Because I haven't learned how to write in Latex yet , my solution is in the attachment

orl wrote: Let $ \mathbb{R}$ denote the set of all real numbers and $ \mathbb{R}^ +$ the subset of all positive ones. Let $ \alpha$ and $ \beta$ be given elements in $ \mathbb{R},$ not necessarily distinct. Find all functions $ f: \mathbb{R}^ + \mapsto \mathbb{R}$ such that \[ f(x)f(y) = y^{\alpha} f \left( \frac {x}{2} \right) + x^{\beta} f \left( \frac {y}{2} \right) \forall x,y \in \mathbb{R}^ + .\] Let $ P(x,y)$ be the assertion that $ f(x)f(y) = y^{\alpha} f \left( \frac {x}{2} \right) + x^{\beta} f \left( \frac {y}{2} \right)$. First the case $ \alpha \neq \beta$: $ P(x,1): f(x)f(1) = y^{\alpha} f \left ( \frac {1}{2} \right ) + f \left ( \frac {x}{2} \right)$ $ P(1,x): f(x)f(1) = y^{\beta} f \left ( \frac {1}{2} \right ) + f \left ( \frac {x}{2} \right)$ Hence $ f \left ( \frac {1}{2} \right ) (x^{\alpha} - x^{\beta} ) = 0 \forall x > 0$. And since $ \alpha \neq \beta: f \left ( \frac {1}{2} \right ) = 0$. So $ f(x)f(1) = f \left ( \frac {x}{2} \right)$ $ P(1,1): (f(1))^2 = 2 f \left ( \frac {1}{2} \right ) \Rightarrow f(1) = 0$. So $ f \left ( \frac {x}{2} \right ) = 0 \forall x > 0$ and hence $ f(x) = 0$ is the only solution when $ \alpha \neq \beta$. (And it is easy to see that it is indeed a solution.) Now assume $ \alpha = \beta$ Let $ h(x) = \left ( \frac {2}{x} \right )^{\alpha} f(x)$ Then $ P(x,y) \iff h(x)h(y) = h \left ( \frac {x}{2} \right ) + h \left ( \frac {y}{2} \right )$. $ P(x,x) \iff (h(x))^2 = 2h \left ( \frac {x}{2} \right )$ So $ 2 h(x)h(y) = 2h \left ( \frac {x}{2} \right ) + 2h \left ( \frac {y}{2} \right ) = (h(x))^2 + (h(y))^2 \iff$ $ (h(x) - h(y))^2 = 0 \iff h(x) = h(y) \forall x,y > 0$. So $ h(x) = c$ for some constant $ c$. $ P(x,x): c^2 = 2c \iff c = 2$ or $ c = 0$. So either $ f(x) = 2 \cdot \left ( \frac{x}{2} \right) ^\alpha \forall x>0$ or $ f(x) = 0 \forall x>0$. It's easy to check that they both are solutions.

Solved with Jeffrey Chen and Nathan Cho. The answer is \(f(x)\equiv c(x^\beta-x^\alpha)\) for \(\alpha\ne\beta\) and \(f\equiv0\) or \(f(x)\equiv x^\alpha/2^{\alpha-1}\) for \(\alpha=\beta\), which work. Now we show these are the only solutions. If \(\alpha\ne\beta\), we clearly have \[y^\alpha f(x/2)+x^\beta f(y/2)=y^\beta f(x/2)+x^\alpha f(y/2),\]which gives \[\frac{f(x/2)}{x^\beta-x^\alpha}=\frac{f(y/2)}{y^\beta-y^\alpha},\]and the result is clear. If \(\alpha=\beta\), we have \(f(x)^2=2x^\alpha f(x/2)\), hence \[f(x)f(y)=y^\alpha\cdot\frac{f(x)^2}{2x^\alpha}+x^\alpha\cdot\frac{f(y)^2}{2y^\alpha}.\]By AM-GM, we should have \(f(x)/x^\alpha=f(y)/y^\alpha\), so \(f(x)\equiv cx^\alpha\) for some \(c\). It is clear \(c=0\) or \(c=1/2^{\alpha-1}\).

Let $P(x,y)$ denote the given assertion. Notice that if $\alpha \neq beta$ then $P(x,y)$ and $P(y,x)$ implies $f(\tfrac{x}{2})=cx^{\alpha}-cx^{\beta}$ where $c$ is constant. Checking gives $f\equiv 0.$ Consider $\alpha = \beta.$ Then $P(x,x)$ gives $f(\tfrac{x}{2})= \tfrac{f(x)^2}{2x^{\alpha}}$ and consequently $P(x,y)$ arranges to $$f(x)f(y)=\frac{x^{\alpha}}{2y^{\alpha}} f(y)^2+\frac{y^{\alpha}}{2x^{\alpha}} f(x)^2$$AM-GM implies $\tfrac{f(x)}{x^{\alpha}}=c$ and checking implies $f(x)\equiv \tfrac{x^{\alpha}}{2^{\alpha}-1}$ and $f\equiv 0$ only works.