Prove that a normed vector space \((X,\|\cdot\|)\) is Banach if and only if every series \[ \sum_{n=1}^\infty x_n \] with \[ \sum_{n=1}^\infty\|x_n\|<\infty \] converges in \(X\).
Solution
1. If \(X\) is Banach, every absolutely convergent series converges.
Suppose \[ \sum_{n=1}^\infty\|x_n\|<\infty \] and let \[ s_N=\sum_{n=1}^N x_n. \] If \(M>N\), then \[ \|s_M-s_N\| =\left\|\sum_{n=N+1}^M x_n\right\| \le\sum_{n=N+1}^M\|x_n\|. \] Because the scalar series of norms converges, the right-hand side tends to \(0\) uniformly in \(M>N\) as \(N\to\infty\). Thus \((s_N)\) is Cauchy. Since \(X\) is complete, \((s_N)\) converges in \(X\).
2. If every absolutely convergent series converges, then \(X\) is Banach.
Let \((y_n)\) be a Cauchy sequence in \(X\). Choose a subsequence \((y_{n_k})\) such that \[ \|y_{n_{k+1}}-y_{n_k}\|<2^{-k} \qquad(k\ge1). \] Then \[ \sum_{k=1}^\infty\|y_{n_{k+1}}-y_{n_k}\| \le\sum_{k=1}^\infty2^{-k}<\infty. \] By hypothesis, the series \[ \sum_{k=1}^\infty (y_{n_{k+1}}-y_{n_k}) \] converges in \(X\). Its partial sums telescope: \[ \sum_{k=1}^{N}(y_{n_{k+1}}-y_{n_k})=y_{n_{N+1}}-y_{n_1}. \] Hence the subsequence \((y_{n_k})\) converges to some \(y\in X\).
Since the original sequence \((y_n)\) is Cauchy, convergence of one subsequence forces the whole sequence to converge to the same limit: given \(\varepsilon>0\), choose \(N\) so that \[ \|y_n-y_m\|<\varepsilon/2 \qquad(n,m\ge N), \] and then choose \(k\) with \(n_k\ge N\) and \[ \|y_{n_k}-y\|<\varepsilon/2. \] For every \(n\ge N\), \[ \|y_n-y\| \le\|y_n-y_{n_k}\|+\|y_{n_k}-y\|<\varepsilon. \] Thus every Cauchy sequence in \(X\) converges, so \(X\) is Banach.