Universal agreement of absolute and unconditional convergence

Suppose $X$ has finite positive dimension with basis $e_1,\dots ,e_d$, [given, L3, L2] and write $x_n={\sum }_j{a}_{j,n}{e}_j$. If ${\sum }_n{x}_n$ is unconditional, then for each $j$ choose the bounded phases ${\lambda }_n=\overline{a_{j,n}}\mathrm{/}\mathrm{\mid }{a}_{j,n}\mathrm{\mid }$ when $a_{j,n}\mathrm{\ne }0$ and zero otherwise. By [L3], ${\sum }_n{\lambda }_n{x}_n$ converges; applying the continuous $j$th coordinate from [L2] shows ${\sum }_n\mathrm{\mid }{a}_{j,n}\mathrm{\mid }<\mathrm{\infty }$.

The triangle inequality gives [given, step 1.1] $\mathrm{\parallel }{x}_n\mathrm{\parallel }\le {\sum }_j\mathrm{\mid }{a}_{j,n}\mathrm{\mid }\mathrm{\parallel }{e}_j\mathrm{\parallel }$. Summing and using step 1.1 over the finite set of coordinates proves ${\sum }_n\mathrm{\parallel }{x}_n\mathrm{\parallel }<\mathrm{\infty }$. If $X=\{0\}$ the claim is immediate. Thus finite dimension implies universal agreement.

Conversely, if $X$ is infinite-dimensional, [A1] and [L1] supply an [given, A1, L1, step 2.1] unconditionally convergent series that is not absolutely convergent. Universal agreement therefore fails. This proves the reverse implication and the equivalence.