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# 微積分證明問題

A function f(x) : [a,b] R satisfies a Lipschitz condition if there exists

a constant L>0 such that 1f(x)-f(y)1<=L1x-y1 whenever x and y in [a,b].

prove that :

(a) a function satisfies a Lipschitz condition on [a,b] is continuous

on [a,b]

(b)if 1f'(x)1<=M for all x屬於[a,b] , then it satisfies a Lipschitz condition on [a,b]

### 1 Answer

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- densityLv 61 decade agoFavorite Answer
(a)It is sufficies t oshow that | f(x) - f(y) |<ε. For all ε>0, take δ=ε/L, then | x - y |<δ. Implies that | f(x) - f(y) | equal and less than L| x - y |<Lδ=L．ε/L=ε. Q.E.D

(b)Using the Mean Value theorem to prove the question.

Source(s): myself

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