9. (16 points) Define $f(x)=f\left(x ; a_{0}, \cdots, a_{2021}, b_{1}, \cdots, b_{2021}\right)=a_{0}+\sum_{k=1}^{2021}\left(a_{k} \cos 2 k \pi+b_{k} \sin 2 k \pi\right)$ where $a_{k}, b_{k} \in \mathbb{R}$, and $f(x)$ satisfies the properties:
(1) $f(x)+f\left(x+\frac{1}{2}\right)=f(2 x)(\forall x \in \mathbb{R})$,
(2) If $f\left(x ; a_{0}, \cdots, a_{2021}, b_{1}, \cdots, b_{2021}\right)=f\left(x ; a_{0}^{\prime}, \cdots, a_{2021}^{\prime}, b_{1}^{\prime}, \cdots, b_{2021}^{\prime}\right)(\forall x \in \mathbb{R})$,
then $\left(a_{0}, \cdots, a_{2021}, b_{1}, \cdots, b_{2021}\right)=\left(a_{0}^{\prime}, \cdots, a_{2021}^{\prime}, b_{1}^{\prime}, \cdots, b_{2021}^{\prime}\right)$

Find all arrays $\left(a_{0}, \cdots, a_{2021}, b_{1}, \cdots, b_{2021}\right)$ that satisfy the conditions.
Answer: $(0, \cdots, 0)$
Difficulty: Easy
Examining: Trigonometric functions, geometric sequences
Analysis:
$$
\begin{array}{l}
f(x)+f\left(x+\frac{1}{2}\right)=f(2 x) \Leftrightarrow \\
2 a_{0}+\sum_{k=1}^{2021}\left[\left(1+(-1)^{k}\right) a_{k} \cos 2 k \pi+\left(1+(-1)^{k}\right) b_{k} \sin 2 k \pi\right]=a_{0}+\sum_{k=1}^{2021}\left(a_{k} \cos 4 k \pi+b_{k} \sin 4 k \pi\right) \\
\Leftrightarrow a_{0}+\sum_{k=1}^{1010}\left(2 a_{2 k} \cos 4 k \pi+2 b_{2 k} \sin 4 k \pi\right)=\sum_{k=1}^{2021}\left(a_{k} \cos 4 k \pi+b_{k} \sin 4 k \pi\right)(4 \text { points })
\end{array}
$$

Considering both sides as $f(x)$, by comparing coefficients, we get:
$$
a_{0}=0, a_{k}=2 a_{2 k}, b_{k}=2 b_{2 k}(k=1, \cdots, 1010), a_{2 k-1}=b_{2 k-1}=0(k=1, \cdots, 1011)(8 \text { points })
$$
For $\forall n \in\{1, \cdots, 2021\}$, let $n=2^{m} l$ (where $l$ is an odd number), then $a_{n}=\frac{1}{2^{m}} a_{l}=0$, similarly $b_{n}=0$ (12 points)
Conversely, $\left(a_{0}, \cdots, a_{2021}, b_{1}, \cdots, b_{2021}\right)=(0, \cdots, 0)$ satisfies the conditions of the problem.
In summary, $\left(a_{0}, \cdots, a_{2021}, b_{1}, \cdots, b_{2021}\right)=(0, \cdots, 0)$ is the only suitable array. (16 points)