= Solution
First, $f(0)=1$. For every $r\geq0$,
$$
f(r)=f(r/2)^2\geq0.
$$
It cannot vanish: if $f(r_0)=0$ for some $r_0>0$, then
$$
0=f(r_0)=f(r_0/n)^n
$$
would imply $f(r_0/n)=0$ for every positive integer $n$, contradicting continuity at zero. Thus $f(r)>0$. Since every characteristic function has modulus at most one,
$$
\boxed{0<f(r)\leq1}.
$$
Now set $h(r)=-\log f(r)$. It is continuous, nonnegative, and additive:
$$
h(r_1+r_2)=h(r_1)+h(r_2).
$$
The classification of a <continuous additive function on the nonnegative real numbers> gives $h(r)=\alpha r$, where $\alpha=h(1)\geq0$. Consequently
$$
\boxed{f(r)=e^{-\alpha r}\qquad(r\geq0)}.
$$
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