Lemma 12.1.3.label Let $E$ be a normed vector space, then the following are equivalent:
- (1)
$E$ is a Banach space.
- (2)
For each $\seq{x_n}\subset E$ with $\sum_{n \in \natp}\norm{x_n}_{E} < \infty$, there exists $x \in E$ such that $x = \sum_{n = 1}^{\infty} x_{n}$.
Proof. (2) $\Rightarrow$ (1): Let $\seq{x_n}\subset E$ be a Cauchy sequence, then there exists a subsequence $\seq{n_k}$ such that $\norm{x_{n_{k+1}} - x_{n_{k}}}_{E} < 2^{-k}$ for all $k \in \natp$. For each $k \in \natp$, let $y_{k} = x_{n_{k+1}}- x_{n_{k}}$, then there exists $y \in E$ such that $y = \sum_{k = 1}^{\infty} y_{k}$. Let $\eps > 0$, then there exists $K \in \natp$ such that:
- (a)
$\norm{y - \sum_{k = 1}^{K-1} y_k}_{E} < \eps$.
- (b)
For each $n \ge n_{K}$, $\norm{x_n - x_{n_K}}_{E} < \eps$.
In which case, for every $n \ge n_{K}$,
Therefore $x_{n} \to y + x_{n_1}$ as $n \to \infty$.$\square$
Post a Comment