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Let a compact interval $[a,b]\subset \mathbb{R}$. If a function $f:[a,b]\to \mathbb{R}$ is continuous and increasing on $[a,b]$, then what can be said about the derivative $f'$? Is it continuous? If it is, then how do we prove it? I came up with this claim, because I want to prove that if $f$ is continuous and increasing on $[a,b]$, then it its derivative is bounded on $[a,b]$.

This is not a homework. I just wanted to fix it. To be honest, I studied an article on Henstock-Stieltjes integral and the problem that I posted is one of the stated statements in that article.

Thanks in advance...

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It seems a bit odd how you state that $f$ is continuous but you don't state that $f$ is differentiable. – wj32 Dec 11 '12 at 10:24
Without the assumption that $f$ is increasing, you cannot even claim that the derivative exists anywhere (the Weierstrass function is an example of a continuous function without derivative anywhere). With the assumption that $f$ is increasing, or more generally of bounded variation, one can prove the derivative exists almost everywhere (i.e. everywhere excluding a set of Lebesgue measure zero). However, it is not necessarily continuous, as explained in the answer below. – Yoni Rozenshein Dec 11 '12 at 10:49
@Yoni Rozenshein Noted. That is why I have deleted my comment. Thanks. :) – juniven Dec 11 '12 at 10:58
up vote 1 down vote accepted

The derivative $f'$ need not be continuous. Let $f$ be a continuous and increasing function on $[0,2]$, $f=x $ when $x\in[0,1]$, and $f=2x-1$ when $x\in (1,2]$.

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Is it because that $f'$ is not continuous at 1? Ah okay, right hand limit is 2 while left hand limit is 1. Am I right? – juniven Dec 11 '12 at 10:37
Because of this I have to give up the claim. So, is it still true that if $f$ is both increasing and continuous on the compact interval $[a,b]$, then its derivative is bounded on $[a,b]$?(We assumed that $f$ is differentiable) – juniven Dec 11 '12 at 10:48
If $f$ is continuous on a compact interval $[a,b]$, then it must be uniformly continuous, so if $f$ is differentiable on $[a,b]$, $f'$ must be bounded. No need to assume its monotonicity. – Chao Chen Dec 11 '12 at 12:44
So, uniform continuity and differentiability of $f$ on $[a,b]$ implies that $f'$ is bounded on $[a,b]$. Thanks. I still have to prove that statement, though at this moment, I don't know how to do it.:) – juniven Dec 11 '12 at 15:20
Sorry I made a mistake, uniform continuity and differentiability of $f$ does not imply $f'$ is bound. Consider $y=-\sqrt{1-x^{2}}$ on $[0,1]$, it's a quarter of a circle. – Chao Chen Dec 12 '12 at 11:07

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