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

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