设f(x)在[0,1]上连续,在(0,1)内可导,且|f’(x)|<1,又f(0)=f(1),证明:对于x1,x2∈[0,1],有 |f(x1)一f(x2)|<.

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问题 设f(x)在[0,1]上连续,在(0,1)内可导,且|f’(x)|<1,又f(0)=f(1),证明:对于x1,x2∈[0,1],有
    |f(x1)一f(x2)|<

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答案联系f(x1)—f(x2)与f’(x)的是拉格朗日中值定理.不妨设0≤x1≤x2≤1.分两种情形: 1)若x2一x1<[*],直接用拉格朗日中值定理得 |f(x1)一f(x2)|=|f’(ξ)(x2一x1)|=|f’(ξ)||x2一x1|<[*]. 2)若x2一x1≥[*],当0<x1<x2<1时,利用条件f(0)=f(1)分别在[0,x1]与[x2,1]上用拉格朗日中值定理知存在ξ∈(0,x1),η∈(x2,1)使得 |f(x1)一f(x2)|=|[f(x1)一f(0)]一[f(x2)一f(1)]| ≤|f(x1)一f(0)|+|f(1)一f(x2)| =|f’(ξ)x1|+|f’(η)(1一x2)| <x1+(1一x2)=1一(x2一x1)≤[*], ①当x1=0且x2≥[*]时,有 |f(x1)一f(x2)|=|f(0)一f(x2)|=|f(1)一f(x2)|=|f’(η)(1一x2)|<[*]. ②当x1≤[*]且x2=1时,同样有 |x(x1)一f(x2)|=|f(x1)一f(1)|=|f(x1)一f(0)|=|f’(ξ)(x1一0)|<[*]. 因此对于任何x1,x2∈[0,1]总有 |f(x1)一f(x2)|<[*].

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