# Prove that $\| \cdot \|_0$ defined by $\| x \|_0=\max\limits_{1\leq i\leq n}|\alpha_i|$ is a norm on $E$.

Given that $$E$$ is a finite dimensional 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} The problem is: I want to prove that $$\| \cdot \|_0$$ defined by $$\| x \|_0=\max\limits_{1\leq i\leq n}|\alpha_i|$$ is a norm on $$E$$.

I, therefore, post the proof in the answer section after it has been approved.

• Your problem makes no sense. What are the $\alpha_i$'s? – José Carlos Santos Dec 26 '18 at 11:58
• @ José Carlos Santos: They are unique scalars such $\{e_i\}^{n}_{i=1}$ is a basis for $E$. – Omojola Micheal Dec 26 '18 at 12:00
• That answer makes no sense either. – José Carlos Santos Dec 26 '18 at 12:01
• Okay, let me edit my post then. Just some time! – Omojola Micheal Dec 26 '18 at 12:02
• @José Carlos Santos: Kindly check, I made some edits. It should be fine now! – Omojola Micheal Dec 26 '18 at 12:07

$$a.\qquad$$ Let $$x\in E,$$ then \begin{align}\|x\|_0=0&\iff \max\limits_{1\leq i\leq n}|\alpha_i|=0 \iff |\alpha_i|=0,\,\forall \,{1\leq i\leq n}\\&\iff \alpha_i=0,\,\forall \,{1\leq i\leq n}\\&\iff x=0,\;\forall\;x\in E\end{align} $$b.\qquad$$ Let $$\lambda \in K$$, then $$\lambda x=\sum_{i=1}^{n}\lambda\, \alpha_i e_i$$ and
\begin{align}\|\lambda x\|_0&= \max\limits_{1\leq i\leq n}|\lambda\,\alpha_i|=|\lambda\,|\max\limits_{1\leq i\leq n}|\alpha_i|=|\lambda\,|\| x\|_0,\;\forall\;x\in E\end{align}
$$c.\qquad$$ Let $$x,y\in E$$, then $$x=\sum_{i=1}^{n}\alpha_i e_i$$ and $$y=\sum_{i=1}^{n}\beta_i e_i,$$ for some $$\alpha_i,\beta_i,\;i\in\{1,2,\cdots,n\}.$$ Thus,
\begin{align}\| x+y\|_0&= \max\limits_{1\leq i\leq n}|\alpha_i+\beta_i|\leq \max\limits_{1\leq i\leq n}|\alpha_i|+\max\limits_{1\leq i\leq n}|\beta_i|=\| x\|_0+\|y\|_0,\;\forall\;x,y\in E\end{align}