# Prove that $f$ is uniformly continuous on $R^n$

Let $f:R^n\to R$ be a continuous function. Suppose $f(x)\to 0$ whenever $||x||\to\infty$. Show that $f$ is uniformly continuous on $R^n$.

Clearly $f$ is uniformly continuous in the closure of any ball. So it suffices to show it is uniformly continuous in the exterior of some ball.

Since $f\to 0$ as $||x||\to \infty,$ for all $\epsilon$ there is $r=r(\epsilon)$ such that $|x| > r\implies |f(x)|<\epsilon/2$. $f$ is uniformly continuous on $\{x:|x|\le r\}$. To see it's uniformly continuous on $A=\{x: |x| > r\}$, we need to prove that for any $\epsilon$ there is $\delta$ such that for all $x,y\in A$ with $|x-y|<\delta$ we have $|f(x)-f(y)|< \epsilon.$ But if $x,y\in A$, then $|x| > r, |y| > r$ and by the above $|f(x)-f(y)|\le |f(x)|+|f(y)|< \epsilon/2+\epsilon/2=\epsilon$ for all $x,y \in A$, not only for those $x,y\in A$ for which $|x-y| < \delta$ for some $\delta$. Is it sufficient to establish continuity, or do I still have to find that $\delta$?

• You're fine having found a non explicit $\delta$, you don't know what values $f$ takes so it's not reasonable to expect a concrete calculation of $\delta$ anyway. However, you have not taken into account the case in which $\|x\| < r$ and $\|y\| \geq r$ – Guido A. Aug 9 '18 at 21:07
• The existence of the $\delta$ is given by Cantor's theorem. To address the problem that Guido A raised, apply Cantor's in the ball $\|x\|\leq r+2\epsilon$ instead of $\|x\|\leq r$. – user582578 Aug 9 '18 at 21:09
• @floodbaharak Why does the existence of $\delta$ follow from that theorem? I don't think I considered a map on a compact metric space. The $\delta$ part deals with a function on $\{|x| > r\}$. – user531232 Aug 9 '18 at 21:25
• @GuidoA. But still the definition says "there exists $\delta$", but I don't have a firm understanding of either because it exists or why we don't need to prove its existence. I don't see where I found a non explicit $\delta$. – user531232 Aug 9 '18 at 21:30
• @user531299 The ball $\|x\|\leq r$ is your compact metric space. – user582578 Aug 9 '18 at 21:59