Question

一道积分证明题

a>0,f'(x)在[0，a]连续，求证f(0)的绝对值<=f(x)绝对值在0到a的积分/a+f'(x)绝对值在0到a的积分

Answer 1

首先存在ξ∈(0,a)，使得∫[0,a]|f(x)|dx=a|f(ξ)|
又|f(ξ)|=|f(0)+∫[0->ξ]f'(x)dx|
≥|f(0)|-|∫[0->ξ]f'(x)dx|
≥|f(0)|-∫[0->ξ]|f'(x)|dx
≥|f(0)|-∫[0->a]|f'(x)|dx
∴|f(0)|≤|f(ξ)|+∫[0->a]|f'(x)|dx=(∫[0,a]|f(x)|dx)/a+∫[0->a]|f'(x)|dx