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Unformatted text preview: 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`aAscoli 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` aAscoli 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 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 )  ≥...
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This note was uploaded on 10/18/2011 for the course MATH S4062Q taught by Professor Staff during the Summer '11 term at Columbia.
 Summer '11
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