6-178 Let $R$ be the set of all real numbers, and $R^{+}$ be the subset of all positive real numbers. $\alpha, \beta$ are given real numbers. Find all functions $f: R^{+} \longrightarrow R$ such that for all $x$ and $y$ in $R^{+}$, we have
$$f(x) f(y)=y^{\alpha} f\left(\frac{x}{2}\right)+x^{\beta} f\left(\frac{y}{2}\right) .$$
[Solution] In the given functional equation, let $y=x$, we get
$$f\left(\frac{x}{2}\right)=\frac{(f(x))^{2}}{x^{\alpha}+x^{\beta}}$$

Rewrite $x$ as $y$ in (1), we get
$$f\left(\frac{y}{2}\right)=\frac{(f(y))^{2}}{y^{\alpha}+y^{\beta}}$$

Substitute (1) and (2) into the original functional equation, we get
$$f(x) \cdot f(y)=\frac{y^{\alpha}}{x^{\alpha}+x^{\beta}}(f(x))^{2}+\frac{x^{\beta}}{y^{\alpha}+y^{\beta}}(f(y))^{2}$$

By swapping $x$ and $y$ in the above equation, we get
$$f(y) f(x)=\frac{x^{\alpha}}{y^{\alpha}+y^{\beta}}(f(y))^{2}+\frac{y^{\beta}}{x^{\alpha}+x^{\beta}}(f(x))^{2}$$
Adding (3) and (4), we get
$$2 f(x) f(y)=\frac{y^{\alpha}+y^{\beta}}{x^{\alpha}+x^{\beta}}(f(x))^{2}+\frac{x^{\alpha}+x^{\beta}}{y^{\alpha}+y^{\beta}}(f(y))^{2},$$

Eliminating the denominators, we get
$$\begin{aligned}
& 2\left(x^{\alpha}+x^{\beta}\right) f(y) \cdot\left(y^{\alpha}+y^{\beta}\right) f(x) \\
= & \left(y^{\alpha}+y^{\beta}\right)^{2}(f(x))^{2}+\left(x^{\alpha}+x^{\beta}\right)^{2}(f(y))^{2}
\end{aligned}$$

That is
$$\frac{f(x)}{x^{\alpha}+x^{\beta}}=\frac{f(y)}{y^{\alpha}+y^{\beta}}$$

In the above equation, let $y=1, \frac{f(1)}{2}=\lambda$, then
$$\begin{array}{l}
\frac{f(x)}{x^{\alpha}+x^{\beta}}=\lambda \\
f(x)=\lambda\left(x^{\alpha}+x^{\beta}\right)
\end{array}$$

If $\lambda=0$, then $f(x)=0$ is clearly a solution to the original functional equation.
If $\lambda \neq 0$, first consider the case $\alpha \neq \beta$.
In this case, we have
$$\begin{array}{l}
f(y)=\lambda\left(y^{\alpha}+y^{\beta}\right) \\
f\left(\frac{x}{2}\right)=\lambda\left(\left(\frac{x}{2}\right)^{\alpha}+\left(\frac{x}{2}\right)^{\beta}\right) \\
f\left(\frac{y}{2}\right)=\lambda\left(\left(\frac{y}{2}\right)^{\alpha}+\left(\frac{y}{2}\right)^{\beta}\right)
\end{array}$$

Substituting into the original functional equation, we get
$$\begin{aligned}
& \lambda^{2}\left(x^{\alpha}+x^{\beta}\right)\left(y^{\alpha}+y^{\beta}\right) \\
= & \lambda y^{\alpha} \lambda\left(\left(\frac{x}{2}\right)^{\alpha}+\left(\frac{x}{2}\right)^{\beta}\right)+\lambda x^{\beta}\left(\left(\frac{y}{2}\right)^{\alpha}+\left(\frac{y}{2}\right)^{\beta}\right) \text {, since } \lambda \neq
\end{aligned}$$
0, multiply both sides by $\frac{1}{\lambda}$ and rearrange to get
$$\left(\lambda-\frac{1}{2^{\alpha}}\right) x^{\alpha} y^{\alpha}+\left(\lambda-\frac{1}{2^{\beta}}-\frac{1}{2^{\alpha}}\right) x^{\beta} y^{\alpha}+\lambda x^{\alpha} y^{\beta}+\left(\lambda-\frac{1}{2^{\beta}}\right) x^{\beta} y^{\beta}=0 .$$

Assume $\alpha>\beta$ (the case $\alpha<\beta$ can be similarly proven).
Multiply both sides by $\frac{1}{x^{\alpha} y^{\alpha}}$, we get
$$\left(\lambda-\frac{1}{2^{\alpha}}\right)+\left(\lambda-\frac{1}{2^{\beta}}-\frac{1}{2^{\alpha}}\right) x^{\beta-\alpha}+\lambda y^{\beta-\alpha}+\left(\lambda-\frac{1}{2^{\beta}}\right) x^{\beta-\alpha} y^{\beta-\alpha}=0 .$$

Let $x \rightarrow+\infty, y \rightarrow+\infty$, we get
$$\lambda=\frac{1}{2^{\alpha}}$$

Substituting the above into (6), we get
$$-\frac{1}{2^{\beta}} x^{\beta} y^{\alpha}+\frac{1}{2^{\alpha}} x^{\alpha} y^{\beta}+\left(\frac{1}{2^{\alpha}}-\frac{1}{2^{\beta}}\right) x^{\beta} y^{\beta}=0$$

Let $x=y$, and multiply both sides by $\frac{1}{y^{2 \beta}}$, we get
$$-\frac{1}{2^{\beta}} y^{\alpha-\beta}+\frac{1}{2^{\alpha}} y^{\alpha-\beta}+\left(\frac{1}{2^{\alpha}}-\frac{1}{2^{\beta}}\right)=0,$$

Let $y \rightarrow 0$, we get $\frac{1}{2^{\alpha}}-\frac{1}{2^{\beta}}=0$, which is a contradiction.
This means that if $\lambda \neq 0$, we must have $\alpha=\beta$.
Now consider the case $\lambda \neq 0$ and $\alpha=\beta$. In this case, we have
$$\begin{array}{l}
f(x)=2 \lambda x^{\alpha} \\
f(y)=2 \lambda y^{\alpha} \\
f\left(\frac{x}{2}\right)=2 \lambda\left(\frac{x}{2}\right)^{\alpha} \\
f\left(\frac{y}{2}\right)=2 \lambda\left(\frac{y}{2}\right)^{\alpha}
\end{array}$$

Substituting the above four equations into the original functional equation, we get
$$4 \lambda^{2} x^{\alpha} y^{\alpha}=2^{1-\alpha} \lambda x^{\alpha} y^{\alpha}+2^{1-\alpha} \lambda x^{\alpha} y^{\alpha},$$

Thus, we have
$$4 \lambda^{2}=2^{2-\alpha} \lambda$$

Since $\lambda \neq 0$, we have
$$\lambda=\frac{1}{2^{\alpha}}$$

That is, in this case, the functional equation has a solution
$$f(x)=2^{1-\alpha} x^{\alpha}$$

In summary, the functional equation has only one solution $f(x)=0$ when $\alpha \neq \beta$; and when $\alpha=\beta$, it has two solutions, $f(x)=0$ and $f(x)=2^{1-\alpha} x^{\alpha}$.