Problem 1298

Proof 1 For all $x \in [0,1]$, we have
\begin{array}{l} |f(x)|=|f(x)-f(0)|\frac{1}{2}$, assume
\frac{1}{2}\frac{1}{2}$,
then let
$$\begin{array}{c} 1 \geqslant x_{1}>x_{2} \geqslant 0 . \\ f(0)=f(1), \end{array}$$

Since
so $\left|f\left(x_{2}\right)-f\left(x_{1}\right)\right|$
$$\begin{array}{l} =\left|f\left(x_{2}\right)-f(0)+f(1)-f\left(x_{1}\right)\right| \\ \leqslant\left|f\left(x_{2}\right)-f(0)\right|+\left|f(1)-f\left(x_{1}\right)\right| \\ <x_{2}-0+1-x_{1} \\ =1-\left(x_{1}-x_{2}\right)<\frac{1}{2} . \end{array}$$

So $\left|f\left(x_{2}\right)-f\left(x_{1}\right)\right|<\frac{1}{2}$.
This is the conventional proof of the problem, using the properties of absolute value inequalities. Deep thinking reveals the geometric meaning contained within, leading to a new proof by constructing a geometric figure:
Proof 3 Because $\left|f\left(x_{2}\right)-f\left(x_{1}\right)\right|<\left|x_{2}-x_{1}\right|$.
So
$$\left|\frac{f\left(x_{2}\right)-f\left(x_{1}\right)}{x_{2}-x_{1}}\right|<1,$$

which means the slope $k$ of the line connecting any two points on the graph of $y=f(x)$ satisfies $|k|<1$. Therefore, all points on the graph of $y=f(x)$ must be confined within a square region with side length $\frac{\sqrt{2}}{2}$ as shown in Figure 5-1. Let the highest point on the graph be $E\left(x_{4}, y_{0}\right)$, and the lowest point be $F\left(x_{b}, y_{b}\right)$. Then $\left|k_{E F}\right|<1$, connect $E F$, intersecting $O B$ at point $P$, extend $F E$ and $E F$ to intersect $O A$ and $B C$ at $G$ and $H$ respectively. Draw a line parallel to $A B$ through point $P$ intersecting $O A$ and $B C$ at $M$ and $N$ respectively, and draw perpendiculars from points $E$, $F$, $G$, $H$, $M$, and $N$ to $O B$, with the feet of the perpendiculars being $E^{\prime}$, $F^{\prime}$, $G^{\prime}$, $H^{\prime}$, $M^{\prime}$, and $N^{\prime}$ respectively, then
$$\begin{aligned} f\left(x_{s}\right)-f\left(x_{n}\right) & <\left|G G^{\prime}\right|+\left|H H^{\prime}\right| \\ & <\left|M M^{\prime}\right|+\left|N N^{\prime}\right| \\ & =\frac{\sqrt{2}}{2}(|P M|+|P N|) \\ & =\frac{\sqrt{2}}{2}|M N|=\frac{1}{2} . \end{aligned}$$

Since $E$ and $F$ are the highest and lowest points on the graph of $y=f(x)$ respectively, we have $\left|f\left(x_{2}\right)-f\left(x_{1}\right)\right| \leqslant f\left(x_{a}\right)-f\left(x_{b}\right)<\frac{1}{2}$.