3. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ so that
(i) $f(2 u)=f(u+v) f(v-u)+f(u-v) f(-u-v)$ for all $u, v \in \mathbb{R}$, and
(ii) $f(u) \geq 0$ for all $u \in \mathbb{R}$.
3. $f$ is either constantly 0 or constantly $1 / 2$.
Clearly, either of the constant functions above satisfies the requirements. Conversely, suppose the given conditions hold. Setting $u=v$, we have
$$
f(2 u)=f(2 u) f(0)+f(0) f(-2 u) .
$$

Case 1. $f(0)=0$.
Then $f(2 u)=0$ for all $u \in \mathbb{R}$ and hence $f$ is constantly 0 .
Case 2. $f(0)=c \neq 0$.
Then
$$
f(-2 u)=\frac{1-c}{c} f(2 u) \quad \text { for all } u \in \mathbb{R} .
$$

Setting $u=0$, we have in particular $c=1-c$. Hence $c=1 / 2$. It follows that $f(-u)=f(u)$ for all $u \in \mathbb{R}$. Therefore,
$$
f(2 u)=f(u+v) f(u-v)+f(u-v) f(u+v)=2 f(u+v) f(u-v) .
$$

Setting $u=0$, we have $1 / 2=f(0)=2 f(v) f(-v)=2(f(v))^{2}$. Thus $f(v)=1 / 2$ for all $v \in \mathbb{R}$ since $f(v) \geq 0$.