关于函数列一致收敛的专题讨论

$f命题：$设$fleft( x ight) in {C^1}left( { - infty , + infty } ight)$，令[{f_n}left( x ight) = nleft[ {fleft( {x + frac{1}{n}} ight) - fleft( x ight)} ight]]

证明：对任意$x in left[ {a,b} ight] subset left( { - infty , + infty } ight)$，有${f_n}left( x ight)$一致收敛于$f'left( x ight)$

参考答案

$f命题：$设$fleft( x ight) in Cleft( { - infty , + infty } ight)$，令[{f_n}left( x ight) = sumlimits_{k = 0}^{n - 1} {frac{1}{n}} fleft( {x + frac{k}{n}} ight)]

证明：对任意$x in left[ {a,b} ight] subset left( { - infty , + infty } ight)$，有${f_n}left( x ight)$一致收敛于$int_0^1 {fleft( {x + t} ight)dt}$

参考答案

$f命题：$$left( f{Dini定理} ight)$设函数列$left{ {{f_n}left( x ight)} ight}$的每一项及其极限函数$fleft( x ight)$

均在$x in left[ {a,b} ight]$上连续，且对每个$ x in left[ {a,b} ight]$有$left{ {{f_n}left( x ight)} ight}$为单调数列，则函数列$left{ {{f_n}left( x ight)} ight}$在$left[ {a,b}  ight]$上一致收敛于${fleft( x ight)}$

方法一

$f命题：$