Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
1  
Ever heard of Lipschitz? – The Chaz 2.0 May 23 '12 at 18:54

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 $\epsilon = 2 \delta$.

share|cite|improve this answer
    
Did you mean $\delta$ = $\frac\epsilon 2$? – Michael May 23 '12 at 19:12

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 $f(x) = 2x$.

share|cite|improve this answer
    
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
1  
@GEdgar: But is it homework?? – The Chaz 2.0 May 24 '12 at 1:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.