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Let $f: \Bbb R → \Bbb R$ be a continuous function such that $f(x)=x$ has no real solution . Then is it true that $f(f(x))=x$ also has no real solution ?

Souvik Dey
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  • I think they mean it doesn't have a fixed point. For example, $f(x) = x^2 + 1$ implies $f(x) = x$ has no real solution. – Suugaku Apr 23 '13 at 08:12
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    @Easy He means a function $f$ so that the function $f(x) - x$ has no real zeros. – Arthur Apr 23 '13 at 08:18
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    @ Arthur: You got it right. – Souvik Dey Apr 23 '13 at 08:20
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    $f(f(x))\not=f(x)\not=x$ – ABC Apr 23 '13 at 09:38
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    @exploringnet: That doesn't work. Being not equal is not a transitive relation. – Fredrik Meyer Apr 23 '13 at 12:26
  • @FredrikMeyer $$f(a)\not=a$$ so, $$f(f(x)\not= f(x)$$ and also $$f(x)\not=x$$ so,from above 2 eq. $$f(f(x))\not=x$$ – ABC Apr 23 '13 at 12:28
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    @exploringnet: Your "proof" doesn't use that this is a real function, nor that this is a function $\mathbb{R} \to \mathbb{R}$. So I can produce a counterexample by letting $f:S^1 \to S^1$ be defined by $f(e^{i\theta})=-e^{i(\theta)}$. Then $f$ has no fixed points, but $f(f(x))=x$ for all $x$. – Fredrik Meyer Apr 23 '13 at 12:39

4 Answers4

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Since $f$ is continuous, saying that $f(x) = x$ has no solution means that either $f(x) < x$ for all $x$ or $f(x) > x$ for all $x$. Let's assume wlog that $f(x) > x$, then $f(f(x)) > f(x) > x$ for all $x$ so $f(f(x))=x$ has no solution either.

roger
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Just a variation of the proof. Suppose $f(f(x))=x$. Define $g:x\mapsto f(x)-x$, which is continuous, and has $g(f(x))=f(f(x))-f(x)=-(f(x)-x)=-g(x)$. Then by the intermediate value theorem $g(y)=0$ for some $y$ in the closed interval bounded by $x$ and $f(x)$, and then of course $f(y)=y$.

Marc van Leeuwen
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Hint: We know that $f(x)\neq x$ for all $x\in\mathbb{R}$. Without loss of generality, let's say that at $x_0\in\mathbb{R}$, we have that $f(x_0)<x_0$. Can it ever be the case that $f(x_1)>x_1$ for any other $x_1\in\mathbb{R}$?

Zev Chonoles
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If $f(f(c))=c$ and $f(c)=r$ then $f(r)=c$.
Since $c\neq r$ it follows that there is an $s$ between $r$ and $c$ such that $f(s)=s$ (in the case $c<r\Rightarrow f(r)-r=c-r>0$ and $f(c)-c=r-c<0$...) $\Rightarrow\Leftarrow$.

P..
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