- no nicknameLv 51 0 年前最佳解答
1. the problem is to show uniformly continuous?
we will do it by contradiction.
if f is not uniform continuous, that means we will have a delta =d ,
such that |x-y| e , think about the behavior close to boundry point. This will get f is not a bounded function. QED.
2. if f is continuous, on a compact set (in R^n the same as close and bounded), then f is uniformly continuous.
say f is periodic => f(x)=f(x+L) , L is the period.
so (0, oo) = [0,L] U [L,2L] U [2L,3L] U ..... infinite many close and bounded interval.
f is uniformly conti on every compact set. because f is periodic in all the interval [nL, (n+1)L] where n is an interger