I have this problem I'm working on:

Problem: Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is a continuous function such that $\lim_{x \to \infty} f(x) = 0$ and $\lim_{x \to -\infty} f(x) = 0.$ Prove that $f$ is uniformly continuous on $\mathbb{R}$.

Attempt: Since $\lim_{x \to \infty} f(x) = 0$, there exists a $N_1 \in \mathbb{R}$ such that $\forall x \in \mathbb{R}$ we have that $|f(x) - 0 | = |f(x)| < 1$ whenever $x > N_1$. Similarly, since $\lim_{x \to - \infty} f(x) = 0$, there exists a $N_2 \in \mathbb{R}$ such that $\forall x \in \mathbb{R}$ it holds that $|f(x)| < 1$ whenever $x < N_2$.

Without loss of generality we can assume that $N_2 < N_1$. Since $[N_2, N_1]$ is a bounded and closed interval, and since $f$ is continuous there, we also know it is uniformly continuous on that interval.

Now it is left to show that $f$ is also uniformly continuous outside that interval. I tried proving this by using the definition of uniform continuity, but I don't know how to pick my $\delta$. Also tried proving Lipschitz continuity, but didn't succeed either.

Any help for this part would be appreciated.

  • $\begingroup$ Start with $\varepsilon$, then define your $N_1$ and $N_2$ (actually you can choose $N_2 = -N_1$ if you want so) and finally find $\delta$. $\endgroup$ Mar 24, 2016 at 21:07
  • $\begingroup$ You could also proceed by contradiction. $\endgroup$ Mar 24, 2016 at 21:14

2 Answers 2


From scratch:

Take $M>0$ so large that $\vert f(x)\vert <\epsilon/3$ whenever $x\leq M$ or $x\leq-M$.

$f$ is uniformly continuous on the compact set $[-M,M]$ so there is a $\delta >0$ s.t. if $x,y\in [-M,M]$ then $\vert x-y\vert<\delta \Rightarrow \vert f(x)-f(y)\vert<\epsilon/2$.

If $x,y\in (-\infty, -M)$, then $\vert f(x)-f(y)\vert\leq \vert f(x)\vert +\vert f(y)\vert<2\epsilon/3<\epsilon$. A similar result holds for for $x,y\in (M,\infty )$.

If $x\in (-\infty, -M), y\in [-M,M]$ and $\vert x-y\vert <\delta,\ $then $\vert-M-y\vert <\delta$ and so

$\vert f(x)-f(y)\vert<\vert f(x)-f(-M)\vert+\vert f(-M)-f(y)\vert\leq \epsilon/3+\epsilon/2<\epsilon$. A similar result holds for $x\in (M,\infty), y\in [-M,M]$


For $e>0$ let $n>0$ where $|x|\geq n\implies |f(x)|<e/3.$

Let $n>d>0$ where, when $x,y\in [-n,n]$ and $|x-y|<d,$ we have $|f(x)-f(x)|<e/3.$

Then if $x\leq -n\leq y<x+d,$ then $y\in [-n,n]$ and $|y-(-n)|<d.$

$$\text {So }\quad |f(x)-f(y)|\leq |f(x)-f(-n)|+|f(-n)-f(y)|<|f(x)-f(-n)|+e/3\leq$$ $$ |f(x)|+|f(-n)|+e/3<e.$$

The rest should be obvious.


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