# Prove whether norm is or is not metric

Let $$(X,d)$$ be a metric space and $$\left\lVert\cdot\right\rVert$$ be a norm on $$\mathbb{R}^2$$. Define $$\tilde{d}$$ by $$\tilde{d}((x_1,x_2),(y_1,y_2)) = \left\lVert(d(x_1,y_1),d(x_2,y_2))\right\rVert$$. Is this $$\tilde{d}$$ a metric on $$X \times X$$? It quite clearly is when e.g. $$\left\lVert\cdot\right\rVert$$ is an $$l^p$$ norm, but how to prove or disprove it in the general case?

Since $$\|\cdot\|$$ is a norm, clearly $$\tilde{d}:X\times X\to [0,+\infty)$$.

$$\tilde{d}((x_1,x_2),(y_1,y_2)) = \left\lVert(d(x_1,y_1),d(x_2,y_2))\right\rVert=0\Leftrightarrow(d(x_1,y_1),d(x_2,y_2))=(0,0)\Leftrightarrow d(x_1,y_1)=0\land d(x_2,y_2)=0\Leftrightarrow x_1=y_1\land x_2=y_2\Leftrightarrow(x_1,x_2)=(y_1,y_2)$$.

$$\tilde{d}((x_1,x_2),(y_1,y_2)) = \left\lVert(d(x_1,y_1),d(x_2,y_2))\right\rVert=\|(d(y_1,x_1),d(y_2,x_2))\|=\tilde{d}((y_1,y_2),(x_1,x_2))$$.

All the previous properties are valid in general. Unfortunately the last property we need, the triangle inequality, can fail:

Take $$X=\Bbb R$$ with $$d(a,b)=|a-b|$$ and consider the norm $$\|(x,y)\|=|x|+|x-y|$$ on $$\Bbb R^2$$. Then $$\tilde{d}((2,0),(0,0))=\|(d(2,0),d(0,0))\|=\|(2,0)\|=4\nleq2=\|(1,1)\|+\|(1,1)\|=\|(d(2,1),d(0,1))\|+\|d(1,0),d(1,0))\|=\tilde{d}((2,0),(1,1))+\tilde{d}((1,1),(0,0))$$.

However, we can require $$\|\cdot\|$$ to be monotone respect the following order: $$(x_1,y_1)\le(x_2,y_2)$$ iff $$|x_1|\le |x_2|\land |y_1|\le |y_2|$$, meaning that whenever $$(x_1,y_1)\le(x_2,y_2)$$ we have $$\|(x_1,y_1)\|\le\|(x_2,y_2)\|$$. The previous partial order can be interpreted as ordering by the distance to the origin, so we are demanding the norm to respect the "farness" of the points of $$\Bbb R^2$$. Here's a link mentioning this monotone norms and a characterisation.

In our case we don't need the absolute values in that partial order of $$\Bbb R^2$$, since $$d$$ is nonnegative, so if your norm $$\|\cdot\|$$ has that property then $$\tilde{d}((x_1,x_2),(y_1,y_2))=\|(d(x_1,y_1),d(x_2,y_2))\|\le\|(d(x_1,z_1)+d(z_1,y_1),d(x_2,z_2)+d(z_2,y_2))\|=\|(d(x_1,z_1),d(x_2,z_2))+(d(z_1,y_1),d(z_2,y_2))\|\le\|(d(x_1,z_1),d(x_2,z_2))\|+\|(d(z_1,y_1),d(z_2,y_2))\|=\tilde{d}((x_1,x_2),(z_1,z_2))+\tilde{d}((z_1,z_2),(y_1,y_2))$$.

So with that additional requirement $$\tilde{d}$$ would be a metric on $$X\times X$$.