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hw2-sol - Modern Analysis Homework 2 Possible solutions 1...

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Modern Analysis, Homework 2 Possible solutions 1. Since { f m } is uniformly bounded on [ a, b ], there exists C R such that for all m N we have sup x [ a,b ] | f ( x ) | ≤ M. This implies that | F n ( x ) | = R x a f n ( t ) dt R x a | f n ( t ) | dt C ( b - a ), and the sequence { F n } is uniformly bounded. Similarly, we have | F n ( x ) - F n ( y ) | ≤ C | x - y | , so the sequence { F n } is equicontinuous. By the Arzel`a-Ascoli theorem, there exists a uniformly convergent subsequence { F n k } . 2. Recall that for metric spaces, compactness is the same thing as sequential compactness, so S is compact if and only if every sequence in S contains a convergent subsequence, with limit in S . ( ) Let { f n } n N be a sequence in S . By the Arzel` a-Ascoli theorem, { f n } contains a uniformly convergent subsequence { f n i } i N . Suppose that f n i f . Since S is uniformly closed, we have f S . Therefore S is compact. ( ) Let’s prove the contrapositive. There are three cases: (a) S is not uniformly closed. This is the same as saying that S is not a closed subset of the metric space C ( K ), but compact subsets of metric spaces are closed (contradiction). (b) If S is not pointwise bounded, then there exists a point x K and a sequence { f n } n N in S such that | f n +1 ( x ) | ≥ | f n ( x ) | + 1 for every n N . Then every subsequence { f n i } i N of { f n } n N satisfies | f n i ( x ) | → ∞ as i → ∞ . It follows that no subsequence of { f n } can converge pointwise, much less uniformly. This shows that there is a sequence in S with no convergent subsequence (contradiction). (c) Suppose that S is not equicontinuous. Then there exists ε 0 > 0 and a sequence { f n } n N , sequences { x n } n N , { y n } n N in K such that | f n ( x n ) - f n ( y n ) | ≥ ε 0 yet | x n - y n | < 1 n .
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