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.
If $f'$ is bounded, then $f$ is uniformly continuous
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0I 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
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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2Of course, this argument establishes Lipschitz continuity, not only uniform continuity. – 2012-04-01
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1This proof seems completely valid except for one significant omission. In order to use the Mean Value Theorem on $f$, $f$ must be continuous on the closed interval $[a, b]$, and differentiable on the open interval $(a, b)$. See [this](http://mathworld.wolfram.com/Mean-ValueTheorem.html). Clearly, $f$ is differentiable on $(a,b)$ and continuous on $(a,b)$, but how do we know that $f$ is continuous on $[a,b]$? In other words, how do we know we can apply MVT? – 2012-11-30
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2In fact, if $f$ is continuous on the closed interval $[a,b]$, $f$ is uniformly continuous on $[a,b]$, see [this](http://planetmath.org/?op=getobj&from=objects&id=3066). – 2012-11-30
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4@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
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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1Thanks. It was nice to see the proof completed! – 2013-12-13
Hint: Use the mean value theorem.