I tried to prove that in non-Banach normed vector space there always exists a series which converges absolutely but which does not converge.

The idea was to consider a Cauchy sequence that doesn't converges and try to construct series with that property, but obvious methods like considering $x_n - x_{n-1}$ didn't work.

So, any hints?


migrated from mathoverflow.net Jun 18 '16 at 17:22

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  • $\begingroup$ This should be moved to MathSE... but, anyway: put the sup-of-differences norm on infinitely-long vectors of complex numbers each with only finitely-many non-zero components. $\endgroup$ – paul garrett Jun 18 '16 at 16:18
  • $\begingroup$ I agree that this should be moved to MathSE, the result holds in every non Banach normed space as the convergence of Cauchy sequences and of absolutely (for an arbitrary Banach, you should say "normal convergence") $\endgroup$ – Duchamp Gérard H. E. Jun 18 '16 at 16:29
  • $\begingroup$ Your example is right, but the point was to prove that in every non-Banach space there is such series. $\endgroup$ – Artur Riazanov Jun 18 '16 at 16:31
  • $\begingroup$ Oh, sorry, I didn't read carefully, ... but, depending on the precise choice of terminology, (see @DuchampGérardH.E.'s comment), this might be the definition of "not complete". $\endgroup$ – paul garrett Jun 18 '16 at 16:36
  • $\begingroup$ It is very easy to check that for normed vector spaces completeness is equivalent to: absolutely convergent series do converge. An immediate consequence: a quotient of a Banach space over a closed linear subspace is complete too. $\endgroup$ – Pietro Majer Jun 18 '16 at 20:30

Let $\{ x_n \}_{n=1}^{\infty}$ be a Cauchy sequence that does not converge. If any subsequence of a Cauchy sequence converges to $x$, then the original Cauchy sequence converges to $x$ as well. Therefore, no subsequence $\{ x_{n_k} \}_{k=1}^{\infty}$ of $\{ x_n \}_{n=1}^{\infty}$ converges. Choose $$ n_1 < n_2 < n_3 < \cdots $$ such that $$ \|x_{n}-x_{m}\| < \frac{1}{2^k},\;\;\; n,m \ge n_k. $$ Then $\|x_{n_i}-x_{n_j}\| < \frac{1}{2^k}$ for $i,j\ge k$. Therefore, the following cannot converge, even though the sum on the right is absolutely convergent: \begin{align} x_{n_k} &= x_{n_1}+(x_{n_2}-x_{n_1})+(x_{n_3}-x_{n_2})+\cdots+(x_{n_k}-x_{n_{k-1}}) \\ &= x_{n_1} +\sum_{j=1}^{k}(x_{n_{j}}-x_{n_{j-1}}). \end{align}


Using the proof of TrialAndError one sees that

Proposition In every normed space (NS) the following properties are equivalent

a) the space is Banach

b) every absolutely (for an arbitrary NS, one should say "normally") convergent series is convergent.


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