A continuous function on $[0,1]$ not of bounded variation

Question

I'm looking for a continuous function $f$ defined on the compact interval $[0,1]$ which is not of bounded variation.

I think such function might exist. Any idea?

Of course the function $f$ such that $$ f(x) = \begin{cases} 1 & \text{if $x \in [0,1] \cap \mathbb{Q}$} \\\\ 0 & \text{if $x \notin [0,1] \cap \mathbb{Q}$} \end{cases} $$ is not of bounded variation on $[0,1]$, but it is not continuous on $[0,1]$.

score 33 · Accepted Answer · answered Dec 27 '14 at 23:04

33

Consider any continuous function passing through the points $(\frac1{2n},\frac1n)$ and $(\frac1{2n+1},0)$, e.g. composed of linear segments. It must have infinite variation because $\sum\frac1n=\infty$.

answered Dec 27 '14 at 23:04

user2345215

16,803

2

Thank you for your answer. Can you draw an example of your function? – Olivier Oloa Dec 28 '14 at 00:35
3

@OlivierOloa I'm a little tired so this will have to do: https://i.sstatic.net/phWFJ.png. It should be easy to imagine anyway. – user2345215 Dec 28 '14 at 01:44
It is OK! Thanks! – Olivier Oloa Dec 28 '14 at 01:54
1

There is no such continuous function over the compact set $[0,1]$ since the construction implies that the function is unbounded. – Muschkopp May 18 '18 at 11:22
1

@Muschkopp it doesn't, the construction implies the function is bounded by $1/2$. – Giafazio Feb 03 '23 at 14:55

Kola B. · Answer 2 · 2014-12-28T21:50:15.873

14

For example $x\sin(1/x)$ (and 0 for $x=0$).

The variation is unbounded, because it is bounded ;-) from below by the sum of absolute values of extrema, but it is the harmonic-like series.

edited Dec 28 '14 at 21:50

answered Dec 27 '14 at 23:02

Kola B.

2,507

Thanks for your answer. +1. – Olivier Oloa Dec 27 '14 at 23:39

score 10 · Answer 3 · edited Aug 12 '21 at 00:54

10

Here is an general idea. A function of bounded variation on $[0,1]$ is necessary differentiable a.e.. In another word, any continuous function which is not differentiable on a positive measure set is not of bounded variation. So, Weierstrass Function is not of bounded variation for sure.

Also, some function which oscillate too much can not be bounded variation as well, for example $$ u(x)=x^a\sin(\frac{1}{x^b}) $$ is not of bounded variation as long as $1 \le a\leq b$.

edited Aug 12 '21 at 00:54

Geoffrey Sangston

2,310

answered Dec 29 '14 at 01:55

spatially

5,720

2

Yes, with $1 < a \le b$ and $u(0)=0$, this also provides an example of a function that is differentiable everywhere but not of bounded variation. – cxseven Mar 25 '16 at 01:12
The property about $x^a \sin(1/x^b)$ is exercise 4.4b on page 205 of Benedetto-Czaja, or exercise 35 of page 119 in Royden-Fitzpatrick. Both exercises assume $a$ and $b$ are positive. Note that if $a$ is negative then $x^a \sin(1/x^b)$ will be unbounded, so will never have bounded variation. – Geoffrey Sangston Aug 12 '21 at 12:59

score 6 · Answer 4 · answered Dec 27 '14 at 23:06

6

The Weierstrass Function restricted to the interval $[0,1]$

answered Dec 27 '14 at 23:06

Kyle Gannon

6,463
2
21
36

A continuous function on $[0,1]$ not of bounded variation

4 Answers4

Linked