How do you show that $f(x) = 2x$ is uniformly continuous on $\mathbb{R}$ ? Or is it not uniformly continuous?
Uniform continuity of $f(x) = 2x$
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calculus
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1Ever heard of Lipschitz? – 2012-05-23
2 Answers
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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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0Did you mean $\delta$ = $\frac\epsilon 2$? – 2012-05-23
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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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1@GEdgar: But *is it* homework?? – 2012-05-24