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Unformatted text preview: Math 116  Problem Set 4 (Thanks to Nick Wage for his writeup. Also, the solution to problem 4c have not been typed up yet. A new version will be posted in the future). 1. (a) Let x (0 , 1] and > 0. Then take 0 < < min { x 2 2 , x 2 , 1 x 2 } . Then if  x x  < we have that 1 x 1 x = x x xx 2( x x ) x 2 < , so f ( x ) = 1 x is continuous. On the other hand, suppose = 1 and > 0. We have that  f (1 / 2) f (1 / 3)  = 1 = , so if > 1 / 6 we are done. Otherwise, take x = and x = / 2. Then  f ( x ) f ( x )  = 1 6 > 1 = , so is not valid. Thus, no choice of works for all x, x (0 , 1] when = 1, and f ( x ) = 1 x is not uniformly continuous on this interval. (b) Suppose f : S X R is a continuous function, and S is compact. Suppose for the sake of contradiction that there exists some > 0 such that for all > 0 there exist x, x S such that  x x  < but  f ( x ) f ( x )  > . Then, for each n N , pick x n , y n S such that  x n y n  < 1 2 n but  f ( x n ) f ( y n )  > . By compactness we know there exists a subsequence of the natural numbers a 1 < a 2 < such that x a 1 , . . . converges to some z S . Further, because the sequence y a 1 , . . . is fully contained in S , we know there exists a subsequence of the natural numbers b 1 < b 2 < such that y a b 1 , y a b 2 , . . . converges to some point w S . Note that x a b 1 , x a b 2 , . . . still converges to z S . For simplicity define z i = x a b i and w i = y a b i . Then for any > 0 we may find an i such that  z w  <  z z i  +  z i w i  +  w i w  < 3 / 3 = , so therefore z = w . Becasue f is continuous at z , we may find a such that if   z  < then  f ( ) f ( z )  < / 2. However then we may find an n such that  z n z  < and  w w n  < . 1 Then because z = w we have that  f ( z n ) f ( w n )   f ( z n ) f ( z )  +  f ( w ) f ( w n )  < 2 / 2 = , which contradicts our choice of z n and w n , and thus shows f is uniformly continuous on S . 2. (a) Theorem 1 (The Weierstrass Approximation Theorem) . If f is a continuous real valued function on the closed interval [ a, b ] , then for any > , there exists a polynomial P ( x ) = n i =0 a i x i for some n N and a i R such that  f ( x ) P ( x )  < for all x [ a, b ] . The function is continuous on a closed and bounded interval in R which we know to be compact, and thus the function is automatically uniformly continuous. Thus, we need not require f to be uniformly continuous, but as a side effect of the other problem assumptions f must end up being uniformly continuous on the interval. (b) If the statement is true when a = 0 and b = 1, then a simple change of variable proves it for all a b , so we may assume we are working on the interval [0 , 1]. Let and n be integers such that 0 n . Then we define the Bernstein polynomial b ,n ( x ) = n x (1 x ) n ....
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This document was uploaded on 08/30/2011.
 Fall '09
 Math

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