需要全部的解答, 请 http://www.cnblogs.com/zhangzujin/p/3527416.html
设 $f(x)$ 是 $[-\pi,\pi]$ 上的凸函数, $f'(x)$ 有界. 求证: $$\bex a_{2n}=\frac{1}{\pi}\int_{-\pi}^\pi f(x)\cos 2nx\rd x\geq 0;\quad a_{2n+1}=\frac{1}{\pi}\int_{-\pi}^\pi f(x)\cos (2n+1)x\rd x\leq 0. \eex$$
证明: $$\beex \bea a_{2n}&=\frac{1}{\pi}\int_{-\pi}^\pi f(x)\cos 2nx\rd x =\frac{1}{2n}\int_{-\pi}^\pi f(x)\rd \sin 2nx\\ &=-\frac{1}{2n}\int_{-\pi}^\pi f'(x)\sin 2nx\rd x =-\frac{1}{(2n)^2} \int_{-2n\pi}^{2n\pi} f'\sex{\frac{t}{2n}} \sin t\rd t\\ &=-\frac{1}{(2n)^2}\sum_{k=-n}^{n-1}\sez{ \int_{2k\pi}^{(2k+1)\pi} +\int_{(2k+1)^\pi}^{(2k+2)\pi} f'\sex{\frac{t}{2n}}\sin t\rd t }\\ &=-\frac{1}{(2n)^2}\sum_{k=-n}^{n-1} \int_{2k\pi}^{(2k+1)\pi}\sez{f'\sex{\frac{t}{2n}}-f'\sex{\frac{t+\pi}{2n}}}\sin t\rd t\\ &\geq 0, \eea \eeex$$ 最后一步是因为 $$\bex -2n\pi \leq t\leq (2n-1)\pi \ra -\pi \leq \frac{t}{2n}\leq \pi -\frac{\pi}{2n} \ra -\pi \leq \frac{t}{2n} \leq \frac{t+\pi}{2n}\leq \pi \eex$$ 及 $f$ 的凸性 $\ra\ f'$ 单调递增. 同理, $$\bex a_{2n+1}=-\frac{1}{(2n+1)^2}\sum_{k=-n-1}^{n-1} \int_{(2k+1)\pi}^{(2k+2)\pi} \sez{f'\sex{\frac{t}{2n+1}}-f'\sex{\frac{t+\pi}{2n+1}}}\sin t\rd t\leq 0, \eex$$ 这里, 我们还需注意 $(2k+1)\pi \leq x\leq (2k+2)\pi\ra \sin x\leq 0$.
