2
$\begingroup$

Prove that any finite dimensional normed linear space over a complete field is complete.

After several comments and corrections, I present the correct proof in the answer section.

$\endgroup$
13
  • $\begingroup$ Looks fine to me. I would probably emphasize why we can choose $N$ uniformly in $i$ (after the line when you say $\{\alpha_{i,r}\}^{n}_{i=1,r\in \Bbb{N}}$ is Cauchy) because this is the key place where the finite dimensionality is required. $\endgroup$ – Pratyush Sarkar Dec 28 '18 at 6:41
  • $\begingroup$ @Pratyush Sarkar: Thanks a lot for your comment! Can you please, elaborate on the use of finite dimension? $\endgroup$ – Omojola Micheal Dec 28 '18 at 6:43
  • $\begingroup$ You are using the fact that $\alpha_{i,r}\to \alpha_{i}\in \Bbb{R}\;\text{or}\;\Bbb{C},\;\text{as}\;r\to \infty$. But there are multiple sequences here -- one for each $i$. For any fixed $i$ and $\epsilon'$, you have a corresponding $N$ (depending on $\epsilon'$ and $i$). But a priori these $N$ may be different for each $i$ (we can write $N_i$ to make the dependence explicit). You want to choose $N = \max\{N_i: i = 1, 2, \dotsc, n\}$ so that the same $N$ works for all $i$. Does that make sense? This also illuminates why things can go wrong in infinite dimensions. $\endgroup$ – Pratyush Sarkar Dec 28 '18 at 6:57
  • $\begingroup$ @Pratyush Sarkar: Oh, now I get you! That makes sense! Is it fine, now? I made some edits! I believe it should be. $\endgroup$ – Omojola Micheal Dec 28 '18 at 7:41
  • 1
    $\begingroup$ I removed some unnecessary parts which you forgot to delete. Looks fine now. $\endgroup$ – Pratyush Sarkar Dec 29 '18 at 1:01
0
$\begingroup$

PROOF

Let $E$ be any finite dimensional normed linear space over a complete field, $\Bbb{R}$ or $\Bbb{C},$ say. Suppose $\dim E=n\geq 1,$ and let $\{e_i\}^{n}_{i=1}$ be a basis for $E.$ Then, there exists scalars $\{\alpha_i\}^{n}_{i=1}$ such that, for arbitrary $x\in E$, \begin{align} x= \sum^{n}_{i=1} \alpha_i e_i .\end{align} Suppose $\{x_r\}_{r\in \Bbb{N}}$ is Cauchy in $E$ w.r.t $\|\cdot\|$ norm and $\epsilon'>0.$ Then, there exists $N$ such that for all $s\geq r\geq N,$ \begin{align} \|x_r-x_s\|<\epsilon'.\end{align} Now, $\|\cdot\|_1$ defined by $\|x\|_1=\sum^{n}_{i=1} |\alpha_i|$ is a norm on $E$ and since All norms defined on a finite dimensional normed linear space are equivalent, we have that $\|\cdot\|_1\sim \|\cdot\|,$ i.e., there exists $\gamma,\beta>0,$ such that \begin{align} \gamma\|x\|_1\leq\|x\|\leq \beta\|x\|_1,\;\forall\,x\in E.\end{align} Then, \begin{align}\gamma|\alpha_{i,r}-\alpha_{i,s}|\leq\gamma\sum^{n}_{i=1}|\alpha_{i,r}-\alpha_{i,s}|=\gamma\|x_r-x_s\|_1\leq\|x_r-x_s\|<\epsilon',\;s\geq r\geq N\end{align} and so the sequence $\{\alpha_{i,r}\}^{n}_{i=1,r\in \Bbb{N}}$ is Cauchy in $\Bbb{R}$ or $\Bbb{C}$. Now, $\alpha_{i,r}\to \alpha_{i}\in \Bbb{R}\;\text{or}\;\Bbb{C},\;\text{as}\;r\to \infty$ by completeness. This implies that for each $i\in\{1,2,\cdots,n\},$ there exists $N_i:=N(i,\epsilon')$ such that \begin{align}|\alpha_{i,r}-\alpha_{i}|<\epsilon',\;r\geq N_i.\end{align} Taking $M=\max\{N_i:1\leq i\leq n\},$ we have that \begin{align}|\alpha_{i,r}-\alpha_{i}|<\epsilon',\;r\geq M.\end{align}

Let $\epsilon>0$ and $n\in\Bbb{N}$, then for $\epsilon'=\dfrac{\epsilon}{n\beta},$ there exists $M$ such that \begin{align}|\alpha_{i}-\alpha_{i,s}|\leq\dfrac{\epsilon}{n\beta},\;s\geq M.\end{align} Taking sums, we have
\begin{align}\|x-x_s\|\leq\beta\|x-x_s\|_1=\beta\sum^{n}_{i=1}|\alpha_{i}-\alpha_{i,s}|\leq\sum^{n}_{i=1}\dfrac{\epsilon}{n},\;s\geq M.\end{align} Hence, \begin{align}\|x_s-x\|\leq\epsilon,\;s\geq M,\end{align} and so, we have that $x\in E,$ which implies that $E$ is a complete finite dimensional normed linear space and we are done!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.