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Math116-PS4

# Math116-PS4 - Math 116 Problem Set 4(Thanks to Nick Wage...

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Math 116 - Problem Set 4 (Thanks to Nick Wage for his write-up. 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 (0 , 1] and > 0. Then take 0 < δ < min { x 2 0 2 , x 0 2 , 1 - x 0 2 } . Then if | x - x 0 | < δ we have that 1 x - 1 x 0 = x 0 - x xx 0 2( x 0 - x ) x 2 0 < , so f ( x ) = 1 x is continuous. On the other hand, suppose = 1 and δ > 0. We have that | f (1 / 2) - f (1 / 3) | = 1 = , so if δ > 1 / 6 we are done. Otherwise, take x = δ and x = δ/ 2. Then | f ( x ) - f ( x ) | = 1 δ 6 > 1 = , so δ is not valid. Thus, no choice of δ 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 δ > 0 there exist x, x S such that | x - x | < δ 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 δ such that if | ξ - z | < δ then | f ( ξ ) - f ( z ) | < / 2. However then we may find an n such that | z n - z | < δ and | w - w n | < δ . 1

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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 > 0 , 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 ν and n be integers such that 0 ν n . Then we define the Bernstein polynomial b ν,n ( x ) = n ν x ν (1 - x ) n - ν . Then for a function f : [0 , 1] R and positive integer n define B n,f ( x ) := n ν =0 f ( ν/n ) b ν,n ( x ) .
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Math116-PS4 - Math 116 Problem Set 4(Thanks to Nick Wage...

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