Let $X$ be a generic Banach space over $\mathbb R$ (also infinite dimensional), and let $B$ be the unit ball in $X$.
I want to prove that the following proposition
$B$ is a fixed-point space if and only if $\partial B$ is not a
retraction of $B$
proof. $(\Rightarrow)$ If $r: B\rightarrow \partial B$ is a retraction, then the function $-r$ can't have fixed points (details are missing), so $B$ is not a fixed-point space.
$(\Leftarrow)$ If $f: B\rightarrow B$ is a continuous map with no fixed points, we can define the map from $B$ to $\partial B$ such that
$$x\longmapsto x+\mu(f(x)-x)$$
where $\mu$ is the unique positive real number such that $\vert\vert x+\mu(f(x)-x)\vert\vert=1$. This define a retraction of $B$ on $\partial B$.
In the above proof I'm not sure about the existence of the number $\mu$; in $\mathbb R^n$ I realize that it exists, but what about the generic case? I need an explicit computation of $\mu$.
addenda: for "unit ball" I mean the closed unit ball.
Best Answer
I assume that by "unit ball" you mean the open unit ball. Consider the continuous convex function $g(t) = \|x + t (f(x) - x)\|$. Since $x \in B$, $g(0) < 1$, while since $f(x) \ne x$, $g(t) > 1$ for $t$ sufficiently large. Now use the Intermediate Value Theorem.
You can't have an "explicit" computation of $\mu$, since you don't specify explicitly what the norm is.