every finite dimensional normed vector space is a Banach space

Theorem 1.

Every finite dimensional normed vector space is a Banach space.

Proof. Suppose $(V,\parallel \cdot \parallel )$ is the normed vector space, and ${(e_i)}_{i=1}^N$ is a basis for $V$. For $x={\sum }_{j=1}^N{\lambda }_j{e}_j$, we can then define

$${\parallel x\parallel }^{\prime }=\sqrt{\sum _{j=1}^N{|{\lambda }_j|}^2}$$

whence $\parallel \cdot {\parallel }^{\prime }:V\to ℝ$ is a norm for $V$. Since all norms on a finite dimensional vector space are equivalent (http://planetmath.org/ProofThatAllNormsOnFiniteVectorSpaceAreEquivalent), there is a constant $C>0$ such that

$$\frac{1}{C}{\parallel x\parallel }^{\prime }\le \parallel x\parallel \le C{\parallel x\parallel }^{\prime },x\in V.$$

To prove that $V$ is a Banach space, let $x_1,x_2,\mathrm{\dots }$ be a Cauchy sequence in $(V,\parallel \cdot \parallel )$. That is, for all $\epsilon >0$ there is an $M\ge 1$ such that

Let us write each $x_k$ in this sequence in the basis $(e_j)$ as $x_k={\sum }_{j=1}^N{\lambda }_{k,j}{e}_j$ for some constants ${\lambda }_{k,j}\in ℂ$. For $k,l\ge 1$ we then have

$\parallel x_k-x_l\parallel$ $\ge$ ${\displaystyle \frac{1}{C}}{\parallel x_k-x_l\parallel }^{\prime }$

$\ge$ ${\displaystyle \frac{1}{C}}\sqrt{{\displaystyle \sum _{j=1}^N}{|{\lambda }_{k,j}-{\lambda }_{l,j}|}^2}$

$\ge$ ${\displaystyle \frac{1}{C}}|{\lambda }_{k,j}-{\lambda }_{l,j}|$

for all $j=1,\mathrm{\dots },N$. It follows that ${({\lambda }_{k,1})}_{k=1}^{\mathrm{\infty }},\mathrm{\dots },{({\lambda }_{k,N})}_{k=1}^{\mathrm{\infty }}$ are Cauchy sequences in $ℂ$. As $ℂ$ is complete, these converge to some complex numbers ${\lambda }_1,\mathrm{\dots },{\lambda }_N$. Let $x={\sum }_{j=1}^N{\lambda }_j{e}_j$.

For each $k=1,2,\mathrm{\dots }$, we then have

$\parallel x-x_k\parallel$ $\le$ $C{\parallel x-x_k\parallel }^{\prime }$

$\le$ $C\sqrt{{\displaystyle \sum _{j=1}^N}{|{\lambda }_j-{\lambda }_{k,j}|}^2}.$

By taking $k\to \mathrm{\infty }$ it follows that $(x_j)$ converges to $x\in V$. $\mathrm{\square }$

Title	every finite dimensional normed vector space is a Banach space
Canonical name	EveryFiniteDimensionalNormedVectorSpaceIsABanachSpace
Date of creation	2013-03-22 14:56:31
Last modified on	2013-03-22 14:56:31
Owner	matte (1858)
Last modified by	matte (1858)
Numerical id	10
Author	matte (1858)
Entry type	Theorem
Classification	msc 46B99