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