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

Question

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.

Answer 1

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