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Suppose that we have a function, $f$, that is continuous on $\mathbb{R}$ and $$\lim_{x \to \pm\infty} f(x) = 0$$ We want to show that $f$ is uniformly continuous.

So, since $\lim_{x \to +\infty} f(x) = 0$, for every $\epsilon > 0$, $\exists a > 0$ such that $x > a$ implies $|f(x)| < \epsilon$. Since $\lim_{x \to -\infty} f(x) = 0$, for every $\epsilon > 0$, $\exists b < 0$ such that $x < b$ implies $|f(x)| < \epsilon$.

So let $\epsilon > 0$. We need to show that $\exists \delta > 0$ such that $\forall x, y \in \mathbb{R}$, if $|x - y| < \delta$, then $|f(x) - f(y)| = \epsilon$.

I got to the point where $|f(x) - f(y)| < \epsilon$ for all $x, y > N_\epsilon$. But I am having a trouble connecting this fact to "$|x - y| < \delta$".

Also, since this function is not explicitly defined, it is very hard to connect between $|x-y|$ and $|f(x) - f(y)|$. I need some help at this moment.

Martin Sleziak
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Steps to follow: 1) Fix $\varepsilon >0$. 2) From the fact that the limits at $\pm \infty $ are $0$, deduce that outside some interval $[a,b]$, $|f(x)-f(y)|<\varepsilon $ (simply since far enough along the $X$-axis the values of $f(x)$ can be made arbitrarily small, in particular smaller than $\varepsilon /2 $). 3) Remember that since $f$ is continuous, it is uniformly continuous on $[a,b]$. 4) Piece things together.

Ittay Weiss
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  • What $\delta$ would make $|f(x) - f(y)| < \epsilon$ on the interval outside $[a, b]$? –  Oct 09 '15 at 19:04
  • You don't need to know! You don't need an explicit $\delta$ or even a way to relate it to $\epsilon$ because you know that, as @IttayWeiss says, $f(x)<\epsilon/2$ once $x$ is close enough to either infinity. So choose both $x$ and $y$ that far out. They'll be $\delta$ apart (and you don't need to know $\delta$) but $|f(x)-f(y)|<\epsilon$ by choice of $x$ and $y$. – postmortes Oct 09 '15 at 19:18
  • @postmortes How do we know that this delta does not depend on $x, y$? –  Oct 09 '15 at 19:33
  • @SeojunHong it probably does, but that doesn't matter. The definition of uniform continuity requires that $\epsilon$ does not depend on the point $x$ chosen, only on the function $f$. – postmortes Oct 10 '15 at 11:45