We know that for a normed vector space $(X, ||\cdot ||)$: X is finite dimensional if and only if the unit ball $K = \left\{x \in X: ||x|| \leq 1 \right\}$ is compact. How is this used in practice? Do we actually use this to prove that a space is finite dimensional? Other than that, are there examples of other theorems where this is used in the proofs?

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    $\begingroup$ In my opinion, it is the other way around. It unfortunately prevents you from using norms and compactness results at the same time in infinite dimensional vector spaces, like spaces of functions, to do functional analysis $\endgroup$
    – Didier
    Feb 4 at 13:07
  • $\begingroup$ @Didier Thank you for your answer. A question that popped up: why is it sufficient to only look at the unit ball? Why does non-compactness of the unit ball prevent you from doing functional analysis? $\endgroup$
    – Si_monster
    Feb 4 at 14:20
  • $\begingroup$ Let' say you have a sequence of functions. If the sequence is bounded in norm, you could expect that it has a convergent subsequence. This is a common trick to show existence of solutions to functional equations such as PDEs, by building a sequence of approximate solutions, whose limit points solve the actual equation. Unfortunately, if your topology is a normed topology, you can't expect to use this trick. One must then find other tolopologies for which the heuristic work $\endgroup$
    – Didier
    Feb 4 at 14:23
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    $\begingroup$ This is the whole point of the notions of Fréchet topology, weak topology, *-topology, weak *-topology, and so on $\endgroup$
    – Didier
    Feb 4 at 14:29
  • $\begingroup$ @Didier I see. If it is a normed topology, how does the non-compactness of the unit ball come into play here? Of course, non-compactness means sequences have no convergent subsequences. $\endgroup$
    – Si_monster
    Feb 5 at 9:12

1 Answer 1


I don't know of any result of the form

Theorem: Let $(V,\|\cdot\|)$ be [$\ldots$]. Then $V$ is finite / infinite dimensional.

Proof: let us show that the unit ball is compact / non-compact. $[\ldots]$

At least, I don't know of any such results where no simpler proof exists. However, the equivalence

$(V,\|\cdot\|)$ is finite dimensional $\iff$ the unit ball is compact.

is of crucial theoretical importance. Note that the fact that the unit ball is compact is equivalent to the fact that closed bounded domains are compact, or to the fact that bounded sequences admit limit points. Thus, the above result tells you that one cannot expect to use compactness arguments in infinite dimensional normed vector spaces to show existence of solutions to some minimising problems or some PDEs. Below is a hand-wavy illustration.

Assume $(V,\|\cdot\|)$ is finite dimensional, and let $F \colon V \to [0,+\infty)$ be some continuous functional. Assume that $F(u) \to \infty$ whenever $\|u\| \to \infty$. Then there exists $u_0 \in V$ such that $F(u_0) = \min_{u\in V} F(v)$. Indeed, one easily builds a sequence $u_k$ such that $F(u_k) \leqslant \inf F + \frac{1}{k}$, then shows that $\{u_k\}$ must be bounded, and exhibits a converging subsequence thanks to a compactness argument. The limit $u_0$ then solves the problem.

Now, assume that $(V,\|\cdot\|)$ is not finite dimensional. Then one can still create a "minimizing" subsequence $\{u_k\}$ such that $F(u_k) \to \inf F$. However, absolutely nothing can ensure the existence of a converging subsequence, and hence, you cannot expect to exhibit a solution to the problem this way.

This example is not artificial: some PDEs are naturally the Euler-Lagrange equations of some suitable functionals as above, and solving those said PDEs boils down to finding minimizers of the functionals. Since function spaces, such as $\mathcal{C}^{\infty}(X)$, $\mathcal{C}^{k,\alpha}(X)$, $\Bbb C[X]$, or $H^p(X)$, are infinite dimensional, you can't expect to use a normed topology to solve many problems in functional analysis, at least not in this way. One has to build appropriate tools to make the above heuristic work. Some of these tools are completeness, compact embeddings of Sobolev spaces, etc. One can also abandon the normed topology and look for a "better" topology that reflects the properties of the problem, such as weak topology, weak-* topology, and so on.

  • $\begingroup$ Thanks a lot! Do you have sources where I can find more examples? $\endgroup$
    – Si_monster
    Feb 5 at 20:13
  • $\begingroup$ @Si_monster More examples of what, exactly? $\endgroup$
    – Didier
    Feb 6 at 10:07
  • $\begingroup$ Sorry, I meant like an actual example of a PDE that is solved by building a sequence of approximate solutions. $\endgroup$
    – Si_monster
    Feb 9 at 11:21
  • $\begingroup$ @Si_monster You can take a look at this quite recent paper by Allen, Lee and Maxwell, (especially section 6), but this is quite a lot tu understand especially if you're not familiar with differential geometry. $\endgroup$
    – Didier
    Feb 9 at 12:32

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