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Viewing Version
6
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'every finite dimensional normed vector space is a Banach space'
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| Title of object: |
every finite dimensional normed vector space is a Banach space |
| Canonical Name: |
EveryFiniteDimensionalNormedVectorSpaceIsABanachSpace |
| Type: |
Theorem |
| Created on: |
2005-01-09 14:50:03 |
| Modified on: |
2005-01-15 09:48:40 |
| Classification: |
msc:46B99 |
Preamble:
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\newcommand{\sR}[0]{\mathbb{R}}
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\newcommand{\Z}{\mathbbmss{Z}}
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Content:
\begin{thm}
Every finite dimensional normed vector space is a Banach space
\end{thm}
\emph{Proof.} Suppose $(V,\Vert\cdot\Vert)$ 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
$$
\Vert x \Vert' = \sqrt{\sum_{j=1}^N |\lambda_j|^2}
$$
whence $\Vert\cdot\Vert'\colon V\to \sR$ is a norm for $V$.
Since
\PMlinkname{all norms on a finite dimensional vector space are equivalent}{ProofThatAllNormsOnFiniteVectorSpaceAreEquivalent},
there is a constant $C>0$ such that
$$
\frac{1}{C} \Vert x \Vert' \le \Vert x \Vert \le C \Vert x \Vert', \quad x\in V.
$$
To prove that $V$ is a Banach space, let $x_1,x_2,\ldots$ be a Cauchy sequence
in $(V,\Vert\cdot \Vert)$. That is,
for all $\varepsilon>0$ there is an $M\ge 1$ such that
$$
\Vert x_j-x_k \Vert <\varepsilon, \ \ \mbox{for all} j,k\ge M.
$$
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
$\lambda_{k,j}\in \C$.
For $k,l\ge 1$ we then have
\begin{eqnarray*}
\Vert x_k-x_l\Vert &\ge& \frac{1}{C} \Vert x_k-x_l \Vert' \\
&\ge& \frac{1}{C} \sqrt{\sum_{j=1}^N |\lambda_{k,j}-\lambda_{l,j}|^2} \\
&\ge& \frac{1}{C} |\lambda_{k,j}-\lambda_{l,j}|
\end{eqnarray*}
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 $\C$. As $\C$ 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
\begin{eqnarray*}
\Vert x-x_k\Vert &\le& C \Vert x-x_k\Vert' \\
&\le& C \sqrt{\sum_{j=1}^N |\lambda_{j}-\lambda_{k,j}|^2}.
\end{eqnarray*}
By taking $k\to \infty$ it follows that $(x_j)$ converges to $x\in V$. $\Box$ |
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