This question was asked earlier Question about proving existence of a function $f$ such that $f \circ f = g$ for an odd function $g$ about the functional square root to an odd function. Now consider an even function $g:\mathbb{R}\rightarrow \mathbb{R}$ where $g(x)>0$ for all $x\in\mathbb{R}$. Does there exist a functional square root $f$ such that $f\circ f=g$? Consider $g(x)=1$ then $f=g$ is a solution. But I suspect in general it does not exist, but I can't seem to reason why.

  • $\begingroup$ I was a bit surprised to see the earlier question. On the whole, we do not expect a functional square root of any function to be defined on the whole line; both local extrema and fixpoints cause trouble. A rare exception is $e^x,$ Helmuth Kneser constructed the thing $\endgroup$
    – Will Jagy
    Dec 4, 2020 at 2:29
  • $\begingroup$ @WillJagy Thanks, that's interesting. I still desire a proof of a counter example. Also I don't require continuity of the functional square root. $\endgroup$
    – Mars
    Dec 4, 2020 at 2:36

1 Answer 1


Consider this even and strictly-positive function $g$:

$$g(x) = \begin{cases} 2&|x|<1.5\\ 1 & |x| \ge 1.5 \end{cases}$$

And assume there exists some $f: \mathbb R\to \mathbb R$, $f(f(x)) = g(x)$ for all $x\in\mathbb R$.

Let $a = f(1)$, then

$$f:1\mapsto a \mapsto 2 \mapsto g(a) \mapsto 1.$$

If $|a| < 1.5$ and $g(a) = 2$, then $f(2)$ is simultaneously equal to both $2$ and $1$.

If $|a| \ge 1.5$ and $g(a) = 1$, then $f(1)$ is simultaneously equal to both $a$ and $1$.

Both cases lead to contradiction, so there is no such $f$ for the function $g$.


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