# Thread: Finite dimensional normed spaces

1. ## Finite dimensional normed spaces

I find this problem interesting since it gives another characterization of finite dim. normed spaces.

It is a standard result that every finite dim. normed space (over $\mathbb{F} = \mathbb{R} , \mathbb{C}$ ) is complete. Prove the converse:

If $V$ is a vector space over $\mathbb{F}$ such that for every norm $\Vert \cdot \Vert : V \rightarrow \mathbb{R}$ we have that $(V,\Vert \cdot \Vert )$ is complete then $\dim (V) < \infty$

I include a hint and the solution for those that want them.

Hint

Spoiler:
Assume $\dim (V) = \infty$, consider the following norm: If $\{ u_a\}_{a\in A}$ is a basis and $x= \sum_{k\in I_x \subset A } \beta _ku_k$ with $\beta _k \in \mathbb{F}$ then $\Vert x \Vert = \sum_{k\in I_x} \vert \beta _k \vert$

Solution

Spoiler:
With assumptions as in the hint assume further that all vectors in the basis have norm $1$ and consider $(u_n)_{n\in \mathbb{N} } \subset \{ u_a \}$ and a sequence $(\alpha _n )_{n\in \mathbb{N} } \subset \mathbb{F}$ such that $\sum_{k=1}^{\infty } \vert \alpha _k \vert < \infty$.

Define $x_n= \sum_{k=1}^{n} \alpha _k u_k$ and assume there exists $x\in V$ such that $x_n \rightarrow x= \sum_{k\in I_x } \beta _k u_k$. Let $m\in \mathbb{N}$ be such that $m> \max \{k\in I_x \cap \mathbb{N} \}$ (I'm taking a little liberty here by taking a countable set in $A$ and identifying it with $\mathbb{N}$ ) then $m$ certainly exists since $I_x$ is finite. We define $N_n=\{ j\in \mathbb{N} : j\leq n \}$. We compute:

$\Vert x_n -x \Vert = \sum_{k\in I_x \cap N_n} \vert \alpha _k - \beta _k \vert + \sum_{k\in I_x \setminus N_n} \vert \beta _k \vert$ $+ \sum_{k\in N_n \setminus I_x } \vert \alpha _k \vert \geq \sum_{k\in N_n \setminus I_x } \vert \alpha _k \vert >0$

This is true for all $n$ but noticing that for $n>m$ we have $\sum_{k\in N_n \setminus I_x } \vert \alpha _k \vert \leq \sum_{k\in N_{n+1} \setminus I_x } \vert \alpha _k \vert$ so we can't have $x_n \rightarrow x$, but $\sum_{k\in \mathbb{N} } \alpha _k u_k$ is an absolutely convergent series. Now just remembering that a normed space is complete iff every absolutely convergent series is convergent we get that indeed $\dim (V) < \infty$

One thing that also came to mind, but haven't given much thought (not posting it as a question but as recreation) if $V$ is such that every two norms are comparable, does it follow $\dim (V) < \infty$ ? (The weaker statement using that two norms are equivalent follows directly from the posed problem).

2. This is well known result. I will update this post with proff which I have in a time, now I'm little buzy. Another interesting question can be putted: Is this valid if we replace norm with metric.
Naimly, $d(x,y)=\|x-y\|$ define metric with folloing properties:

$1) d(\lambda x,\lambda y)=|\lambda|d(x,y)
\ 2)d(x+z,y+z)=d(x,y), \forall z\in X
$

And conversally, if metric have this two additional properties, then $d(x,0)=\|x\|$ define a norm.
But, not all metrics have this addicional properties. I believe the answer is affirmative.

3. Originally Posted by ns1954
This is well known result. I will update this post with proff which I have in a time, now I'm little buzy. Another interesting question can be putted: Is this valid if we replace norm with metric.
Naimly, $d(x,y)=\|x-y\|$ define metric with folloing properties:

$1) d(\lambda x,\lambda y)=|\lambda|d(x,y)
\ 2)d(x+z,y+z)=d(x,y), \forall z\in X
$

And conversally, if metric have this two additional properties, then $d(x,0)=\|x\|$ define a norm.
But, not all metrics have this addicional properties. I believe the answer is affirmative.
Well, if V is complete with every metric it's complete with respect to every norm and so $\dim (V)< \infty$ . I don't think the converse is true, for example if one were to prove this http://www.mathhelpforum.com/math-he...t-metrics.html then since V is obviously not compact (with any norm) there would exist an equivalent non-complete metric.

4. And here must be noted that metric space is compact iff is complete and totally bounded.