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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.

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  • $\begingroup$ 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 $n.$ $\endgroup$ – Will 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
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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.

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