# finding the functional square root of an even function

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.

• 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 Commented Dec 4, 2020 at 2:29
• @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.
– Mars
Commented Dec 4, 2020 at 2:36

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 $$g$$.