设 $f(x)$ 是 $[0,2\pi]$ 上的凸函数, $f'(x)$ 有界. 求证: $$\bex a_n=\frac{1}{\pi}\int_0^{2\pi} f(x)\cos nx\rd x\geq 0. \eex$$

证明: $$\beex \bea a_n&=\frac{1}{n\pi}\int_0^{2\pi} f(x)\rd \sin nx =-\frac{1}{n\pi}\int_0^{2\pi} f'(x)\sin nx\rd x\\ &=-\frac{1}{n^2\pi}\int_0^{2n\pi}f'\sex{\frac{t}{2n}} \sin t\rd t\\ &=-\frac{1}{n^2\pi}\sum_{k=0}^n \sez{\int_{2k\pi}^{(2k+1)\pi} +\int_{(2k+1)\pi}^{(2k+2)\pi} f'\sex{\frac{t}{2n}} \sin t \rd t}\\ &=-\frac{1}{n^2\pi}\sum_{k=0}^n \int_{2k\pi}^{(2k+1)\pi} \sez{f'\sex{\frac{t}{2n}} -f'\sex{\frac{t+\pi}{2n}}}\sin t\rd t\\ &\geq 0. \eea \eeex$$