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.

  • $\begingroup$ First, do you see why the conditions are necessary? Which tools do you want to be used? $\endgroup$ – Davide Giraudo Oct 5 '12 at 16:50
  • $\begingroup$ @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. $\endgroup$ – blindman Oct 5 '12 at 16:55
  • $\begingroup$ 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. $\endgroup$ – Davide Giraudo Oct 5 '12 at 17:01
  • $\begingroup$ @Davide Giraudo: Dear Sir. Hamel basis is not familiar with me. Can you use a simpler tool? Thank you for your helping. $\endgroup$ – blindman Oct 5 '12 at 17:03
  • $\begingroup$ @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. $\endgroup$ – Christopher A. Wong Oct 5 '12 at 17:52

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.


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