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Here's a theorem: A normed linear space X is a Banach space iff every absolutely convergent series in X is convergent. How is this possible? I need the proof.

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    Show that that condition is equivalent to "every Cauchy sequence is convergent". Construct a Cauchy sequence from the terms of an absolutely convergent series, and extract an absolutely convergent series from (a subsequence of) a Cauchy sequence. – Daniel Fischer Oct 05 '13 at 23:46

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$\implies$: Suppose $\sum_n a_n$ is absolutely convergent, i.e. $\sum_n\|a_n\|$ is convergent in $\Bbb R$. Then use triangle inequality to show that the corresponding sequence $$s_n:=\sum_{k\le n}a_k$$ is Cauchy.

$\Longleftarrow$: If $(a_n)$ is a Cauchy sequence, then select a subsequence $a_{n_k}$ such that $\ \|a_{n_{k+1}}-a_{n_k}\|< \displaystyle\frac1{2^n}$, and define $b_k:=a_{n_{k+1}}-a_{n_k}$ with $b_0:=a_{n_0}$. Then $\sum_n b_n$ is an absolute convergent series, so it converges, and then by the Cauchy property conclude that its limit point must be the limit of $(a_n)$.

Berci
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  • Ok. Pls check this theorem out, is there anything like: Let X be a normed linear space. Then the normed conjugate X' of X is complement. Is that suppose to be complement or complete? –  Oct 06 '13 at 01:06