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 ) is complete. Prove the converse:

If is a vector space over such that for every norm we have that is complete then

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

Hint

Spoiler:

Assume , consider the following norm: If is a basis and with then

Solution

Spoiler:

With assumptions as in the hint assume further that all vectors in the basis have norm and consider and a sequence such that .

Define and assume there exists such that . Let be such that (I'm taking a little liberty here by taking a countable set in and identifying it with ) then certainly exists since is finite. We define . We compute:

This is true for all but noticing that for we have so we can't have , but 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

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

January 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, define metric with folloing properties:

And conversally, if metric have this two additional properties, then define a norm.
But, not all metrics have this addicional properties. I believe the answer is affirmative.

January 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, define metric with folloing properties:

And conversally, if metric have this two additional properties, then 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 . 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.

January 23rd 2010, 08:17 AM

ns1954

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