2. Let the function $f(x)(f(x) \neq 0)$ have the domain $(0,+\infty)$, and for $x \in \mathbf{R}_{+}, y \in \mathbf{R}$, it always holds that $f\left(x^{y}\right)=$ $y f(x)$. If $a>b>c>1$, and $a, b, c$ form an arithmetic sequence, then the relationship between $f(a) f(c)$ and $(f(b))^{2}$ is ( ).
(A) $f(a) f(c)(f(b))^{2}$
(D) Cannot be determined
2. A.

Let $a=b^{r}, c=b^{s}$, then $r, s$ are positive numbers.
Also, $a+c=2b, a>c$, then
$$
b>\sqrt{ac}=\sqrt{b^{r+s}}>b^{\sqrt{rs}} \text{.}
$$

Thus, $rs<1$.
Therefore, $f(a) f(c)=f\left(b^{r}\right) f\left(b^{s}\right)=rs(f(b))^{2}<(f(b))^{2}$