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设 $f(x)$ 是 $[-pi,pi]$ 上的凸函数, $f'(x)$ 有界. 求证: $$ex a_{2n}=frac{1}{pi}int_{-pi}^pi f(x)cos 2nx d xgeq 0;quad a_{2n+1}=frac{1}{pi}int_{-pi}^pi f(x)cos (2n+1)x d xleq 0. eex$$
证明: $$eex ea a_{2n}&=frac{1}{pi}int_{-pi}^pi f(x)cos 2nx d x =frac{1}{2n}int_{-pi}^pi f(x) d sin 2nx\ &=-frac{1}{2n}int_{-pi}^pi f'(x)sin 2nx d x =-frac{1}{(2n)^2} int_{-2npi}^{2npi} f'sex{frac{t}{2n}} sin t d t\ &=-frac{1}{(2n)^2}sum_{k=-n}^{n-1}sez{ int_{2kpi}^{(2k+1)pi} +int_{(2k+1)^pi}^{(2k+2)pi} f'sex{frac{t}{2n}}sin t d t }\ &=-frac{1}{(2n)^2}sum_{k=-n}^{n-1} int_{2kpi}^{(2k+1)pi}sez{f'sex{frac{t}{2n}}-f'sex{frac{t+pi}{2n}}}sin t d t\ &geq 0, eea eeex$$ 最后一步是因为 $$ex -2npi leq tleq (2n-1)pi a -pi leq frac{t}{2n}leq pi -frac{pi}{2n} a -pi leq frac{t}{2n} leq frac{t+pi}{2n}leq pi eex$$ 及 $f$ 的凸性 $ a f'$ 单调递增. 同理, $$ex a_{2n+1}=-frac{1}{(2n+1)^2}sum_{k=-n-1}^{n-1} int_{(2k+1)pi}^{(2k+2)pi} sez{f'sex{frac{t}{2n+1}}-f'sex{frac{t+pi}{2n+1}}}sin t d tleq 0, eex$$ 这里, 我们还需注意 $(2k+1)pi leq xleq (2k+2)pi a sin xleq 0$.