Let $a, b>0$. Find all functions $f: \mathbb{R}_{+} \rightarrow \mathbb{R}$ such that for all positive real numbers $x, y$:

$$
f(x) f(y)=y^{a} f\left(\frac{x}{2}\right)+x^{b} f\left(\frac{y}{2}\right)
$$
The left-hand member is symmetric in $x$ and $y$. Therefore, the right-hand member must also be symmetric, thus

$$
x^{a} f(y / 2)+y^{b} f(x / 2)=y^{a} f(x / 2)+x^{b} f(y / 2)
$$

for all positive real numbers $x, y$. Therefore, if $a \neq b$, there exists $c$ such that $f(x / 2) /\left(x^{a}-x^{b}\right)=c$ for all $x \neq 0,1$. Thus,

$$
f(x)=c\left(2^{a} x^{a}-2^{b} x^{b}\right)
$$

for all $x>0$ different from $1 / 2$ and 1. We rewrite the functional equation with this expression, and we get

$$
c^{2}\left(2^{a} x^{a}-2^{b} x^{b}\right)\left(2^{a} y^{a}-2^{b} y^{b}\right)=c\left(y^{a} x^{a}-y^{b} x^{b}\right)
$$

so $c^{2}\left(2^{a} x^{a}-2^{b} x^{b}\right)\left(2^{a} y^{a}-2^{b} y^{b}\right)-c\left(y^{a} x^{a}-y^{b} x^{b}\right)=0$. We treat this expression as a polynomial in $x$: the coefficients of the same degree must be equal on both sides of the equation (this is true even if $a$ and $b$ are not necessarily integers). The term in $x^{a}$ gives $c^{2} 2^{a}\left(2^{a} y^{a}-2^{b} y^{b}\right)-c y^{a}=0$. Treating this as a polynomial in $y$, and looking at the term of degree $b$, we find $c=0$, so $f=0$ on $\mathbb{R}^{*} \backslash\{1 / 2,1\}$. By substituting $x=y=1 / 2$ into the initial equation, we have $f(1 / 2)=0$. Then $x=y=1 / 2$ gives $f(1)=0$, so $f$ is the zero function, which is indeed a solution to the problem.

Let's handle the case $a=b$: with $x=y$ in the initial equation, we get $f(x)^{2}=2 x^{a} f(x / 2)$, so $f(x / 2)$ is positive for all $x$, so $f$ never takes strictly negative values. Therefore, $f(x)=\sqrt{2 x^{a} f(x / 2)}$. By replacing $f(x)$ and $f(y)$ with this new expression in the initial equation, we get

$$
4 x^{a} f\left(\frac{x}{2}\right) y^{a} f\left(\frac{y}{2}\right)=x^{2 a} f\left(\frac{y}{2}\right)^{2}+y^{2 a} f\left(\frac{x}{2}\right)^{2}
$$

which can be factored as

$$
\left(x^{a} f\left(\frac{y}{2}\right)-y^{a} f\left(\frac{x}{2}\right)\right)^{2}=0
$$

The left-hand member must therefore be zero, which implies that the ratio $f(x / 2) / x^{a}$ is constant (for non-zero $x$). Therefore, $f(x)=c x^{a}$ for some $c \in \mathbb{R}$ (which is also true at $x=0$, because by setting $x=y=0$ in the initial equation, we find $f(0)=0$). Conversely, we see that these functions are solutions to the problem.