Finite Dimensional Normed Linear Spaces are Banach Spaces

# Finite Dimensional Normed Linear Spaces are Banach Spaces

Theorem 1: Let $(X, \| \cdot \|_X)$ be a finite-dimensional normed linear space. Then $(X, \| \cdot \|_X)$ is a Banach space. |

**Proof:**Let $X$ be a finite-dimensional normed space with $\mathrm{dim}(X) = n$. Then by the theorem referenced above there exists an isomorphism $T : X \to \mathbb{R}^n$.

- Now let $(x_n)_{n=1}^{\infty}$ be a Cauchy sequence in $X$. We want to show that $(T(x_n))_{n=1}^{\infty}$ is a Cauchy sequence in $\mathbb{R}^n$. Let $\epsilon > 0$ be given and let $\displaystyle{\delta = \frac{\epsilon}{\| T \|} > 0}$ (Note that $\| T \| \neq 0$ since $\| T \| = 0$ if and only if $T = 0$ and the zero linear operator is not bijective). Since $(x_n)_{n=1}^{\infty}$ is Cauchy in $X$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then:

\begin{align} \quad \| x_m - x_n \|_X < \delta = \frac{\epsilon}{\| T \|} \end{align}

- So if $m, n \geq N$ we have that:

\begin{align} \quad \| T(x_m) - T(x_n) \| \leq \| T \| \| x_m - x_n \|_X < \| T \| \delta = \| T \| \cdot \frac{\epsilon}{\| T \|} = \epsilon \end{align}

- So $(T(x_n))_{n=1}^{\infty}$ is a Cauchy sequence in $\mathbb{R}^n$. But $\mathbb{R}^n$ is complete. So $(T(x_n))_{n=1}^{\infty}$ converges to some $y \in \mathbb{R}^n$, that is:

\begin{align} \quad \lim_{n \to \infty} \| T(x_n) - y \| = 0 \end{align}

- Since $T$ is a bijection, for $y \in Y$ there exists $x \in X$ such that $T(x) = y$. So $x = T^{-1}(y)$, and:

\begin{align} \quad \lim_{n \to \infty} \| x_n - x \|_X = \lim_{n \to \infty} \| T^{-1}(T(x_n)) - T^{-1}(y) \| = \lim_{n \to \infty} \| T^{-1} \| \| T(x_n) - y \| = \| T^{-1} \| \cdot \lim_{n \to \infty} \| T(x_n) - y \| = 0 \end{align}

- Therefore $(x_n)_{n=1}^{\infty}$ converges to $x \in X$. Hence every Cauchy sequence in $X$ converges in $X$, and so $(X, \| \cdot \|_X)$ is a Banach space. $\blacksquare$

Corollary 1: Let $(X, \| \cdot \|)$ be a normed linear space. If $Y \subseteq X$ is a finite-dimensional subspace of $X$ then $Y$ is closed. |