Re: [分析] uniformly continuous

[PaulErdos (My brain is open)]

→ h2o1125     :微分有界→unicont
→ jimmy780331 :不懂"微分有界→uni.cont."
→ h2o1125     :應該是微分→0才對

Definition od Lipschitz condition:

http://planetmath.org/encyclopedia/Lipschitzcondition.html

http://mathworld.wolfram.com/LipschitzCondition.html




Theorem:

 If f satisfies Lipschitz condition on I,then f is uniformly continuous

on I.

http://www.proofwiki.org/wiki/Lipschitz_Condition_Implies_Uniform_Continuity

http://planetmath.org/encyclopedia/UniformContinuityOfLipschitzFunctions.html




Theorem:

 If f is differentiable on I and there exists M＞0

such that│f'(x)│＜M on I.

(in other words, f has bounded derivative on I)

Then f satisfies Lipschitz condition, and hence uniformly continuous on I.



proof:
                         f(x)－f(y)
　By Mean Value Theorem, ─────＝f'(c)＜M  for some c in (x,y)⊂ I
                           x－y




NOTE: The converse is NOT true

            ＿
Let f(x)＝√x on(0,1) ,then f'is not bounded but uniformly continuous on (0,1)




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