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 spaceMathworldPlanetmath, say X, with the norm, say 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 sequenceMathworldPlanetmath (xn) be in X such that the series


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




is a Cauchy sequenceMathworldPlanetmathPlanetmath in X. Since X is Banach, sk converges in X.

Conversely, suppose that absolute convergenceMathworldPlanetmath implies convergence. Let (xn) be a Cauchy sequence in X. Then for all m1 there exists Nm such that for all k,kNm we have xk-xk<1/m2. We’ll conveniently choose Nm so that Nm is an increasing sequence in m. Then in particular, xNm-xNm+1<1/m2. 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 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 limM(xN1-xNm)=s. Since s and xN1 are both in X, so is the limit of xNm, which is a subsequence of the Cauchy sequence (xn). Hence (xn) converges in X. So X is Banach.

This completesPlanetmathPlanetmathPlanetmathPlanetmath the proof.

Title proof of necessary and sufficient conditions for a normed vector space to be a Banach spaceMathworldPlanetmath
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