ia2sp10h4s

# However it should be clear that x k 1 a nk a kk k for

This preview shows pages 3–4. Sign up to view the full content.

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 0 n (0) and | f 0 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?

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern