裴礼文 一元积分学 370页 练习4.3.26 解答

castelu

练习4.3.26：

　　设$f(x)$是$[-\pi,\pi]$上凸函数，$f'(x)$有界。求证：
$$a_{2n}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos 2nxdx \ge 0$$
$$a_{2n+1}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos (2n+1)xdx \le 0$$

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解：
　　由于$f(x)$是$[-\pi,\pi]$上凸函数，$f'(x)$有界，故$f'(x)$在$[-\pi,\pi]$上单调递增且可积
　　当
$$-2n\pi \le t \le (2n-1)\pi$$
　　时
$$-\pi \le \frac{t}{2n} \le \frac{t+\pi}{2n} \le \pi$$
$$\begin{eqnarray*}
a_{2n}&=&\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos 2nxdx = \frac{1}{2n\pi}\int_{-\pi}^{\pi}f(x) d (\sin 2nx)\\
&=&-\frac{1}{2n\pi}\int_{-\pi}^{\pi}f'(x)\sin 2nxdx=-\frac{1}{4n^2\pi}\int_{-2n\pi}^{2n\pi}f'\left(\frac{t}{2n}\right)\sin tdt\\
&=&-\frac{1}{4n^2\pi}\sum\limits_{k=-n}^{n-1}\left[\int_{2k\pi}^{(2k+1)\pi}f'\left(\frac{t}{2n}\right)\sin tdt+\int_{(2k+1)\pi}^{(2k+2)\pi}f'\left(\frac{t}{2n}\right)\sin tdt\right]\\
&=&-\frac{1}{4n^2\pi}\sum\limits_{k=-n}^{n-1}\int_{2k\pi}^{(2k+1)\pi}\left[f'\left(\frac{t}{2n}\right)-f'\left(\frac{t+\pi}{2n}\right)\right]\sin tdt\\
&\ge 0&
\end{eqnarray*}$$
　　同理
$$\begin{eqnarray*}
a_{2n+1}&=&-\frac{1}{(2n+1)^2\pi}\sum\limits_{k=-n-1}^{n-1}\int_{(2k+1)\pi}^{(2k+2)\pi}\left[f'\left(\frac{t}{2n+1}\right)-f'\left(\frac{t+\pi}{2n+1}\right)\right]\sin tdt\\
&\le 0&
\end{eqnarray*}$$