# A generalized norm function in $\mathbb{R}^n$ [closed]

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.

• As the answer states, you are looking at a particular class of Minkowski functionals. But note that your conditions on $P$ are not sufficient to get a norm. Consider the case $n = 2$ and $P = \{ -1 < x < 1\}$. This set is convex and centrally symmetric, but $N_p((0,1)) = 0$. Similarly, you can consider the case where $P = \{ x = 0, |y| < 1\}$. In this case $N_p((1,0) = \infty$. Finally, if you allo your $P$ to be open (as I did in the first example), you really want to use $\inf$ instead of $\min$. Apr 12, 2021 at 13:24
• In convex geometry this is the reciprocal of the radial function of a convex body. Apr 12, 2021 at 15:30