$f\circ f$ is linear implies $f$ is linear

Question

I saw a question that asked for the possible values of $f(-1)$ if $(f\circ f)(x)=4x-12$.

The solution uses the fact that if the degree of $f$ is $n$, the degree of $f\circ f$ is $n^2$, and concludes that $f$ must be linear i.e. be of the form $ax+b$. The rest is easy.

I understand that this is the case if $f$ is a polynomial, but the question doesn’t state that. What if there exists a weird function with logarithms, trigonometric functions etc. that is not linear but its composition with itself is linear? In other words, maybe the composition “cancels out” the weirdness. Obviously, involutions are a trivial example since the composition is $x$.

My question is: Is there a function $f$ that is not linear, but $f\circ f$ is linear and is not the identity function?

Would a self-inverse function like $f(x)=\dfrac {1-x}{1+x}$ do? — player3236, Sep 13 '20 at 16:19
I assume that you want the domain to the the whole $\mathbb R$? — player3236, Sep 13 '20 at 16:31
I think there is the tacit assumption that the domain of $f$ is all real numbers (since the domain of $f\circ f$ can be $\Bbb R$ but is limited to whatever the domain of $f$ is). — Clayton, Sep 13 '20 at 16:31
@player3236 : This question is explicitly about missing assumptions... — Eric Towers, Sep 13 '20 at 16:38

score 4 · Answer 1 · answered Sep 13 '20 at 16:35

4

For $a,b>0$, let $$f(x)=-\frac{a+b}2x+\frac{b-a}2|x|=\begin{cases}-ax&x\ge0\\-bx&x\le 0\end{cases} $$ Then $$ f(f(x))=abx,$$ so $f\circ f$ is linear. If $a\ne b$ and $ab\ne1$, then $f$ is not linear and $f\circ f$ is not the identity.

answered Sep 13 '20 at 16:35

Hagen von Eitzen

374,180

$f\circ f$ is linear implies $f$ is linear

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