设f′(χ)在[0,1]上连续,且f(1)=f(0)=1.证明:∫01f′2(χ)dχ≥1.

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问题 设f′(χ)在[0,1]上连续,且f(1)=f(0)=1.证明:∫01f′2(χ)dχ≥1.

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答案由1=f(1)-f(0)=∫01f′(χ)dχ, 得12=1=(∫01f′(χ)dχ)2≤∫0112dχ∫01f′2(χ)dχ=∫01f′2(χ)dχ,即∫01f′2(χ)dχ≥1.

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