1

*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?

Daschin
  • 675
  • Without any additional assumption of $f$, the claim is false. Take $f(x) = 1$ if $x\in\mathbb{Q}$, $f(x) = 2$ if $x\in\mathbb{R}\setminus\mathbb{Q}$. – Rigel May 27 '17 at 11:30
  • @Rigel additional condition "$f: [a,b] \to \Bbb R$ be of bounded variation." is attached to OP. – Daschin May 27 '17 at 11:46
  • 1
    You can't prove that $1/f$ is of bounded variation in this way. To show that a function has bounded variation, you need an upper bound for the sum that is independent of the partition. Yes, in the (usual) definition, only finite partitions are considered, but these can have arbitrarily many points, so $\frac{2}{c} n$ is not bounded. – Daniel Fischer May 27 '17 at 12:02

2 Answers2

1

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

Gio67
  • 20,905
1

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

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

duncan
  • 395