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Let $f_1$ and $f_2$ be arbitrary self-mappings on $C([0,1])$ with $f_2 > f_1$. Define set $F = \{f \in (C[0,1])| f_1 \leq f \leq f_2 \mbox{ and } f \mbox{ is increasing}\}$. Is it true that every continuous self-mapping $\Phi: F \rightarrow F$ has a fixed point? For concreteness one could set $f_1(x) = 0.5x$ and $f_2(x) = 0.5 + 0.5x$.

P.S.: Note that $F$ is not compact and therefore I could not apply Schauder's fixed point theorem. In the same time I was not able to construct a mapping $\Phi$ that does not have a fixed point for the case when $f_1(x) = 0.5x$ and $f_2(x) = 0.5 + 0.5x$.

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Incidentally, any closed convex non-compact subset of a Banach space admits a continuous self-map with no fixed points (check the references in this similar question… ) – Pietro Majer Nov 4 '13 at 11:04
up vote 5 down vote accepted

I think the following is an example of a map $\Phi$ that does not have a fixed point.

For arbitrary $f_1<f_2$ and $f \in F$ define $$ \Phi(f)(x) = \begin{cases} \cos(\pi x)f_1(x) + (1-\cos(\pi x)) f(x) &{\rm for\;} x \le 1/2,\\ \cos(\pi (1-x))f_2(x) + (1-\cos(\pi (1-x)))f(x) &{\rm for\;} x \ge 1/2. \end{cases} $$ This function pointwise in $x$ takes a weigthed average of $f$ and $f_1$ for $x<1/2$ and of $f$ and $f_2$ for $x>1/2$. Hence a fixed point $f$ must satisfy $f=f_1$ for $x<1/2$ and $f=f_2$ for $x<1/2$, which is clearly not continuous.

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What if I add the requirement for all functions in $F$ to be increasing? – Sergii Nov 3 '13 at 1:08
My example would work just as well: (by coincidence) I chose the function $f$ to converge to the smaller $f_1$ on the first half of the interval $[0,1]$. – Jaap Eldering Nov 3 '13 at 21:00

I will give an idea similar to that of Jaap Eldering. However, I find the "retraction to $[0,\infty)$" approach quite useful in general, so I decided to post this answer.

Consider the family $\{g_t\}_{t\in [2,\infty)}$ given by:

  • $g_t(x)=f_1(x)$ for $t\leq 0.5-\frac{1}{t}$,
  • $g_t(x)=f_2(x)$ for $t\geq 0.5+\frac{1}{t}$,
  • $g_t(x)$ linear on $[0.5-\frac{1}{t},0.5+\frac{1}{t}]$.

This family should form (correct me if I am wrong) a closed subset in $C([0,1])$ that is homeomorphic to the half-line $[0,\infty)$. Since $[0,\infty)$ is an AR (absolute retract), the subset $\{g_t\}_{t\in [2,\infty)}$ is a retract of $C([0,1])$. But $[0,\infty)$ does not have the fixed point property, so $C([0,1])$ also does not have this property.

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