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 , > bonkie 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