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201A_F10_hw3_sol

# 201A_F10_hw3_sol - 201A Fall 10 Thomases Homework 3 Mihaela...

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201A, Fall ’10, Thomases Homework 3 Mihaela Ifrim 1. Let X be a normed linear space. A series in x n in X is absolutely convergent if bardbl x n bardbl converges to a finite value in R . Prove X is a Banach space if and only if every absolutely convergent series converges. Proof. Let X be a Banach space and suppose that ( x n ) is absolutely convergent; i. e., bardbl x n bardbl = y for some finite y . For any ε > 0, let N be sufficiently large so that n = N bardbl x n bardbl < ε . If m, n N , then bardbl i = m x i i = m ( x i ) bardbl = bardbl m i = n ( x i ) bardbl < m i = n bardbl x i bardbl < ε (applying the triangle inequality). Therefore, n x n is Cauchy and hence convergent. We conclude that every absolutely convergent series in X converges. Conversely, let X be a normed linear space in which every absolutely convergent series converges, and suppose that { x n } n is a Cauchy se- quence. For each k N , choose n k such that bardbl ( x m x n ) bardbl < 2 - k for m, n n k . In particular, bardbl x n k +1 x n k bardbl < 2 - k . If we define y 1 = x n 1 and y k +1 = x n k +1 x n k for k 1, it follows that ( bardbl y n bardbl ) ≤ bardbl x n 1 bardbl + 1 : i. e., ( y n ) is absolutely convergent, and hence convergent. More explicitly n bardbl y n bardbl ) = k bardbl x n k +1 x n k bardbl + bardbl x n 1 bardbl ≤ bardbl x n 1 bardbl + k 1 2 k = bardbl x n 1 bardbl + 1. However, this implies that bardbl y n bardbl ≤ ∞ . But we considered that every absolute convergent series is convergent, which means that y n ≤ ∞ . But the sequence of partial sums of y n is a sub- sequence of x n . Since x n is Cauchy and has a convergent subsequence, then it will aslo converge to the same limit as the subsequence. Hence we have started with a random Cauchy sequence in X and we got that it is a convergent sequence, so we can conclude that X

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201A_F10_hw3_sol - 201A Fall 10 Thomases Homework 3 Mihaela...

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