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Unformatted text preview: 2 Exchanging infinite processes is a delicate operation; it has to be done with care. For example, suppose we take a nk = 1 n 2 k if k 6 = n , k k=n For a fixed k , once n > k the sequence is the same as 1 / (2 k n ) → 0 as n → ∞ , thus a k = lim n →∞ a nk = 0 for all k . However, it should be clear that ∞ X k =1 a nk > a kk = k for all n . So we have here a situation where each series ∑ ∞ k =1 a nk converges (absolutely since all terms are positive), the series coming from the limit sequence { a k } converges even more so since all terms are 0, but (1) is quite false. 3. Several of you got 5 for this exercise for using a correct proof incorrectly. It gave me the impression that you were not quite sure of what you were doing. Before using that proof, a lot of things have to be proved. Here is a sketch of how it works. Let X be a normed vector space. Because it is a vector space, ∑ n k =1 a k makes sense for any finite number a 1 ,...,a k ∈ X . If a n ∈ X for n = 1 , 2 , 3 ,... , one can talk now of the series ∑ ∞ n =1 a n and say that it converges iff and only if the sequence { ∑ n k =1 a k } ∞ n =1 converges, and write a = ∑ ∞ n =1 a n iff lim n →∞ ∑ n k =1 a k = a ; i.e., if lim n →∞ k ∑ n k =1 a k a k = 0. We say the series converges absolutely iff the series of nonnegative real terms ∑ ∞ n =1 k a n k converges. Then one has: Theorem 1 If X is a complete normed space; i.e., a Banach space, absolute convergence implies convergence. The proof is similar (or identical) to the proof in the complex case, but it requires a proof. Another result needed for this other proof of completeness, a result that cannot be used so freely, that requires a proof, is one that is valid in all metric spaces, namely: Lemma 2 Let { p n } be a Cauchy sequence in a metric space M . Then there is a subsequence { p n k } such that d ( p n k 1 ,p n k ) < 2 k . The proof is easy, but at our level it is needed. The main result involved is the converse of Theorem 1, namely: Theorem 3 If X is a normed vector space in which every absolutely convergent sequence converges, then X is a Banach space. People who used this approach provided what amounts to a moderately incorrect proof of this result. All in all, I felt there were too many gaps, and several errors, in the proofs i saw, to warrant more points than I gave. 3. Let { f n } be a sequence of continuously differentiable functions on [0 , 1] with f n (0) = f n (0) and  f n ( x )  ≤ 1 for all x ∈ [0 , 1] and n ∈ N . Show that if lim n →∞ f n ( x ) = f ( x ) for all x ∈ [0 , 1], then f is continuous on [0 , 1]. Must the sequence converge? Must there be a convergent subsequence?...
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 Spring '11
 Speinklo
 Metric space, Limit of a sequence, Hilbert space, Cauchy sequence, Banach space

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