The given solution I have is that $f(x) = 0.5 + 0.5x$ but I am not entirely how these results are derived.

A function $f : X \to X $ is a contraction if $d(f(x),f(y)) \leq \gamma d(x,y)$

so here,

$$|f(x) - f(y)| \leq \gamma (x-y)$$ $$|0.5 + 0.5x - 0.5 -0.5y| \leq 0.5(x-y)$$

So obviously the solution checks out since $\gamma = 0.5 < 1$

We could have also chosen $f(x) = c + cx$ where $0<c<1$ , correct?

How would my approach have changed if we had to find

$a)$ A contraction $f:(0,1) \to (0,1)$ with at least 1 fixed point

$b)$ A contraction $f:(0,5) \to (0,5)$ with no fixed points

  • 1
    $\begingroup$ Hmmm... What are the values of $c$ such that $x\mapsto c+cx$ is a function $(0,1)\to(0,1)$ with no fixed point, already? // Re $f(x)=(x+1)/2$, it works, yes, but a simpler example would be $f(x)=ax$, for any $0<a<1$. $\endgroup$
    – Did
    Commented Jun 22, 2015 at 7:16
  • $\begingroup$ @Didf can' it be anything between $0$ and $1$? For instance, if $c = 0.01$ $$f(x) = 0.01 + 0.01x$$ $$|f(x)-f(y)| = |0.01x - 0.01y| \leq 0.01|x-y|$$ is a contraction. edit: my main concern is how the problem changes with a different domain and range, and the requirement of a fixed point $\endgroup$
    – elbarto
    Commented Jun 22, 2015 at 7:18
  • 1
    $\begingroup$ No, $f(x)=(x+1)/100$ has a fixed point in $(0,1)$. And no, $f(x)=ax$ has no fixed point in $(0,1)$. $\endgroup$
    – Did
    Commented Jun 22, 2015 at 7:21
  • 1
    $\begingroup$ The problem for functions $f:(0,5)\to(0,5)$ is entirely reducible to the problem for functions $g:(0,1)\to(0,1)$, simply consider $g(x)=f(5x)/5$ and note that $f$ has some fixed points if and only if $g$ has. $\endgroup$
    – Did
    Commented Jun 22, 2015 at 7:22
  • 1
    $\begingroup$ Just solve the equation $f(x)=x$, and you find $x=1/99$. $\endgroup$
    – mickep
    Commented Jun 22, 2015 at 7:31

1 Answer 1


First, is there any result that guarantees the existence of an application without fixed points?

Yes, fixed-point theorems requires a compact set, while (0,1) is bounded but not closed. Thus, there exists a continuous application without fixed points.

Then, you must adhere to some requirements:

  • Your application should be a contraction, the definition is the one you said;
  • It must return values in $(0,1$), with no fixed points.

Having not to deal with $0$ and $1$ (the toughest ones), you can say that $f(x) = \gamma x$, such that $0 < \gamma < 1$ satisfies the requests. Also, shifting by a constant $\phi$ such that $\phi + \gamma \le 1$ gives you acceptable results. So, every function written as $f(x) = \phi + \gamma x$ is a contraction with no fixed points in $(0,1)$.

How about a contraction with at least one fixed point?

Identity doesn't work well here, since $\gamma = 1$ doesn't give a contraction. What if you try to contract the unit open interval to a point which belongs to $(0,1)$?

What changes if we search a contraction $(0,5) \to (0,5)$ without fixed points?

You can partly recycle the previous reasonings. However, while $0 < \gamma < 1$ still holds, does the condition about $\phi$ change?


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