I have a problem where I have to show that every $n$-dimensional normed space $E$ has the same norm as the euclidean space $E_n$.

Here's what I've got:

Since $E$ is $n$-dimensional then for the basis $(e_1, ..., e_n)$ of $E$ there's a unique representation for every $x \in E$ i.e. $x=\sum_{i=1}^{n}{\lambda_i e_i}$.

First I show for every $x \in E$ the function $g(x)=(\lambda_1, ... \lambda_n) \in E_n$ is an isomorphism.

Then I show that $\lVert x\rVert_{E} \le C\lVert v\rVert_{E_n}$ and thus the image is continuous.

After that I show that$\lVert x\rVert_{E_n} \le C_2\lVert v\rVert_{E}$ and thus the image is homeomorphic.

I also conclude that $m\lVert v\rVert_{E_n} \le \lVert x\rVert_{E} \le M\lVert v\rVert_{E_n}$ for every $x$.

I'm not clear on how to conclude that the norms are equivalent. How can I use the last inequality to show that?

Thanks in advance!

  • $\begingroup$ There is no such thing as a finite normed linear space and the title is wrong. What is your definition of equivalent norms? $\endgroup$ Jan 9, 2019 at 12:26

1 Answer 1


What you want to prove is not true. For instance take $E_2$, and consider the norms $$ \|(a,b)\|_1=|a|+|b|,\ \ \|(a,b)\|_\infty=\max\{|a|,|b|\}. $$ Then $$\|(1,0)\|_1\ne\|(1,1)\|_1,$$ while $$ \|(1,0)\|_\infty=\|(1,1)\|_\infty.$$


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