This is a problem I have encountered on my exam. It is probably an easy one but after an hour of trying I could not come with an answer. If you could give me a small hint (just a starting point) that would be great.
Question is; prove or disprove that $$g(v)\colon\mathbb{R}^n\to\mathbb{R}^n\\ g(v) = \frac{v}{\sqrt{1 + \|v\|^2}}$$ is uniformly continuous.
Thanks in advance.
Let us write $v=r\xi, w =s\eta$ where $r = \|v\|, s=\|w\|\geq 0$ and $\xi,\eta \in \mathbb{S}^{n-1}$. Then we have by triangle inequality $$ \|g(v)-g(w)\| = \|\frac{r}{\sqrt{1+r^2}}\xi - \frac{s}{\sqrt{1+s^2}}\eta\|\leq ||\frac{r}{\sqrt{1+r^2}}\xi - \frac{s}{\sqrt{1+s^2}}\xi\|+\frac{s}{\sqrt{1+s^2}}\|\xi-\eta\|. $$ Since $r\in[0,\infty)\mapsto \frac{r}{\sqrt{1+r^2}}$ is a continuous function with limit, it is uniformly continuous. Hence we know that for arbitrary $\epsilon>0$ there is $\delta'>0$ such that $$|r-s|<\delta', \|\xi-\eta\|<\delta'\Rightarrow \|g(v)-g(w)\|<\epsilon.$$
Try to establish that there is $\delta>0$ such that $$ \|v-w\|<\delta \Rightarrow |r-s|<\delta', \|\xi-\eta\|<\delta', $$ if $\|v\|,\|w\| \geq 1.$