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$f命题1：$$（f{Bendixon判别法}）$设$sumlimits_{n = 1}^infty {{u_n}left( x ight)} $为$left[ {a,b} ight]$上的可微函数项级数，且$sumlimits_{n = 1}^infty {{u_n}^prime left( x ight)} $的部分和函数列在$left[ {a,b} ight]$上一致有界

证明：如果$sumlimits_{n = 1}^infty {{u_n}left( x ight)} $在$left[ {a,b} ight]$上收敛，则必在$left[ {a,b} ight]$上一致收敛

证明：由一致有界的定义知，存在$C>0$，使得对每个正整数$n$和每个$x in left[ {a,b} ight]$，有
[left| {sumlimits_{k = 1}^n {{u_k}^prime left( x ight)} } ight| le C]
对任给$varepsilon > 0$，取区间$left[ {a,b} ight]$的等距分划$left{ {{x_0},{x_1}, cdots ,{x_m}} ight}$，使得当$m$充分大时，分划的细度[Delta {x_i} = frac{{b - a}}{m} < frac{varepsilon }{{4C}}]
由于$sumlimits_{n = 1}^infty {{u_n}left( x ight)} $在$left[ {a,b} ight]$上处处收敛，则由$f{Cauchy收敛准则}$知，对任给$varepsilon >0$，存在$N>0$，使得当$n>N$时，对任意正整数$p$和分划的每个分点${x_i}$，同时成立

[left| {sumlimits_{k = n + 1}^{n + p} {{u_k}left( {{x_i}} ight)} } ight| < frac{varepsilon }{2},i = 0,1, cdots ,m]
于是对任意$x in left[ {a,b} ight]$，不妨设$x in left[ {{x_{i - 1}},{x_i}} ight]$，由微分中值定理知，存在${xi _i} in left( {x,{x_i}} ight)$，使得
egin{align*}
left| {sumlimits_{k = n + 1}^{n + p} {{u_k}left( x ight)} } ight| &= left| {sumlimits_{k = n + 1}^{n + p} {{u_k}left( {{x_i}} ight)} + sumlimits_{k = n + 1}^{n + p} {left( {{u_k}left( x ight) - {u_k}left( {{x_i}} ight)} ight)} } ight|\&
le left| {sumlimits_{k = n + 1}^{n + p} {{u_k}left( {{x_i}} ight)} } ight| + left| {sumlimits_{k = 1}^n {{u_k}^prime left( {{xi _i}} ight)} } ight|left| {x - {x_i}} ight|\&
< frac{varepsilon }{2} + 2Cleft| {x - {x_i}} ight| le frac{varepsilon }{2} + frac{varepsilon }{2} = varepsilon 
end{align*}
从而由函数项级数一致收敛的$f{Cauchy准则}$即证