Absolutely Convergent Series is Convergent iff Normed Vector Space is Banach/Sufficient Condition

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Theorem

Let $\struct {X, \norm {\, \cdot \,}}$ be a normed vector space.

Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ be an absolutely convergent series in $X$.

Suppose $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ is convergent.


Then $X$ is a Banach space.


Proof

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:

$\displaystyle n > n_k \implies \norm {x_n - x_{n_k} } < \frac 1 {2^k}$


Basis for the Induction

Let $\displaystyle \epsilon = \frac 1 2$.

Choose $n_1$ such that:

$\displaystyle n, m \ge n_1 \implies \norm {x_n - x_m} < \frac 1 2$

In particular, when $m = n_1$:

$\displaystyle n \ge n_1 \implies \norm {x_n - x_{n_1}} < \frac 1 2$


Induction Step

Suppose $x_{n_1}, \dots x_{n_k}$ have been constructed.

Let $\displaystyle \epsilon = \frac 1 {2^{k + 1}}$.

Choose $n_{k + 1}$ such that $n_{k + 1} > n_k$ and:

$\displaystyle n, m \ge n_{k + 1} \implies \norm {x_n - x_m} < \frac 1 {2^{k + 1}}$

In particular, when $m = n_{k + 1}$:

$\displaystyle n \ge n_{k + 1} \implies \norm {x_n - x_{n_{k + 1}}} < \frac 1 {2^{k + 1}}$

$\Box$

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 $\displaystyle \sum_{k \mathop = 1}^\infty \norm {u_k}$:

\(\displaystyle \sum_{k \mathop = 1}^\infty \norm {u_k}\) \(=\) \(\displaystyle \norm {u_1} + \sum_{k \mathop = 2}^\infty \norm {u_k}\)
\(\displaystyle \) \(=\) \(\displaystyle \norm {u_1} + \sum_{k \mathop = 2}^\infty \norm {x_{n_k} - x_{n_{k - 1} } }\)
\(\displaystyle \) \(=\) \(\displaystyle \norm {u_1} + \sum_{k \mathop = 1}^\infty \norm {x_{n_{k + 1} } - x_{n_k} }\) Relabeling: $k \to k + 1$
\(\displaystyle \) \(\le\) \(\displaystyle \norm {x_{n_1} } + \sum_{k \mathop = 1}^\infty \frac 1 {2^k}\)
\(\displaystyle \) \(\le\) \(\displaystyle \norm {x_{n_1} } + 1\)
\(\displaystyle \) \(<\) \(\displaystyle \infty\)

Thus, $\displaystyle \sum_{k \mathop = 1}^\infty u_k$ is absolutely convergent.

By assumption in the theorem, $\displaystyle \sum_{k \mathop = 1}^\infty u_k$ is convergent.

In other words:

$\displaystyle \lim_{k \mathop \to \infty} \sum_{j \mathop = 1}^k u_j = u$.

On the other hand:

\(\displaystyle \sum_{j \mathop = 1}^k u_j\) \(=\) \(\displaystyle x_{n_1} + \sum_{j \mathop = 2}^k \paren {x_{n_j} - x_{n_{j \mathop - 1} } }\)
\(\displaystyle \) \(=\) \(\displaystyle x_{n_k}\) Telescoping Series

Therefore:

$\displaystyle \lim_{k \mathop \to \infty} x_{n_k} = u =: x$

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.

$\blacksquare$


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