4
$\begingroup$

Does someone know if the following is true:

Let $\mathbb{X}$ be some arbitrary Banach space. $\{x_k \}_{k=1}^{\infty} \in \mathbb{X}$ is a sequence chosen from $\mathbb{X}$.

Now, if the series $$\sum_{k=1}^\infty \|x_k\|_X$$ converges, would the "more generic" series
$$\sum_{k=1}^\infty x_k$$ also converge?

If yes, could you please give the proof (or just mark the proof steps) ?

Thank you.

  • 2
    If you remember how in calculus the absolute convergence of a series implies the convergence of the series without absolute values, then this is the same and I believe it is also called absolute convergence.2010-12-20
  • 2
    For everyone who is interested - the full proof can be found here: http://planetmath.org/?method=l2h&from=objects&name=ProofOfNecessaryAndSufficientConditionsForANormedVectorSpaceToBeABanachSpace&op=getobj2010-12-20

2 Answers 2

4

Yes. You can use the hypothesis that $\sum\|x_k\|$ converges to show that the sequence of partial sums of the series $\sum x_k$ is a Cauchy sequence.

Conversely, if this property holds in a normed space, then the space is complete; that direction isn't as straightforward.

4

It is indeed true. It is in fact almost the same as the statement of the Weierstrass M-test.