Date: Wed, 27 Nov 1996 22:48:46 -0800
From: bruck@pacificnet.net (Ronald Bruck)
To: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: contraction on a metric space
Newsgroups: sci.math.research

(A copy of this message has also been posted to the following newsgroups:
sci.math.research)

In article <57fe3i$mi8@gannett.math.niu.edu>, rusin@vesuvius.math.niu.edu
(Dave Rusin) wrote:

> In article <pbonckae-1911961603020001@hyacinth.luc.ac.be>,
> bonkie <pbonckae@luc.ac.be> wrote:
> >Question: characterize the statement:'a map f:V -> V of a metric space is
> >a contraction up to a decent choice of the metric (=equivalent metric)'
> 
> I'm not sure what you want here. If you mean
>         "For every map  f: V-> V  of a metric space there is an
>         equivalent metric relative to which  f  is a contraction"
> then this is clearly false, since every contraction has a unique fixed point,
> and every contraction is continuous -- these are properties which can
> expressed with reference to the metric [at least within the equivalence
> class of metrics]. (If your metric space  V  is not complete replace
> "a unique" with "at most one").
> 
> If you mean
>         "Characterize the functions  f: V-> V  of a metric space
>         for which there is an equivalent metric relative to which
>         f  is a contraction"
> then I don't know an answer but as noted above there would have to be
> some restrictions on  f.
> 
> (I'm interpreting "contraction" in the strong sense here:
> d( f(x), f(y) ) <= r * d( x, y)  for some  r < 1. Obviously a weaker
> definition would allow more functions to be contractions relative
> to some metric.)

I don't remember the reference, or even the author, but as I recall there
is a theorem which runs roughly:  given a metric space (M,d) and a
self-mapping f from M to M, a necessary and sufficient condition that there
be an equivalent metric \rho (complete metric?) such that f is a strict
contraction with respect to \rho is:  for each x in M, the sequence
{f^n(x)} converges to the same point u of M.  The NECESSITY is obvious.  I
may be forgetting some details.

Here's a teaser:  suppose a metric space (M,d) has the property that every
strict contraction f:M \to M has a fixed-point (necessarily unique).  Must
M be complete?  (The answer is due to W. A. Kirk.)

--Ron Bruck