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

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问题 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

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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

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