设f(x)在(一∞,+∞)上是导数连续的有界函数,|f(x)一f’(x)|≤1.证明:|f(x)|≤1.

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问题 设f(x)在(一∞,+∞)上是导数连续的有界函数,|f(x)一f’(x)|≤1.证明:|f(x)|≤1.

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答案因为f(x)有界,所以[*] 于是e-xf(x)|x+∞=∫x+∞[e-xf(x)]’dx, 即一e-xf(x)=∫x+∞一e-x[f(x)一f’(x)]dx,两边取绝对值得e-x|f(x)|≤∫x+∞e-x|f(x)一f’(x)|dx≤∫x+∞e-xdx=e-x,故|f(x)|≤1.

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