

I am proving the claim below:

Let $f: [a,b] \to \Bbb R$ be of bounded variation.

$f(x) \ge c \gt 0$ for all $x \in [a, b]$ where $c$ is a constant

$\Rightarrow$ $h(x)$ = $1 \over f(x)$ is of bounded variation on $[a, b]$.

To prove it, I had made up inequality such as - $\mid {1 \over h(x_i)} - {1 \over h(x_{i-1})} \mid \le \mid {1 \over h(x_i)}\mid + \mid {1 \over h(x_{i-1)}}\mid \le {2 \over c}$

Then I want to derive from above, the fact that $\sum_{i=1}^{n}\mid {1 \over h(x_i)} - {1 \over h(x_{i-1})} \mid \le {2 \over c}n$(*)

but from (*) to derive the fact that $\sum_{i=1}^{n}\mid {1 \over h(x_i)} - {1 \over h(x_{i-1})} \mid \le {2 \over c}n \lt\infty$, I need a guarantee that partition of definition of Bounded of Variation is $\lt \infty$.

so Is it true for the defnition of Bounded Variation, it only requires the finite partition?

Answer 1

$$\sum_{i=1}^n\left\vert\frac1{f(x_i)}-\frac1{f(x_{i-1})}\right\vert=\sum_{i=1}^n\left\vert\frac{f(x_{i-1})-f(x_{i})}{f(x_i)f(x_{i-1})}\right\vert\le\sum_{i=1}^n\frac{\vert f(x_{i-1})-f(x_{i})\vert}{c^2}\le\frac1{c^2}\text{Var}f$$

Answer 2

if f is of bounded variation, it's easy:

$|\frac{1}{f(x_i)} - \frac{1}{f(x_{i-1})}| = |\frac{f(x_{i-1}) - f(x_{i})}{f(x_i)f(x_{i-1})}| = \frac{1}{f(x_i)f(x_{i-1})}|f(x_{i-1}) - f(x_{i})| \leq \frac{1}{c^2}|f(x_{i-1}) - f(x_{i})|$