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 $f\_2(x) = 0.5 + 0.5x$. 



Fixed point for a self-mapping on subset of C[0,1]
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.


