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