# every finite dimensional normed vector space is a Banach space

###### Theorem 1.

Every finite dimensional normed vector space is a Banach space.

Proof. Suppose $(V,\|\cdot\|)$ 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

 $\|x\|^{\prime}=\sqrt{\sum_{j=1}^{N}|\lambda_{j}|^{2}}$

whence $\|\cdot\|^{\prime}\colon V\to\mathbb{R}$ 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

 $\frac{1}{C}\|x\|^{\prime}\leq\|x\|\leq C\|x\|^{\prime},\quad x\in V.$

To prove that $V$ is a Banach space, let $x_{1},x_{2},\ldots$ be a Cauchy sequence in $(V,\|\cdot\|)$. That is, for all $\varepsilon>0$ there is an $M\geq 1$ such that

 $\|x_{j}-x_{k}\|<\varepsilon,\ \ \mbox{for all}j,k\geq M.$

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 $\lambda_{k,j}\in\mathbbmss{C}$. For $k,l\geq 1$ we then have

 $\displaystyle\|x_{k}-x_{l}\|$ $\displaystyle\geq$ $\displaystyle\frac{1}{C}\|x_{k}-x_{l}\|^{\prime}$ $\displaystyle\geq$ $\displaystyle\frac{1}{C}\sqrt{\sum_{j=1}^{N}|\lambda_{k,j}-\lambda_{l,j}|^{2}}$ $\displaystyle\geq$ $\displaystyle\frac{1}{C}|\lambda_{k,j}-\lambda_{l,j}|$

for all $j=1,\ldots,N$. It follows that $(\lambda_{k,1})_{k=1}^{\infty},\ldots,(\lambda_{k,N})_{k=1}^{\infty}$ are Cauchy sequences in $\mathbbmss{C}$. As $\mathbbmss{C}$ is complete, these converge to some complex numbers $\lambda_{1},\ldots,\lambda_{N}$. Let $x=\sum_{j=1}^{N}\lambda_{j}e_{j}$.

For each $k=1,2,\ldots$, we then have

 $\displaystyle\|x-x_{k}\|$ $\displaystyle\leq$ $\displaystyle C\|x-x_{k}\|^{\prime}$ $\displaystyle\leq$ $\displaystyle C\sqrt{\sum_{j=1}^{N}|\lambda_{j}-\lambda_{k,j}|^{2}}.$

By taking $k\to\infty$ it follows that $(x_{j})$ converges to $x\in V$. $\Box$

Title every finite dimensional normed vector space is a Banach space EveryFiniteDimensionalNormedVectorSpaceIsABanachSpace 2013-03-22 14:56:31 2013-03-22 14:56:31 matte (1858) matte (1858) 10 matte (1858) Theorem msc 46B99