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Let $f\colon(a,b) \to \mathbb R$ be differentiable and suppose that there exists an $M>0$ such that |f'(x)| \leq M for all $x$ in $(a,b)$. Prove that $f$ is uniformly continuous.

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    I tried to fix up some of the formatting and give the question a more descriptive title. Consider adding some thoughts on what you've tried and where you're stuck. Questions consisting of merely a command to prove something are not popular here. Welcome, by the way!2012-04-01

3 Answers 3

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Expanding on Azreal's hint: The Mean Value Theorem implies that for any distinct $x$, $y$ in $(a,b)$ we have $\tag{1} \biggl|{f(x)-f(y)\over x-y}\biggr|\le M. $

Keep in mind what you need to show:

Given $\epsilon>0$, there is a $\delta>0$ such that $ |f(x)-f(y)|\lt \epsilon\quad\text{whenever}\quad |x-y|\lt\delta. $

Can you see how to use $(1)$ to find the required $\delta$?

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    @Dan We apply the Mean Value Theorem to $f$ on the interval determined by $x$ and $y$, not on the interval $[a,b]$. Since $x$ and $y$ are elements of $(a,b)$, the Mean Value Theorem indeed applies.2012-11-30
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Let us chose some $\epsilon>0$. We need to prove the existence of $\delta>0$ such that if $|x-x'|<\delta$ then $|f(x)-f(x')|<\epsilon$. Let us now write Lagrange's theorem on closed interval $[x,x']$:

$\left|\frac{f(x)-f(x')}{x-x'}\right|=|f'(c)|\leq M ,\qquad c\in(x,x').$

Therefore, $ |f(x)-f(x')|\leq M|x-x'|$. Now, demanding $M|x-x'|<\epsilon$. We'll get $|x-x'|<\frac{\epsilon}{M}$.

So, for $\delta=\frac{\epsilon}{M}$ we'll get that:

$|x-x'|<\delta\Longrightarrow|f(x)-f(x')|<\epsilon$

(Notice that our $\delta$ value is not depnding on $x$ and $x'$ )

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    Thanks. It was nice to see the proof completed!2013-12-13
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Hint: Use the mean value theorem.