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, 2019 at 22:46
  • $\begingroup$ By "equivalence of norms" you mean that they induce the same topology, correct? $\endgroup$
    – Math1000
    Nov 22, 2019 at 23:12

2 Answers 2

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, 2021 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, 2021 at 5:42
  • $\begingroup$ @WillG The question starts with 'Assume that all finite dimensional subspaces of a normed space closed'. $\endgroup$ May 28, 2021 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, 2021 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 $\endgroup$ May 28, 2021 at 5:59
0
$\begingroup$

Just to point out, what the answer above actually does is to show that if $X$ is a finite-dimensional normed vector space, then any linear map $\phi\colon X \to Y$ is continuous: exactly as it points out for the identity map, the continuity for a general linear map $\phi$ follows by, say, picking a basis $\{f_1,\ldots, f_k\}$ for $\phi(X)\leq Y$ (the image of $X$ is finite-dimensional because $X$ is) and writing $\phi(v) = \sum_{i=1}^k \phi_i(v).f_i$, where the $\phi_i$ are linear functionals on $X$.

You can then use induction to prove the continuity of linear functionals on a finite-dimensional vector space: if $f\colon X \to \mathbb R$ is a linear functional, then $H=\ker(f)$ is a subspace of dimension $n-1$ where $n = \dim(X)$. Now it is enough to show that $H$ is complete, because a complete subspace of a metric space has to be closed. To see that $H$ is complete, pick a basis of $H$, $\{e_1,\ldots,e_{n-1}\}$ and let $\psi \colon \ell_2^n \to H$ (where $\ell_2^n$ denotes the $n$-dimensional Euclidean vector space/inner product space) be given by $\psi((a_1,\ldots,a_{n-1}) \mapsto \sum_{i=1}^{n-1} a_ie_i$. Now $\psi$ is an isomorphism of linear spaces, and by the induction hypothesis, $\psi$ is bounded. But since $\ell_2^n$ is complete and any bounded linear map is uniformly continuous, it follows that $H$ is complete as required!

[Just for completeness, the continuity of any linear map $f\colon X\to Y$ when $\dim(X)=1$ really is trivial: if we pick a vector $e \in X$ with $\|e\|=1$, then $\|f\| = |f(e)|$.]

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .