Suppose I is a non-degenerate interval, and f : I→R is
differentiable. If there is an M∈R such that |f ′(x)|≤M for any
x∈I, show that f is uniformly continuous on I.
- SamLv 68 年前最佳解答
Ref WIKI:Definition of Uniform continuity: Given metricspaces (X, d1) and (Y, d2),a function f : X → Y is called uniformlycontinuous if for every real number ε > 0 there exists δ > 0such that for every x, y ∈ X with d1(x, y) < δ,we have that d2(f(x), f(y)) < ε.If X and Yare subsets of the real numbers, d1 and d2can be the standard Euclidean norm, || · ||, yielding the definition:for all ε > 0 there exists a δ > 0 suchthat for all x, y ∈ X, |x − y| < δimplies |f(x) − f(y)| < ε.[PF]Sincef is differentiable in I, for any x,y in I, by the mean value theorem,wehave [f(y)-f(x)]/(y-x)=f`(z) where z is between x and y,and z is in I.So| [f(y)-f(x)]/(y-x)| =|f`(z)| <= M. => |f(y)-f(x)| < = M |(y-x) |.Nowfor any e(epsilon) >0, take d(delta)=e/M，Wehave |f(y)-f(x)| < = M |(y-x) |<=M*d=M*e/M=efor all |(y-x) |<=d.Hence by definition, f is uniformly continuous on I.
2012-10-31 02:17:02 補充：
if a,b in R, a < b, and 00=infinity;
then (a,b), (a,b],[a,b),[a,b].
(-00,00)=R are all non-degenerate interval.
- 教書的Lv 68 年前