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.

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