# Hw11sol - Honors Introduction to Analysis I Homework XI Solution May 4 2009 Problem 1 Suppose f n → f and the functions f n all satisfy the

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Unformatted text preview: Honors Introduction to Analysis I Homework XI Solution May 4, 2009 Problem 1 Suppose f n → f and the functions f n all satisfy the Lipschitz condition | f n ( x )- f n ( y ) | ≤ M | x- y | for some constant M independent of n . Prove that f also satisfies the same Lipschitz condition. Solution. First observe that: | f ( x )- f ( y ) | = | f ( x )- f n ( x ) + f n ( x )- f n ( y ) + f n ( y )- f ( y ) | ≤ | f ( x )- f n ( x ) | + | f n ( x )- f n ( y ) | + | f n ( y )- f ( y ) | Since f n → f (in this case it is enough pointwise), we have that for every x , for any m there is an n x such that | f ( x )- f n ( x ) | ≤ 1 / 2 m , for every n ≥ n x . Let n x,y = max { n x ,n y } . We therefore have that for x and y , for every m : | f ( x )- f n ( x ) | + | f n ( y )- f ( y ) | ≤ 1 2 m + 1 2 m = 1 m for n ≥ n x,y . The first inequality becomes: | f ( x )- f ( y ) | ≤ | f n ( x )- f n ( y ) | + 1 m ≤ M | x- y | + 1 m for any m ∈ N . Hence we can take m → ∞ and get: | f ( x )- f ( y ) | ≤ M | x- y | So f satisfies the same Lipschitz condition. Problem 2 Give an example of a sequence of continuous functions on a compact domain converging pointwise but not uniformly to a continuous function. Solution. Consider the sequence of “tent” functions f n : [0 , 100] → R : f n ( x ) = n 2 x x ∈ [0 , 1 /n ]- n 2 x + 2 n x ∈ [1 /n, 2 /n ] x ≥ 2 /n One can check that f n → f pointwise, where f : [0 , 100] → R , f ( x ) = 0 for all x . But since R 100 f n ( x ) dx = 1, for all n and R 100 f ( x ) dx = 0, we don’t have uniform convergence.= 0, we don’t have uniform convergence....
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## This note was uploaded on 09/22/2010 for the course MATH 413 at Cornell University (Engineering School).

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Hw11sol - Honors Introduction to Analysis I Homework XI Solution May 4 2009 Problem 1 Suppose f n → f and the functions f n all satisfy the

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