Proof of the Completeness of Finite Dimensional Normed Spaces
Theorem 1
Finite-dimensional normed spaces are complete.
Description
Accordingly, finite-dimensional vector spaces become Banach spaces merely by the definition of a norm. This is particularly useful because of the frequent use of $\mathbb{R}^{n}$ or $\mathbb{C}^{n}$.
Proof
Strategy: Utilize the fact that we are dealing with a finite-dimensional vector space to break down every vector into basis units and define a convenient norm. Transform abstract calculations into direct calculations by leveraging the equivalence of norms.
For a finite-dimensional normed space $(X, \| \cdot \|_{0} )$, there exists a basis $\left\{ e_{1} , \cdots , e_{n} \right\}$.
Part 1
Let’s define the Cauchy sequence $\left\{ x_{k} \right\}_{k \in \mathbb{N} }$ of $X$ as follows.
$$ x_{1} = \lambda_{1}^{(1)} e_{1} + \cdots + \lambda_{n}^{(1)} e_{n} $$
$$ x_{2} = \lambda_{1}^{(2)} e_{1} + \cdots + \lambda_{n}^{(2)} e_{n} $$
$$ \vdots $$
$$ x_{k} = \lambda_{1}^{(k)} e_{1} + \cdots + \lambda_{n}^{(k)} e_{n} $$
Part 2
$$ \| \lambda_{1} e_{1} + \cdots + \lambda_{n} e_{n} \| : = \sum_{k=1}^{n} | \lambda_{k} | $$
Let’s define the norm $\| \cdot \|$ as the sum of the absolute values of the coefficients of the linear combination.
Since $X$ is a finite-dimensional normed space, $\| \cdot \|_{0} \sim \| \cdot \|$ and for some $m , M > 0$
$$ m \| x_{k} - x_{m} \| \le \| x_{k} - x_{m} \|_{0} \le M \| x_{k} - x_{m} \| $$
Part 3
By the definition of $\| \cdot \|$
$$ m \| x_{k} - x_{m} \| \le \| x_{k} - x_{m} \|_{0} $$
$$ \implies m \left( \left| \lambda_{1}^{(k)} - \lambda_{1}^{(m)} \right| + \cdots + \left| \lambda_{n}^{(k)} - \lambda_{n}^{(m)} \right| \right) \le \| x_{k} - x_{m} \|_{0} $$
Since $\left\{ x_{k} \right\}_{k \in \mathbb{N} }$ is a Cauchy sequence, when $k,m \to \infty$, then $\| x_{k} - x_{m} \|_{0} \to 0$. Thus, for all $1 \le i \le n$
$$ \lim_{k,m \to \infty } m \left( \left| \lambda_{i}^{(k)} - \lambda_{i}^{(m)} \right| \right) \to 0 $$
Therefore, $\left\{ \lambda_{i}^{(k)} \right\}_{k \in \mathbb{N}}$ are Cauchy sequences, and since $\mathbb{C}$ is a complete space, they converge to some $\lambda_{i} \in \mathbb{C}$.
Part 4
If we set $x := \lambda_{1} e_{1} + \cdots + \lambda_{n} e_{n}$
$$ \| x_{k} - x \|_{0} \le M \| x_{k} - x \| $$
$$ \implies \| x_{k} - x \|_{0} \le M \left( \left| \lambda_{1}^{(k)} - \lambda_{1} \right| + \cdots + \left| \lambda_{n}^{(k)} - \lambda_{n} \right| \right) $$
$$ \implies 0 \le \lim_{k \to \infty} \| x_{k} - x \|_{0} \le \lim_{k \to \infty} M \left( \left| \lambda_{1}^{(k)} - \lambda_{1} \right| + \cdots + \left| \lambda_{n}^{(k)} - \lambda_{n} \right| \right) $$
$$ \implies \lim_{k \to \infty} \| x_{k} - x \|_{0} = 0 $$
Therefore, the Cauchy sequence $\left\{ x_{k} \right\}_{k \in \mathbb{N} }$ of $X$ converges to $x \in X$.
■
Kreyszig. (1989). Introductory Functional Analysis with Applications: p73. ↩︎