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?
Best Answer
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 $D$.