m425b-hw1soln-s08

# m425b-hw1soln-s08 - MATH 425b ASSIGNMENT 1 SOLUTIONS SPRING...

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MATH 425b ASSIGNMENT 1 SOLUTIONS SPRING 2008 Prof. Alexander Chapter 7 (1) Suppose f n f uniformly. Let > 0. There exists an N such that || f N - f || < . Since f N is bounded, there exists an M N such that | f N ( x ) | ≤ M N for all x . Then for all x , | f ( x ) | ≤ | f N ( x ) | + | f N ( x ) - f ( x ) | < M N + , so f is uniformly bounded. (2) Suppose f n f and g n g , both uniformly. Let > 0. There exist N 1 , N 2 such that n N 1 = ⇒ | f n ( x ) - f ( x ) | < 2 for all x, n N 2 = ⇒ | g n ( x ) - g ( x ) | < 2 for all x. Then for n max( N 1 , N 2 ), ( f n ( x ) + g n ( x )) - ( f ( x ) + g ( x )) f n ( x ) - f ( x ) + g n ( x ) - g ( x ) < 2 + 2 = , for all x . Thus f n + g n f + g uniformly. Now suppose also that each f n and g n is bounded. By Problem 1, { f n } and { g n } are uniformly bounded, that is, there exists M such that | f n ( x ) | ≤ M and | g n ( x ) | ≤ M for all x . Let > 0. There exist N 3 , N 4 such that n N 3 = ⇒ | f n ( x ) - f ( x ) | < 2 M for all x, n N 4 = ⇒ | g n ( x ) - g ( x ) | < 2 M for all x. Then for n max( N 3 , N 4 ), for all x , | f n ( x ) g n ( x ) - f ( x ) g ( x ) | = | f n ( x )( g n ( x ) - g ( x )) + ( f n ( x ) - f ( x )) g ( x ) | ≤ | f n ( x ) | | g n ( x ) - g ( x ) | + | f n ( x ) - f ( x ) | | g ( x ) | < M · 2 M + 2 M · M = .

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