(AMM. Problems and Solutions. 2015. 01) Let $f$ be a twice continuously differentiable function from $[0,1]$ into $bR$. Let $p$ be an integer greater than $1$. Given that $$ex sum_{k=1}^{p-1} fsex{frac{k}{p}}=-frac{1}{2}[f(0)+f(1)], eex$$ prove that $$ex sez{int_0^1 f(x) d x}^2leq frac{1}{5!p^4} int_0^1 [f''(x)]^2 d x. eex$$
证明: By Newton-Leibniz formula and Fubini's theorem, we have $$eex ea int_0^1 f(x) d x&=sum_{k=1}^p int_frac{k-1}{p}^frac{k}{p}f(t) d t =sum_{k=1}^p int_{frac{k-1}{p}}^frac{k}{p} sez{fsex{frac{k-1}{p}}+int_{frac{k-1}{p}}^t f'(s) d s} d t\ &=frac{1}{p} sum_{k=1}^p fsex{frac{k-1}{p}} +sum_{k=1}^p int_{frac{k-1}{p}}^frac{k}{p} int_{frac{k-1}{p}}^tf'(s) d s d t\ &=frac{1}{p} sez{f(0)-frac{1}{2}(f(0)+f(1))} +sum_{k=1}^p int_{frac{k-1}{p}}^frac{k}{p} sex{frac{k}{p}-t}f'(t) d t\ &=-frac{1}{2p}[f(1)-f(0)] +sum_{k=1}^p int_{frac{k-1}{p}}^frac{k}{p} sex{frac{k}{p}-t}f'(t) d t\ &=-frac{1}{2p}sum_{k=1}^p sez{fsex{frac{k}{p}}-fsex{frac{k-1}{p}}} +sum_{k=1}^p int_{frac{k-1}{p}}^frac{k}{p} sex{frac{k}{p}-t}f'(t) d t\ &=sum_{k=1}^p int_{frac{k-1}{p}}^frac{k}{p} f'(t)sex{frac{k}{p}-t-frac{1}{2p}} d t\ &=sum_{k=1}^p int_{frac{k-1}{p}}^frac{k}{p} sex{frac{k}{p}-t-frac{1}{2p}}sez{f'sex{frac{k-1}{p}}+int_{frac{k-1}{p}}^t f''(s) d s} d t\ &=sum_{k=1}^p f'sex{frac{k-1}{p}}int_{frac{k-1}{p}}^frac{k}{p} sex{frac{k}{p}-t-frac{1}{2p}} d t\ &quad +sum_{k=1}^p int_{frac{k-1}{p}}^frac{k}{p} sex{frac{k}{p}-t-frac{1}{2p}}int_{frac{k-1}{p}}^t f''(s) d s d t\ &=sum_{k=1}^p int_{frac{k-1}{p}}^frac{k}{p} sex{frac{k}{p}-t-frac{1}{2p}}int_{frac{k-1}{p}}^t f''(s) d s d t\ &=sum_{k=1}^p int_{frac{k-1}{p}}^frac{k}{p}f''(s)int_s^frac{k}{p} sex{frac{k}{p}-t-frac{1}{2p}} d t d s, eea eeex$$ Then invoking the Cauchy-Schwarz inequality, we obtain $$eex ea sez{int_0^1 f(x) d x}^2 &=sez{sum_{k=1}^p int_{frac{k-1}{p}}^frac{k}{p}f''(s)int_s^frac{k}{p} sex{frac{k}{p}-t-frac{1}{2p}} d t d s}^2\ &leq psum_{k=1}^p sez{ int_{frac{k-1}{p}}^frac{k}{p}f''(s)int_s^frac{k}{p} sex{frac{k}{p}-t-frac{1}{2p}} d t d s}^2\ &leq psum_{k=1}^p int_{frac{k-1}{p}}^frac{k}{p} [f''(s)]^2 d s cdot int_{frac{k-1}{p}}^frac{k}{p} sez{int_s^frac{k}{p} sex{frac{k}{p}-t-frac{1}{2p}} d t}^2 d s\ &=psum_{k=1}^p int_{frac{k-1}{p}}^frac{k}{p} [f''(s)]^2 d scdot frac{1}{120p^5}\ &=frac{1}{5!p^4} int_0^1 [f''(x)]^2 d x. eea eeex$$