Show Existence of Fixed Point in $\mathbb{R}^n$ with Euclidean Metric

Question

Consider the closed unit ball in $\mathbb{R}^n$, $B = \{ x \in \mathbb{R}^n : \|x\| \leq 1 \}$ with the Euclidean metric. Then let $g : B \rightarrow B$ be a function such that $$\|g(x) - g(y)\| \leq \|x - y\|$$ for all $x$, $y \in B$. Then show that $g$ has a fixed point.

I have considered breaking $B$ down into two sets. Let $A$ is the set of all points in $B$ such that $g$ is an isometry. Then we have that if $B = A$ then the zero point is fixed. Otherwise, $B \setminus A$ under $g$ is a contraction map. It would therefore have a fixed point if $B \setminus A$ was closed (hence complete), but I cannot seem to show this. Any advice would be appreciated! Do I need to do something with compactness perhaps?

Answer 1

If $g:R^n\rightarrow R^n$, this is not true, consider a non trivial translation.

Since you have edited your question, If you say that $g:B\rightarrow B$, it is an application of the Brouwer fixed-point theorem.

https://en.wikipedia.org/wiki/Brouwer_fixed-point_theorem

Answer 2

Hint: Here are three steps to prove it

1) If there is a continuous map $ g: B \to B$ with no fixed points then there will be a smooth map $h : B \to B$, without fixed points

Use the fact that we can approximate a continuous function by a differentiable one, and consider the positive continuous function $x \mapsto |g(x) - x|$.

2) If there is a smooth map $h: B \to B $ with no fixed points then the unitary sphere $S^{n-1}$ is a smooth retract of the unitary ball $B \subset \mathbb R^n$.

Consider the retraction defined by $\rho: B \to S^{n-1}$ where $\rho (x) = $ intersection of the semi-line $\overline{g(x)x}$ with the sphere $S^{n-1}$.

3) There is no smooth retraction $\rho : B \to S^{n-1}$.

Suppose there is a retraction $\rho: B \to \partial B$ of class $C^2$. Let $\omega$ be a $C^1$ form of maximum degree in $\partial B$, such that $\displaystyle \int_{\partial B} \omega \neq 0$ (take the volume element, for instance). Now $\omega$ is closed, that is, $d\omega = 0$. By Stokes Theorem, plus the fact that $\rho|_{\partial B} = id$, we have $$0 \neq \int_{\partial B} \omega = \int_{\partial B} \rho^*\omega = \int_{B} d(\rho^*\omega) = \int_{B} \rho^*(d\omega) = \int_{B} \rho^*0 = 0$$

which yields a contradiction.