Some_sols_Homework5_typeset - 1 Solution to Problem...

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1 Solution to Problem 4.1.5.15. By uniform continuity, we have that for every ²> 0, there exists δ> 0 such that a<x<y<a + δ = ⇒| f ( x ) - f ( y ) | <². Choose ² n =1 /n , and let δ n be a corresponding δ . We may take δ 0 n = min( δ n , 1 /n ), and choose x n = a + δ n / 2. Then n, m N = x n ,x m ( a, a + δ n )= f ( x n ) - f ( x m ) | < 1 n , so the sequence n 7→ f ( x n ) is Cauchy, hence convergent. Call the limit A , and do a similar construction near b to find a number B . Then the function g :[ a, b ] R biven by g ( x A if x = a f ( x )i f a<x<b B if x = b is a continuous map which coincides with f on ( a, b ). To see that it is continuous at x , we need to show that for any 0 there exists 0 such that if a x<ea + δ then f ( x ) - A | , and we have verified this above. The construction at b is analogous. Solution to Problem 4.2.4.5. This is immediate. If | f ( x ) - f ( y ) |≤ M 1 | x - y | and | g ( x ) - g ( y ) M 2 | x - y | , then | ( f + g )( x ) - ( f + g )( y ) | = f ( x ) - f ( y ) · + g ( x ) - g ( y ) ·fl ( M 1 + M 2 ) | x - y | .
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Some_sols_Homework5_typeset - 1 Solution to Problem...

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