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Old 2007 October 25th, 10:05 AM   #1 (permalink)
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problem with function

Suppose that f is a continuous real-valued function defined on the closed interval [0, 1]. Which of the following must be true?
I. There is a constant C>0 such that |f(x)-f(y)|<=C for all x and y in [0, 1]
II. There is a constant D>0 such that |f(x)-f(y)|<=1 for all x and y in [0, 1] that satisfy |x-y|<=D
III. There is a constant E>0 such that |f(x)-f(y)|<=E|x-y| for all x and y in [0, 1]

(A) I only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II, and III

I chose (E) but the answer is (C). My argument is that |f(x)-f(y)|/|x-y|<=E. Where is my mistake?
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Old 2007 October 27th, 04:24 PM   #2 (permalink)
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Help me, por favor, El Seņor.
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Old 2007 October 27th, 07:20 PM   #3 (permalink)
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the answer is c because III requires to be differential at all points..which may not be necessary for a continous function...
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Old 2007 October 28th, 01:49 PM   #4 (permalink)
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Thanks sagiari!
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Old 2007 December 12th, 11:03 PM   #5 (permalink)
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The Answer is C since (III) is wrong. f is a continuous on the closed interval [0,1] then f is bounded (I) and uniformly continuous, but f is not Liptchitz (III).
Eg: f = sqrt(x)
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Old 2007 December 13th, 01:38 AM   #6 (permalink)
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Quote:
Originally Posted by sagiari View Post
the answer is c because III requires to be differential at all points..which may not be necessary for a continous function...
With C, It maybe not differentiable.
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