real-analysis-hw4-2007-10-2 - f erence | f x − f y | is...

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Real Analysis Homework 4, due 2007-10-9 in class 1. (20 points) (a) (10 points) Use de fi nition (do not use Theorem 3.33) to show that the Cantor- Lebesgue function f ( x ) : [0 , 1] [0 , 1] is not a Lipschitz continuous function. (b) (10 points) Show that the Cantor-Lebesgue function f ( x ) : [0 , 1] [0 , 1] satis fi es the following | f ( x ) f ( y ) | 2 | x y | α , x, y [0 , 1] where α (0 , 1) is a constant given by α = log 2 / log 3 . (Hint: Use the fact that if x, y [0 , 1] with | x y | 3 k for some k N , then the di
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Unformatted text preview: f erence | f ( x ) − f ( y ) | is at most 2 − k . For arbitrary x, y ∈ [0 , 1] one can choose an unique k ∈ N such that 3 − k − 1 < | x − y | ≤ 3 − k , which implies | f ( x ) − f ( y ) | ≤ 2 − k . Rewrite the es-timate without involving k. ) 2. (10 points) Do Exercise 20 in p. 48. 3. (10 points) Do Exercise 21 in p. 48. 4. (10 points) Do Exercise 23 in p. 49. 1...
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