"Different" way of proving every norm is equivalent in finite dimensional vector space?

Question

It certainly is easy to find as many proofs as you want for the equivalence of any two norms in a finite dimensional vector space.

I was wondering what is the relationship between homeomorphisms and the equivalence of any two norms in a finite dimensional vector space.

This question urged to me since I am doing a worksheet that asked to to show that the mapping $T:(\Bbb K^n, \| \cdot \|_\infty) \rightarrow (X,\| \cdot \|)$ is a homeomorphism (more details can be seen here) and afterwards to prove that any two norms on a finite dimensional space $X$ are equivalent, based on this result.

Formal exercise. Show that in a finite-dimensional vector space $X$, every two norms are equivalent, using this result.

My attempt of resolution. Let $X$ be a finite-dimensional vector space and let $\|.\|_\beta$ and $\| \cdot \|_\gamma$ be two norms defined on $X$. Since $X$ is finite-dimensional, let $\{e_1\dots,e_n\}$ be a basis for $X$ and define a third norm on $X$ as follows:

Given $x \in X$ we can write: $$ x = \sum_{j=1}^n \alpha_je_j,$$ for some scalars $\alpha_j \in \Bbb K, \forall j \in \{1,\dots,n\}.$ Set $$ \| x \|_\infty = \max\{|\alpha_j|: j \in \{1,\dots,n\}\}$$ This is the exact same norm used for $\Bbb K^n$ in the previous result I linked. By transitivity, it suffices to show that both $\| \cdot \|_\beta$ and $\| \cdot \|_\gamma$ are equivalent to $\| \cdot \|_\infty$. This is easy to do (I will omit it here) using norm properties (triangle inequality and homogeneity).

Now, one could finish this proof in the usual way - i.e. - considering the unit sphere $S = \{x \in X : \| x \|_\infty = 1\}$ and applying some topological arguments.

I just don't see where I am applying the result I linked... Does it allow me to make some kind of shortcut? I am not really familiar with homeomorphisms so maybe this is way easier than what I think. So, basically, what I want to do is to use the linked result to somehow help me proving this. I believe that the infinite norm definition has to stick around, but I don't see how to relate it with the homeomorphism.

Thanks for any help in advance.

Answer 1

The result you want to use is: for any $n$-dimensional $\Bbb K$-vector space $X$ ($\Bbb K=\Bbb R$ or $\Bbb C$) any linear bijection $\Bbb K^n\to X$ is an homeomorphism $(\Bbb{K}^n, \| \cdot \|_\infty) \rightarrow (X, \| \cdot \|),$ where $\| \cdot \|$ is an arbitrary norm on $X.$

Let $T:\Bbb K^n\to X$ be a linear bijection. Then, using this result twice, $T:(\Bbb{K}^n, \| \cdot \|_\infty)\to(X, \| \cdot \|_\beta)$ and $T:(\Bbb{K}^n, \| \cdot \|_\infty)\to(X, \| \cdot \|_\gamma)$ ar both homeomorphims hence by composition, $\mathrm{id}_X:(X, \| \cdot \|_\beta)\to(X, \| \cdot \|_\gamma)$ is a homeomorphism, i.e. $ \|_\gamma$ and $\cdot \|_\beta$ are equivalent.

Answer 2

Consider teh identity map $I:(X,\lvert \lvert \rvert \rvert_{\infty})\to (X,\lvert \lvert \rvert \rvert_{\alpha})$,it is a linear map which is continous,this is inverse is also is continous,but the continuity is exactly the inequality giving equivalence of norms .$$ \lvert \lvert I(x)\rvert \rvert_{\alpha} \leq \lvert \lvert x \rvert \rvert_{\infty} \sum_{i=1}^{n} \lvert \lvert e_{i}\rvert \rvert_{\alpha}$$ So $$\lvert \lvert x \rvert \rvert_{\alpha} \leq M \lvert \lvert x\rvert \rvert_{\infty}$$Where $M=\sum_{i=1}^{n} \lvert \lvert e_{i}\rvert \rvert_{\alpha}$,Now the continuity of the inverse means ,there is some $N$ such that $$\lvert \lvert I^{-1}(x) \rvert \rvert_{\infty}=\lvert \lvert x \rvert \rvert_{\infty} \leq N \lvert \lvert x\rvert \rvert_{\alpha}$$ Combinig both inequalities gives us the equivalence .