Suppose that $f$ is a real-valued function on $\Bbb R$ whose derivative exists at each point and is bounded. Prove that $f$ is uniformly continuous.

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    $\begingroup$ Could you provide any work you've done on this problem already? $\endgroup$ – Ian Coley Apr 30 '13 at 19:40
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    $\begingroup$ This is a trivial consequence of the mean value theorem. It is more interesting to note that $f$ can be uniformly continuous, even if its derivative is unbounded. You'll see examples here. One of them is easy to extend to $\mathbb{R}$. $\endgroup$ – Julien Apr 30 '13 at 20:22

Since $f'$ is bounded then there's $M>0$ s.t. $$|f'(x)|\leq M\quad \forall x\in\mathbb{R}$$ hence by mean value theorem we find $$|f(x)-f(y)|\leq M|x-y|\quad \forall x,y\in\mathbb{R}$$ so $f$ is a lipschitzian function on $\mathbb{R}$ and therefore it's s uniformly continuous on $\mathbb{R}$.


Hint: $f(a)-f(b)=f'(\xi)(a-b)$.


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