Converse to Banach's fixed point theorem? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T04:14:55Z http://mathoverflow.net/feeds/question/26119 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26119/converse-to-banachs-fixed-point-theorem Converse to Banach's fixed point theorem? Xandi Tuni 2010-05-27T08:35:40Z 2011-06-23T10:09:24Z <p>Let $(X,d)$ be a metric space. Banach's fixed point theorem states that if $X$ is complete, then every contraction map $f:X\to X$ has a unique fixed point. A contraction map is a continuous map for which there is an real number $0\leq r &lt; 1$ such that $d(f(x),f(y))\leq rd(x,y)$ holds for all $x,y\in X$.</p> <blockquote> <p>Suppose $X$ is a metric space such that every contraction map $f:X\to X$ has a unique fixed point. Is $X$ complete?</p> </blockquote> http://mathoverflow.net/questions/26119/converse-to-banachs-fixed-point-theorem/26122#26122 Answer by Gjergji Zaimi for Converse to Banach's fixed point theorem? Gjergji Zaimi 2010-05-27T08:47:47Z 2010-05-27T08:47:47Z <p>The answer is no, for example look at the graph of $\sin(1/x)$ on $(0,1]$. But for more information and related questions check out "On a converse to Banach's Fixed Point Theorem" by Márton Elekes.</p> http://mathoverflow.net/questions/26119/converse-to-banachs-fixed-point-theorem/26126#26126 Answer by Pete L. Clark for Converse to Banach's fixed point theorem? Pete L. Clark 2010-05-27T09:46:14Z 2011-05-24T18:20:34Z <p>As has been pointed out, the putative converse to the Contraction Mapping Theorem suggested in the question is not true. But there <em>is</em> a result which may reasonably be viewed as the converse of CMT. </p> <p>Theorem (Bessaga, 1959): Let $X$ a set and $f: X \rightarrow X$ a function such that for all $n \in \mathbb{Z}^+$, the iterate $f^n$ has a unique fixed point. Then there exists a complete metric on $X$ with respect to which $f$ is a contraction mapping (for any preassigned constant $c \in (0,1)$). </p> <p>[<b>Addendum</b>: I just looked at Elekes' paper and saw that it cites this result of Bessaga and says that it is seemingly the earliest converse. So I guess this post is not exactly exciting news. Oh well.] </p> <p>I learned about this result from a talk that Keith Conrad gave in the Undergraduate Math Club at UGA. For more information, see his "blurb"</p> <p><a href="http://www.math.uconn.edu/~kconrad/blurbs/analysis/contractionshort.pdf" rel="nofollow">http://www.math.uconn.edu/~kconrad/blurbs/analysis/contractionshort.pdf</a></p> <p>In later correspondence with him I pointed out the following result, which is now included in his writeup:</p> <p>Theorem: For a function $f: X \rightarrow X$, the following are equivalent:<br> (i) Every iterate $f^n$ has at most one fixed point.<br> (ii) There is a metric (not necessarily complete!) with respect to which $f$ is a contraction.</p> <p>This is not an earth-shattering result but it has a nice, crisp statement and afterwards I decided that it was too good to be true that I was the first to think of it. And I was right -- after a quick internet search I found the result in a published paper. (Unfortunately I didn't take note of the reference. Sorry, K.)</p> http://mathoverflow.net/questions/26119/converse-to-banachs-fixed-point-theorem/26129#26129 Answer by Pietro Majer for Converse to Banach's fixed point theorem? Pietro Majer 2010-05-27T10:31:43Z 2010-05-27T11:41:57Z <p>If you want a sort of positive result, this comes to my mind, for what is worth: </p> <blockquote> <p>If X is a metric space such that any contraction map T:Y&rarr;Y on any nonempty closed subset of X has a fixed point, then X is complete.</p> </blockquote> <p>Indeed, a Cauchy sequence (x<sub>n</sub>) converges if and only if some subsequence converges. Up to extracting a subsequence, a Cauchy sequence is either stationary (thus convergent) or injective, and verifies d(x<sub>p+1</sub>,x<sub>q+1</sub>)&le;d(x<sub>p</sub>,x<sub>q</sub>)/2. If the set {x<sub>n</sub>: n&isin;<strong>N</strong>} were closed in X, T(x<sub>n</sub>):=x<sub>n+1</sub> defines a contraction with no fixed points there. Hence it's not closed, thus it's closure contains exactly one more point, the limit of the sequence.</p> http://mathoverflow.net/questions/26119/converse-to-banachs-fixed-point-theorem/68599#68599 Answer by Stefan Geschke for Converse to Banach's fixed point theorem? Stefan Geschke 2011-06-23T10:02:42Z 2011-06-23T10:09:24Z <p>This is now an old thread, but I just came across it. The question you are asking was asked by Behrends a couple of years ago, and he knew about the counter example on the plane that Gjergji points out. Behrends was specifically asking for an incomplete metric space that is a subset of the line and has the fixed point property.</p> <p>I came up with a partition of $\mathbb R$ into two dense sets $X_0$ and $X_1$ such that for all $i\in{0,1}$, every map $f:X_i\to X_i$ that satisfies $$\forall x,y\in X_i(x\not=y\rightarrow|f(x)-f(y)|&lt;|x-y|)$$ is actually constant (and in particular has a fixed point). The $X_i$ can be constructed by an easy transfinite recursion and there is no control of how complicated the sets are. This is the huge advantage of Elekes' examples which are Borel.</p> <p>The notes with my proof are here: <a href="http://www.hausdorff-center.uni-bonn.de/people/geschke/papers/Behrends.pdf" rel="nofollow">http://www.hausdorff-center.uni-bonn.de/people/geschke/papers/Behrends.pdf</a></p>