Completeness of a normed linear space is equivalent to the condition that absolute summability implies summability.

Definition: A space X is complete if every Cauchy sequence converges to a point in the space. Note that by "converges in X" we mean

$$ \left\| f_n - f\right\| \to 0 $$

A sequence \(\{f_n\}\subseteq X\) is called summable if the sum \(\sum f_n\) converges in X to a point \(f\in X\).

The sequence is absolutely summable if \(\sum \|f_n\|<\infty\) as a series of real numbers.

Theorem: Let X be a normed linear space. X is complete if and only if every absolutely summable sequence in X is summable.

Proof: Suppose X is complete, and let \(\{f_n\}\) be an absolutely summable sequence. We will first prove that the sequence \(\left\{\sum_{k=1}^m f_k\right\}\) as a sequence in m is a Cauchy sequence.

$$ \left|\sum_{k=a}^b f_k\right| \le \sum_{k=a}^b \|f\| < \infty $$

Therefore the sequence is Cauchy, and since the space is complete, there is a \(f\in X\) such that \(\sum f_k = f\). We now need to show that the convergence is "in X". That is to say we need to show \( \|\sum_{k=1}^m f_k - f \| \to 0\). But this is just the tail of the sequence, and

$$ \left|\sum_{k=a}^\infty f_k \right| \le \sum_{k=a}^\infty \|f_k\| \to 0 $$

This completes the first half of the proof. For the converse we assume that the space, X, has the stated property and we proceed now to prove that the space is complete. To begin with we assume that \(\{f_k\}\) is a Cauchy sequence. We will build an absolutely summable sequence \(\{z_k\}\) in the following way. First, define a subsequence of \(\{f_k\}\), which we call \(\{f_{n_k}\}\), such that for all \(a,b \ge n_k \)

$$ |f_{n_a}-f_{n_b}| < 1/2^k $$

We then define \(z_k = f_{n_k}-f_{n_{k-1}}\) except for \(z_1=f_1\). It then follows that this sequence is absolutely summable

$$ \sum \|z_k\| = \sum \|f_{n_k}-f_{n_{k-1}}\| \le \sum 1/2^k = 1 $$

Ok, the above ignores \(f_1\), but honestly \(f_1\) deserves to be ignored (because the sum will still be finite even when it's included). Since the sum is absolutely summable then it is summable and therefore there exists an \(f\in X\) such that

$$ f = \sum z_k = \lim f_k $$

It now remains to show that the convergence is in X. But as before \(\left\| f- \sum_{k=1}^m z_k\right\|\) is just the tail of the sum \(\sum z_k\), because this is absolutely summable, the tail goes to 0. (The explicit check is just like last time.)