Are there functions $f$ of one real variable $x\in X$ that are not contracting maps on the set $X$ but for which, given the starting point $x_0$, the fixed-point iteration $x_n=f(x_{n-1})$, for $n=1,2,3,\dots$ will still converge to a fixed-point?

  • $\begingroup$ Not if the fixed point, call it $a,$ so $f(a) = a,$ has also $|f'(a)| > 1.$ $\endgroup$ – Will Jagy Jan 14 '13 at 3:56
  • $\begingroup$ So at some point we have: if the fixed-point iteration $x_n=f(x_{n−1})$ converges in the vicinity of the fixed point $a$, then $f$ is a contracting map in a neighbourhood of $a$? $\endgroup$ – pluton Jan 14 '13 at 4:06
  • 3
    $\begingroup$ Consider $f(x)=x^2$ on the set $[0,1]$: it is not a contraction, but the iterates do converge to a fixed point. $\endgroup$ – user53153 Jan 14 '13 at 4:22
  • 1
    $\begingroup$ @5PM I was commenting, because pluton had added that question in a comment right before yours. $\endgroup$ – JSchlather Jan 14 '13 at 5:06
  • 2
    $\begingroup$ @Jacob Oh, I see. Well, how about $f=\chi_{\mathbb Q}$ then. $\endgroup$ – user53153 Jan 14 '13 at 5:10

Consider the function $f=\chi_{\mathbb Q}$ (i.e., $f(x)=1$ if $x$ is rational and $0$ otherwise). It is nowhere continuous, let alone contracting. On the other hand, $f(f(x))=1$ for all $x$.

  • $\begingroup$ This is a quite problematic function but the answer is helpful. Then, for which class of functions the most basic fixed-point theorem ($f$ continuous, $f(X)\subset X$ and $|f'|<1$ on $X$) provides necessary and sufficient conditions instead of the usual sufficient conditions only? $\endgroup$ – pluton Jan 14 '13 at 14:15
  • $\begingroup$ @pluton Seeing that such a class must exclude the quadratic polynomial $x\mapsto x^2$, I'm not optimistic that you'll find anything more interesting than degree $1$ polynomials. $\endgroup$ – user53153 Jan 14 '13 at 14:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.