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If $f:[0,1]\rightarrow[0,1]$ has continuous derivative. Can we say that $f$ is absolutely continuous?

Ittay Weiss
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mojipapa
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2 Answers2

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Is the derivative continuous?

If so, is $[0,1]$ compact?

Would this imply that $f$ has a bounded derivative?

If so, might this imply that $f$ is Lipschitz?

Udo Klein
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    Be carefull, if $C^1 [0,1]$ is defined as continuous on $[0,1]$ and differentiable on $(0,1)$ the derivate only exists on $(0,1)$, so $\sqrt{x}$ is not lipschitz. – Dominic Michaelis Feb 23 '13 at 21:57
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    Excellent point :) In this case the argument needs a little massage for 0 and 1. The functions $$f_n:= f((1-1/n)x+1/n))$$ look very nice. – Udo Klein Feb 23 '13 at 22:05
  • Hm. I think that for most people $C^1[0,1]$ is the space of functions which can be extended to a $C^1$ function on an open set containing $[0,1]$. It is a matter of definition, though, and the OP should probably tell us what is his. – Mariano Suárez-Álvarez Feb 24 '13 at 01:03
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Yes, you can say that f is absolutely continuous, since [0,1] is compact, and f is continuos (is $C^1$ therefore is $C^0$).

Affirmation: Every continuous function $f:[a,b] \rightarrow \mathbb R$ is absolutely continuos.

Proof:

$$\forall \epsilon > 0, \forall x \in [a,b], \exists\; \delta_x > 0;\\ y \in I_x\cap[a,b]=(x - \delta_x, x+\delta_x)\cap[a,b] \rightarrow \!|f(y) - f(x)| < \frac{\epsilon}{2}$$

It's easy to see that $[a,b] \subset \bigcup_{x\in[a,b]}I_x$. Since [a,b] is a compact there exists a finite number of $I_x$ that covers [a,b] (Borel-Lebesgue theorem). Therefore, $[a,b] \subset \bigcup_{\lambda = 1}^{n}I_\lambda$. $\delta = \min(\delta_{\lambda_1}, ..., \delta_{\lambda_n}) \rightarrow \forall x, y \in [a,b]; |y-x| < \frac{\delta}{2}. x, x_\lambda\in I_\lambda,$ for some $0<\lambda\leq n$, where $|x-x_\lambda| < \frac{\delta}{2} \rightarrow |f(x) - f(x_\lambda)| \leq \frac{\epsilon}{2}.$

$$|y - x_\lambda| \leq |y-x| + |x - x_\lambda| \leq \frac{\delta}{2} + \frac{\delta}{2} = \delta \rightarrow |f(y)-f(x)| \leq |f(y) - f(x_\lambda)| + |f(x_\lambda) - f(x)| \leq \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon$$

Sorry for the poor formating. Hope that helped.

Asaf Karagila
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Pedro Amorim
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    This only shows uniform continuity, not absolute continuity. The affirmation you state is actually false: the cantor function is continuous everywhere on $[0,1]$, but not absolutely continuous. – user6247850 Jun 26 '19 at 15:57