I am having some troubles solving a question on my homework sheet:
Prove that $\,\,f:\Bbb{R}^n\setminus\{0\}\to \Bbb{R}^n$, defined as $$f(x) = \frac{x}{||x||}$$ is continuous but not uniformly continuous.
I have thought about the case when $\,n=1$, a map sending $\,\Bbb{R}^-$ to $-1$ and $\,\Bbb{R}^+$ to $1$, which is easy to prove. However, what about the cases when $n>1$? I have no idea how to start the $\delta-\epsilon$ statement. Many thanks for your help.
Best Answer
$f$ is continuous since $x\mapsto x$ and $x\mapsto\|x\|$ are continuous maps.
Note that for $\delta>0$ we can look at the image of $U_{\delta}=B_{\delta}(0)\setminus\{0\}$ under $f$ and see that $x\in U_{\delta}$ and $-x$ satisfy $\|f(x)-f(-x)\|=2$. Thus $f$ can not be uniformly continuous.
Additional comment: if we let $g_{\delta}$ to be the restriction of $f$ to $\mathbb{R}^n\setminus B_{\delta}(0)$, then one can show that $g_{\delta}$ is uniformly continuous. This intuitively shows that the problem about uniform continuity of $f$ is at neighborhoods of the origin and only there.