2
$\begingroup$

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

  • 1
    Ever heard of Lipschitz?2012-05-23

2 Answers 2

5

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 $\epsilon = 2 \delta$.

  • 0
    Did you mean $\delta$ = $\frac\epsilon 2$?2012-05-23
0

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 $f(x) = 2x$.

  • 1
    @GEdgar: But *is it* homework??2012-05-24