1
$\begingroup$

I have seen and understand the proof of equivalence of norms when we assume the compactness of the unit ball. However, I have been asked

Assume that all finite-dimensional subspaces of a normed space are closed. Prove that all norms on a finite-dimensional normed space are equivalent without assuming the unit ball is compact. You may start by showing that all norms on a 1-dimensional normed space are equivalent. Then apply mathematical induction.

I am very lost on how to prove this. I can easily get the upper (or lower bound depnding on the norm used) but I am having a very hard time with the induction part of the argument.

$\endgroup$
2
  • $\begingroup$ Maybe this helps: if $X$ is a finite dimensional subspace of dimension $n+1$ then $X = Y+Z$ (topological sum, i.e. $X$ is homeomorphic to $Y \times Z$) with $Y$ a vector space of dimension $n.$ $\endgroup$
    – William M.
    Nov 22 '19 at 22:46
  • $\begingroup$ By "equivalence of norms" you mean that they induce the same topology, correct? $\endgroup$
    – Math1000
    Nov 22 '19 at 23:12
2
$\begingroup$

Consider any normed linear space $X$. If the kernel of a linear functional is closed then the functional is continuous. Proof: let $f$ be a non-zero functional and choose $y$ such that $f(y)=1$. If $f$ is not continuous then there exists $(x_n)$ such that $|f(x_n)| >n$ and $\|x\|_n=1$ for all $n$. But then $\frac {x_n} {f(x_n)}-y$ is a sequence in the kernel of $f$ converging to $-y$ and $f(-y) \neq 0$ leading to a contradiction.

Now your result follows immediately: if $X$ is finite dimensional space and every subspace is assumed to be closed then every linear functional is continuous. Hence the identity map from $(X,\|.\|_1) \to (X,\|.\|_2)$ is continuous for any two norms $\|.\|_1$ and $\|.\|_2$ proving that any two norms are equivalent.

$\endgroup$
5
  • $\begingroup$ But can we prove that every subspace is closed without relying on the equivalence of norms? $\endgroup$
    – WillG
    May 28 '21 at 5:38
  • $\begingroup$ To prove that every subspace of a finite-dimensional vector space is closed, you can argue that it is complete. But completeness depends on the norm, unless all norms are equivalent—but this is the result we are trying to prove. $\endgroup$
    – WillG
    May 28 '21 at 5:42
  • $\begingroup$ @WillG The question starts with 'Assume that all finite dimensional subspaces of a normed space closed'. $\endgroup$
    – Kavi Rama Murthy
    May 28 '21 at 5:44
  • $\begingroup$ Ah, fair enough. But out of curiosity, do you know if there is a proof of this statement which still does not rely on compactness of the unit ball (or equivalence of norms)? $\endgroup$
    – WillG
    May 28 '21 at 5:50
  • $\begingroup$ Yes, you can prove this using just the fact that any bounded seqeunce of real numbers has a convergent subsequence. @WillG $\endgroup$
    – Kavi Rama Murthy
    May 28 '21 at 5:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.