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I am very bad with proofs that ask you to show that "$\exists$ something..." because in most of them you have to explicitly show the something. The following question is one such. Any help will be greatly appreciated!

Let $\{u_n\}$ a sequence in a Banach space $X$. Suppose that $\sum_{n=1}^{\infty} \|u_n\| < \infty$. Prove that there exists some $x \in X$ such that

$$\lim_{n \to \infty} \sum_{k=1}^{n} u_k = x$$

Thank you very much in advance!

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marked as duplicate by 23rd, egreg, Davide Giraudo, Lord_Farin, Johannes Kloos Nov 23 '13 at 23:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

You meant $\sum_{k=1}^nu_k$. – 1015 Feb 21 '13 at 3:45
Precisely! I will correct it. – user44069 Feb 21 '13 at 3:47

1 Answer 1

up vote 5 down vote accepted

Let $ m,n \in \mathbb{N} $ satisfy $ m \leq n $. By the Triangle Inequality for the norm $ \| \cdot \|_{X} $, we have $$ \left\| \sum_{k=m}^{n} u_{k} \right\|_{X} \leq \sum_{k=m}^{n} \| u_{k} \|_{X}. $$ As $ \displaystyle \sum_{k=1}^{\infty} \| u_{k} \|_{X} < \infty $, it follows from the Cauchy Criterion for Series that for every $ \epsilon > 0 $, there exists an $ N \in \mathbb{N} $ sufficiently large so that $$ \forall m,n \in \mathbb{N}_{\geq N}: \quad m \leq n ~ \Longrightarrow ~ \sum_{k=m}^{n} \| u_{k} \|_{X} < \epsilon. $$ Consequently, $$ \forall m,n \in \mathbb{N}_{\geq N}: \quad m \leq n ~ \Longrightarrow ~ \left\| \sum_{k=m}^{n} u_{k} \right\|_{X} < \epsilon. $$ We thus see that the sequence $ \displaystyle \left( \sum_{k=1}^{n} u_{k} \right)_{n \in \mathbb{N}} $ is Cauchy in $ X $. As $ X $ is a Banach space, every Cauchy sequence in $ X $ converges; in particular, the sequence $ \displaystyle \left( \sum_{k=1}^{n} u_{k} \right)_{n \in \mathbb{N}} $ converges to some $ x \in X $: $$ \sum_{k=1}^{\infty} u_{k} ~ \stackrel{\text{def}}{=} ~ \lim_{n \to \infty} \sum_{k=1}^{n} u_{k} = x. $$

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Thank you very much! – user44069 Feb 21 '13 at 3:21
You’re welcome! :) – Haskell Curry Feb 21 '13 at 3:23
Do you think you can have a look on this question too? Its another one of mine. – user44069 Feb 21 '13 at 3:25
Which problem is that? – Haskell Curry Feb 21 '13 at 3:27… – user44069 Feb 21 '13 at 3:27