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 $\\|\\cdot\\|$ to beBanach, 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}{\\|x_{n}\\|}

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

\\left\\|\\sum_{n+1}^{m}{x_{n}}\\right\\|\\leq\\sum_{n+1}^{m}{\\|x_{n}\\|}<\\epsilon

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 sequencein $X$ . Then for all $m\\geq 1$ there exists $N_{m}$ such that for all $k,k^{\\prime}\\geq N_{m}$ we have $\\|x_{k}-x_{k^{\\prime}}\\|<1/m^{2}$ .We’ll conveniently choose $N_{m}$ so that $N_{m}$ is an increasing sequence in $m$ . Then in particular, $\\|x_{N_{m}}-x_{N_{m+1}}\\|<1/m^{2}$ . Hence we have,

\\sum_{m=1}^{M}{\\|x_{N_{m}}-x_{N_{m+1}}\\|}<\\sum_{m=1}^{M}{\\frac{1}{m^{2}}}

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 telescopingseries, however, the sum above converges to $\\lim_{M\\rightarrow\\infty}({x_{N_{1}}-x_{N_{m}}})=s$ . 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.