A fixed point free ball map produces a boundary retraction

Put $b=f(x)$, $v=x-b$, $a=⟨b,v⟩$, and $q(s)=\mathrm{\mid }b+sv{\mathrm{\mid }}^2-1$. Since $v\mathrm{\ne }0$, this is an upward quadratic with $q(0)\le 0$ and $q(1)\le 0$. Its larger root is $t(x)=\frac{-a+\sqrt{a^2+(1-\mathrm{\mid }b{\mathrm{\mid }}^2)\mathrm{\mid }v{\mathrm{\mid }}^2}}{\mathrm{\mid }v{\mathrm{\mid }}^2}\mathrm{.}$ The discriminant is nonnegative and $q(1)\le 0$ implies $t(x)\ge 1$.

Set $r(x)=b+t(x)v$. The nonzero denominator and the continuous square root of a nonnegative continuous function make $t$ and $r$ continuous. The root equation gives $\mathrm{\mid }r(x)\mathrm{\mid }=1$, so $r$ has the required target. No differentiability or uniform lower bound on the denominator is needed.

If $\mathrm{\mid }x\mathrm{\mid }=1$, then $q(1)=0$ and $q^{\mathrm{\prime }}(1)=2⟨x,x-b⟩=2(1-⟨x,b⟩)>0$: equality in $⟨x,b⟩\le \mathrm{\mid }b\mathrm{\mid }\le 1$ would force $b=x$, excluded by hypothesis. Thus $1$ is the larger root and $r(x)=x$. This proves the retraction property, also on the two endpoints when $n=1$.