946

已知定义在 ${\displaystyle [0,1]}$ 上的函数 ${\displaystyle f(x)}$ 满足:

1. ${\displaystyle f(0)=f(1)=0}$;
2. 对所有 ${\displaystyle x,y\in [0,1]}$, 且 ${\displaystyle x\ne y}$, 有 ${\displaystyle |f(x)-f(y)|<{\displaystyle \frac{1}{2}|x-y|}}$.

若对所有 ${\displaystyle x,y\in [0,1]}$, ${\displaystyle |f(x)-f(y)|<k}$, 求 ${\displaystyle k}$ 的最小值.

一方面，取${\displaystyle {\displaystyle \{\begin{array}{rl}& tx,0\le x\le \frac{1}{2},\\ & t(1-x),\frac{1}{2}<x\le 1\end{array}(0<t<\frac{1}{2})}}$
满足题意，则 ${\displaystyle |f(x)-f(y)|\le {\displaystyle \frac{t}{2}\to \frac{1}{4}(t\to \frac{1}{2})}}$, 故 ${\displaystyle k}$ 可以取到 ${\displaystyle {\displaystyle \frac{1}{4}}}$.
另一方面，如果存在 ${\displaystyle x_0,y_0}$ 使得 ${\displaystyle {\displaystyle |f(x_0)-f(y_0)|>\frac{1}{4}}}$, 则不妨设 ${\displaystyle y_0>x_0,f(y_0)>f(x_0)}$. 则$$\begin{array}{r}\frac{1}{4}<f(y_0)-f(x_0)<\frac{1}{2}(y_0-x_0)\Rightarrow y_0-x_0>\frac{1}{2}.\end{array}$$
此外$$\begin{array}{r}|f(x_0)-0|<\frac{1}{2}(x_0-0),|f(y_0)-0|<\frac{1}{2}(1-y_0),\end{array}$$
两式相加得$$\begin{array}{r}\frac{1}{4}>\frac{1}{2}(x_0+1-y_0)>|f(x_0)|+|f(y_0)|\ge |f(x_0)-f(y_0)|>\frac{1}{4},\end{array}$$
矛盾！因此 ${\displaystyle k}$ 无法继续减小，这样 ${\displaystyle k}$ 的最小值是 ${\displaystyle {\displaystyle \frac{1}{4}}}$.