📂Banach Space

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.

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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 $x \in X$.

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