Absolute Summability Criterion for Completeness

# Absolute Summability Criterion for Completeness

Recall from the Absolutely Summable Series page that if $(X, \| \cdot \|)$ is a normed space then a series $\sum_{i=1}^{\infty} x_i$ is said to be absolutely summable if the numerical series $\sum_{i=1}^{\infty} \| x_i \|$ converges. We can use this definition to come up with a criterion to determine when certain normed spaces are complete.

 Theorem 1 (Absolute Summability Criterion for Completeness): Let $(X, \| \cdot \|)$ be a normed linear space. Then $(X, \| \cdot \|)$ is complete if and only if every absolutely summable series in $X$ converges.
• Proof: $\Rightarrow$ Suppose that $(X, \| \cdot \|)$ is complete. Let $\sum_{i=1}^{\infty} x_i$ be an absolutely summable series in $X$. Then the numerical series $\displaystyle{\sum_{i=1}^{\infty} \| x_i \|}$ converges. Let $(s_n)$ be the sequence of partial sums of $\sum_{i=1}^{\infty} x_i$ and let $(a_n)$ be the sequence of partial sums of $\sum_{i=1}^{\infty} \| x_i \|$.
• Let $\epsilon > 0$ be given. Since $(a_n)$ converges and is hence Cauchy, there exists an $N \in \mathbb{N}$ such that if $m \geq n \geq N$ then:
(1)
\begin{align} \quad |a_m - a_n| = \biggr | \sum_{i=1}^{m} \| x_i \| - \sum_{i=1}^{n} \| x_i \| \biggr | = \biggr | \sum_{i=n+1}^{m} \| x_i \| \biggr | < \epsilon \end{align}
• For this same $N$, if $m \geq n \geq N$ we have that:
(2)
\begin{align} \quad \| s_m - s_n \| = \left \| \sum_{i=1}^{m} x_i - \sum_{i=1}^{n} x_i \right \| = \left \| \sum_{i=n+1}^{m} x_i \right \| \leq \sum_{i=n+1}^{m} \| x_i \| < \epsilon \end{align}
• Therefore $(s_n)$ is Cauchy. Since $(X, \| \cdot \|)$ is complete, this implies that $(s_n)$ converges. So $\sum_{i=1}^{\infty} x_i$ converges in $X$.
• $\Leftarrow$ Suppose that every absolute summable series in $X$ converges in $X$. Let $(x_i)$ be a Cauchy sequence in $X$. Then for each $k = 1, 2, ...$ there exists an $N_k \in \mathbb{N}$ such that if $m, n \geq N_k$ then:
(3)
\begin{align} \quad \| x_m - x_n \| < \frac{1}{2^k} \end{align}
• Without loss of generality, assume that $N_1 < N_2 < ...$. Then for each $k \in \mathbb{N}$ we have that:
(4)
\begin{align} \quad \| x_{N_k} - x_{N_{k+1}} \| < \frac{1}{2^k} \end{align}
• Observe that the series $\sum_{k=1}^{\infty} [x_{N_k} - x_{N_{k+1}}]$ is absolutely summable since:
(5)
\begin{align} \quad \sum_{k=1}^{\infty} \| [x_{N_k} - x_{N_{k+1}} \| <\sum_{k=1}^{\infty} \frac{1}{2^k} = 1 < \infty \end{align}
• So by assumption, the series $\sum_{k=1}^{\infty} [x_{N_k} - x_{N_{k+1}}]$ converges to some $x \in X$. So the $(x_{N_k})$ converges to $x_{N_1} - x$. But $(x_{N_k})$ is a convergent subsequence of the Cauchy sequence $(x_n)$. So $(x_n)$ converges. Hence $(X, \| \cdot \|)$ is complete. $\blacksquare$