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?

Since $f(a)f(b)=f(b)$ for all $b\ge a$, either $f(b)=0$ or $f(a)=1$. If $f(a)=0$, then $f(b)=0$ for all $b\ge a$. If $f(b)=1$, then $f(a)=1$ for all $a\le b$. 
Thus, it appears that for any $a\ge0$, the functions
$$
f_a^+(x)=\left\{\begin{array}{}
1&\text{for }x\le a\\
0&\text{for }x\gt a
\end{array}\right.
$$
and for any $a\gt0$, the functions
$$
f_a^-(x)=\left\{\begin{array}{}
1&\text{for }x\lt a\\
0&\text{for }x\ge a
\end{array}\right.
$$
and the function
$$
f_\infty(x)=1\quad\text{for all }x
$$
all satisfy the conditions, and these should be all.
