Given that $E$ is a finite dimensional normed linear space. Let $\dim E=n\geq 1$ 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 $\| x \|_0=\max\limits_{1\leq i\leq n}|\alpha_i|$ is a norm on $E$. that $\| \cdot \|_0$ defined by $\| x \|_0=\max\limits_{1\leq i\leq n}|\alpha_i|$ 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 $x\in E,$ \begin{align}\|x\|=\|\sum_{i=1}^{n}\alpha_i e_i\|\leq \max\limits_{1\leq i\leq n}|\alpha_i|\big\|\sum_{i=1}^{n}e_i\big\| \leq \max\limits_{1\leq i\leq n}|\alpha_i|\sum_{i=1}^{n}\big\|e_i\big\|=\beta\| x \|_0,\end{align} where $\beta:=\sum_{i=1}^{n}\big\|e_i\big\|.$ Now, define $S=\{x\in E: \| x \|_0=1\}.$ Clearly, $S$ is compact. Let \begin{align}\psi:(E&,\| \cdot \|_0)\longrightarrow \Bbb{R},\\& x\mapsto \psi(x)=\|x\|\end{align}
Let $\epsilon>0$ and $ x,y\in E$ be arbitrary such that $\Big\Vert x-y\Big\Vert_0<\delta,$ then \begin{align}\left|\psi\left(x \right)-\psi\left(y \right)\right|&=\left|\Vert x\Vert-\Big\Vert y \Big\Vert\right| \\&\leq \Big\Vert x-y\Big\Vert \\&\leq \beta\,\Big\Vert x-y\Big\Vert_0 \\&<\beta \delta. \end{align} So, given any $\epsilon>0$, choose $\delta=\dfrac{\epsilon}{\beta+1}>0,$ then \begin{align}\left|\psi\left(x \right)-\psi\left(y \right)\right|&<\beta \delta=\beta\left(\frac{\epsilon}{\beta+1}\right)<\epsilon. \end{align} Thus, $\psi$ is uniformly continuous on $E$ and is automatically continuous on $E$. Since $S\subseteq E$, then $\psi$ is continuous on $S$, and the minimum is attained in the set, i.e. there exists $t_0\in S$ such that $\psi(t_0)=\min\limits_{t\in S} \psi(t)$ and \begin{align}0<\psi(t_0)\leq \psi(t)=\|t\|,\;\;t\in S.\end{align} Let $u=\frac{x}{\| x\|_0}$, then $u\in S$ and \begin{align}\gamma\leq \psi(u)=\Big\|\frac{x}{\| x\|_0}\Big\|\implies \gamma \| x\|_0\leq \| x\|,\;\;\gamma:=\psi(t_0).\end{align} Finally, we have \begin{align}\gamma\leq \psi(t)=\Big\|\frac{x}{\| x\|_0}\Big\|\implies \gamma \| x\|_0\leq \| 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 $\psi(t_0)>0?$