# Equivalence of norms on a finite dimension space without using compactness of the unit ball

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.

• 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 $n.$ – Will M. Nov 22 '19 at 22:46
• By "equivalence of norms" you mean that they induce the same topology, correct? – Math1000 Nov 22 '19 at 23:12

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.