$f,g\colon [a,b]\to [a,b]$ 2 continuous functions for which holds $f(g(x))=g(f(x))$. Prove that there exist $x$ such that $f(x)=g(x)$.
If such $x$ does not exist, then $f(x)-g(x)$ has constant sign. Assuming it is positive, then $f(x)>g(x)$ for all $x$, and then $f(f(x))>g(g(x))$, but I can't find a contradiction. Please, help me.
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Jean Marie
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I don't see why $f(x)>g(x) \implies f(f(x))>g(g(x))$. – Jean Marie Mar 30 '22 at 06:16
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@JeanMarie because $f(g(x)) > g(g(x))$ and $f(f(x)) > g(f(x)) = f(g(x))$ – Mar 30 '22 at 06:17
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@User Thanks !... – Jean Marie Mar 30 '22 at 06:18
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It's certainly not necessarily true that $g=f^{-1}$ (for example, on $[0,1]$ both $f(x)$ and $g(x)$ can be arbitrary positive powers of $x$). – Greg Martin Mar 30 '22 at 06:23
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@Greg Martin You are right. I erase my erroneous comment. – Jean Marie Mar 30 '22 at 07:44
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An interesting answer here about the particular case where $f$ and $g$ are polynomials. – Jean Marie Mar 30 '22 at 07:59
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Hint: Consider the infimum of the set of fixed points of $f$ (that this set is nonempty follows by IVT, which gives us a special case of the Brouwer fixed point theorem for compact intervals).
ligtningTORM
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You can argue using the intermediate value theorem that $g$ has a fixed point on $[a,b]$. Pick some $c$ in $[a,b]$ such that $g(c) = c$. Then $f(c)$ is also a fixed point of $g$ since
$$f(c) = f(g(c)) = g(f(c)).$$
Since $f$ and $g$ commute, you can also get that
$$f^n(c) = f^n(g(c)) = g(f^n(c))$$
Thus $f^n(c)$ is also a fixed point of $g$. Define a sequence $x_n = f^n(c)$.
Use that $f$ and $g$ commute and your assumption that $f>g$ to show that $x_n$ is increasing and converges to a fixed point of both $f$ and $g$. There's your contradiction.
Philip Hoskins
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