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Given that $E$ is a finite dimensional normed linear space. Let $dim E=ngeq 1$ and ${e_i}^{n}_{i=1}$ be a basis for $E.$ Then, there exists unique scalars ${alpha_i}^{n}_{i=1}$ such that
begin{align}x=sum_{i=1}^{n}alpha_i e_i.end{align}
PROOF

I proved here Prove that $| cdot |_0$ defined by $| x |_0=maxlimits_{1leq ileq n}|alpha_i|$ is a norm on $E$. that $| cdot |_0$ defined by $| x |_0=maxlimits_{1leq ileq n}|alpha_i|$ is a norm on $E$. So, the next thing to do, is to prove that any norm $| cdot |$ defined on $E,$ is equivalent to $| cdot |_0$.

So, for any $xin E,$
begin{align}|x|=|sum_{i=1}^{n}alpha_i e_i|leq maxlimits_{1leq ileq n}|alpha_i|big|sum_{i=1}^{n}e_ibig| leq maxlimits_{1leq ileq n}|alpha_i|sum_{i=1}^{n}big|e_ibig|=beta| x |_0,end{align}
where $beta:=sum_{i=1}^{n}big|e_ibig|.$ Now, define $S={xin E: | x |_0=1}.$ Clearly, $S$ is compact. Let
begin{align}psi:(E&,| cdot |_0)longrightarrow Bbb{R},\& xmapsto psi(x)=|x|end{align}

Let $epsilon>0$ and $ x,yin E$ be arbitrary such that $BigVert x-yBigVert_0<delta,$ then
begin{align}left|psileft(x right)-psileft(y right)right|&=left|Vert xVert-BigVert y BigVertright| \&leq BigVert x-yBigVert \&leq beta,BigVert x-yBigVert_0 \&<beta delta. end{align}
So, given any $epsilon>0$, choose $delta=dfrac{epsilon}{beta+1}>0,$ then
begin{align}left|psileft(x right)-psileft(y right)right|&<beta delta=betaleft(frac{epsilon}{beta+1}right)<epsilon. end{align}
Thus, $psi$ is uniformly continuous on $E$ and is automatically continuous on $E$. Since $Ssubseteq E$, then $psi$ is continuous on $S$, and the minimum is attained in the set, i.e. there exists $t_0in S$ such that $psi(t_0)=minlimits_{tin S} psi(t)$ and begin{align}0<psi(t_0)leq psi(t)=|t|,;;tin S.end{align}
Let $u=frac{x}{| x|_0}$, then $uin S$ and begin{align}gammaleq psi(u)=Big|frac{x}{| x|_0}Big|implies gamma | x|_0leq | x|,;;gamma:=psi(t_0).end{align}
Finally, we have begin{align}gammaleq psi(t)=Big|frac{x}{| x|_0}Big|implies gamma | x|_0leq | x|leq beta| x |_0, ;;text{for some} ;;gamma,beta>0.end{align}
Therefore, any norm $| cdot |$ defined on $E,$ is equivalent to $| cdot |_0$ and we are done!

Kindly help check if the proof is correct.

QUESTION:

What gives the assurance that $psi(t_0)>0?$

The key of the argument is the fact that $S$ is compact, and you are glossing over that. It's not very hard, but it is not the right part of the proof to say "clearly": it's precisely the part of the proof where you have to use that $E$ is finite-dimensional.

Once you know that $S$ is compact, you are done. You have already proven that $psi$ is uniformly continuous, as your argument started with $psi(x)leqbeta|x|_0$. And you don't even need uniform continuity, which is automatic for a continuous function on a compact set. Once you know that $S$ is compact and $psi$ is continuous, it is standard that it attains a max and a min on $S$, and you are done.

Finally, your (unneeded) argument to show that $psi$ is uniformly continuous starts by saying that $x/|x-y|_0in S$, which is not the case. And the argument cannot be right because it doesn't use that $S$ is compact, so it doesn't use that $E$ is finite-dimensional; nor it uses what $|cdot|_0$ is (which you did use in your original proof of $psi(x)leqbeta|x|$): it is a "proof" that any two norms on a vector space are comparable, something that is not true.

Finally, you have $psi(t_0)>0$ because $psi$ is a norm; since $t_0in S$, you have $|t_0|_0=1$, so $t_0ne0$ and then $psi(t_0)>0$.

Since the minimum is attained in the set $S$, it's pretty straightforward. Recall that $S$ is defined as :

$$S = {x in E : |x|_0 = 1}$$

You defined $psi(t)$ as :

$$begin{align}psi:(E&,| cdot |_0)longrightarrow Bbb{R},\& xmapsto psi(x)=|x|end{align}$$

But, note that for any norm, it is :

$$|x| = 0 Leftrightarrow x = 0$$

The importance of $Leftrightarrow$ as an if and only if operator is noted here.

Specifically, it is : $psi(t_0) = |t_0|$ with $t_0 in S$. But $|t_0|_0 = 1 > 0 Leftrightarrow t_0 > 0 Leftrightarrow |t_0| equiv psi(t_0) >0$.

We know that if and only if $||x||=0$ then $x=0$. And we know the same for $||x||_0$.
Since $||t_0||_0=1$ we get that $||t_0||neq 0$. And since $||x||geq0$ we get positivity of $||t_0||=psi(t_0)$