What are the properties of continuous functions satisfying $f(x)=f(x^2+c)$?

[Ackbach]

Here is this week's POTW:

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Let $c>0$ be a constant. Give a complete description, with proof, of the set of all continuous functions $f:\mathbb{R}\to\mathbb{R}$ such that $f(x)=f(x^2+c)$ for all $x\in\mathbb{R}$.

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[Ackbach]

No one answered this week's POTW. This was a Putnam problem from 1996. I post an Internet solution below:

Spoiler

We first consider the case $c<1/4$; we shall show in this case that $f$ must be constant. The relation
$$f(x)=f(x^2+c)=f((-x)^2+c)=f(-x)$$
proves that $f$ is an even function. Let $r_1\le r_2$ be the roots of $x^2+c-x,$ both of which are real. If $x>r_2,$ define $x_0=x$ and $x_{n+1}=\sqrt{x_n-c}$ for each positive integer $x$. By induction on $n$, $r_2<x_{n+1}<x_n$ for all $n$, so the sequence $\{x_n\}$ tends to a limit $L$ which is a root of $x^2+c=x$ not less than $r_2$. Of course this means $L=r_2$. Since $f(x)=f(x_n)$ for all $n$ and $x_n\to r_2$, we conclude $f(x)=f(r_2)$, so $f$ is constant on $x\ge r_2$.

If $r_1<x<r_2$ and $x_n$ is defined as before, then by induction, $x_n<x_{n+1}<r_2$. Note that the sequence can be defined because $r_1>c$; the latter follows by noting that the polynomial $x^2-x+c$ is positive at $x=c$ and has its minimum at $1/2>c$, so both roots are greater than $c$. In any case, we deduce that $f(x)$ is also constant on $r_1\le x\le r_2$.

Finally, suppose $x<r_1$. Now define $x_0=x, \; x_{n+1}=x_n^2+c$. Given that $x_n<r_1$, we have $x_{n+1}>x_n$. Thus if we had $x_n<r_1$ for all $n$, by the same argument as in the first case we deduce $x_n\to r_1$ and so $f(x)=f(r_1)$. Actually, this doesn't happen; eventually we have $x_n>r_1$, in which case $f(x)=f(x_n)=f(r_1)$ by what we have already shown. We conclude that $f$ is a constant function.

Now suppose $c>1/4$. Then the sequence $x_n$ defined by $x_0=0$ and $x_{n+1}=x_n^2+c$ is strictly increasing and has no limit point. Thus if we define $f$ on $[x_0,x_1]$ as any continuous function with equal values on the endpoints, and extend the definition from $[x_n,x_{n+1}]$ to $[x_{n+1},x_{n+2}]$ by the relation $f(x)=f(x^2+c)$, and extend the definition further to $x<0$ by the relation $f(x)=f(-x)$, the resulting function has the desired property. Moreover, any function with that property clearly has this form.