TITLE: a problem on uniform continuity
QUESTION [5 upvotes]: Let $f : \mathbb{R} → \mathbb{R}$ be a continuous function. Which of the following imply
that it is uniformly continuous?
(a) $f$ is $2\pi$-periodic.
(b) $f$ is differentiable and its derivative is bounded on $\mathbb{R}$.
(c) $f$ is absolutely continuous.   
my thoughts:
(a) is a periodic function so uniformly continuous.
(b) false.example $\sqrt[3]{x}$.here i am confused because the answer is given that it is also true.
(c) i am not sure.

REPLY [3 votes]: (a) You can consider a compact subset $[a,b]$ s.t. $b-a=T$, where $T$ is the period. Then you have uniform continuity on $[a,b]$ by Heine-Cantor and hence uniform continuity over the whole $\mathbb R$. 
(b) If a differentiable function has bounded derivative it is Lipschitz continuous (you can take as Lipschitz constant $L= \sup_{x}\vert f' (x) \vert$) and hence also uniformly continuous. 
(c) Yes, it is well known that absolutely continuous is much stronger than continuity and uniform continuity.