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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.

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Your example for (b) is not quite right. The derivative of $\sqrt[3]{x}$ is $\frac{1}{3x^{3/2}}$ which is not bounded. It tends to infinity when $x$ tends to $0$. In fact, the derivative is not defined at $x=0$. – Paul Dec 31 '12 at 12:48
up vote 4 down vote accepted

(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.

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(b) Assume $|f'(x)|<M$ for all $x$. Then $f(y)-f(x)=f'(\xi)(y-x)$ with some $\xi$ between $x$ and $y$, hence $|y-x|<\delta:=\frac \epsilon M$ implies $|f(y)-f(x)|<\epsilon$.

(c) should be seen as trivially true (consider sums with only one summand in the definition of absolute continuity)

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