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I would like to find a short proof for the following theorems:

Theorem 1. A normed space is finite dimensional iff all of its linear functional is continuous.

Theorem 2. A normed space is finite dimensional iff its unit ball is compact.

Thank you in advance.

blindman
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  • First, do you see why the conditions are necessary? Which tools do you want to be used? – Davide Giraudo Oct 05 '12 at 16:50
  • @Davide Giraudo: Dear Sir. One of the direction in both theorems is easy to prove. How to prove that the given space is finite dimensional. I would like to use the simplest tool. – blindman Oct 05 '12 at 16:55
  • For the first theorem, I know a proof which uses Zorn lemma to get a Hamel basis. Then with that, we can construct a non-continuous linear functional. – Davide Giraudo Oct 05 '12 at 17:01
  • @Davide Giraudo: Dear Sir. Hamel basis is not familiar with me. Can you use a simpler tool? Thank you for your helping. – blindman Oct 05 '12 at 17:03
  • @blindman, A hamel basis is simply the "usual" basis that you're familiar with: a linearly independent set such that every vector in the space can be written as a finite linear combination of elements from that set. – Christopher A. Wong Oct 05 '12 at 17:52
  • Maybe the shortest proof isnt so easy to understand. – Tomás Oct 05 '12 at 18:09
  • I wouldn't expect "short proofs" of those facts. One is deceived by the way they are written, because they only talk about finite-dimensional spaces. But the two assertions say exactly this:
    1. A normed space is infinite-dimensioan if and only if it has a non-continuous linear functional.
    2. A normed space is infinite-dimensional if and only if its unit ball is not compact.

    Usual proofs are of the form "finite-dimensional implies continuous linear functional and compact unit ball", and "infinite-dimensional implies a discontinuous linear functional and non-compact unit ball".

     – Martin Argerami Oct 05 '12 at 18:36
  • Ok. For the second theorem take a look on the page 160 of "Brezis - Functional Analysis, Sobolev Spaces and PDE". For the first one is like Davide Giraudo said. – Tomás Oct 05 '12 at 19:01

1 Answers1

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The direction you asked for in the comments:

Let $X$ be an infinite dimensional normed space.

  1. Pick a countable independent collection $(e_n)_{n\in\mathbb{N}}$, pick $(y_i)_i$ such that $(e_n,y_i)_{n,i}$ is a basis. Let $f$ be the functional determined by $f(e_n)=n\|e_n\|$ and $f(y_i)=0$. Then $f$ is unbounded.

  2. By Riesz's lemma one easily constructs a sequence of independent vectors in the unit ball without a converging subsequence.

wildildildlife
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