Is $f$ increasing

Question

The Cantor-Lebesgue function gives an example of a function which is increasing but whose derivative (where it exists) is $0$. But is the following true?

Let $f$ is a continuous function on a compact set, $K \subset \mathbb{R}$. Suppose $f'$ exists and $f'>0$ on a dense subset of $K$, then $f$ is increasing.

Does it make a difference if $K$ is an interval or not?

I suppose K is a subset of $\mathbb R$. To define derivative f at a point it has to be defined in a neighborhood of the point. Some clarification on what kind of K you want to consider is needed. Perhaps you want to take K to be a compact interval. — Kavi Rama Murthy, Mar 16 '18 at 05:40
@KaviRamaMurthy The spaces I had in mind were indeed in $\mathbb{R}$ or $\mathbb{C}$. I shall make the edit. — NPH, Mar 16 '18 at 05:41
You are still of the mark!. A set like ${0,1/2,1/3,...}$ is compact in $\mathbb R$. How do you define derivative of a function defined on this set? — Kavi Rama Murthy, Mar 16 '18 at 08:39

score 2 · Accepted Answer · answered Mar 16 '18 at 06:45

Minkowski's question mark function is a strictly increasing continuous function $?$ on $[0,1]$, with $?(0)=0$ and $?(1)=1$, whose derivative exists and is $0$ on the rationals. Thus $f(x) = x - ?(x)$ is a continuous function whose derivative is $+1$ on a dense subset of $[0,1]$, but $f(1) = f(0)=0$ so $f$ is not increasing.

Is $f$ increasing

1 Answers1