📝 题目
例 7 设 $f\left( x\right) \in R\left\lbrack {a,b}\right\rbrack$ ,求证: ${\mathrm{e}}^{f\left( x\right) } \in R\left\lbrack {a,b}\right\rbrack$ .
💡 答案与解析
证 作分划 $\Delta : a = {x}_{0} < {x}_{1} < {x}_{2} < \cdots < {x}_{n} = b$ . 设 ${x}^{\prime },{x}^{\prime \prime } \in$ $\left\lbrack {{x}_{k},{x}_{k + 1}}\right\rbrack$ ,则根据微分中值定理, $\exists \xi$ ,使得
$$ \left| {{\mathrm{e}}^{f\left( {x}^{\prime }\right) } - {\mathrm{e}}^{f\left( {x}^{\prime \prime }\right) }}\right| = {\mathrm{e}}^{\xi }\left| {f\left( {x}^{\prime }\right) - f\left( {x}^{\prime \prime }\right) }\right| , \tag{2.2} $$
其中 $\xi$ 介于 $f\left( {x}^{\prime }\right)$ 与 $f\left( {x}^{\prime \prime }\right)$ 之间. 因为可积函数一定有界,所以可设 $\left| {f\left( x\right) }\right| \leq M$ . 于是由 (2.2) 式得
$$ \left| {{\mathrm{e}}^{f\left( {x}^{\prime }\right) } - {\mathrm{e}}^{f\left( {x}^{\prime \prime }\right) }}\right| = {\mathrm{e}}^{M}\left| {f\left( {x}^{\prime }\right) - f\left( {x}^{\prime \prime }\right) }\right| . \tag{2.3} $$
设 ${\omega }_{k}$ 与 ${\widetilde{\omega }}_{k}$ 分别表示 $f\left( x\right)$ 与 ${\mathrm{e}}^{f\left( x\right) }$ 在 $\left\lbrack {{x}_{k},{x}_{k + 1}}\right\rbrack$ 上的振幅,在公式 (2.3) 中,让 ${x}^{\prime },{x}^{\prime \prime }$ 在 $\left\lbrack {{x}_{k},{x}_{k + 1}}\right\rbrack$ 上变化,两边取上确界得到
$$ {\widetilde{\omega }}_{k} \leq {\mathrm{e}}^{M}{\omega }_{k}\;\left( {k = 0,1,\cdots ,n - 1}\right) , $$
由此推出
$$ \mathop{\sum }\limits_{{k = 0}}^{{n - 1}}{\widetilde{\omega }}_{k}\Delta {x}_{k} \leq {\mathrm{e}}^{M}\mathop{\sum }\limits_{{k = 0}}^{{n - 1}}{\omega }_{k}\Delta {x}_{k} \tag{2.4} $$
令 $\displaystyle{\lambda = \mathop{\max }\limits_{{0 \leq k \leq n - 1}}\Delta {x}_{k}}$ ,因为 $f\left( x\right) \in R\left\lbrack {a,b}\right\rbrack$ ,所以 $\displaystyle{\mathop{\lim }\limits_{{\lambda \rightarrow 0}}\mathop{\sum }\limits_{{k = 0}}^{{n - 1}}{\omega }_{k}\Delta {x}_{k} = 0}$ . 由此, 令 $\lambda \rightarrow 0$ ,对 (2.4) 式取极限得
$$ 0 \leq \mathop{\lim }\limits_{{\lambda \rightarrow 0}}\mathop{\sum }\limits_{{k = 0}}^{{n - 1}}{\widetilde{\omega }}_{k}\Delta {x}_{k} \leq {\mathrm{e}}^{M}\mathop{\lim }\limits_{{\lambda \rightarrow 0}}\mathop{\sum }\limits_{{k = 0}}^{{n - 1}}{\omega }_{k}\Delta {x}_{k} = 0 \Rightarrow \mathop{\lim }\limits_{{\lambda \rightarrow 0}}\mathop{\sum }\limits_{{k = 0}}^{{n - 1}}{\widetilde{\omega }}_{k}\Delta {x}_{k} = 0. $$
因此 ${\mathrm{e}}^{f\left( x\right) } \in R\left\lbrack {a,b}\right\rbrack$ .