I have a problem from Carother's Real Analysis, page 116.

Suppose that $f : \mathbb{R} \to \mathbb{R}$ is continuous and $f(x) \to 0$ as $x \to \pm\infty$. Prove that $f$ is uniformly continuous.

I would like to do it myself, but I have tried starting in a few ways and haven't solved it.

I figure I could start by showing that $f(x)$ is bounded. So I assume to derive a contradiction that it is not - i.e. for any $M \in \mathbb{R}$, $\exists x : |f(x)| > M$. With this I would like to contradict the continuity of $f$, which says that $\forall \epsilon > 0 \,\, \exists \delta $ such that $$|x-y| < \delta \implies |f(x)-f(y)|<\epsilon$$ However even on this I am having brain-block. So really two quetsions: is proving that $f$ is bounded the correct way forward, and how do I prove boundedness anyway?! (I know that just showing $f$ bounded would not suffice in any case).


A direct proof is much easier. Since $f(x) \to 0$ as $x \to \pm \infty$, given $\epsilon > 0$, there exists $K(\epsilon)$ such that for all $x > K(\epsilon)$, we have $\vert f(x) \vert < \epsilon/2$. Now on the compact interval $[-K,K]$, the function is continuous and therefore uniformly continuous. On the interval $(-\infty,-K] \cup [K,\infty)$, we have $\vert f(x) -f(y) \vert < \epsilon$ for any $x,y \in (-\infty,-K] \cup [K,\infty)$.

| cite | improve this answer | |
  • $\begingroup$ Very well written :) $\endgroup$ – user860374 Nov 16 '14 at 18:31
  • $\begingroup$ Nice answer, +1. $\endgroup$ – Swapnil Tripathi Nov 16 '14 at 18:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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