杂题 - Eufisky - The lost book
含奇点的第二型曲面积分计算
曲线积分的计算

杂题

Eufisky posted @ 2017年9月13日 19:55 in 数学分析 with tags 函数 , 1054 阅读

设函数$~f(x)$ 在$~[a,b]$ 上连续,但不为常数.求证:$~\exists \xi\in(a,b)$,使$~f(x)$ 在$~\xi$ 不取极值.

注意到 $f$ 不是常数函数, 并且 $f\in C[a,b],$ 所以 $f$ 的值域 $R$ 是一个有限闭区间. 

先将 $f$ 开拓定义到整个 $\mathbf R$ 上: 对 $x\leq a,$ 让 $f(x)=f(a);$ 对 $x\geq b,$ 让 $f(x)=f(b).$ 让 $$C=\{f(x);\ \hbox{$x$ 是 $f$ 在 $\mathbf R$ 中的极小值点}\},$$ 下面来证明 $C$ 是一个至多可列集. 

任给 $c\in C,$ 存在 $x\in\mathbf R$ 以及 $u_x,v_x\in\mathbf Q,$ 使得 $$u_x<x<v_x,\ f(y)\geq f(x)=c,\ \forall y\in(u_x,v_x).$$这样就得到了一个从 $C$ 到 $\mathbf Q\times\mathbf Q$ 的单射 $c\mapsto(u_x,v_x)$, 故 $C$ 是至多可列集.

类似可证 $f$ 极大值的全体也是至多可列集. 从而 $f$ 的极值的全体是 $R$ 的至多可列子集. 这就完成了证明. 

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