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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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  • $\begingroup$ 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! $\endgroup$ – Dylan Moreland Apr 1 '12 at 16:43
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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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    $\begingroup$ Of course, this argument establishes Lipschitz continuity, not only uniform continuity. $\endgroup$ – t.b. Apr 1 '12 at 16:58
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    $\begingroup$ In fact, if $f$ is continuous on the closed interval $[a,b]$, $f$ is uniformly continuous on $[a,b]$, see this. $\endgroup$ – user51304 Nov 30 '12 at 16:53
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    $\begingroup$ This 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. 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? $\endgroup$ – user51304 Nov 30 '12 at 16:53
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    $\begingroup$ @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. $\endgroup$ – David Mitra Nov 30 '12 at 19:15
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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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    $\begingroup$ Thanks. It was nice to see the proof completed! $\endgroup$ – user66360 Dec 13 '13 at 3:24
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Hint: Use the mean value theorem.

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