We defined a new norm. The norm of $x \in \mathbb{R}^n$ is defined as $$ N_P(x) = \min \{t \geq 0 : x \in t\cdot P\} \enspace,$$ where $P$ is a centrally symmetric and convex body centered at the origin point.

We prove that it is a norm.

1.Identity of indiscernibles.

Obviously, $N_P(x)=0 \Leftrightarrow x=0$.

2.Absolutely scalable.

Because of centrally symmetric property, $N_P(ax)=|a|N_P(x)$.

3.Triangle inequality.

Denote $N_P(x+y), N_P(x),N_P(y)$ as $t_0,t_1,t_2$ respectively. And let vectors $x+y,x,y$ go from the origin point and hit the border of $P$ at $a,b,c$ respectively.

Therefore $(x+y)=x+y$ implies $t_0\vec{a}=t_1\vec{b}+t_2\vec{c}$, implies $\vec{a}=\frac{t_1}{t_0}\vec{b}+\frac{t_2}{t_0}\vec{c}$

Suppose $t_0 > t_1+t_2$, thus $0\leq \frac{t_1}{t_0}+\frac{t_2}{t_0}<1$.

However, this contradict to the convex property because border $bac$ is not convex. QED

We realized this new norm consists all possible norms in $\mathbb{R}^n$, including $\ell_p$.

Because for any norm $N(\cdot)$, define $P=\{ x : N(x)\leq 1\}$, one can verify that $N_P=N$. It shows a simple fact: a norm is equivalent to the space which has unit norm.

Our question is, did anyone discover it before? What is the name? We googled it but did not get answers.