Let’s prove the completeness of . The argument consists of two steps.

**Claim 1.** Suppose is a normed space in which every absolutely convergent series converges; that is, converges whenever are such that converges. Then the space is complete.

**Proof.** Take a Cauchy sequence . For find an integer such that as long as . (This is possible because the sequence is Cauchy.) Also let and consider the series . By the hypothesis this series converges. Its partial sums simplify (telescope) to . Hence the subsequence has a limit. It remains to apply a general theorem about metric spaces: if a Cauchy sequence has a convergent subsequence, then the entire sequence converges. This proves Claim 1.

**Claim 2.** Every absolutely convergent series in Â converges.

**Proof.** The elements of are functions from to , so let’s write them as such: . (This avoids confusion of indices.) Suppose the series converges. Then for any the series also converges, by Comparison Test. Hence converges (absolutely convergent implies convergent for series of real or complex numbers). Let . So far the convergence is only pointwise, so we are not done. We still have to show that the series converges in , that is, its tails have small norm: as .

What we need now is a dominating function, so that we can apply the Dominated Convergence Theorem. Namely, we need a function such that

(1) , and

(2) for all .

Set . Then (2) follows from the triangle inequality. Also, is the increasing limit of functions , for which we have

using the triangle inequality in . Therefore, by the Monotone Convergence Theorem.