Let $X$ be a metric space. Recall that a function $f: X \rightarrow X$ is contractive if there exists $C \in (0,1)$ such that for all $x,y \in X$, $d(f(x),f(y)) \leq C d(x,y)$; a function $f$ is weakly contractive if for all $x \neq y \in X$, $d(f(x),f(y)) < d(x,y)$.

Also let us say that an attracting point for $f: X \rightarrow X$ is a point $L \in X$ such that for every $x_0 \in X$, the sequence of iterates $\{ (f \circ \cdots \circ f)(x_0)\}_{n=0}^{\infty}$ converges to $L$. (Since $f$ is continuous, an attracting point is necessarily a fixed point.) Finally, we say $f: X \rightarrow X$ is attractive if it has an attracting point.

Recall the following two Attraction Theorems:

  1. Banach Attraction Theorem: If $X$ is complete, then every contractive mapping $f: X \rightarrow X$ is attractive.

  2. Edelstein Attraction Theorem: If $X$ is compact, then every weakly contractive mapping $f: X \rightarrow X$ is attractive.

(Side Remark: Of course these theorems imply that $f$ has a fixed point and are usually called "fixed point theorems". In the case of Banach's Theorem the stronger conclusion is well known: in fact it is well known that the sequences of iterates converge with exponential speed. But in the case of Edelstein's Theorem, most of the literature I have found states it as a fixed point theorem only, despite the fact that this weaker conclusion is at the level of a homework or exam problem: consider $\min_{x \in X} d(x,f(x))$...One exceptional reference which treats this topic very nicely is Keith Conrad's note on contraction mappings.)

I started thinking about these results in the context of teaching a "Spivak Calculus course" this past academic year. In the context of that course, functions were defined on intervals, not metric spaces, but not necessarily intervals that are closed and/or bounded. Thus one is not always working in a complete metric space. It is clear that in an incomplete case a contraction mapping need not have a fixed point: one need look no further than

$f: (0,1) \rightarrow (0,1), \ x \mapsto \frac{x}{2}$.

However this counterexample doesn't look "serious", because $f$ continuously extends to $[0,1)$ and has $0$ as an attracting point there. Thinking over this phenomenon, I realized it was rather general:

$3$. Let $I \subset \mathbb{R}$ be an(y) interval, and let $f: I \rightarrow I$ be continuous.
a) At least one of the following occurs:
(i) $f$ has a fixed point in $I$.
(ii) $\sup I$ is an attracting point for $f$. (If $I$ is unbounded above, this means that every sequence of iterates diverges to $\infty$.)
(iii) $\inf I$ is an attracting point for $f$. (If $I$ is unbounded below...)
b) If $f$ is moreover weakly contractive, then it has an attracting point in $[\inf I,\sup I]$.

I expect many readers will find a proof immediately, but see $\S 10.4$ of these notes if you like.

The case $I = \mathbb{R}$ of Theorem 3a) was the subject of a short Monthly note of A.F. Beardon.

Is there an extension of Theorem 3 to $\mathbb{R}^n$? If you look at the proof of Theorem 3, you see that it really uses the order-theoretic, rather than the metric properties of $\mathbb{R}^n$: when $I$ is a bounded interval the order-theoretic completion and the metric completions coincide, but not in the unbounded case. So let's restrict to bounded sets and ask the following question:

Question 1: Let $X \subset \mathbb{R}^n$ be a bounded subset. Does every weakly contractive map $f: X \rightarrow X$ have an attracting fixed point in $\overline{X}$ (closure in $\mathbb{R}^n$ = metric completion)?

Here is a reformulation / generalization of the question using the concepts of metric spaces. If I have a contraction mapping $f$ on an incomplete metric space $X$, I can't expect it to have a fixed point, but I can ask for a fixed point in the unique extension of $f$ to a continuous mapping $\overline{f}: \overline{X} \rightarrow \overline{X}$ on the metric completion $\overline{X}$. But it is easy to see that if $C$ is a contraction constant for $f$, it is also a contraction constant for $\overline{f}$, so the following result is not really any more general than Theorem 1.

$1'$ Every contractive map on a metric space extends to an attractive map on the metric completion.

As for 2., it is clear that the compactness hypothesis cannot just be thrown out entirely: for instance $f: \mathbb{R} \rightarrow \mathbb{R}, \ x \mapsto \sqrt{x^2+1}$ is weakly contractive but has no (finite!) fixed point. However, recalling that a metric space is compact iff it is complete and totally bounded there is a way to "remove only the completeness hypothesis". So I ask:

Question 2: Let $X$ be a totally bounded metric space. Must every weakly contractive map on $X$ have an attracting point (or even a fixed point) in the compact space $\overline{X}$?

Note that bounded sets in $\mathbb{R}^n$ are totally bounded, so an affirmative answer to Question 2 indeed implies an affirmative answer to Question 1.

Finally, is a priori conceivable -- although I am rather doubtful about it -- that Question 2 may follow trivially from Theorem 2 in the same way that Theorem 1'. follows trivially from Theorem 1. That is:

Question 3: Let $f$ be a weakly contractive map on a metric space $X$, and let $\overline{f}$ be the unique continuous extension to the metric completion $\overline{X}$. Must $\overline{f}$ be weakly contractive?


Q1. Take $X=\{\\,z\in\mathbb C\mid 1<|z|<2\\,\}$ and $f(z)=-\tfrac{z}{\sqrt{|z|}}$.

It seems to answer the other questions.

You may take a spiral in X which is $f$-invariant. This produce a simply connected example.

  • 1
    $\begingroup$ I guess you mean $\sqrt{|z|}$ intead of $\sqrt{z}$. Anyway, thanks: this gives a simple negative answer to all three questions. $\endgroup$ – Pete L. Clark Aug 1 '12 at 23:31
  • $\begingroup$ Can you give an example in which there is a fixed point which is not attracting? $\endgroup$ – Pete L. Clark Aug 1 '12 at 23:36
  • $\begingroup$ @Pete, $f(z)=\bar z/\sqrt{|z|}$. $\endgroup$ – Anton Petrunin Aug 2 '12 at 9:57

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