Re: [分析] uniformly continuous
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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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