Let $F$ be the set of real polynomials $f(x)$ satisfying
(1) the degree of $f(x)$ is less than or equal to 3;
(2) for any $x \in [0,1]$, $|f(x)| \leqslant 1$.
Find $\max _{f \in F} f(2)$.
Solution: Let $f(x)=a x^{3}+b x^{2}+c x+d$. Denote
$$
\left.\begin{array}{l}
A=f(1)=a+b+c+d, \\
B=f\left(\frac{3}{4}\right)=\frac{27}{64} a+\frac{9}{16} b+\frac{3}{4} c+d, \\
C=f\left(\frac{1}{4}\right)=\frac{1}{64} a+\frac{1}{16} b+\frac{1}{4} c+d, \\
D=f(0)=d .
\end{array}\right\}
$$

By condition (2), we know $A, B, C, D \in[-1,1]$.
Let $f(2)=8 a+4 b+2 c+d=x A+y B+z C+w D$.
Substituting (1) into the above equation and comparing the coefficients on both sides, we get
$$
\left\{\begin{array}{l}
x+\frac{27}{64} y+\frac{1}{64} z=8, \\
x+\frac{9}{16} y+\frac{1}{16} z=4, \\
x+\frac{3}{4} y+\frac{1}{4} z=2, \\
x+y+z+w=1 .
\end{array}\right.
$$

Solving, we get $(x, y, z, w)=\left(\frac{70}{3},-\frac{112}{3}, \frac{80}{3},-\frac{35}{3}\right)$.
Thus, $f(2)=\frac{70}{3} A-\frac{112}{3} B+\frac{80}{3} C-\frac{35}{3} D$
$$
\leqslant \frac{70}{3}+\frac{112}{3}+\frac{80}{3}+\frac{35}{3}=99 \text {. }
$$

On the other hand, let $f(x)=32 x^{3}-48 x^{2}+18 x-1$ (i.e., if (2) holds with equality, then $A=C=1, B=D=-1$, solving for $a, b, c, d$ gives $f(x))$, then $f(2)=99$.
Clearly, $f(x)$ satisfies condition (1). Next, we prove that $f(x)$ satisfies condition (2).
Since $f^{\prime}(x)=96 x^{2}-96 x+18=6(4 x-1)(4 x-3)$, the maximum and minimum values of $f(x)$ must be attained at $0, \frac{1}{4}, \frac{3}{4}, 1$.
And $f(0)=f\left(\frac{3}{4}\right)=-1, f\left(\frac{1}{4}\right)=f(1)=1$,
so, $f(x)$ satisfies condition (2) (or directly verify $f\left(\frac{1+\cos x}{2}\right)=\cos 3 x$ to show that $f(x)$ satisfies condition (2)).
Therefore, $f \in F$.
Hence $\max _{f \in F} f(2)=99$.
(Zhu Qing San, Peking University School of Mathematical Sciences, Class 2, Grade 04, 100871)