设f(χ)在[0,+∞)上连续,在(0,+∞)内可导且满足f(0)=0,f(χ)≥0,f(χ)≥f′(χ)(χ>0),求证:f(χ)≡0.

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问题 设f(χ)在[0,+∞)上连续,在(0,+∞)内可导且满足f(0)=0,f(χ)≥0,f(χ)≥f′(χ)(χ>0),求证:f(χ)≡0.

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答案由f′(χ)-f(χ)≤0, 得 e-χ[f′(χ)-(χ)]=[e-χ(χ)]′≤0. 又f(χ)e-χχ=0=0, 则f(χ)e-χ≤f(χ)e-χχ=0=0.进而f(χ)≤0(χ∈[0,+∞)), 因此f(χ)≡0([*]χ∈[0,+∞)).

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