Mathematics Prelims

July 16, 2008

Every Finite Dimensional Normed Space is Banach

Filed under: Analysis,Functional Analysis — cjohnson @ 2:45 pm

Every finite dimensional normed space (over a complete field, namely $\mathbb{R}$ or $\mathbb{C}$) is Banach (complete in the metric induced by the norm).

Proof: Let $Y$ be an n-dimensional normed space with basis $\{ e_1, ..., e_n \}$ and let $(y_n)_{n \in \mathbb{N}}$ be any Cauchy sequence in $Y$.  As $(y_n)$ is Cauchy, for any given $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that for all $m, n > N$ we have $\| y_n - y_m \| < \epsilon$.  However, by a previous lemma,there exists a $c > 0$ such that

$\displaystyle \| y_n - y_m \| = \left\| \sum_{i=1}^n \left( \alpha_i^{(n)} - \alpha_i^{(m)} \right) e_i \right\| \geq c \sum_{i=1}^n \left| \alpha_i^{(n)} - \alpha_i^{(m)} \right|$

Which means that $\sum_{i=1}^n \left| \alpha_i^{(n)} - \alpha_i^{(m)} \right| \leq \epsilon/c$, which in turn implies for a fixed $i$, the sequence $(\alpha_i^{(m)})$ is Cauchy in a complete space, so it converges.  Let the limit of this sequence be $\alpha_i$ and let $y = \alpha_1 e_1 + ... \alpha_n e_n$.  Now we have

$\displaystyle \| y_m - y \| = \left\| \sum_{i=1}^n \left( \alpha_i^{(m)} - \alpha_i \right) e_i \right\| \leq \sum_{i=1}^n \left| \alpha_i^{(m)} - \alpha_i \right| \|e_i\|$.

Since $\alpha_i^{(m)} \to \alpha_i$, we can make this right-hand side arbitrarily small by picking a large enough $N \in \mathbb{N}$ and considering $m > N$.  This means that $\| y_m - y \| \to 0$, so $y_m \to y$, and the space is complete.