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FJU 發問時間： 科學數學 · 1 0 年前

高等微積分（Uniform Continuity)

1 個解答

• 1 0 年前
最佳解答

1. the problem is to show uniformly continuous?

if yes,

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