Example 1.1.6 Consider the function $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ given by $f(x)=x /(1+\|x\|)$, where $\|x\|=\left(\sum x_i^2\right)^{1 / 2}$. We have $$ \begin{aligned} \|f(y)-f(x)\| & =\frac{\|(y-x)+x(\|x\|-\|y\|)+\| x\|(y-x)\|}{(1+\|x\|)(1+\|y\|)} \\ & \leq \frac{(1+2\|x\|)}{(1+\|x\|)(1+\|y\|)}\|y-x\| . \end{aligned} $$
Therefore, $f$ maps the open ball $B(x ; \delta)$ into the open ball $B(f(x) ; \epsilon)$ if $\delta=\epsilon$ for $x=0$, and $\delta=\min \{1,(1+\|x\|) \in /(1+2\|x\|)\}$ for $x \neq 0$. Thus $f$ is continuous.
For above process, I can understand the part of $\delta=(1+\|x\|) \epsilon /(1+2\|x\|)$, but I don't under stand where the min {1, ...} comes from.
My process is below:
I need to calculate a $\delta$ , so that when $\|y-x\|<\delta$, $\|f(y)-f(x)\|<\epsilon$.
\begin{aligned} \|f(y)-f(x)\| & =\frac{\|(y-x)+x(\|x\|-\|y\|)+\| x\|(y-x)\|}{(1+\|x\|)(1+\|y\|)} \\ & \leq \frac{(1+2\|x\|)}{(1+\|x\|)(1+\|y\|)}\|y-x\| \leq \frac{(1+2\|x\|)}{(1+\|x\|)}\|y-x\|\leq \epsilon \end{aligned}
So $\|y-x\|\leq\epsilon(1+\|x\|)/(1+2\|x\|)$, and then if I set $\delta=(1+\|x\|) \epsilon /(1+2\|x\|)$, then for $\|y-x\|<\delta$, $\|f(y)-f(x)\|<\epsilon$.
But the textbook answer is above. and I don't understand why the need to consider when x=0, and the min{1,...}.
