 Solution

A norm on a complex vector space is a map $\Vert \cdot \Vert :X\to [0,\infty )$ that vanishes only at zero, satisfies $\Vert \lambda x\Vert =|\lambda |\Vert x\Vert$, and obeys the triangle inequality. The space is Banach when every Cauchy sequence in the resulting metric converges in $X$.

If $X$ is Banach and ${\sum }_j\Vert x_j\Vert <\infty$, the estimate $\Vert {\sum }_{j=m}^n{x}_j\Vert \le {\sum }_{j=m}^n\Vert x_j\Vert$ shows that its partial sums are Cauchy, hence converge. Conversely let every such series converge and take a Cauchy sequence $(y_n)$. Choose a subsequence with $\Vert y_{n_{k+1}}-y_{n_k}\Vert <2^{-k}$. The series consisting of $y_{n_1}$ followed by these successive differences is absolutely summable in norm, so converges by hypothesis; its partial sums are exactly this subsequence. A Cauchy sequence with a convergent subsequence converges to the same limit, proving the absolute-series criterion for completeness.

On the absolutely summable sequence space ${\ell }^1$, coordinatewise addition and scalar multiplication preserve membership, and summing the scalar triangle inequality verifies its norm axioms. To prove completeness using the criterion, let $x^{(k)}=(x_j^{(k)})\in {\ell }^1$ with ${\sum }_k\Vert x^{(k)}{\Vert }_1<\infty$. For each $j$ the sum $x_j={\sum }_k{x}_j^{(k)}$ converges absolutely, and the permitted interchange of nonnegative double sums givesThus $x\in {\ell }^1$. MoreoverEvery absolutely norm-summable series therefore converges in ${\ell }^1$, proving ${\ell }^1$ is a Banach space.