Tao Analysis 2 Lemma 6.6.6 (lemma for inverse function theorem)

Question

Let $B(0,r)$ be a ball in $\mathbb{R}^n$ centered at the origin, and let $g: B(0,r) \to \mathbb{R}^n$ be a map such that $g(0) = 0$ and $$||g(x) - g(y)|| \le \frac12 ||x-y||$$ for all $x,y \in B(0,r)$ (here $||x||$ denotes the length of $x$ in $\mathbb{R}^n$). Then, the functions $f : B(0,r) \to \mathbb{R}^n$ defined by $f(x) : = x + g(x)$ is one to one, and furthermore the image $f(B(0,r))$ of this map contains ball $B(0,r/2)$.

I understand how to prove one to one. The proof for the second part is on the picture below.

It is not clear to me how the same argument shows that $F$ maps $\overline{B(0, r-\epsilon)}$ to itself. If I use the same argument when $x \in \overline{B(0, r-\epsilon)}$, $$||F(x)|| \le ||y|| + ||g(x)|| \le \frac{r}2 + ||g(x) - g(0)|| \le \frac{r}2 + \frac{r-\epsilon}2 = r- \frac{\epsilon}2$$ This is not sufficient to show that it maps to $\overline{B(0, r-\epsilon)}$. I think I am missing something here. I would appreciate if you give some help.

Answer 1

What happens is that $y$ is in the open ball of radius $r/2$. So, if $\epsilon$ is small enough, you can get $$\|y\|<r-\epsilon/2.$$

Answer 2

In the argument above $y$ is fixed, and $\|y\|<\frac{r}{2}$. So say $\|y\|=\frac{r-\varepsilon}{2}$.

Then $F(x)=y-g(x)$ maps $\overline{B(0,r-\varepsilon)}$ into itself, since $$ \|F(x)\|\le \|y\|+\|g(x)-g(0)\|\le \frac{r-\varepsilon}{2}+\frac{1}{2}\|x-0\|\le r-\varepsilon. $$