Copy of s13 - Solutions to homework 13 1 First of all for...

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Solutions to homework # 13. 1. First of all, for every function f n there exists a bound M n such that | f n ( x ) | ≤ M n for all x . Now pick an arbitrary ε > 0. Then there exists N = N ( ε ) such that | f n ( x ) - f N ( x ) | < ε for all x whenever n > N. Then | f n ( x ) | ≤ ε + | f N ( x ) | ≤ M N + ε for all n N . Let M := max { M 1 , . . . , M N - 1 , M N + ε } . Then | f n ( x ) | ≤ M for all n and all x . 2. Let f denote the limit of ( f n ) and let g denote the limit if ( g n ). For any ε > 0, there exists N = N ( ε ) such that | f n ( x ) - f ( x ) | ≤ ε/ 2, | g n ( x ) - g ( x ) | ≤ ε/ 2 for all x . Then | f n ( x ) + g n ( x ) - f ( x ) - g ( x ) | ≤ | f n ( x ) - f ( x ) | + | g n ( x ) - g ( x ) | ≤ ε forall x. If now both ( f n ) and ( g n ) are sequences of bounded functions, we know from Exercise 1 that they are uniformly bounded. So there exists a constant such that | f n ( x ) | ≤ M , | g n ( x ) | ≤ M for all x and all n IN. Then the limits
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This document was uploaded on 11/17/2010.

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