设函数f0(x)在(-∞,+∞)内连续,fn(x)=∫0xfn-1(t)dt(n=1,2,…). 证明:fn(x)绝对收敛.

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问题 设函数f0(x)在(-∞,+∞)内连续,fn(x)=∫0xfn-1(t)dt(n=1,2,…).
证明:fn(x)绝对收敛.

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答案对任意的x∈(-∞,+∞),f0(t)在[0,x]或[x,0]上连续,于是存在M>0(M与x有关),使得|f0(t)|≤M(t∈[0,x]或t∈[x,0]),于是 |fn(x)|≤[*]|∫0x(x-t)n-1dt|M/n!|x|n [*] 根据比较审敛法知[*]fn(x)绝对收敛.

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