Norms on a vector space over $\mathbb{R}$ In an exercise in the book "Topology and groupoids" the following is asked:
Let V be a finite dimensional right vector space over $\mathbb{R}$ ($dim_RV=n)$. Show that any $2$ norms on $V$ are equivalent. (My understanding is that equivalent means that they induce the same metric topology on $V$).
I would like someone to give me help me with this. I prefer to take hints instead of complete answers
Thank you

My attempt: I reduced the problem to the following statement which I am not sure if its true. Let $||\,||$ be norm on $V$. There exists a basis $\{v_1,v_2,...,v_n\}$ of $V$ such that for any $a_1,a_2,...,a_n\in R$ we have:
$$\max\{|a_i|(||v_i||):1\leq i\leq n\}\leq \ ||\sum_{i=1}^nv_ia_i||$$

Now Is there a similar result for infinite dimensional spaces ?
 A: $\left\|x\right\|_\infty=\max\limits_{1\le k\le n}\left|x_k\right|$
Take $(e_1,\dots,e_n)$ basis of $\Bbb R^n$
$x=\sum\limits_{k=1}^nx_ke_k$

$\|x\|=\left\|\sum\limits_{k=1}^nx_ke_k\right\| \le \sum\limits_{k=1}^n\left\|x_ke_k\right\|=\sum\limits_{k=1}^n\left|x_k\right|\left\|e_k\right\|\le \left(\max\limits_{1\le k\le n} \left|x_k\right|\right)\left(\sum\limits_{k=1}^n\left\|e_k\right\|\right) = \left(\sum\limits_{k=1}^n\left\|e_k\right\|\right)\left\|x\right\|_\infty$ where $\sum\limits_{k=1}^n\left\|e_k\right\|$ is a constant

Let $S=\left\{x\in \Bbb R^n, \left\|x\right\|_\infty = 1\right\}$

 Try defining a function on $S$

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Let $\varphi:\begin{array}{ll}S&\to&\Bbb R\\ x &\mapsto& \left\|x\right\|\end{array}$



 Try to prove it is continuous with respect to $\|.\|_\infty$

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$\forall x, y \in \Bbb R^n, \left|\left\|x\right\|-\left\|y\right\|\right| \le \left\|x-y\right\| \le \left(\sum\limits_{k=1}^n\left\|e_k\right\|\right) \left\|x-y\right\|_\infty$ so $\varphi$ is continuous with respect to with $\|.\|_\infty$ norm.



 Since $\varphi$ is continuous with respect to $\|.\|_\infty$, you should be able to find useful properties of $\varphi\left(S\right)$ coming from properties of $S$

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$\Bbb R ^n$ is finite dimensional, $S$ is closed and bounded so it is compact (all of that with respect to $\|.\|_\infty$). So $\varphi$ is bounded on $S$, $\exists \alpha \in S, \varphi(\alpha)=m=\inf\limits_{x\in S}\varphi(x)$ and $\exists \beta \in S, \varphi(\beta)=M=\sup\limits_{x\in S}\varphi(x)$.



 Try to prove that $0<m$ and $0<M$

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Since $\alpha,\beta\in S, \left\|\alpha\right\|_\infty=\left\|\beta\right\|_\infty=1\not= 0$ so $\alpha\not=0$ and $\beta\not=0$ so $0<m=\varphi(\alpha)=\left\|\alpha\right\|$ and $0<M=\varphi(\beta)=\left\|\beta\right\|$



 Try to apply those inequalities to vectors outside of $S$

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$\forall x \in \Bbb R^n\setminus \left\{0\right\},\cfrac{x}{\left\|x\right\|_\infty}\in S$ so $m \le \left\|\cfrac{x}{\left\|x\right\|_\infty}\right\| \le M$. That is $\forall x \in \Bbb R^n\setminus \left\{0\right\}, m\left\|x\right\|_\infty\le \left\|x\right\|\le M \left\|x\right\|_\infty$



 Try to extend those inequalities to all vectors of $\Bbb R^n$

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 For $x=0$, the equality holds so the inequality does too.


$\boxed{\forall x \in \Bbb R^n, m\left\|x\right\|_\infty\le \left\|x\right\|\le M \left\|x\right\|_\infty \text{ where } m > 0 \text{ and } M > 0}$
