Is there a uniformly continuous function $f:\mathbb R \to \mathbb R$ such that $a_{n+1}=f(a_n)$? 
Let $a_1=\frac 2 3 , \ a_{n+1}=a_n-a^2_n$ for $n\ge 1$, $a_n$ is monotonically decreasing and bounded: $0\le a_{n+1} \le 1$ and $\displaystyle\lim_{n\to\infty} a_n=0$.
Is there a uniformly continuous function $f:\mathbb R \to \mathbb R$ such that $a_{n+1}=f(a_n), \ \forall n\in \mathbb N$ ?

I don't know how to start here but I'm pretty sure there is such function since $a_n$ is converging then from the definition of uniform continuity, we'll have $\forall \epsilon >0 :\exists \delta >0: \forall n,m\in \mathbb N: |a_n-a_m|<\delta\Rightarrow |f(a_n)-f(a_m)|<\epsilon$
$a_{n+1}=f(a_n)=f(a_{n-1}-a_{n-1}^2)$ so I need to find a function that will "push the index" by 1, like $f(x)=x+1$ and is uniformly continuous, but I have no idea how to find such function.
 A: Isn't $f(x)=x-x^2$ what you are looking for?
Edit:
This function is uniformly continuous on $[0,1]$ as a continuous function on the compact interval $[0,1]$. As noticed by DanZimm, this function is unfortunately not uniformly continuous on the unbounded set $\mathbb{R}$. But as proposed by Daw and Ihf you may just define piecewise your function outside of the interval of interest to solve this little problem. That is considering for $a<b$, $a,b \in \mathbb{R}$ the function
$g(x) = \left\{ \begin{array}{l l} f(a) & \text{if } x \leq a\\ f(x) & \text{if } a \leq x \leq b \\ f(b) & \text{if } x \geq b \end{array} \right.$
In particular you said that $0 \leq a_n \leq 1$ in your question so it makes sens to choose $a=0$ and $b=1$ which leads to the answers of the other posts. It is now clear that the function $g$ is continuous on the interval $[a-1,b+1]$ and constant everywhere else thus uniformly continuous on $\mathbb{R}$.
A: What about
$$
f(x) = \max(x-x^2,\ 0)?
$$
It is globally Lipschitz continuous, hence uniformly continuous.
A: How about $f(x)=x-x^2$ for $x\in [0,1]$ and $0$ otherwise?
