If $f: \mathbb R \to \mathbb R$ and $f$ is continuous and for every real value of $x : f(f(x))=x$ , prove that there is a $c$ where $f(c)=c$

Question

If $f: \mathbb R \to \mathbb R$ and $f$ is continuous and for every real value of $x : f(f(x))=x$ , prove that there is a $c$ where $f(c)=c$.

I know how to prove that a continuous function like $f: [a,b] \to [a,b]$ has a fixed point, but I don't know how to distinguish intervals in this question.

Hint: take $a$ and $b$ to be $0$ and $f(0)$ in some order. – Greg Martin Dec 29 '19 at 16:33 — Greg Martin, Dec 29 '19 at 16:33
I actually didn't get what you mean... – Emily Dec 29 '19 at 17:00 — Emily, Dec 29 '19 at 17:00

score 3 · Answer 1 · answered Dec 29 '19 at 16:50

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Let $g(x)=f(x)-x$, suppose that it doesn't have any root. Then, by continuity of $g$, either it is always positive, or always negative. Suppose that it is always positive, then $f(x)>x\forall x\in\Bbb R$. Hence, $f(f(x))>f(x)>x$, which is a contradiction. The case when $g(x)$ is always negative is similar.

answered Dec 29 '19 at 16:50

Martund

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you're right, I was trying to solve this problem with intermediate value theorem, but this way is probably better. thanks! – Emily Dec 29 '19 at 17:06

score 1 · Answer 2 · answered Dec 29 '19 at 16:47

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if we always had $f(x) > x,$ we would then have $f(f(x)) > f(x) > x,$ so $f(f(x)) > x$ always, finally $f(f(x)) \neq x.$ That's a contradiction of hypotheses, so we must sometimes have $f(x) \leq x.$

Similar for $f(x) < x$ always...

answered Dec 29 '19 at 16:47

Will Jagy

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I see! that's right. I thought I have to distinguish a subset of R as domain, but this contradiction is a good way – Emily Dec 29 '19 at 17:07

score 1 · Answer 3 · answered Dec 29 '19 at 16:50

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Clearly $A=(a,f(a))$ and $B=(f(a),a)$ lies on a graph of function $f$.

Then if $f(a)=a$ we are done else $f(a)\ne a$ so $A,B$ are on different sides of a line $y=x$, so the graph must cut $y=x$ (since $f$ is contiuous) and we are done.

answered Dec 29 '19 at 16:50

nonuser

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this way is interesting, but i didn't get how you declare A and B? – Emily Dec 29 '19 at 17:09
Just take any real $a$... – nonuser Dec 29 '19 at 17:11
ok, this is a really new approach – Emily Dec 29 '19 at 17:51

score 1 · Accepted Answer · answered Dec 29 '19 at 17:06

I know how to prove that a continuous function like f:[a,b]→[a,b] has a fixed point, but I don't know how to distinguish intervals in this question.

Then Make an interval.

Let $x_0$ be any point. If $f(x_0) =x_0$ we can let $c = x_0$ and we are done.

If $x_0 < f(x_0)$ let $a = x_0$ and $b = f(x_0)$. In this case you have i) $a < b$ and ii) $f(a) = b$. And iii) $f(b) = f(f(a)) = a$.

If $x_0 > f(x_0)$ then let $a=f(x_0)$ and let $b =x_0$. Again in this case you have i)$a < b$ and ii) $f(a) = f(f(b))=b$ and ii) $f(b) = a$.

and thus you know that $f$ is continuous on $[a,b]$ and that $f(a) = b$ and $f(b) = a$ and .... now go to town.

....

Let $g(x)=f(x) -x$ which is continuous. $g(a) = f(a) - a = b-a >0$ and $g(b) = f(b) - b=a-b < 0$ so there is a $c: a < c < b$ so that $g(c) = 0$ so $f(c) - c = 0$ and $f(c) = c$.

wow, that's genius!!! that's what I exactly looking for – Emily Dec 29 '19 at 17:23 — Emily, Dec 29 '19 at 17:23

If $f: \mathbb R \to \mathbb R$ and $f$ is continuous and for every real value of $x : f(f(x))=x$ , prove that there is a $c$ where $f(c)=c$

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