0
$\begingroup$

Let $f\colon X\times \mathbb{R}\to X, (x,\varepsilon)\mapsto y$, with $X$ open, be a continuous function in both arguments. Consider the following fixed-point iteration \begin{align} x_{k+1} = f(x_k,\varepsilon),\quad x_0\in X.\quad (\star) \end{align} Assume that the above iteration converges to a (possibly different, depending on $\varepsilon$) point $x_{\varepsilon}^*\in X$ contained in an open and connected set consisting of (a subset of) fixed points of $f(\cdot,0)$, say $\mathcal{E}$, for all $\varepsilon\in(0,1]$ and for all $x_0\in X$.

My question. If I know that for $\varepsilon=0$, $(\star)$ also converges to a fixed point of $f(\cdot,0)$ for all $x_0\in X$, can I conclude that, in this case, $(\star)$ converges to the closure of $\mathcal{E}$, i.e. $\overline{\mathcal{E}}$, for all $x_0\in X$? If not, which kind of assumptions do I need to arrive at such a conclusion?

Any comment and/or reference is very welcome.

A different setting.

Assume that, for all $x_0\in X$ and for all $\varepsilon\in(0,1]$, iteration $(\star)$ converges to the set of fixed points of $f(x_k,\varepsilon)$, say $\mathcal{F}_{\varepsilon}$, which corresponds to a closed and connected subset of the set of fixed points of $f(\cdot,0)$. Assume moreover that $\mathcal{F}_{\varepsilon_1}\subset \mathcal{F}_{\varepsilon_2}$ for every $\varepsilon_1> \varepsilon_2$.

My question (revised). If I know that for $\varepsilon=0$, $(\star)$ also converges to a fixed point of $f(\cdot,0)$ for all $x_0\in X$, can I conclude that $(\star)$ converges to the closure of $\mathcal{F}_{0^+}$ for all $x_0\in X$?

Improve this question
$\endgroup$
6
  • 2
    $\begingroup$ Certainly not. Take $\mathcal E=(-1,1)$, $f(x,\varepsilon)=x-\varepsilon x$. As to assumptions, it would be nice to specify the terms in which you want to make them. Otherwise, the best answer may turn out to be "it happens if and only if it happens". $\endgroup$
    – fedja
    Commented Sep 29, 2016 at 12:07
  • $\begingroup$ @fedja: Right! I've just added a revised version of my question, in which there is an additional assumption that could be interesting for me. $\endgroup$
    – Ludwig
    Commented Sep 29, 2016 at 14:55
  • $\begingroup$ Erm... $\mathcal F_\epsilon$ is also always closed (as the set of all fixed points of a continuous mapping). Now you also want it open and connected. Of course, this is still possible (if all these sets are exactly the same and the rest is far away) but is it really what you want? $\endgroup$
    – fedja
    Commented Oct 1, 2016 at 7:26
  • $\begingroup$ @fedja: My point is that if I know that $\mathcal F_{\varepsilon_1}$ is strictly contained in $\mathcal F_{\varepsilon_2}$, for every $\varepsilon_1>\varepsilon_2$, $\varepsilon_1,\varepsilon_2>0$, is the answer still negative? $\endgroup$
    – Ludwig
    Commented Oct 1, 2016 at 8:17
  • $\begingroup$ Alas, it is: just make it $x-\varepsilon\max(0,x-1+\varepsilon)$ on the positive semi-axis and extend by (odd) symmetry. $\endgroup$
    – fedja
    Commented Oct 1, 2016 at 12:09

0

Reset to default

You must log in to answer this question.