# Is the Minkowski distance actually the $p$-norm function?

The Minkowski distance of order $$p$$ between two points

$${\displaystyle X=(x_{1},x_{2},\ldots ,x_{n}),Y=(y_{1},y_{2},\ldots ,y_{n})\in \mathbb {R} ^{n}}$$

is defined as

$${\displaystyle D\left(X,Y\right)=\left(\sum _{i=1}^{n}|x_{i}-y_{i}|^{p}\right)^{1/p}}.$$

The p-norm of a vector $${\displaystyle \mathbf {x} =(x_{1},\ldots ,x_{n})}$$ is

$$\left\|\mathbf {x} \right\|_{p}:={\bigg (}\sum _{i=1}^{n}\left|x_{i}\right|^{p}{\bigg )}^{1/p}.$$

It seems that the Minkowski distance is actually the $$p$$-norm function, is it?

You are right in the sense that the Minkowski distance is indeed the metric that is induced by the $$p$$-norm, i.e. $$D(X,Y)=\|X-Y\|_p \quad \text{for all X,Y\in\Bbb R^n}.$$ Note however that, strictly-speaking, the statement "the Minkowski distance is the $$p$$-norm function" does not make sense, as $$D$$ has the domain $$\Bbb R^n\times\Bbb R^n$$ (it has two arguments, so to speak), while the $$p$$-norm is defined for elements in $$\Bbb R^n$$ (it has only one argument).
PS: You might want to include the dependence on $$p$$ in your notation for the Minkowski-distance, i.e. write $$D_p$$ instead of just $$D$$.