Every Finite-Dimensional Normed Linear Space is a Banach Space

Every Finite-Dimensional Normed Linear Space is a Banach Space

Recall from the Two Finite-Dimensional Normed Linear Spaces of the Same Dimension are Isomorphic page that if $X$ and $Y$ are finite-dimensional normed linear spaces such that $\mathrm{dim} (X) = \mathrm{dim} (Y)$ then $X$ and $Y$ are isomorphic, that is, there exists a bijective linear operator $T : X \to Y$ that is bounded and such that $T^{-1} : Y \to X$ is bounded.

We will now use this result to show that every finite-dimensional normed linear space $X$ is also a Banach space.

Theorem 1: Let $X$ be a finite-dimensional normed linear space. Then $X$ 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 $T : X \to \mathbb{C}^n$.
  • 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 $\mathbb{C}^n$. 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:
(1)
\begin{align} \quad \| x_m - x_n \| < \delta = \frac{\epsilon}{\| T \|} \end{align}
  • So if $m, n \geq N$ we have that:
(2)
\begin{align} \quad \| T(x_m) - T(x_n) \| \leq \| T \| \| x_m - x_n \| < \| T \| \delta = \| T \| \cdot \frac{\epsilon}{\| T \|} = \epsilon \end{align}
  • So $(T(x_n))_{n=1}^{\infty}$ is a Cauchy sequence in $\mathbb{C}^n$. But $\mathbb{C}^n$ is complete. So $(T(x_n))_{n=1}^{\infty}$ converges to some $y \in \mathbb{C}^n$, that is:
(3)
\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:
(4)
\begin{align} \quad \lim_{n \to \infty} \| x_n - 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 $X$ is a Banach space. $\blacksquare$
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