| Version 6 |
Version 5 |
| \begin{thm} |
\begin{thm} |
| Every finite dimensional normed vector space is a Banach space |
Every finite dimensional normed vector space is a Banach space |
| \end{thm} |
\end{thm} |
|
|
| \emph{Proof.} Suppose $(V,\Vert\cdot\Vert)$ is the normed vector space, |
\emph{Proof.} Suppose $(V,\Vert\cdot\Vert)$ is the normed vector space, |
| and $(e_i)_{i=1}^N$ is a basis for $V$. |
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 |
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} |
\Vert x \Vert' = \sqrt{\sum_{j=1}^N |\lambda_j|^2} |
| $$ |
$$ |
| whence $\Vert\cdot\Vert'\colon V\to \sR$ is a norm for $V$. |
whence $\Vert\cdot\Vert'\colon V\to \sR$ is a norm for $V$. |
| Since |
Since |
| \PMlinkname{all norms on a finite dimensional vector space are equivalent}{ProofThatAllNormsOnFiniteVectorSpaceAreEquivalent}, |
\PMlinkname{all norms on a finite dimensional vector space are equivalent}{ProofThatAllNormsOnFiniteVectorSpaceAreEquivalent}, |
| there is a constant $C>0$ such that |
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. |
\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 |
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, |
in $(V,\Vert\cdot \Vert)$. That is, |
| for all $\varepsilon>0$ there is an $M\ge 1$ such that |
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. |
\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 $(e_j)$ |
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 |
as $x_k=\sum_{j=1}^N \lambda_{k,j} e_j$ for some constants |
|
$\lambda_{k,j}\in \C$.
|
$\{ \lambda_{k,j}\in \C \mid k=1,2,\ldots, j=1,\ldots, N\}$.
|
|
For $k,l\ge 1$ we then have
|
For $k,l\ge 1$ we have
|
| \begin{eqnarray*} |
\begin{eqnarray*} |
| \Vert x_k-x_l\Vert &\ge& \frac{1}{C} \Vert x_k-x_l \Vert' \\ |
\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} \sqrt{\sum_{j=1}^N |\lambda_{k,j}-\lambda_{l,j}|^2} \\ |
| &\ge& \frac{1}{C} |\lambda_{k,j}-\lambda_{l,j}| |
&\ge& \frac{1}{C} |\lambda_{k,j}-\lambda_{l,j}| |
| \end{eqnarray*} |
\end{eqnarray*} |
| for all $j=1,\ldots, N$. |
for all $j=1,\ldots, N$. |
| It follows that |
It follows that |
| $(\lambda_{k,1})_{k=1}^\infty, \ldots, (\lambda_{k,N})_{k=1}^\infty$ |
$(\lambda_{k,1})_{k=1}^\infty, \ldots, (\lambda_{k,N})_{k=1}^\infty$ |
| are Cauchy sequences in $\C$. As $\C$ is complete, these converge to |
are Cauchy sequences in $\C$. As $\C$ is complete, these converge to |
| some complex numbers $\lambda_1, \ldots, \lambda_N$. |
some complex numbers $\lambda_1, \ldots, \lambda_N$. |
| Let $x=\sum_{j=1}^N \lambda_j e_j$. |
Let $x=\sum_{j=1}^N \lambda_j e_j$. |
|
|
| For each $k=1,2,\ldots$, we then have |
For each $k=1,2,\ldots$, we then have |
| \begin{eqnarray*} |
\begin{eqnarray*} |
| \Vert x-x_k\Vert &\le& C \Vert x-x_k\Vert' \\ |
\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}. |
&\le& C \sqrt{\sum_{j=1}^N |\lambda_{j}-\lambda_{k,j}|^2}. |
| \end{eqnarray*} |
\end{eqnarray*} |
| By taking $k\to \infty$ it follows that $(x_j)$ converges to $x\in V$. $\Box$ |
By taking $k\to \infty$ it follows that $(x_j)$ converges to $x\in V$. $\Box$ |
