Is there any distance function such that $g(\mathbf{b}+\mathbf{c}) = g(\mathbf{b}) + g(\mathbf{c})$?

Question

Suppose that we have $$\mathbf{a}_i - \mathbf{a}_j = \mathbf{b} + \mathbf{c},$$ where all variables are of the same dimension (i.e., $n$ dimensional). If we let $g(\mathbf{a}_i-\mathbf{a}_j)$ denote the squared Euclidean distance between $\mathbf{a}_i$ and $\mathbf{a}_j$, then the following is not true $$g(\mathbf{b}+\mathbf{c}) = g(\mathbf{b}) + g(\mathbf{c})\;$$ because $g(\mathbf{b}+\mathbf{c})=(b_1+c_1)^2+...+(b_n+c_n)^2 \ne b_1^2+c_1^2+...+b_n^2+c_n^2=g(\mathbf{b}) + g(\mathbf{c}).$

Is there any distance function $g(\cdot)$ such that $$g(\mathbf{b}+\mathbf{c}) = g(\mathbf{b}) + g(\mathbf{c})\;?$$

Answer 1

If you are considering a normed space $(X,\|\cdot\|)$ and $X$ consists of more than $0$, then $\|x+y\| = \|x\| + \|y\|$ cannot hold for every $x,y$. Take for example $x\neq y$ and $y = -x$, then you get $0 = \|x-x\| = \|x\| + \|x\|$, which implies $\|x\| = 0 \Leftrightarrow x = 0$, which is a contradiction.

Answer 2

Actually, if $X$ is any set with at least $2$ elements then you can't define any metric on it such that the triangle inequality is always an equality. Indeed, if $x\ne y$ then $d(x,x)=0$ is strictly smaller than $d(x,y)+d(y,x)=2d(x,y)$.

In particular, a norm like you want exists only for a trivial vector space.