方企勤 第三章 一元函数积分学 第18题
📝 题目
例 18 设 $f\left( x\right)$ 是在 $\left( {-\infty , + \infty }\right)$ 上的周期函数,周期为 $T$ ,并满足:
(1) $\left| {f\left( x\right) - f\left( y\right) }\right| \leq L\left| {x - y}\right| \left( {\forall x,y \in \left( {-\infty , + \infty }\right) }\right)$ ,其中 $L$ 为常数;
(2) $\displaystyle{\int }_{0}^{T}f\left( x\right) \mathrm{d}x = 0$ .
求证: $\mathop{\max }\limits_{{x \in \left\lbrack {0,T}\right\rbrack }}\left| {f\left( x\right) }\right| \leq \frac{1}{2}{LT}$ .
💡 答案解析
证 由条件 (1) 成立推出 $f\left( x\right)$ 连续,进而知存在 ${x}_{M} \in \left\lbrack {0,T}\right\rbrack$ ,使得
$$ f\left( {x}_{M}\right) = \mathop{\max }\limits_{{x \in \left\lbrack {0,T}\right\rbrack }}\left| {f\left( x\right) }\right| ; $$
又由条件 (2) 及积分中值定理可知,存在 ${x}_{0} \in \left( {0,T}\right)$ ,使得
$$ f\left( {x}_{0}\right) = \frac{1}{T}{\int }_{0}^{T}f\left( x\right) \mathrm{d}x\overset{\text{ 条件 ( 2 ) }}{ = }0. $$
以下分三种情况讨论:
(1) 当 ${x}_{M} = {x}_{0}$ 时,
$$ f\left( {x}_{M}\right) = f\left( {x}_{0}\right) = 0 \Rightarrow \mathop{\max }\limits_{{x \in \left\lbrack {0,T}\right\rbrack }}\left| {f\left( x\right) }\right| = 0, $$
显然要证的不等式成立.
(2)当 ${x}_{M} > {x}_{0}$ 时,这时由 $f\left( x\right)$ 的周期性,有
$$ 2\left| {f\left( {x}_{0}\right) - f\left( {x}_{M}\right) }\right| = \left| {f\left( {x}_{0}\right) - f\left( {x}_{M}\right) }\right| $$
$$ + \left| {f\left( {{x}_{0} + T}\right) - f\left( {x}_{M}\right) }\right| $$
$$ \leq L\left( {{x}_{M} - {x}_{0}}\right) + L\left( {{x}_{0} + T - {x}_{M}}\right) $$
$$ = {LT}\text{ , } $$
即要证的不等式成立.
(3)当 ${x}_{M} < {x}_{0}$ 时,这时还由 $f\left( x\right)$ 的周期性,有
$$ 2\left| {f\left( {x}_{0}\right) - f\left( {x}_{M}\right) }\right| = \left| {f\left( {x}_{M}\right) - f\left( {x}_{0}\right) }\right| $$
$$ + \left| {f\left( {{x}_{M} + T}\right) - f\left( {x}_{0}\right) }\right| $$
$$ \leq L\left( {{x}_{0} - {x}_{M}}\right) + L\left( {{x}_{M} + T - {x}_{0}}\right) $$
$$ = {LT}\text{ , } $$
即要证的不等式也成立.