Example 6 Let $f(x)$ be an even function defined on $\mathbf{R}$, whose graph is symmetric about the line $x=1$. For any $x_{1}, x_{2} \in \left[0, \frac{1}{2}\right]$, we have $f\left(x_{1}+x_{2}\right)=f\left(x_{1}\right) \cdot f\left(x_{2}\right)$. Also, $f(1)=a>0$.
(1) Find $f\left(\frac{1}{2}\right)$ and $f\left(\frac{1}{4}\right)$;
(2) Prove that $f(x)$ is a periodic function;
(3) Let $a_{n}=f\left(2 n+\frac{1}{2 n}\right)$, find $\lim _{n \rightarrow \infty}\left(\ln a_{n}\right)$.
(1) Solution: Since for $x_{1}, x_{2} \in\left[0, \frac{1}{2}\right]$,
we have $f\left(x_{1}+x_{2}\right)=f\left(x_{1}\right) \cdot f\left(x_{2}\right)$,
thus $f(x)=f\left(\frac{x}{2}\right) \cdot f\left(\frac{x}{2}\right) \geqslant 0$, $x \in[0,1]$.
$$
\begin{array}{c}
\because f(1)=f\left(\frac{1}{2}+\frac{1}{2}\right)=f\left(\frac{1}{2}\right) \cdot f\left(\frac{1}{2}\right) \\
=\left[f\left(\frac{1}{2}\right)\right]^{2}, \\
f\left(\frac{1}{2}\right)=f\left(\frac{1}{4}+\frac{1}{4}\right)=f\left(\frac{1}{4}\right) \cdot f\left(\frac{1}{4}\right)= \\
{\left[f\left(\frac{1}{4}\right)\right]^{2} \cdot f(1)=a>0,} \\
\therefore f\left(\frac{1}{2}\right)=a^{\frac{1}{2}}, f\left(\frac{1}{4}\right)=a^{\frac{1}{4}} .
\end{array}
$$
(2) Proof: According to the problem, $y=f(x)$ is symmetric about the line $x=1$,
thus $f(x)=f(1+1-x)$,
i.e., $f(x)=f(2-x), x \in \mathbf{R}$.
Also, since $f(x)$ is an even function, we have $f(-x)=f(x)$,
$$
x \in \mathbf{R}, \therefore f(-x)=f(2-x), x \in \mathbf{R},
$$

Substituting $-x$ with $x$, we get $f(x)=f(x+2), x \in \mathbf{R}$.

This shows that $f(x)$ is a periodic function on $\mathbf{R}$, and 2 is one of its periods.
(3) Solution: From (I), we know $f(x) \geqslant 0, x \in[0,1]$.
$$
\begin{array}{l}
\because f\left(\frac{1}{2}\right)=f\left(n \cdot \frac{1}{2 n}\right) \\
=f\left(\frac{1}{2 n}+(n-1) \cdot \frac{1}{2 n}\right) \\
=f\left(\frac{1}{2 n}\right) \cdot f\left((n-1) \cdot \frac{1}{2 n}\right) \\
=f\left(\frac{1}{2 n}\right) \cdot f\left(\frac{1}{2 n}\right) \cdot \cdots \cdot f\left(\frac{1}{2 n}\right) \\
=\left[f\left(\frac{1}{2 n}\right)\right]^{n}, \\
\because f\left(\frac{1}{2}\right)=a^{\frac{1}{2}}, \therefore f\left(\frac{1}{2 n}\right)=a^{\frac{1}{2 n}} .
\end{array}
$$
Since one period of $f(x)$ is 2,
$$
\therefore f\left(2 n+\frac{1}{2 n}\right)=f\left(\frac{1}{2 n}\right), \text{ thus } a_{n}=a^{\frac{1}{2 n}},
$$
$$
\therefore \lim _{n \rightarrow \infty}\left(\ln a_{n}\right)=\lim _{n \rightarrow \infty}\left(\frac{1}{2 n} \ln a_{n}\right)=0 .
$$