Is there a uniformly continuous function such that $a_{n+1} = f(a_n)$?

Question

Let $a_{n+1} = a_n - a_n^2$ and $a_1 = \frac{2}{3}$. I already proved that $a_n \to 0$

Now I was asked, is there a uniformly continuous function such that $a_{n+1} = f(a_n)$?

All I can think of is the obvious $f(x) = x-x^2$ which answers the demand but it doesn't look uniformly continuous to me, because it goes dramatically to negative infinity.

Answer 1

$a_{n+1}-a_n\leq -a_n^2\leq 0$, hence $a_n$ is a decreasing sequence , since $a_n\to 0$ then $0\leq a_n\leq a_0=\frac{2}{3}$. Now $f:[0,\frac{2}{3}]\to \Bbb R$ by $f(x)=x-x^2$ we have $a_{n+1}=f(a_n)$. it is a continuous function on a compact $[0,\frac{2}{3}]$, hence it is uniformly continuous.
EDIT: According the remark below we can extend $f$ to an uniformly continuous function in $\Bbb R$, $$f(x)=f(0)\ \ \text{if}\ x\leq 0 ,\ \ f(x)=x-x^2\ \quad \text{if}\ x\in[0,\frac{2}{3}]\ \quad \text{and} \ f(x)=f(\frac{2}{3})\quad \text{if}\ x\geq \frac{2}{3}$$