I'm interested in the solvability of the functional equation (which we would call it $(*)$) $$f\circ f=g$$ where $g$ is some functions to be defined.

Certainly, even if $(*)$ has a solution, it would not be expected to be simple. For example:

Example 1.1 Consider $g(x)=0$, it easy to show that $$f_1(x)=0$$ $$f_2(x)=D(x)-1\;\text{where $D(x)$ is the Dirichlet function.}$$ $$f_3(x)=\begin{cases}0\;\text{if $x\le0$.}\\-x^2\;\text{if $x>0$}\end{cases}$$ Are all solutions. In fact, by extending $f_2$ and $f_3$, we can get some general form like $$f_2^*=\begin{cases}0\;\text{if $P(x)$.}\\a\;\text{if $\neg P(x)$.}\end{cases}$$ where $P$ is some proposition such that $P(0)$ is true and $a$ is a constant such that $P(a)$ is true. Or: $$f_3^*=\begin{cases}0\;\text{if $x\in B$.}\\h(x)\;\text{if $x\notin B$}\end{cases}$$ where $0\in B$ and $h:\mathbb R\to B$. Notice that $f_3^*$ implies a sad fact that $f$ could nearly have any local property you want (i.e. differentiable or not, continuous or not, etc.). So, it appears to be not simple to find the general form of $(*)$, but how about the existence problem?

To ensure that our work worth itself, the following example of non existence of the solution of $(*)$ by suitably chosen $g$ is presented below:

Example 1.2 Consider the following functional equation $$f(f(x))=g(x)=x^2-2$$ We are going to prove that it has no solution.


Consider the following two sets $$\mathcal A:=\{x\mid g(x)=x\}=\{-1,2\}$$ $$\mathcal B:=\{x\mid g(g(x))=x\}=\{-1,2,\frac{-1+\sqrt 5}{2},\frac{-1-\sqrt 5}{2}\}$$ Now, assume that $f$ exists.

To be convenient, let $$a:=-1,b:=2,c:=\frac{-1+\sqrt 5}{2},d:=\frac{-1+\sqrt 5}{2}$$ As $g\circ f=f\circ f\circ f=f\circ g$, $g(f(a))=f(a)$ and $g(f(b))=f(b)$, which implies that $f(a),f(b)\in\mathcal A$. By a similar approach we could show that $f(c),f(d)\in\mathcal B$.

If $f(c)=a$, then $g(c)=f(f(c))=f(a)=a$, but as $g(c)=d$, it is a contradiction. Similarly $f(c)\neq b$. And if $f(c)=c$, then $g(c)=f(f(c))=c\neq d$

So $f(c)=d$. But it turns out that$$c=g(d)=f(f(d))=f(f(f(c)))=f(g(c))=f(d)=f(f(c))=g(c)=d$$, a contradiction. And we are done. $\square$

Remark As we could actually show $g(c)=d$ and $g(d)=c$ without using $g$ explicitly (i.e. By considering $g(g(g(c)))$, we could show that $g(c)\in\mathcal B$, followed by $g(c)\neq a,\neq b,\neq c$, giving $g(c)=d$), for every function $g$ with $\#\mathcal A=2$ and $\#\mathcal B=4$, $(*)$ is unsolvable.

So, is there actually a criterion for $g$ such that we could know the solvability of $(*)$? If no, could we extend the above remark to get a class of functions $g$ such that $(*)$ is unsolvable?

Thanks in advance.

  • $\begingroup$ If you are going to insist on real valued functions, any fixed point $g(a) = a$ with $g'(a) < 0$ will fail. For example, $g(x) = -x.$ Easy over the complexes, $f(x) = ix.$ $\endgroup$ – Will Jagy Aug 23 '17 at 3:56
  • $\begingroup$ The sentence "And if $f(c)=c,$ then $g(c)=f(f(c))=c\ne d$" would be clearer if you emphasized that $g(c)=d.$ E.g. "And if $f(c)=c,$ then $d=g(c)=f(f(c))=c\ne d.$" $\endgroup$ – DanielWainfleet Aug 24 '17 at 7:47
  • $\begingroup$ There is a theory of fractional iterates (f is a "half-iteration" of g) and even fractional derivatives. I DK anything about it. $\endgroup$ – DanielWainfleet Aug 24 '17 at 7:50

There are some necessary conditions that don't require continuity. For example, if $(a,b)$ is a $2$-cycle of $g$ (i.e. $g(a) = b$ and $g(b) = a$), then $(f(a), f(b)$ is also a $2$-cycle, and thus $g$ can't have an odd number of $2$-cycles.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.