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Let $f : I^2 \to I$ be a continuous map, where $I := [0,1]$ is the unit interval. It is a basic fact that for each $y\in I$, the function $x \mapsto f(x,y)$ admits a fixed point. I want to ask whether one can always choose those fixed points as a continuous function of $y$.

Question: Does there always exist a continuous path $\gamma : I \to I$ such that $f(\gamma(y),y) = \gamma(y)$ for every $y\in I$?

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  • $\begingroup$ Do you mean that $f : [0,1] \times [0,1] \to [0,1]$ and searching for a continuous $g : [0,1] \to [0,1]$ such that $f(g(y), y) = g(y)$ for all $y \in [0,1]$, is it? $\endgroup$
    – Daniele Zuddas
    Jan 24, 2014 at 4:05
  • $\begingroup$ Yes. Exactly. I forgot mentioning the range of $f(x,y)$. $\endgroup$
    – LSZ
    Jan 24, 2014 at 5:11
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    $\begingroup$ You may find arxiv.org/abs/1210.6496 interesting. $\endgroup$
    – Michał Kukieła
    Jan 25, 2014 at 10:54

1 Answer 1

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Not necessarily. Let $f(x,0)=0$, $\,f(x,1/2)=x$, $\,f(x,1)=1$, and define $f(x,y)$ for $0<y<1/2$ and $1/2<y<1$ by linear interpolation (that is, $f(x,y) = 2yx$ and $f(x,y)=1-2(1-y)(1-x)$ respectively). Then $g(y)$ would have to be $0$ for $y<1/2$ and $1$ for $y>1/2$, so $g$ cannot be continuous at $y=1/2$.

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    $\begingroup$ Good example... is there an more strange example such that the fixed point at $y=0$ and the fixed point at $y=1$ stay in different path components of the set of such fixed points? $\endgroup$
    – Daniele Zuddas
    Jan 24, 2014 at 5:41

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