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

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!

• 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 $f(0)=f(1)$.
– user958916
Sep 16, 2021 at 7:57

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.

• Dear, but this problem means we need to find all functions $f(x),$ $f(\sqrt{x-1})$ is not as clear function. Sep 16, 2021 at 7:49
• @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). Sep 16, 2021 at 7:51