# 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 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

 $\sum_{n}{\|x_{n}\|}$

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 sequence in $X$. 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 telescoping series, 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.

Title proof of necessary and sufficient conditions for a normed vector space to be a Banach space ProofOfNecessaryAndSufficientConditionsForANormedVectorSpaceToBeABanachSpace 2013-03-22 17:35:11 2013-03-22 17:35:11 willny (13192) willny (13192) 4 willny (13192) Proof msc 46B99