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