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Math Help - Finite dimensional normed spaces

  1. #1
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    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).
    Last edited by Jose27; January 5th 2010 at 06:31 PM.
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  2. #2
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    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)<br />
\ 2)d(x+z,y+z)=d(x,y), \forall z\in X<br />
    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.
    Last edited by ns1954; January 22nd 2010 at 07:57 AM.
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  3. #3
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    Quote Originally Posted by ns1954 View Post
    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)<br />
\ 2)d(x+z,y+z)=d(x,y), \forall z\in X<br />
    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.
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  4. #4
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    And here must be noted that metric space is compact iff is complete and totally bounded.
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