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 $x \in \mathbb R$.
A contradiction can be easily obtained by letting $f=g$ without $f$ being linear.
As an alternative example $f(x)=x^2$ and $g(x)=x^3$
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.
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)) $.
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.
