Prove that relations beneath have conditions for being a norm function
A) for $C^{n}: \left \| x \right \|= \left (\sum_{j=1}^{n}\left | \xi _{j} \right |^{2} \right )^{1/2}$
B) for $C\left[a,b\right]: \left \| x \right \|=\displaystyle{\max_{t\in [a,b]}} \:\left | x(t) \right |$
(B) we need to show:
(i) $\|x\|\ge 0,\forall x\in C[a,b]$ and $\|x\|=0\iff x=0$
(ii)$\|\alpha x\|=|\alpha|\|x\|,\forall\alpha\in\Bbb R,\forall x\in C[a,b]$
(iii) $\|x+y\|\le\|x\|+\|y\|,\forall x,y\in C[a,b]$
Proof:
(i)$\|x\|=\max_{t\in[a,b]}|x(t)|\ge 0$ since $|\cdot|$ is a norm.
$\|x\|=0\Rightarrow \max_{t\in[a,b]}|x(t)|=0\Rightarrow |x(t)|\le 0\Rightarrow x=0$
$x=0\Rightarrow \|x\|=\|0\|= \max_{t\in[a,b]}|0|=0$.
(ii)$\|\alpha x\|=\max_{t\in[a,b]}|\alpha x(t)|=|\alpha|\max_{t\in[a,b]}|x(t)|=|\alpha|\|x\|$
(iii)|x+y|\le|x|+|y|\le \max_{t\in[a,b]}|x(t)|+\max_{t\in[a,b]}|y(t)|=\|x\|+\|y\|$
$\Rightarrow \max_{t\in[a,b]}|x(t)+y(t)|\le \max_{t\in[a,b]}|x(t)|+\max_{t\in[a,b]}|y(t)|$
$\Rightarrow \|x+y\|\le\|x\|+\|y\|$