# $f(x)=f(x^2+ 1/4)$ , $f$ is continuous from $\mathbb{R}$ to $\mathbb{R}$

Find all continous functions $f\colon\mathbb{R}\rightarrow\mathbb{R}$ satisfying $f(x)=f(x^2+ 1/4)$

What I've tried so far:

• suppose that $f$ is one-one thus $x=x^2+1/4$ ... $x=1/2$

• then $f(x)=f(1/2)$ is constant

That's it, I'm stuck here.

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$f$ can't be one-to-one, since for example $f(0) = f(\frac14)$. – Javier Sep 8 '13 at 15:40
i missed that part , thanks well the situation just got worse then – Steve Sep 8 '13 at 15:42
@Steve It didn't really get worse. You now have less to worry about. – Git Gud Sep 8 '13 at 15:42

Step 1: Pick $a \in [-\frac{1}{2}, \frac{1}{2}]$. Define

$$x_1=a ; x_{n+1}= x_n^2+\frac{1}{4} \,.$$

Prove that $x_n$ is convergent. What is the limit?

Deduce from here that $f(x_1)=..=f(x_n)=...=f(l)$.

Step 2: Pick $a \in (\frac{1}{2}, \infty)$. Define

$$x_1=a ; x_{n+1}= \sqrt{x_n-\frac{1}{4}} \,.$$

Prove that $x_n$ is convergent. What is the limit?

Deduce from here that $f(x_1)=..=f(x_n)=...=f(l)$.

This proves that $f$ is constant on $[-\frac{1}{2}, \infty)$.

To complete the problem, just observe that for $a < -\frac{1}{2}$ we have $a^2+\frac{1}{4} > -\frac{1}{2}$

and $f(a)=f(a^2+\frac{1}{4})$.

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Was this inspiration or a known technique? – Git Gud Sep 8 '13 at 15:47
wow i have no idea how you thought about this , amazing ! – Steve Sep 8 '13 at 15:51
@GitGud The idea of the proof is to study the convergence of $x_{n+1}=x_n^2+\frac{1}{4}$. As soon as I realized that, there is a standard technique for that. We do that by solving $l=l^2+\frac{1}{4}$ and then see what happens with the monotony of $x_{n}$ inside and outside the roots, which is easy to see since that is a quadratic function. – N. S. Sep 9 '13 at 2:20
@N.S.Thank you. – Git Gud Sep 9 '13 at 10:11