{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

real-analysis-hw4-2007-10-2

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Real Analysis Homework 4, due 2007-10-9 in class 1. (20 points) (a) (10 points) Use de f 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 f 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 =log2 / log 3 . (Hint: Use the fact that if x, y [0 , 1] with | x y | 3 k for some k N , then the di
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online