Proving $f$ is uniform continuous

Question

I am a little off my game today, so I can't immediately see a "way out" out of this question.

If $f$ is continuous on $\Bbb R$ and $\lim_{x \to \pm \infty} f(x) = 0$, $f$ must be uniformly continuous.

I think I am supposed to do this by contradiction.

Pick $\varepsilon > 0$. Consider the set ${ x : \lvert f(x)\rvert \geqslant \varepsilon/2}$. — Daniel Fischer, Nov 17 '14 at 16:01

score 2 · Accepted Answer · answered Nov 17 '14 at 16:06

2

No need for contradiction.Given $\epsilon >0\exists G>0$ such that forall $x>G |f(x)|<\epsilon$ Also $x<-G$ will imply $|f(x)|<\epsilon$ Since$ f$ is continuous on $\mathbb R$ thus continuous on $[-G,G]$ and thus uniformly continuous there

answered Nov 17 '14 at 16:06

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Oh I see now, we split up the region. – Lemon Nov 17 '14 at 16:07

score 1 · Answer 2 · answered Nov 17 '14 at 16:14

Define a continuous contraction map $\pi$ from $\mathbb{R}$ to $S^1$ and extend it by mapping $\pm \infty$ to one point on the circle. Extend your map by $f(\pm\infty)=0$. Then $f\pi^{-1}$ is a continuous function on the circle, hence is uniformly continuous: furthermore, $d(\pi^{-1}(x),\pi^{-1}(y))>d(x,y)$ so you may use the same $\delta(\epsilon)$ function on $\mathbb{R}$ as you used on $S^1$. I hope it makes sense: I know it's not the elementary solution but you may find it interesting.

Proving $f$ is uniform continuous

2 Answers2