计算题
13.设f(x)在[0,1]上具有连续导数,证明:当x∈[0,1],有
|f(x)|≤∫01(|f(t)|+|f'(t)|)dt
【正确答案】由中值定理可知:∫01|f(t)|dt=|f(ξ)|,0≤ξ≤1
f(x)一f(ξ)=∫ξxf'(t)dt
则:f(x)=f(ξ)+∫ξxf'(t)dt
则:|f(x)|≤|f(ξ)|+|∫ξxf(t)dt|≤|f(ξ)|+∫ξx|f'(t)|dt≤|f(ξ)|+∫01|f(t)|dt
即:|f(x)|≤∫01[|f(t)|+|f'(t)|]dt
【答案解析】