This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View
Full
Document
This note was uploaded on 09/22/2010 for the course MATH 413 at Cornell University (Engineering School).
 '08
 PROTSAK

Click to edit the document details