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How do you show that $f(x) = 2x$ is uniformly continuous on $\mathbb{R}$ ? Or is it not uniformly continuous?

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Ever heard of Lipschitz? –  The Chaz 2.0 May 23 '12 at 18:54
    
@Michael I have taken the liberty to add homework tag, which is to be used for all the homework problem. Feel free to remove it if the question is not, in fact, a homework question. –  Sasha May 23 '12 at 19:02

2 Answers 2

You check the definition.

For all $\epsilon > 0$, there exists a $\delta > 0$, such that for all $x, y \in \mathbb{R}$ the $|x-y| < \delta$ implies $|f(x)-f(y)| < \epsilon$.

For $f(x) = 2 x$, take $\epsilon = 2 \delta$.

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Did you mean $\delta$ = $\frac\epsilon 2$? –  Michael May 23 '12 at 19:12
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Is your definition of uniform continuity correct? –  Michael May 23 '12 at 19:28
    
@Michael Yes, I misquoted the definition. I will edit. –  Sasha May 23 '12 at 19:48

This is related to the comment left by @The Chaz

A differentiable real-valued function on $\mathbb{R}$ with bounded derivative is uniformly continuous on $\mathbb{R}$.

This is clearly(?) satisfied by your function $f(x) = 2x$.

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But a homework problem, show $f(x) = 2x$ is uniformly continuous, would not be done by this method! –  GEdgar May 24 '12 at 1:01
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@GEdgar: But is it homework?? –  The Chaz 2.0 May 24 '12 at 1:13

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