3
$\begingroup$

This question is related to solving $f(f(x))=g(x)$.

Assume that $g$ is a bijective function $g:\mathbb{R}\to \mathbb{R}$. If there is no continuous function $f : \mathbb R \to \mathbb R\,$ for which $f\circ f=g,\, $ does this imply also that there is no continuous function $f : \mathbb R \to \mathbb R\,$ for which $f\circ f=g+g^{-1}\ $(where $\,g^{-1}$ is the compositional inverse of $\,g$)? What if we replace $\mathbb R$ by $\mathbb C$?

Note:Assume $g$ is continuous in both cases $\mathbb{R}$ and also $\mathbb{C}$ and analytic in $\mathbb{C}$

Improve this question
$\endgroup$
4
  • $\begingroup$ You quantified each of your variables several times (for example, you began by fixing $f$, then asked about the existence of $f$, which I think resulted in some confusion in an answer). I edited so that each variable is quantified only once, hopefully without changing the meaning. $\endgroup$
    – LSpice
    Commented Apr 28, 2020 at 4:56
  • 1
    $\begingroup$ What motivates this question? Do you mean to assume that $g$ is continuous? Do you mean to assume any additional regularity in the complex case? $\endgroup$
    – LSpice
    Commented Apr 28, 2020 at 4:57
  • $\begingroup$ OP (user14204) is welcome to edit their post again. I did what is, in my opinion, nice and mathematically clearer but I will not edit this question any further. $\endgroup$
    – Wlod AA
    Commented Apr 30, 2020 at 3:26
  • $\begingroup$ Too bad, I would improve the formulation still further (but a word is a word). $\endgroup$
    – Wlod AA
    Commented Apr 30, 2020 at 4:12

2 Answers 2

Reset to default
3
$\begingroup$

I address the real case. $g$ may be either increasing or decreasing (as it is continuous). In the second case neither $g$ nor $g+g^{-1}$ has a "functional root". Let us show that in the first case $g$ always has a functional root $f$, by constructing one such.

For any $a$ with $g(a)=a$, set $f(a)=a$.

Assume that $g(a)=c>a$. Choose any $b\in(a,c)$ and define $f$ monotonously and continuously on $[a,b]$ so that $f(a)=b$, $f(b)=c$. Thus, $[a,b]\mapsto[b,c]$. This automatically extends to a function $[b,c]\mapsto [c,g(b)]$ as $f(x)=g(f^{-1}(x))$. Proceed further in this way. The segments will either cover $ [a,+\infty)$ or converge to the smallest fixed point of $g$ to the right of $a$.

Act similarly to define $f$ on a sequence of segments to the left of $a$, using the relation $f(x)=f^{-1}(g(x))$.

Thus $f$ becomes defined on any interval between fixed points (one-side nearest to each other). This way, $f$ becomes a monotone bijection of $\mathbb R$, hence continuous, and $f^2=g$.

Improve this answer
$\endgroup$
6
  • $\begingroup$ @Rafikmathz, why not accept it? $\endgroup$
    – LSpice
    Commented Apr 29, 2020 at 23:24
  • $\begingroup$ Why are fixed points of $g$ isolated? $\endgroup$
    – LSpice
    Commented Apr 29, 2020 at 23:25
  • 1
    $\begingroup$ @LSpice: They are not! But any non-fixed point has a right-nearest and a left-nearest ones (if there are any on a specific side), so it will be covered. BTW: I do not pretend for acceptance, since the complex setup still remains. $\endgroup$
    – Ilya Bogdanov
    Commented Apr 30, 2020 at 0:12
  • $\begingroup$ Oh, sorry! I mixed up which $a$ was which, and thought you were looking at the right-nearest fixed point to a fixed point. $\endgroup$
    – LSpice
    Commented Apr 30, 2020 at 0:44
  • $\begingroup$ There were MO-threads mentioning solving totally $\ f\circ f = g\ $ for $\ f\ $ for arbitrarily given non-decreasing homomorphism $\ \Bbb R\to\Bbb R. $ Furthermore, the was a paper before 1961 which has described the group of such homeomorphisms completely. (It is equivalent but epsilon-easier the describe non-decreasing homomorphism of the [0;1]). $\endgroup$
    – Wlod AA
    Commented Apr 30, 2020 at 3:13
-1
$\begingroup$

EDIT: The below counterexample is for a fixed $f$. As for the question asked, it still remains open.

Consider $f(x)=(x-1)^2+1.8$ and $g(x)=x$. Then $f(f(x)) = ((x-1)^2+1.8-1)^2+1.8$ and $g^{-1}(x)=x$.

(1) One can compute that $f(f(x))-g(x) = ((x-1)^2+1.8-1)^2+1.8-x = 0$ has no real roots (in particular roots are $\frac{15\pm i\sqrt{55}}{10}$ and $\frac{5\pm i\sqrt{155}}{10}$).

(2) Where as, $f(f(x))-g(x)-g^{-1}(x) = ((x-1)^2+1.8-1)^2+1.8-2x=0$ has two real roots (in particular, 1.2917, 1,6546).

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ This doesn't really seem to be an answer to the question, which asks about the existence of a function $f$ such that $f(f(x)) = h(x)$ (for fixed $h$) identically in $x$, not about the existence of solutions $x$ to $f(f(x)) = h(x)$ (for fixed $h$ and $f$). $\endgroup$
    – LSpice
    Commented Apr 28, 2020 at 4:53
  • 2
    $\begingroup$ @LSpice: Guess I got the question completely wrong, before your last edit. Thanks for the clarification. $\endgroup$
    – DSM
    Commented Apr 28, 2020 at 5:02
  • $\begingroup$ It's $\ g\ $ which is given, and one has to solve it (if possible) for $\ f$ $\endgroup$
    – Wlod AA
    Commented Apr 30, 2020 at 3:16

You must log in to answer this question.