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$f命题1：$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$f'left( x ight)$

证明：由$fleft( x ight) in {C^1}left( { - infty , + infty } ight)$知，$f'left( x ight) in Cleft[ {a,b}
ight]$，则

由$f{Cantor定理}$知，$f'left( x ight)$在$left[ {a,b} ight]$上一致连续，即对任意$varepsilon > 0$，存在$delta > 0$，使得对任意的$x,y in left[ {a,b} ight]$满足$left| {x - y} ight| < delta $时，有[left| {f'left( x ight) - f'left( y ight)} ight| < varepsilon ]

由微分中值定理知，存在${xi _n} in left( {x,x + frac{1}{n}} ight)$，使得
[{f_n}left( x ight) = nf'left( {{xi _n}} ight)frac{1}{n} = f'left( {{xi _n}} ight)]
取$N = frac{1}{delta }$，则当$n > N$时，对任意$x in left[ {a,b} ight]$，有[left| {x - {xi _n}} ight| < delta ]

从而有[left| {f'left( x ight) - f'left( {{xi _n}} ight)} ight| < varepsilon ]
所以对任意$varepsilon > 0$，存在$N = frac{1}{delta } > 0$，使得当$n > N$时，对任意$x in left[ {a,b} ight]$，有[left| {f'left( x ight) - {f_n}left( x ight)} ight|{ m{ = }}left| {f'left( x ight) - f'left( {{xi _n}} ight)} ight| < varepsilon ]
从而由函数列一致收敛的定义即证

$f注1：$$N$的取值由不等式${left| {x - {xi _n}} ight| < delta }$放缩得到

$f注2：$由于$f'left( x ight) in Cleft( { - infty , + infty } ight)$，所以[ lim limits_{n o infty } {f_n}left( x ight) = mathop {lim }limits_{n o infty } frac{{fleft( {x + frac{1}{n}} ight) - fleft( x ight)}}{{frac{1}{n}}} = f'left( x ight)]
即${f_n}left( x ight)$在$left( { - infty , + infty } ight)$上处处收敛于$f'left( x ight)$