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Is the following statement true? In case not, what's a counterexample? Thank you.

If $f,g: \mathbb R \to \mathbb R$ are two continuous functions satisfying $f \circ g = g \circ f$, then either $f$ is linear or $g$ is linear (Where neither $f$ nor $g$ is invertible and $f \neq g$).

Def: $f : \mathbb R \to \mathbb R$ is said to be linear if, there exists $a \in \mathbb R$ such that $f(x) = ax$ for all $x \in \mathbb R$.

Fly by Night
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4 Answers4

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A contradiction can be easily obtained by letting $f=g$ without $f$ being linear.

As an alternative example $f(x)=x^2$ and $g(x)=x^3$

Asinomás
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  • Or $f(x)=\ln\ln x$ and $g(x)=\ln\ln\ln x$. – Akiva Weinberger Aug 10 '15 at 00:09
  • I appreciate the answer, though that's not what I was looking for. I'll ask another one. –  Aug 10 '15 at 00:10
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    @BolzWeir what are you looking for? – Asinomás Aug 10 '15 at 00:11
  • I should have worded my question more properly. I am looking for general sufficient conditions for that equality to hold. I am interested in a non-trivial condition. –  Aug 10 '15 at 00:14
  • I.e., a condition which does not look like $f = g^n$, etc. Something that is analytical in flavor. –  Aug 10 '15 at 00:15
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    This appears to be related http://math.stackexchange.com/questions/11431/when-functions-commute-under-composition – Asinomás Aug 10 '15 at 00:16
  • I had not thought about these cases and that's why I put the condition as: $f$ is linear, thinking that this would prevent the attempts to answer the question at face value. The link you gave is what I was looking for, thanks. –  Aug 10 '15 at 00:20
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Look at what happens if $f=g$. Or $f=g^2$ - or, in general $f=h^a$ and $g=h^b$ for some function $h$. We only need to use the associativity of composition to prove that these commute.

Milo Brandt
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Two sets of functions such that $h_n(h_m(x)) =h_{mn}(x) $ are the powers ($h_n(x) =x^n $) and the Chebychev polynomials (look them up) $h_n(x) =T_n(x) $.

For distinct $m$ and $n$ set $f(x) = h_n(x)$ and $g(x) = h_m(x)$.

Then $f(g(x)) =h_n(h_m(x)) =h_{nm}(x) =h_{mn}(x) =h_m(h_n(x)) =g(f(x)) $.

marty cohen
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    Indeed, I believe that there’s a theorem of Ritt that says that these are essentially the only examples of polynomials with the property $f\circ g=g\circ f$ other than pairs that are iterates of some polynomial. – Lubin Aug 10 '15 at 02:27
  • Sounds familiar. Nice when I can be exhaustive rather than exhausted. – marty cohen Aug 10 '15 at 03:43
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In general, let

$$ f(x) = h^{[m]}(x) \quad \textrm{and} \quad g(x) = h^{[n]} h(x), $$

where

$$ h^{[n+1]}(x) = h \circ h^{[n]}(x), $$

then

$$ f \circ g(x) = g \circ f(x), $$

and $f(x)$ nor $g(x)$ is linear.

johannesvalks
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