Example 19 (Adapted from the 2002 National High School Competition) Let the quadratic function $f(x)=a x^{2}+b x+c(a, b, c \in$ R) satisfy the following conditions:
(1) For $x \in \mathbf{R}$, $f(x-4)=f(2-x)$, and $f(x) \geqslant x$;
(2) For $x \in(0,2)$, $f(x) \leqslant\left(\frac{x+1}{2}\right)^{2}$;
(3) The minimum value of $f(x)$ on $\mathbf{R}$ is 0.
Find $f(x)$.
Given the conditions: when $x \in \mathbf{R}$, $f(x-4)=f(2-x)$, i.e., $f(-4+x)=f(2-x)$. By Conclusion 10, the axis of symmetry of the function $f(x)$ is $x=-1$, thus $-\frac{b}{2 a}=-1$, which means $b=2 a$.
Since the minimum value of $f(x)$ is 0 and the axis of symmetry is $x=-1$, then $f(-1)=0$, i.e., $a-b+c=0$.
From the given conditions, when $x \in(0,2)$, $x \leqslant f(x) \leqslant\left(\frac{x+1}{2}\right)^{2}$.
Taking $x=1$ in the above inequality, we get $1 \leqslant f(1) \leqslant 1$.
Thus, $f(1)=1$, i.e., $a+b+c=1$.
Solving equations (1), (2), and (3), we get $a=c=\frac{1}{4}, b=\frac{1}{2}$.
Therefore, $f(x)=\frac{1}{4}(x+1)^{2}$.
Note: (*) is a sandwich inequality. If the condition $x \leqslant f(x) \leqslant\left(\frac{x+1}{2}\right)^{2}$ is replaced by $x-4 \leqslant f(x) \leqslant x^{2}-3 x$, the same sandwich effect can be achieved. Solving $x-4=x^{2}-3 x$ gives $x=2$, then $-2 \leqslant f(2) \leqslant-2$, hence $f(2)=-2$.