# Bounded variation

1. Jul 11, 2011

### Rasalhague

Am I right in thinking that the statement "f:[a,b]-->R is of bounded variation" is equivalent to the statements "f:[a,b]-->R has bounded range" and "f;[a,b]-->R is a bounded function".

2. Jul 11, 2011

### micromass

Staff Emeritus
Hi Rasalhague!

No, you are not correct in thinking that. Being of bounded variation is much more strict than being bounded. For example, the function on [0,1]

$$f(x)=\sin(1/x)$$

is bounded between -1 and 1, but it is not of bounded varietion. Another example is the Dirichlet function on [0,1]:

$$f(x)=\left\{\begin{array}{c} 1~\text{if}~x\in \mathbb{Q}\\ 0~\text{if}~x\notin \mathbb{Q}\\ \end{array}\right.$$

This is bounded, but not of bounded variation. It can be shown that the following must hold for functions of bounded variation:

• The function is continuous everywhere except possibly in a countable set.
• The function has one-sided limits everywhere.
• The function has a derivative almost everywhere (i.e. except in a set of measure 0).

so you see that being of bounded variation is pretty strict.

3. Jul 11, 2011

### Rasalhague

Argh, thanks micromass, I see now where my confusion came from. For some reason my mind was just blanking out the summation sign!! Oopsh. Ack. Blushy-faced emoticon. I think it's asking for a rest :)