📝 题目
例 25 设函数 $f\left( x\right) \in C\left\lbrack {a,b}\right\rbrack ,f\left( x\right) \geq 0$ ,且 $\displaystyle{\int }_{a}^{b}f\left( x\right) \mathrm{d}x = 0$ . 求证: 在 $\left\lbrack {a,b}\right\rbrack$ 上, $f\left( x\right) \equiv 0$ .
💡 答案与解析
证法 1 因为对于区间(a, b)中的每一个点 $x$ ,
$$ 0 \leq {\int }_{a}^{x}f\left( t\right) \mathrm{d}t \leq {\int }_{a}^{b}f\left( t\right) \mathrm{d}t = 0 \Rightarrow {\int }_{a}^{x}f\left( t\right) \mathrm{d}t \equiv 0, $$
所以
$$ f\left( x\right) = {\left( {\int }_{a}^{x}f\left( t\right) \mathrm{d}t\right) }^{\prime } \equiv 0 \Rightarrow f\left( x\right) \equiv 0\;\left( {\forall x \in \left( {a,b}\right) }\right) . $$
又函数 $f\left( x\right)$ 在 $\left\lbrack {a,b}\right\rbrack$ 上连续,故有 $f\left( x\right) \equiv 0\left( {\forall x \in \left\lbrack {a,b}\right\rbrack }\right)$ .
证法 2 首先证明 $f\left( x\right) \equiv 0\left( {\forall x \in \left( {a,b}\right) }\right)$ . 用反证法.
假设 $f\left( x\right) ≢ 0\left( {\forall x \in \left( {a,b}\right) }\right)$ ,则存在 ${x}_{0} \in \left( {a,b}\right)$ ,使得 $f\left( {x}_{0}\right) \neq$ 0,不妨设 $f\left( {x}_{0}\right) > 0$ . 取 $\varepsilon = \frac{f\left( {x}_{0}\right) }{2} > 0$ ,由 $\mathop{\lim }\limits_{{x \rightarrow {x}_{0}}}f\left( x\right) = f\left( {x}_{0}\right)$ ,存在 $\delta > 0$ $\left( {a < {x}_{0} - \delta ,{x}_{0} + \delta < b}\right)$ ,使得对 $\forall x\left( {\left| {x - {x}_{0}}\right| < \delta }\right)$ ,有
$$ \left| {f\left( x\right) - f\left( {x}_{0}\right) }\right| < \varepsilon , $$
由此得
$$ f\left( x\right) > f\left( {x}_{0}\right) - \left| {f\left( x\right) - f\left( {x}_{0}\right) }\right| > f\left( {x}_{0}\right) - \varepsilon > 0. $$
这样有
$$ 0 = {\int }_{a}^{b}f\left( x\right) \mathrm{d}x $$
$$ = {\int }_{a}^{{x}_{0} - \delta }f\left( x\right) \mathrm{d}x + {\int }_{{x}_{0} - \delta }^{{x}_{0} + \delta }f\left( x\right) \mathrm{d}x + {\int }_{{x}_{0} + \delta }^{b}f\left( x\right) \mathrm{d}x $$
$$ \geq 0 + \frac{f\left( {x}_{0}\right) }{2} \cdot {2\delta } + 0 = {\delta f}\left( {x}_{0}\right) > 0. $$
于是产生 $0 > 0$ 的矛盾. 故反证法假设不成立,即结论成立,这表明 $f\left( x\right) \equiv 0$ ,所以 $f\left( x\right) \equiv 0\left( {\forall x \in \left( {a,b}\right) }\right)$ . 再因为 $f\left( x\right)$ 在 $\left\lbrack {a,b}\right\rbrack$ 上连续,所以
$$ f\left( a\right) = \mathop{\lim }\limits_{{x \rightarrow a + 0}}f\left( x\right) = 0,\;f\left( b\right) = \mathop{\lim }\limits_{{x \rightarrow b - 0}}f\left( x\right) = 0, $$
即得在 $\left\lbrack {a,b}\right\rbrack$ 上, $f\left( x\right) \equiv 0$ .
引申 本例的逆否命题: 如果在 $\left\lbrack {a,b}\right\rbrack$ 上,函数 $f\left( x\right) \in C\left\lbrack {a,b}\right\rbrack$ , 且 $f\left( x\right) ≢ 0$ ,则 $\displaystyle{\int }_{a}^{b}f\left( x\right) \mathrm{d}x > 0$ . 这是一个非常有用的结果.