18.100pset12

# 18.100pset12 - 18.100B/C Fall 2008 Solutions to Homework 12...

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18.100B/C: Fall 2008 Solutions to Homework 12 1.(a) (Rudin page 165 problem 2) Let f and g be the functions such that f n f and g n g . Choose ε > 0. Then there is some N 1 N such that n N 1 implies | f n ( x ) - f ( x ) | ≤ ε/ 2 for all x and some N 2 N such that n N 2 implies | g n ( x ) - g ( x ) | ≤ ε/ 2 for all x . Then with N = max { N 1 , N 2 } we have n N implies | ( f n + g n )( x ) - ( f + g )( x ) | ≤ | f n ( x ) - f ( x ) | + | g n ( x ) - g ( x ) | ε 2 + ε 2 = ε for all x , so ( f n + g n ) converges uniformly to f + g . If f n and g n are bounded for all n then we have by an argument similar to problem 4b from the last problem set that there is some M R such that | f ( x ) | ≤ M and | g ( x ) | ≤ M and | f n ( x ) | ≤ M and | g n ( x ) | ≤ M for all x . Fix ε > 0, then we can choose N N such that n N implies | g n - g | ≤ ε / 2 M and | f n - f | ≤ ε/ 2 M for all x . Then | f n g n - fg | = | ( f n - f ) g n + f ( g n - g ) | ≤ | f n - f | · | g n | + | f | · | g n - g | ε 2 M + ε 2 M = ε for all x , so ( f n g n ) converges uniformly to fg . (b) (Rudin page 165 problem 3) Let f n ( x ) = x and g n ( x ) = 1 + 1 n for all n N , x R . These converge uniformly to f ( x ) = x (since f n = f is actually independent of n ) resp. g n ( x ) = 1 (since g n converges pointwise and does not actually depend on x ). Also, we clearly have pointwise convergence f n · g n x . Now ﬁx any ε > 0 and any n , and choose some x > nε . Then f n ( x ) g n ( x ) = x · (1 + 1 /n ) > f ( x ) g ( x ) + ε , so convergence is not uniform. 2.

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18.100pset12 - 18.100B/C Fall 2008 Solutions to Homework 12...

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