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.


2 Answers 2


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.

  • 1
    $\begingroup$ I think that we must consider the closed unit ball because necessarily $\partial B\subset B$ if we are considering retractions. $\endgroup$
    – Dubious
    Jun 17, 2013 at 9:37
  • $\begingroup$ If it's the closed unit ball, and the norm is not strictly convex, you might have $\|x\| = \|y\| = 1$ and $\|x + \mu (y - x)\| = 1$ for all $\mu \in [0,1]$. $\endgroup$ Jun 17, 2013 at 10:10

Let's begin with the theorem. I'll give an answer that is correct but that may not be what you want. Then, after the proof, I'll give you the answer that I think you are looking for. Recall that a metric space $X$ has the fixed point property if for every continuous function $f\colon X\to X$ has a fixed point.

Theorem: The closed unit ball $B$ has the fixed point property if and only if the unit sphere $S$ is not a retraction of B.

proof: The closed unit sphere in any infinite dimensional normed space is a retract of the closed unit ball. Such a Ball does not have the fixed point property because it is norm compact.

Now suppose that we are talking about a finite dimensional Banach space. The closed unit ball is convex and compact. Thus, it has the fixed point property.

The sphere in the finite dimensional space is not a retract of the unit ball because if there is a retract $f$, then letting $\psi$ be a rotation function from the sphere to itself we see $\psi\circ f$ is continuous and does not have a fixed point.

That proves the result you need. QED

But the idea I'm guessing (the idea behind the question) is if that the statement

A1. there is no retract from $B$ to $S$.

implies (quickly and straightforwardly; no such thing as two equivalent theorems! Axioms can be equivalent but not theorems.)

A2. Every continuous function from $B$ to $B$ has a fixed point.

Thus, a proof of A1 is a proof of Brouwer's fixed point theorem.

Assume A1. Suppose that $f$ is continuous with domain and range being the closed unit ball $B$. If $f$ does not have a fixed point, then the function that takes $x$ into the point in the sphere that lies on the ray starting from $f(x)$ and going through $x$, is a retract. This is impossible. So A2 holds.

That A2 implies A1: we have already shown this.


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