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 $X$, with the norm, say $\parallel \cdot \parallel$ to be Banach, it is necessary and sufficient that convergence of every absolutely convergent series in $X$ implies convergence of the series in $X$.

Suppose that $X$ is Banach. Let a sequence $(x_n)$ be in $X$ such that the series

$$\sum _n\parallel x_n\parallel$$

converges. Then for all $ϵ>0$ there exists $N$ such that for all $m>n>N$ we have

Hence

$$s_k=\sum _{n=1}^k{x}_n$$

is a Cauchy sequence in $X$. Since $X$ is Banach, $s_k$ converges in $X$.

Conversely, suppose that absolute convergence implies convergence. Let $(x_n)$ be a Cauchy sequence in $X$. Then for all $m\ge 1$ there exists $N_m$ such that for all $k,k^{\prime }\ge N_m$ we have . We’ll conveniently choose $N_m$ so that $N_m$ is an increasing sequence in $m$. 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

$$\sum _{m=1}^M(x_{N_m}-x_{N_{m+1}})$$

converges as $M$ tends to infinity. So there is an $s$ in $X$ which is the limit of the sum above. As a telescoping series, however, the sum above converges to ${lim}_{M\to \mathrm{\infty }}(x_{N_1}-x_{N_m})=s$. Since $s$ and $x_{N_1}$ are both in $X$, so is the limit of $x_{N_m}$, which is a subsequence of the Cauchy sequence $(x_n)$. Hence $(x_n)$ converges in $X$. So $X$ is Banach.

This completes the proof.

Title	proof of necessary and sufficient conditions for a normed vector space to be a Banach space
Canonical name	ProofOfNecessaryAndSufficientConditionsForANormedVectorSpaceToBeABanachSpace
Date of creation	2013-03-22 17:35:11
Last modified on	2013-03-22 17:35:11
Owner	willny (13192)
Last modified by	willny (13192)
Numerical id	4
Author	willny (13192)
Entry type	Proof
Classification	msc 46B99