下凸函数在开区间内闭 Lip 连续 ¶

2020 微积分甲期末考试 11 题

设函数 \(f\) 在 \((0,1)\) 上连续且满足 \(\forall x_{1} \in (0,1), x_{2} \in (0,1), x_{3} \in (0,1)\)，且 \(x_{1} < x_{2} < x_{3}\)，都有 \(\dfrac{f(x_{2}) - f(x_{1})}{x_{2}-x_{1}} \leq \dfrac{f(x_{3}) - f(x_{2})}{x_{3}-x_{2}}\) 成立。若设 \(a < b\) 且 \([a,b] \subset (0,1)\)，试证明：

\(\exists L > 0\) 使得 \(\forall t_{1}, t_{2} \in [a,b]\)，有 \(\vert f(t_{2}) - f(t_{1})\vert \leq L\vert t_{2} - t_{1}\vert\) 成立。

不妨设 \(0 < a \leq t_{1} < t_{2} \leq b < 1\)，则由条件有

\(\dfrac{f(a) - f(\dfrac{a}{2})}{a - \dfrac{a}{2}} \leq \dfrac{f(t_{1}) - f(a)}{t_{1} - a} \leq \dfrac{f(t_{2}) - f(t_{1})}{t_{2} - t_{1}} \leq \dfrac{f(b) - f(t_{2})}{b - t_{2}} \leq \dfrac{f(\dfrac{b + 1}{2}) - f(b)}{\dfrac{b + 1}{2}-b}\)

由此 \(\dfrac{f(t_{2}) - f(t_{1})}{t_{2} - t_{1}}\) 有界。取 \(L=\max\left\{\left|\dfrac{f(a) - f(\dfrac{a}{2})}{\dfrac{a}{2}}\right|, \left|\dfrac{f(\dfrac{b + 1}{2}) - f(b)}{\dfrac{1 - b}{2}}\right|\right\}\) 即可。