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 $V$.Sinceall 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 sequencein $(V,\parallel \cdot \parallel )$. 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 tosome 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 $x\in V$. $\mathrm{\square }$