Prove a continuous function on the unit ball vanishes, using Brouwer's fixed point theorem

Question

Let $B^n\in\mathbb R^n$ be the closed unit ball, and $f:B\to\mathbb R^n$ be continuous. Suppose further that $f$ takes the boundary to itself, i.e., $|x|=1\implies f(x)=x$.

I would like to use Brouwer's fixed point theorem to show that there must exist a point $p\in B^n$ such that $f(p)=0$, where $0$ is the $0$ vector in $\mathbb R^n.$

Answer 1

Assume by contradiction that $f:B^n \to \mathbb{R}^n$ has no zero. Set $$ g:B^n \to B^n, \quad g(x)=-\frac{f(x)}{\|f(x)\|}. $$ Then $g$ is continuous and by Brouwer's FPT it has a fixed point $y$, say. Now $-f(y)=\|f(y)\|y$. Taking the norm yields $\|f(y)\| = \|f(y)\| \|y\|$, hence $\|y\|=1$. Thus $f(y)=y$, so $-y=y$, a contradiction.

Answer 2

If the aim is to use Brouwer's theorem, then you want to find a function $g$ such that:

1. $g: B^n \rightarrow B^n$ (because we need to map a compact convex set to itself, so $\mathbb{R}^n$ is too big);
2. $g$ is continuous; and
3. A fixed point of $g$ is somehow related to a zero of $f$, i.e. $g(p) = p \implies f(p') = 0$ (and probably we'll want those points to be the same).

To satisfy 2 and 3, $g(x) = f(x) - x$ looks promising, but in that case we can't guarantee 1. If you could maybe do something to shrink that function so that $|g(x)| \leq 1$ without breaking the other parts, that would be nice - is there a scaling factor that might make sense (maybe using some important implication of $f$ being continuous)?