functional relation I need to find functions $f : \mathbb{R}_{+} \rightarrow \mathbb{R}$ which satisfies
$$ f(0) =1 $$
$$ f(\max(a,b))=f(a)f(b)$$
For each $a,b \geq 0$.
I have found two functions which satisfy my criteria. 
$$ f_1(x)=1$$
$$f_2(x) = \begin{cases}
0, & \text{if } x>0 \\
1, & \text{if } x=0
\end{cases}
$$
Is there another function which satisfies my criteria?

For any $x>0$, from $f(x)=f(x)^2$ it follows that $f(x)=0$ or $1$. Prove that, if $f(a)=1$ then $f(x)=1 \ \forall x\leq a$.
