📝 题目
例 8 设 $f\left( x\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }n{\mathrm{e}}^{-n}\cos {nx}$ ,求证:
(1) $\mathop{\max }\limits_{{0 \leq x \leq {2\pi }}}\left| {f\left( x\right) }\right| \geq \frac{2}{\mathrm{e}}$ ； (2) ${f}^{\prime }\left( x\right)$ 存在;
(3) $\mathop{\max }\limits_{{0 \leq x \leq {2\pi }}}\left| {{f}^{\prime }\left( x\right) }\right| \geq \frac{2}{\pi \mathrm{e}}$ .
💡 答案与解析
证 $\left( 1\right) f\left( 0\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }n{\mathrm{e}}^{-n} = \frac{1}{\mathrm{e}} + \mathop{\sum }\limits_{{n = 2}}^{\infty }n{\mathrm{e}}^{-n}$
$$ \geq \frac{1}{\mathrm{e}} + 2\mathop{\sum }\limits_{{n = 2}}^{\infty }{\mathrm{e}}^{-n} = \frac{1}{\mathrm{e}}\left( {1 + \frac{2}{\mathrm{e} - 1}}\right) > \frac{2}{\mathrm{e}}, $$
因此
$$ \mathop{\max }\limits_{{0 \leq x \leq {2\pi }}}\left| {f\left( x\right) }\right| \geq \left| {f\left( 0\right) }\right| > 2/\mathrm{e}. $$
(2)因为级数
$$ \mathop{\sum }\limits_{{n = 1}}^{\infty }{\left( n{\mathrm{e}}^{-n}\cos nx\right) }^{\prime } = - \mathop{\sum }\limits_{{n = 1}}^{\infty }{n}^{2}{\mathrm{e}}^{-n}\sin {nx} $$
在实轴上一致收敛,所以 ${f}^{\prime }\left( x\right)$ 存在,并且连续,可表示为
$$ {f}^{\prime }\left( x\right) = - \mathop{\sum }\limits_{{n = 1}}^{\infty }{n}^{2}{\mathrm{e}}^{-n}\sin {nx}. $$
(3)证法 1 用贝塞尔不等式,
$$ {\int }_{0}^{2\pi }{\left| {f}^{\prime }\left( x\right) \right| }^{2}\mathrm{\;d}x = \mathop{\sum }\limits_{{n = 1}}^{\infty }\pi {n}^{2}{\mathrm{e}}^{-{2n}} > \frac{\pi }{{\mathrm{e}}^{2}}. $$
又设 ${\left| {f}^{\prime }\left( {x}_{0}\right) \right| }^{2} = \mathop{\max }\limits_{{0 \leq x \leq {2\pi }}}{\left| {f}^{\prime }\left( x\right) \right| }^{2}$ ,则
$$ {\left| {f}^{\prime }\left( {x}_{0}\right) \right| }^{2} \geq \frac{1}{2\pi }{\int }_{0}^{2\pi }{\left| {f}^{\prime }\left( x\right) \right| }^{2}\mathrm{\;d}x \geq \frac{1}{2{\mathrm{e}}^{2}}. $$
从而
$$ \mathop{\max }\limits_{{0 \leq x \leq {2\pi }}}\left| {{f}^{\prime }\left( x\right) }\right| \geq \frac{1}{\sqrt{2}\mathrm{e}} > \frac{2}{\pi \mathrm{e}} $$
证法 2 由 ${f}^{\prime }\left( x\right)$ 的傅氏系数公式,
$$ \frac{1}{\pi }{\int }_{0}^{2\pi }{f}^{\prime }\left( x\right) \sin x\mathrm{\;d}x = - \frac{1}{\mathrm{e}}, $$
所以
$$ \frac{1}{\mathrm{e}} = \frac{1}{\pi }\left| {{\int }_{0}^{2\pi }{f}^{\prime }\left( x\right) \sin x\mathrm{\;d}x}\right| \leq \frac{1}{\pi }{\int }_{0}^{2\pi }\left| {{f}^{\prime }\left( x\right) }\right| \left| {\sin x}\right| \mathrm{d}x $$
$$ \leq \mathop{\max }\limits_{{0 \leq x \leq {2\pi }}}\left| {{f}^{\prime }\left( x\right) }\right| \cdot \frac{1}{\pi }{\int }_{0}^{2\pi }\left| {\sin x}\right| \mathrm{d}x = \frac{4}{\pi }\mathop{\max }\limits_{{0 \leq x \leq {2\pi }}}\left| {{f}^{\prime }\left( x\right) }\right| , $$
由此即得
$$ \mathop{\max }\limits_{{0 \leq x \leq {2\pi }}}\left| {{f}^{\prime }\left( x\right) }\right| \geq \frac{\pi }{4\mathrm{e}} > \frac{2}{\pi \mathrm{e}} $$