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Let $(x_k)$ be a sequence in a Banach space $X$. if $\sum_{k=1}^\infty \|x_k\|$ converges , then $\sum_{k=1}^\infty x_k$ also converges.

$\textbf{My attempt}$

Let $(e_k)$ be the orthonormal basis of $X$. since $\sum_{k=1}^\infty \|x_k\|$ converges there exists $ 0< M <\infty$ such that $\|x_k\|<M$, then we can write $$|\sum_{k=1}^\infty x_k| \le \sum_{k=1}^\infty |x_k e_k| \le \|x_k\|\|e_k\| < \infty$$

since the series converges absolutely then it converges in ordinary sense.

domath
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  • Is this useful? https://math.stackexchange.com/questions/298587/about-banach-spaces-and-absolute-convergence-of-series – 1.414212 Mar 17 '20 at 04:34
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    A Banach space (as opposed to Hilbert space) need not have an orthonormal basis. – GEdgar Mar 17 '20 at 14:20

1 Answers1

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You cannot write $\sum x_n$ before proving that the series converges. So your attempt fails. Prove that $\left\|\sum\limits_{k=n}^{m} x_n\right\| \to 0$ as $n,m \to \infty$ which makes the partial sum sequence a Cauchy sequence. Then use completeness to finish the proof.

Davide Giraudo
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Kavi Rama Murthy
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  • would this work? Choose $N$ such that $\sum_{k=1}^j |x_k| \le \frac{\epsilon}{2}$ for any $j>N$. Let $v_n = \sum_{k=1}^n x_k$, then to show that $v_n$ is cauchy we have that , for any $n,m>N$ $$|v_n-v_m| =| \sum_{k=1}^n x_k- \sum_{k=1}^m x_k| \le \sum_{k=1}^n |x_k|+ \sum_{k=1}^m |x_k| \le \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$ so the sequence $v_n$ is cauchy and by the completeness of the space it converges. – domath Mar 17 '20 at 17:47
  • @stat No. This doesn't work. There is such $N$ unless all the $x_k$'s are $0$. Choose $N$ such that $ \sum\limits_{k=N}^{\infty} |x_k| <\epsilon$. – Kavi Rama Murthy Mar 17 '20 at 23:13