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Find all functions $f:\mathbb{R} \to \mathbb{R},$ which is continuous in $\mathbb{R}$ then $f(x)=f(x^2+1).$

My tried:

If $x \le 1,$ we consider the sequence

$x_0=a\le 1$

$x_{n+1}=x_n^2 +1$

$\Rightarrow f(x_n) = f(x_{n-1}^2 +1) =f(x_{n-1}).$

$$\therefore f(x_n) =f(x_0)=f(a) \forall n, x_{n+1}-x_n=x_n^2 -x_n+1>0\forall x_n$$ I don't know what to do next, help me?

Thanks for a real lot!

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    $\begingroup$ By $f(x)=f(x^2+1)$, if $f$ is known in $[0,1]$, it is also known in $f[1,2]$, then in $[2,5]$... and also in $[-1,0]$, $[-2,-1]$... So $f$ can be chosen arbitrarily in $[0,1]$, provided it is continuous and $f(0)=f(1)$. $\endgroup$
    – user958916
    Sep 16, 2021 at 7:57

1 Answer 1

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Note that $f(-x)=f(x^2+1)=f(x)$. So $f(x)$ is even. We just need to solve for the non-negative part.

Consider the sequence $a_n=0,1,2,5,26,\dots$ generated by $a_0=0, a_{n+1}=a_n^2+1$. Clearly, $g(x)=x^2+1$ is a monotonic bijection from $[a_n,a_{n+1})$ to $[a_{n+1}, a_{n+2})$ for each $n$.

Choose any continuous function $\phi(x)$ with domain $[0,1)$, with $\phi(1^-) = \phi(0)$. It uniquely generates a solution $$f(x)=\begin{cases}\phi(x)&0\le x<1\\f(\sqrt{x-1}) &\text{otherwise}\end{cases}$$ Also, it is clear that all solutions can be generated in this way. Therefore, this is the simplest possible form of the answer.

We can derive another form of the answer though.

$$f(x) = \phi(h^{(n)}(x)) \qquad (a_n \le x < a_{n+1})$$

where $h(x) = \sqrt{x-1}$, and $h^{(n)}$ stands for the $n$-th iteration.


And don't forget to define the negative part by using evenness: $f(-x) = f(x) \qquad (x < 0)$. That makes the answer complete.

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  • $\begingroup$ Dear, but this problem means we need to find all functions $f(x),$ $f(\sqrt{x-1})$ is not as clear function. $\endgroup$
    – NKellira
    Sep 16, 2021 at 7:49
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    $\begingroup$ @tthnew As I said, you can't get any better than that. The function is determined by it's value in the 'fundamental region' [0,1). $\endgroup$
    – Trebor
    Sep 16, 2021 at 7:51

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