# Finite dimensional normed spaces

• Jan 5th 2010, 05:26 PM
Jose27
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 $\displaystyle \mathbb{F} = \mathbb{R} , \mathbb{C}$ ) is complete. Prove the converse:

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

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

Hint

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

Solution

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

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

$\displaystyle \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$$\displaystyle + \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 $\displaystyle n$ but noticing that for $\displaystyle n>m$ we have $\displaystyle \sum_{k\in N_n \setminus I_x } \vert \alpha _k \vert \leq \sum_{k\in N_{n+1} \setminus I_x } \vert \alpha _k \vert$ so we can't have $\displaystyle x_n \rightarrow x$, but $\displaystyle \sum_{k\in \mathbb{N} } \alpha _k u_k$ 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 $\displaystyle \dim (V) < \infty$

One thing that also came to mind, but haven't given much thought (not posting it as a question but as recreation) if $\displaystyle V$ is such that every two norms are comparable, does it follow $\displaystyle \dim (V) < \infty$ ? (The weaker statement using that two norms are equivalent follows directly from the posed problem).
• Jan 22nd 2010, 07:17 AM
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, $\displaystyle d(x,y)=\|x-y\|$ define metric with folloing properties:

$\displaystyle 1) d(\lambda x,\lambda y)=|\lambda|d(x,y) \ 2)d(x+z,y+z)=d(x,y), \forall z\in X$
And conversally, if metric have this two additional properties, then $\displaystyle d(x,0)=\|x\|$ define a norm.
But, not all metrics have this addicional properties. I believe the answer is affirmative.
• Jan 23rd 2010, 07:33 AM
Jose27
Quote:

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, $\displaystyle d(x,y)=\|x-y\|$ define metric with folloing properties:

$\displaystyle 1) d(\lambda x,\lambda y)=|\lambda|d(x,y) \ 2)d(x+z,y+z)=d(x,y), \forall z\in X$
And conversally, if metric have this two additional properties, then $\displaystyle d(x,0)=\|x\|$ 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 $\displaystyle \dim (V)< \infty$ . 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.
• Jan 23rd 2010, 08:17 AM
ns1954
And here must be noted that metric space is compact iff is complete and totally bounded.