Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach
Theorem
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be an absolutely convergent series in $X$.
Then $\ds \sum_{n \mathop = 1}^\infty a_n$ is convergent if and only if $X$ is a Banach space.
Proof
Necessary Condition
That $\ds \sum_{n \mathop = 1}^\infty a_n$ is absolutely convergent means that $\ds \sum_{n \mathop = 1}^\infty \norm {a_n}$ converges in $\R$.
Hence by Convergent Sequence in Normed Vector Space is Cauchy Sequence:
- the sequence of partial sums is a Cauchy sequence.
Now let $\epsilon > 0$.
Let $N \in \N$ such that for all $m, n \in \N$, $m \ge n \ge N$ implies that:
- $\ds \sum_{k \mathop = n + 1}^m \norm {a_k} = \size {\sum_{k \mathop = 1}^m \norm {a_k} - \sum_{k \mathop = 1}^n \norm {a_k} } < \epsilon$
This $N$ exists because the sequence is Cauchy.
Now observe that, for $m \ge n \ge N$, one also has:
\(\ds \norm {\sum_{k \mathop = 1}^m a_k - \sum_{k \mathop = 1}^n a_k}\) | \(=\) | \(\ds \norm {\sum_{k \mathop = n + 1}^m a_k}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \sum_{k \mathop = n + 1}^m \norm {a_k}\) | Triangle inequality for $\norm {\, \cdot \,}$ | |||||||||||
\(\ds \) | \(<\) | \(\ds \epsilon\) |
It follows that the sequence of partial sums of $\ds \sum_{n \mathop = 1}^\infty a_n$ is Cauchy.
As $X$ is a Banach space, this implies that $\ds \sum_{n \mathop = 1}^\infty a_n$ converges.
$\Box$
Sufficient Condition
Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $X$.
We have that:
- $\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N: \norm {x_n - x_m} < \epsilon$
We will prove the existence of a subsequence $\sequence {x_{n_k} }_{k \mathop \in \N}$ such that:
- $n > n_k \implies \norm {x_n - x_{n_k} } < \dfrac 1 {2^k}$
Basis for the Induction
Let $\epsilon = \dfrac 1 2$.
Choose $n_1$ such that:
- $n, m \ge n_1 \implies \norm {x_n - x_m} < \dfrac 1 2$
In particular, when $m = n_1$:
- $n \ge n_1 \implies \norm {x_n - x_{n_1} } < \dfrac 1 2$
Induction Step
Suppose $x_{n_1}, \dots x_{n_k}$ have been constructed.
Let $\epsilon = \dfrac 1 {2^{k + 1} }$.
Choose $n_{k + 1}$ such that $n_{k + 1} > n_k$ and:
- $n, m \ge n_{k + 1} \implies \norm {x_n - x_m} < \dfrac 1 {2^{k + 1} }$
In particular, when $m = n_{k + 1}$:
- $n \ge n_{k + 1} \implies \norm {x_n - x_{n_{k + 1} } } < \dfrac 1 {2^{k + 1}}$
$\Box$
Define:
- $u_1 := x_{n_1}$
- $u_{k + 1} := x_{n_{k + 1} } - x_{n_k}$
Now we have a sequence $\sequence {u_k}_{k \mathop \in \N}$.
Consider the series $\ds \sum_{k \mathop = 1}^\infty \norm {u_k}$:
\(\ds \sum_{k \mathop = 1}^\infty \norm {u_k}\) | \(=\) | \(\ds \norm {u_1} + \sum_{k \mathop = 2}^\infty \norm {u_k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {u_1} + \sum_{k \mathop = 2}^\infty \norm {x_{n_k} - x_{n_{k - 1} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {u_1} + \sum_{k \mathop = 1}^\infty \norm {x_{n_{k + 1} } - x_{n_k} }\) | Relabeling: $k \to k + 1$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {x_{n_1} } + \sum_{k \mathop = 1}^\infty \frac 1 {2^k}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {x_{n_1} } + 1\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds \infty\) |
Thus, $\ds \sum_{k \mathop = 1}^\infty u_k$ is absolutely convergent.
By assumption in the theorem, $\ds \sum_{k \mathop = 1}^\infty u_k$ is convergent.
In other words:
- $\ds \lim_{k \mathop \to \infty} \sum_{j \mathop = 1}^k u_j = u$.
On the other hand:
\(\ds \sum_{j \mathop = 1}^k u_j\) | \(=\) | \(\ds x_{n_1} + \sum_{j \mathop = 2}^k \paren {x_{n_j} - x_{n_{j \mathop - 1} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x_{n_k}\) | Telescoping Series |
Therefore:
- $\ds \lim_{k \mathop \to \infty} x_{n_k} = u =: x$
So $\sequence {x_{n_k} }_{k \mathop \in \N}$ converges in $X$.
We have that $\sequence {x_{n_k} }_{k \mathop \in \N}$ is a convergent subsequence of a Cauchy sequence$\sequence {x_n}_{n \mathop \in \N}$.
By Convergent Subsequence of Cauchy Sequence in Normed Vector Space, $\sequence {x_n}_{n \mathop \in \N}$ is convergent with the same limit $x$.
By definition, the underlying space is Banach.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis: $\S 1$: Normed and Banach spaces