设 $a>0$, 函数 $f(x)$ 在 $[0,a]$ 上连续可微, 证明: $$\bex |f(0)|\leq \frac{1}{a}\int_0^a |f(x)|\rd x+\int_0^a |f'(x)|\rd x. \eex$$ (华中师范大学)
解答: 由 $$\beex \bea \int_0^a f(x)\rd x&=\int_0^a \sez{f(0)+\int_0^x f'(t)\rd t}\rd x\\ &=af(0)+\int_0^a \int_t^a f'(t)\rd x\rd t\\ &=af(0)+\int_0^a f'(t)(a-t)\rd t \eea \eeex$$ 知 $$\beex \bea |f(0)|&=\sev{\frac{1}{a}\int_0^a f(x)\rd x -\int_0^a f'(t)\sex{1-\frac{t}{a}}\rd t}\\ &\leq \frac{1}{a}\int_0^a |f(x)|\rd x +\int_0^a |f'(t)|\rd t. \eea \eeex$$
