Let $X$ be a Banach space and $g:B_r\to X$ a continuous function, where $B_r:=\{x\in X\mid \|x\|\leq r\}$.
Suppose that
- $g(x)\neq 0$, for all $x\in B_r$.
- $g(B_r)$ is relatively compact.
Define $f:B_r\to B_r$ by $f(x)=-r\frac{g(x)}{\|g(x)\|}$.
I want to show that $f(B_r)$ is relatively compact.
I'm trying to apply the "sequential criterion". For this, we need to show that every sequence in $f(B_r)$ has a subsequence which converges in $X$.
Let $y_n=-r\frac{g(x_n)}{\|g(x_n)\|}$ be a sequence in $f(B_r)$. Since $g(B_r)$ is relatively compact, $z_n=g(x_n)$ has a convergent subsequence $z_{n'}=g(x_{n'})$. Let us write $$\lim_{n'\to\infty} z_{n'}=\lim_{n'\to\infty} g(x_{n'})=z.$$
If $z\neq 0$, then $y_n$ has the convergent subsequence $y_{n'}=-r\frac{g(x_{n'})}{\|g(x_{n'})\|}$, which converges to $-r\frac{z}{\|z\|}$.
Question: How to deal with the case $z=0$?
Different arguments (without sequences) would also be appreciated.
Edit
The assertion that $f(B_r)$ is relatively compact came from the image below (page 22 in this link). Theorem 1.18 is the Schauder–Tychonoff fixed point theorem.
