proof of necessary and sufficient conditions for a normed vector space to be a Banach space
We prove here that in order for a normed space, say , with the norm, say to be
Banach, it is necessary and sufficient that convergence of every absolutely convergent series in
implies convergence of the series in .
Suppose that is Banach. Let a sequence be in such that the series
converges. Then for all there exists such that for all we have
is a Cauchy sequence in . Since is Banach, converges in .
Conversely, suppose that absolute convergence implies convergence. Let be a Cauchy sequence in . Then for all there exists such that for all we have . We’ll conveniently choose so that is an increasing sequence in . Then in particular, . Hence we have,
The sum on the right converges, so must the sum on the left. Since absolute convergence implies convergence, we must have
converges as tends to infinity. So there is an in which is the limit of the sum above. As a telescoping
series, however, the sum above converges to . Since and are both in , so is the limit of , which is a subsequence of the Cauchy sequence . Hence converges in . So is Banach.
This completes the proof.
|Title||proof of necessary and sufficient conditions for a normed vector space to be a Banach space|
|Date of creation||2013-03-22 17:35:11|
|Last modified on||2013-03-22 17:35:11|
|Last modified by||willny (13192)|