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Let $f \colon [ 0, 1] \to [0, 1]$ be a continuous map such that $$ f\big( f(x) \big) = x \ \mbox{ for each } x \in [0, 1], $$ and $$ f(x) \neq x \ \mbox{ for at least one } x \in [0, 1], $$ then how to show that $f$ has exactly one fixed point in $[0, 1]$?

I know (how to show) that $f$ does have at least one fixed point in $[0, 1]$.

Suppose that $u \in [0, 1]$ is such that $f(u) \neq u$. Suppose further that $p, q \in [0, 1]$ satisfy $f(p) = p$ and $f(q) = q$.

What next? Can we arrive at a contradiction?

Of course, $$ f\big( f(u) \big) = u, $$ so that $$ f \big( f(u) \big) \neq f(u). $$

Context:

This is Prob. 1.2 in the book An Introduction To Metric Spaces And Fixed Point Theory by Mohamed A. Khamsi and William A. Kirk.

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    $\begingroup$ For some intuition on self-inverse functions: they symmetric to the line $f(x)=x$ as explained here. Having this symmetry makes them strictly decreasing. Such functions can have only a single intersection with any strictly increasing function as $f(x)=x$. This intersection is the fixed point. $\endgroup$ – M. Winter Aug 19 '18 at 16:30
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$f$ is injective, since $f(x)=f(y) \implies f(f(x))=f(f(y)) \implies x=y$. Now, every injective continuous function is monotone. Now, by Brouwer's fixed point theorem, there is at least one fixed point. If there are two fixed points $x_1 > x_2$, then $f(x_1)>f(x_2)$ i.e. $f$ is monotone increasing. Now, there exists $x_0$ such that $f(x_0) \neq x_0$. Suppose, $f(x_0)>x_0$. Then $f(f(x_0))>f(x_0) \implies x_0>f(x_0)$. Contradiction.

If $f(x_0)<x_0$ then we arrive at the same contradiction.

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  • $\begingroup$ @Empty Well, I never said "strictly". It's increasing because there is a pair where $f$ preserves their order i.e. $x>y$ and $f(x)>f(y)$ and $f$ is monotone. $\endgroup$ – SinTan1729 Aug 19 '18 at 15:29
  • $\begingroup$ @Empty You have two options. Monotone increasing or decreasing. Now, as there is a pair which shows it's increasing, then it's increasing. $\endgroup$ – SinTan1729 Aug 19 '18 at 15:30
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    $\begingroup$ Question1: Why $f(x_0)>x_0$ always ? If I take $f(x)=x^2$ then $f$ is strictly increasing. $f(1/2)=1/4<1/2$. !! So you have to specify two different cases. $\endgroup$ – Empty Aug 19 '18 at 15:48
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    $\begingroup$ Question 2 You have to specify which fixed point Theorem you are using to say $f$ has at least one fixed point. $\endgroup$ – Empty Aug 19 '18 at 15:50
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    $\begingroup$ @CameronBuie Yeah..It is not a question, but the thing is not clear enough !! $\endgroup$ – Empty Aug 19 '18 at 15:52
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Quite obviously $f$ is onto and one-to-one, so either (a) $f(0) =0 $ and $f(1)=1 $ or (b) $f(0)=1$ and $f(1) =0$

Also, because $f$ is continuous and a bijection, $f$ is strictly monotonic.

In case $(b)$ it's not difficult to see that $f$ has then exactly one fixed point (I leave this as a task to you).

It remains to show that $(a)$ cannot happen. So assume $f(0)=0, f(1)=1$ and suppose $f(x_0) \neq x_0$ for some $x_0$, assume $f(x_0) < x_0$. Then $f(f(x_0))= x_0 > f(x_0)$. and $f$ is not increasing. A similar reaoning applies if $f(x_0)>x_0$

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    $\begingroup$ Why is it obvious that $f$ is onto? $\endgroup$ – DanielWainfleet Aug 19 '18 at 15:18
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    $\begingroup$ @DanielWainfleet The image of $f$ has to be $[0,1]$ or else $f(f(x)) = x$ would be false for some $x$. $\endgroup$ – eyeballfrog Aug 19 '18 at 15:48
  • $\begingroup$ @eyeballfrog. You're right. It IS obvious. Now...:) $\endgroup$ – DanielWainfleet Aug 19 '18 at 16:27
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It is not hard to see that if

  • if $f(x)$ is not injective, then $f(f(x))$ is not injective.
  • if $f(x)$ is not surjective, then $f(f(x))$ is not surjective.

So since $f(f(x))=x$ is bijective (on $[0,1]$), so is $f(x)$. This makes it either strictly increasing, or strictly descreasing on $[0,1]$.

It cannot be strictly increasing, because then

$$x>f(x) \implies f(x)>\underbrace{f(f(x))}_x \implies f(x)>x.$$

But a strictly decreasing function has obviously a single intersection with any strictly increasing function, e.g. the function $g(x)=x$. But intersections with $g$ are exactly the fixed points.

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