Question

数学分析 一致连续 证明题

Answer 1

证明 g(x)一致连续，对任意的b>0,存在c_1>0,使得当|x_2-x_1|<c时有
|g(x_2)-g(x_1)|<b/4
又存在M>0,使得当x>=M时，|f(x)-g(x)|<b/4
当a<=x<=M时，f(x)一致连续。存在c_2>0使得当,a<=x_1,x_2<=M,0<|x_2-x_1|<c_2时有
|f(x_2)-f(x_1)|<b/4<b,
当x_2,x_1>=M时,若|x_2-x_1|< c_1
|f(x_2)-f(x_1)|=|f(x_2)-g(x_2)+g(x_1)-f(x_1)+g(x_2)-g(x_1)|
<=3b/4<b
如果x_1<=M,x_2>M, x_2-x_1<c=min(c_1,c_2)
|f(x_2)-f(x_1)|=|f(x_2)-f(M)+f(M)-f(x_1)|<=|f(x_2)-f(M)|+b/4
<=|f(x_2)-f(M)|+b<=3b/4+b/4=b
总之，任意的b>0,存在c>0使得|x_2-x_1|<c时，|f(x_2)-f(x_1)|<b.